s:: ~ ~' ~-'t ~ ~ i~~, PROBLEMS IN ILLUSTRATION OF THE PRINCIPLES OF PLANE COORDINATE GEOMETRY. BY WILLIAM WALTON, 1M.A., TRINITY COLLEGE, CAMBRIDGE. XcXE.roTv fir 7TrapaaEFti#j/aa'C XpwLUvov iKavoS E'vELKVvaC'ai L 'r wv /.ELtOVWV. II X d~'., v. CAMBRIDGE: JOHN DEIGHTON; LONDON: SIMPKIN & CO.; G. BELL, FLEET STREET. M.DCCC.LI. CAMB RIDGE: PRINTED BY MIETCALFE AND PALMER. PREFACE. WITHIN the last few years have appeared many valuable works on Coordinate Geometry, involving a variety of new and elegant methods of Algebraical investigation. Several of these treatises are enriched with numerous incidental problems, and to the solution of problems of a miscellaneous character some of them are principally devoted. The tendency of even detached and desultory problems is no doubt to sharpen the intellect of the learner, and to give him a true conception of the signification of corresponding and more abstract propositions: the classification however of problems by some law of affinity is, in my opinion, calculated not only to produce the same result in a still higher degree, but also to qualify him by an enlarged comparison for becoming skilful and judicious in effecting his own solutions. In this conviction has originated the composition of this work. In preparing this collection of Problems for publication, I have been guided by two principles of arrangement, the one depending upon similarity of geometrical properties, the other upon analogy between methods of solution. In Algebraic Geometry the latter principle seems to be usually the more important, and, excepting where its application would have iv PREFAiCE. been disadvantageously at variance with the former, has been in this treatise adopted as the basis of classification. The obligations under which I am placed to various authors, for so much of what is comprised in this volume, have been particularly acknowledged throughout its pages. The methods of solution, however, which I have given, are rarely even similar to those which are developed in the works where the geometrical properties were originally enunciated. The greater number or all of the solutions contained in many of these works are purely geometrical. The advantage, it is proper to add, which I have derived from the various Examination papers published in the University of Cambridge and its various Colleges, is very considerable. Cambridge, 30th January, 1851, CONTENTS, STRAIGHT LINE. Section 1. Elementary problems. Rectangular axes 2. Elementary problems. Oblique axes 3. Polar equation 4. Rectilinear loci.. 5. Transversals. Explicit parameters 6. Transversals. Implicit parameters 7. Rectilinear areas CIRCLE. 1. Referred to two perpendicular diameters. Tangents 2. Referred to two perpendicular diameters. Chords 3. Referred to two perpendicular diameters. Points 4. Referred to any rectangular axes 5. Referred to two tangents as axes of coordinates 6. Referred to any oblique axes 7. Polar coordinates 8. Polar equations to tangents and chords 9. Poles and polars.. 10. Radical axes, centres of similitude, &c. 11. Inscribed and circumscribed polygons 12. Circular loci Page 1 7 14 15 24 40 57 66 68 71 72 80 82 83 87 89 90 96 101 PARABOLA. 1. Referred to the axis and its tangent. Ordinates 2. Referred to the axis and its tangent. Tangents. 3. Referred to the axis and its tangent. Magical equation to the tangent.. 4. Referred to the axis and its tangent. Normals 5. Referred to the axis and its tangent. Chords. 6. Referred to the axis and its tangent. Focal properties 7. Referred to a tangent and its diameter as axes 8. Referred to two tangents as axes 112 114 116 121 124 127 130 132 v1 CONTENTS. Section 9. Referred to any rectangular axes whatever. Reduction 10. Polar equation. Focus the pole 11. Polar equation. Vertex the pole 12. Polar equation. Pole a point in the axis 13. Polar equation. Pole anywhere 14. Linear equation 15. Polar equation to the tangent 16. Poles and polars 17. Intersection of parabolas 18. Parabolic loci.. 19. Parabolic envelops 20. Miscellaneous problems ELLIPSE. Page 1 39 141 145 146 147 148 149 153 156 159 166 169 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24, 25. 26. 27. Referred to its axes. Ordinates Referred to its axes. Tangents Referred to its axes. Magical equation to the tangent Referred to its axes. Normals. Referred to its axes. Chords Referred to its axes. Focal properties Referred to its axes. Conjugate diameters Referred to axes parallel to the axes of the curve Polar equation. Centre the pole Polar equation. Focus the pole Polar equation. End of the axis major the pole Polar equation. End of the axis minor the pole Polar equation. Point in the axis the pole Polar equation. Pole anywhere Referred to conjugate diameters Referred to any two diameters Referred to any rectangular axes whatever Linear equation. Intersections of Ellipses Polar equation to the tangent Polar equation to a chord Polar equation to the normal Poles and polars Inscribed and circumscribed polygons Elliptic loci. Elliptic envelops Miscellaneous problems 172 174 178 185 188 192 194 201 203 207 210 211 212 213 215 219 221 224 224 228 231 232 233 234 242 251 253 CONTENTS. vii HYPERBOLA. Section Page I. Referred to its axes. Ordinates... 262 2. Referred to its axes. Tangents.. 263 3. Referred to its axes. Magical equation to the tangent. 264 4. Referred to its axes. Focal properties... 265 5. Referred to its axes. Conjugate diameters. Conjugate hyperbola 267 6. Referred to its axes. Asymptotes... 269 7. Referred to its transverse axis and the tangent at its vertex 272 8. Referred to conjugate diameters. Asymptotes.. 273 9. Referred to conjugate diameters. Conjugate hyperbola. 276 10. Referred to any two diameters. Conjugate hyperbola. 278 11. Referred to its asymptotes... 280 12. Referred to any rectangular axes... 292 13. Referred to any rectangular axes. Reduction.. 294 14. Polar equation. Centre the pole... 296 15. Polar equation. Focus the pole... 297 16. Polar equation. Point in the axis the pole.. 299 17. Polar equation. Pole anywhere... 300 18. Poles and Polars..... 301 19. Hyperbolic loci..... 305 20. Miscellaneous problems.... 317 LINES OF THE SECOND ORDER. 1. Referred to a principal diameter and its tangent. Normals 324 2. Referred to a principal diameter and its tangent. Chords. 325 3. Referred to a principal diameter and its tangent. Focal properties.....326 4. Referred to any oblique diameters... 328 5. Referred to two tangents as axes... 329 6. Referred to a tangent and normal... 334 7. Referred to any axes whatever. Centres.. 338 8. Referred to any axes whatever. Tangents.. 338 9. Referred to any axes whatever. Chords.. 341 10. Referred to any axes whatever. Directrix.. 345 11. Referred to any axes whatever. Conjoint lines and circles 346 12. Passing through given points... 348 13. Passing through given points and touching given straight lines 351 14. Determination of their equations from given conditions. 355 15. Poles andpolars..... 357 16. Polar equations..... 364 17. Linear equation..... 366 viii CONTENTS. Section 18. Polar equation to the tangent 19. Polar equation to the chord of a conic section 20. Inscribed polygons 21. Circumscribed polygons 22. Problems relating to several curves 23. Intersections of conic sections. Common chords 24. Double contact 25. Conical loci 26. Envelops. 27. Similar curves 28. Miscellaneous problems Appendix Page 368 375 383 392 398 403 405 407 415 420 424 428 Page 5, 82, 108, 109, 119, ERRATA. Line Error. 5 from bottom, y = - x, y = x - 2a, 7, 4a, 7 from bottom, + 6ay cot/3 = 9c2. Correction. y = - 2x, y =x- 3a. 2a. + 2ay tan / = c2. - or +. 1, 4 or -. 5 from bottom, -2, ADDENDUM. Page 40. If u = 0, v = 0, be the equations to two straight lines, then, x cos e + y sin e = 6 being the general type of the equation to a straight line, the two bisecting lines will be denoted by the equations u + v = 0. This useful proposition is due to Mr. Rutherford, Annals of Philosophy for May, 1843, p. 353. It is intended to publish an additional table of Errata, should it be found desirable, in January 1852. ALGEBRAIC GEOMETRY. STRAIGHT LINE. SECTION I. Elementary Problems. Rectangular Axes. 1. THE distances of a certain point from the three lines x cosa + y sina =, x cosa'+ y sin' = 8', xcos + y sina"= '", being respectively c, c', c", to find the relation between the quantities c, a', a, a, 8', S", c, c', c". By the conditions of the problem, we have x cos a + y sin a - = c, x cos a + y sin a - 8' = c' x cos a" + y sin a" -"=; or, putting 8 + c = X, 8' + c' = ', 8" + c" = X", and eliminating x and y by cross multiplication, X (cos a" si' ossin a' - cos a sin a") + (c os a sin os a" sin a) + X" (cos a' sin a - cos a sin a') = 0, sin (a' - a") sin (a" - a) sin (a - a') or + + =0. 2. To find the coordinates of a point, the distances of which from two given straight lines are assigned. Let X, Y, be the coordinates of the required point, and let the equations to the given straight lines be Ix + my = 8, l'x + n'y = 8', B 2 STRAIGHT LINE. (1, m) being the direction-cosines of the former, and (1', m') of the latter line. Then, the algebraical symbols of the distances of the point (X, Y) from these lines being respectively X, X', we have X + mY- =, = 'X X+ m'Y- 8' = X' X, X', being positive or negative accordingly as the point (X, Y) is, relatively to the origin, on the more remote or on the nearer side of the corresponding lines. From these two equations we get, for the coordinates of the required point, m'- mi' y (8 + X) 1' - (8' + x') 1 I'm - Im' 3. To find the equation to a straight line such that, if perpendiculars be drawn to it from n proposed points, the sum of the perpendiculars on one side of it may be equal to the sum of the perpendiculars on the other. Let (x, y,), (x,, y2), (x,, y3),... (x, y), be the n points, and suppose the equation to the straight line to be Xx + y =........................ (1), X, /, being its direction-cosines. Then the algebraical expressions for the distances of the n points from this line will be XX + /-Y1 - Xx2 +Y2- x + 8Y - + 3-.. XX + /y,-8. But, since the geometrical sums of the two opposite groups of perpendiculars are, by the condition of the problem, equal, it follows that the algebraical sum of all the perpendiculars must be zero. Hence X (xi + x2 + 3 +... +,1) + I (Y1 + y+Y3 Y " +. + y-) = n *... (2). Eliminating 8 between (1) and (2), we get X (x +x2 + X... +x-nx) + I ( +y + y +...+y,,-ny)=0, as the equation to a line satisfying the hypothesis. ELEMENTARY PROBLEMS. RECTANGULAR AXES. 3 Since the ratio of X to,u is not determinate, it follows that there is an infinite number of such straight lines, all passing through a point of which the coordinates are (x, + x2+ x,3+ +X) (Y1~ +Y) 2 (1 3 +..+ Yt) + n n Puissant: Recueil de diverses propositions de Geometrie, p. 192, troisieme edition. 4. To determine the position of a point in a given straight line, such that the difference between the squares of its distances from two given points may be equal to a given area. Let the origin of coordinates coincide with one of the given points, and let the axis of x contain the other. Let a be the abscissa of the latter point, and let the equation to the given line be x cosa + y sin a =.................. (1). The coordinates of the required point being (x, y), we have, by the condition of the problem, c2 denoting the given area, x2 + Y2 = c2 + (x - a)' + y2, 2 + 2 or x = 2..................... (2). 2a.(2) The ordinate y is known from (1) and (2), the intersection of the two lines, which they represent, being the required point. COR. Suppose the lines represented by (1) and (2) to be coincident: then cos a 2a sin a = 0, +.- a= + C2 a2 + c2 whence a=O, 8- a; 2a and the value of y cannot be ascertained. If then we restrict a and 8 respectively to the values 0, a2 + c2 and a 2+, the problem becomes indeterminate, every point in the given straight line satisfying the condition of the problem. B2 4 STRAIGHT LINE. This example affords an instance of the degeneration of a determinate into an indeterminate problem, in consequence of our restricting the data by particular conditions, and belongs therefore, according to Playfair's interpretation of the ancient signification of the term, to a class of propositions called, by the old geometricians, Porisms. Playfair* says that" a Porism may be defined a proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate, or capable of innumerable solztions." 5. A straight line, inclined to the axis of x at an angle of 30~, cuts the positive axes of coordinates in A and B. To find tle equation to a straight line bisecting AB and passing through the origin. The required equation is x= /3.y. 6. To find the equations to the four sides of a square, the coordinates of two of the angular points being (2, 3) and (3, 4). The required equations are x = 2, y=3, x=3, y = 4. 7. To find the equation to a line which cuts the two lines y= 2, y = 3, at distances 3, 4, respectively, from the axis of y. The required equation is x = y+ 1. 8. To find the equations to the straight lines which pass through the intersection of the lines y = 2x+4, y = 3x+6, and bisect the supplementary angles which they include. The required equations are y = (1 + 42). (x + 2). * Transactions of the Royal Society of Edinburgh, Vol. iii. p. 170; 1794. ELEMENTARY PROBLEMS. RECTANGULAR AXES. 5 9. To find the distance of the point of intersection of the lines denoted by the equations 3x + 2y + 4 = 0, 2x + 5y + 8 =0 from the line denoted by the equation y = 5x + 6. 62 Required distance = 11 \(26)' 10. To find the tangent of the angle between the lines represented by the equations 2x + 3y + 4 = 0, 3x + 4y + 5 = 0. If b denote the angle between them, tan ( = I. 11. To find the distance of the origin of coordinates from the line, the equation of which is - =1. 2 3 6 Required distance = -(. V(13' 12. To find the equation to a line passing through the intersection of the lines x=a, x+y+a=O, and through the origin: also to the line passing through the point of intersection of the same lines and perpendicular to the latter of them. The required equations are respectively y = - x y = x - 2a. 13. To find the length of the perpendicular from the vertex upon the base of a triangle, the coordinates of the vertex being (3, 3), and those of the extremities of the base (0, 2), (2, 1). Required length = /5. 6 STRAIGHT LINE. 14. To form the equation for determining the abscissa of a point, in the straight line H l> a b the distance of which from a given point (a, /3) shall be 8, and thence to shew that there are in general two such points and that, in the particular case in which those two points coincide, 82. (a2 + b2) = (a/3 + ba - ab)2 15. To find the abscissa of a point in the axis of x, such that its distance from a point (0, k') may exceed its distance from a point (h, k), by a given length c. If x represent the required abscissa, (c2 - h2) x = e2h + c {e4 - (c2 - A2) K2), where e2 = - (c2 + k2 - h2- k2). Newton: Arithmetica Universalis, Prob. XI. 16. If ABCDE is a regular pentagon; to find the equations to its several sides, AB being the axis of x, and a line through A, at right angles to AB, the axis of y. To find also the equation to the line joining E, B, and to the line joining A, C. The equations to the lines AB, BC, CD, DE, EA, EB, A C are, respectively, 27w y=0, y=tan- (x-a), y -tan (x- a) + 2a sinn n y=-tanW y ta(x-a), tn 5 27r 7 \ 7r -tan. y - tan (- a), y = tan -. x. g, 17. To prove that, if the distances of a point from two points (a, b), (a', '), are c, c', respectively, the point will lie in a straight line, the intercepts of which on the coordinate axes are (b"2 - b 2) (C 2 _ c2) (an' +2(a a)) and (b + b) + 2'" - (a2) - (C'2 2) 2 '/i"2 (b'- b) ELEMENTARY PROBLEMS. OBLIQUE AXES. 7 18. If the ordinates of a straight line, drawn perpendicularly to the axis of x, be all turned about their intersections with this axis, so as to make an angle 0 with it; to shew that, the equation to the original line being y = x tan a, the angle 0 between this line and the rectilinear locus of the extremities of the inclined ordinates, may be found from the equation tan sin a + tan a. sin (a- 0) tan 6 = cosa + tan a. cos (a- 0) and that the new line will coincide with the original one, provided that = -7r + 2a. 19. To prove that the equation 2y2 - 3xy - 2x2 - y + 2x = 0 represents two straight lines perpendicular to each other. 20. To shew that the equation y2 - xy - 2x2 + 5x - y - 2 = 0 represents two straight lines inclined to each other at an angle tan-' 3. 21. To prove that the equation 2 - 2xy. sec a + x2 = 0 represents two straight lines, including an angle a. SECTION II. Elementary Problems. Oblique Axes. 1. To determine the geometrical signification of the equation 4y - 8xy + xo + 4y + 2x - 2 = 0. Completing the square with regard to y, we have (2y + 1 - 2)2 = 3 (1 - x)'2 8 STRAIGHT LINE. and therefore 2y + 1 - 2x = + /3 (1 - x)7 or 2y + (~ 3 - 2) x = /3 - 1, the equations to two straight lines. 2. To determine the geometrical signification of the equation x2 + y2 + xy = 0. Multiplying the equation by 4, we have (x + 2y)2 + 32 = 0. This equation cannot be true, unless each of the squares be separately equal to zero. Hence X = 0, and x + 2y = 0, and therefore x and y must be simultaneously equal to zero. The proposed equation therefore represents the origin of coordinates. 3. The axes of coordinates being inclined to each other at an angle of 60~, to determine the equation to a line parallel to a line, the equation of which is x + y = 3a, and at a distance from it equal to ~ a/3. Let OA = 3a = OB; then AB is the line represented by the equation x + y = 3a. Draw OPQ at right angles to AB, and take QP- = aV3. Draw HPK parallel to A QB. Then PQ = AH cos 30~, a^V3 = 1 4V3. HA, o HA = a =KB. Hence OH = 2a = OK, and the equation to HK is therefore x + y = 2a. ELEMENTARY PROBLEMS. OBLIQUE AXES. 9 If we had taken a point P' on the other side of Q, so that P'Q = -a V3, we should evidently have got as the equation to the required line x + y = 4a. 4. To determine the conditions that the equation ax" + by2 + 2cxy + 2ax + 2b'y + c' = 0 may represent two parallel straight lines. If the equation represent two parallel straight lines, its lefthand member must be resolvable into two factors of the forms Xx + py + v, Xx + py + v. Hence, identically, ax2 + by2 + 2cxy + 2dx + 2b'y +c = (Xx + Py + v) (Xx +,y + y'). Equating the coefficients of x2 y2, 2xy, x, y, on both sides, we have X2 = a.........(1), 2 =........(2), Xp = c.........(3), X ( + v) =2a......(4), (v+ v') =2b'.......(5). From (1) and (4), we have a X 2a v + Vf and, from (3) and (5), c X a c Similarly, from (2) and (5), combined with (3) and (4), there is -,.............................. (7). b ac The equations (6) and (7) constitute the required conditions. 5. To prove that the lines of which the equations are y=2x + 3, y=3x + 4, y=4x + 5, all pass through one point. 10 STRAIGHT LINE. 6. To find the value of a that the line y = ax + 3 may pass through the intersection of the two lines y = x + 1, y = 2x + 2. The required value of a is 3. 7. To find the equations to the two straight lines which, passing through the origin of coordinates, divide into three equal parts the distance between the points, in which the axes of coordinates are intersected by the line x +y =1. The required.equations are 2x = y and 2y = x. 8. To find the equations to the diagonals of the parallelogram formed by the four lines x =a, x = a, y= b, y = b'. The required equations are (a - a) y - (V - b) x = ab - ab', and (a - a') y - ('- b) x = ab - a'b'. 9. The equations to two straight lines referred to oblique axes, inclined to each other at an angle co, being y - mx = 0, and my + x = 0, to find the angle between them. The angle between them is equal to tan1 ( + 1 tan ). 10. If the straight lines denoted by the equations Ix + my = 8, I'x + m'y = 8', ELEMENTARY PROBLEMS. OBLIQUE AXES. 11 be perpendicular to each other, to prove that, co being the inclination of the axes, It + nmm = (Im + 'm) cos co. 11. To prove that the group of straight lines defined by the equation a'b - ab' b- b a -a a ( a-a where X is arbitrary, are identical with the group of straight lines defined by the equation y - ax - b = -t (y - a'x - b'), where /u is arbitrary. 12. What are the geometrical significations of the equations x2+y2=o, xy=o? The former represents the origin, and the latter the axes of coordinates. 13. A straight line passes through a given point; to determine the magnitudes of its intercepts on the axes of coordinates, the product of the intercepts being given. If a, b, be the coordinates of the given point, and c2 the given product; then, m, n, representing the required intercepts, n =- - {c + ('2 - 4ab)}, = 2b n = c T (c2 - 4ab)i). Puissant: Recueil de diverses propositions de Geometrie, p. 141, troisieme edition. 14. To determine the position of a straight line, which passes through a given point, and cuts two given straight lines so that its intercepted portion is of a given length. If the two given lines be taken as axes of coordinates; then, o denoting their included angle; A, k, the coordinates of the given point, and c the given length; the intercepts a, b, of the 12 STRAIGHT LINE. straight line on the axes of coordinates will be determined by the two equations 4 - 2 (h + k cosco) a3 + (A2 + k2 - c2 + 2hk coso) a2+ 2c2ha-c2h2=0, b -2 (k + h cos o) b3 + (k2 + h2 - c + 2kh cos w) b2 + 2c2kb - 7' = 0. Newton: Universal Arithmetic, prob. 24. Gergonne: Annales de Mathematiques, tom. 10, p. 205. 15. Through a given point in a given angle, equidistant from its two sides, to draw a straight line terminated at the two sides of the angle, so that this point may divide the straight line into two parts the sum of the squares of which shall be of a given value. Let a, a, be the coordinates of the given point referred to the sides of the angle as axes; let co denote the given angle. Then the equation to the required line will be y - a = m (x - a), m being determined by the equation ( 2 c2 S m + -+-cos + cos co the four values of m corresponding to four different lines which satisfy the conditions of the problem. " Ce probleme est un de ceux qui furent proposes en 1819 au concours general des colleges royaux de Paris." Gergonne: Annales de Mathematiqzues tom. x. p. 73. Puissant: Recueil de diverses propositions de Geome'trie, p. 142, troisieme e'dition. 16. To ascertain the geometrical signification of the equation y - xy + 2 -2x + 1 = 0. It represents the point of intersection of two straight lines x=, y =. 17. To determine the geometrical meaning of the equation x3 - 4a2x == 0. ELEMENTARY PROBLEMS. OBLIQUE AXES. 13 It represents three parallel straight lines, viz. the axis of y, and two equidistant lines on each side of it. 18. To interpret the equation xy - 2ay + 3ax - 6a2 = 0. It represents two straight lines, one parallel to the axis of y, cutting the axis of x at a distance + 2a from the origin, and the other parallel to the axis of x, cutting the axis of y at a distance - 3a from the origin. 19. To assign the geometrical signification of the equations (x +y+ c).(x - y + c) = O, (x + y + c)2 + (x - y + C)2 = 0. The former equation denotes two straight lines, the intercepts of which on the axes of coordinates are respectively (- c, - c) and (- c, c). The latter equation denotes a point, viz. the point of intersection of the two lines. 20. To ascertain the geometrical signification of the equation \a 2 \a b ) 2 The equation represents two straight lines, the equations of which are x +y + 1 --+ +1=0,.-...l=0. a b a 2b 21. If the equation ax2 + by2 + 2cxy = c' represent two different straight lines; to prove that, either ' is zero and c2 > ab, or c' is finite, and c2 = ab. 22. To prove that the equation (2 +2 1) (/2 +" ) =(a + ~ -1) represents two straight lines, and to find their equations. 14 STRAIGHT LINE. The equations to the two lines are included in the single equation x-_ c y- B m the two values of the ratio between I and m being determinable from the quadratic 12 (2 _ 32) + 2lmca/ + m2 (a - 2) = 0. SECTION III. Polar Equation. 1. AB and BC are two right lines perpendicular to each other; A is a fixed point, B moves along a given right line EF, and AB is to BC in a given ratio; to determine the locus B of C. From A draw AE at right angles to EF. Let LCAE=, L CAB=a; // a being a constant quantity, because the A ratio of AB to BC is constant. Let AE=a, AC=r. Then r cos a = AB =a sec (O - a), r = a sec a sec (0 - a), the equation to the locus of C, which is therefore a straight line. 2. To find the polar coordinates of the point of intersection of the straight lines represented by the equations r = 2a sec (0 - 7r), r = a sec (0 - -r), and the angle between them. The polar coordinates required are (~r%, 2a), and ~7r is the angle between the two lines. RECTILINEAR LOCI. 15 3. To find the polar equation to a straight line which passes through the two points (r,, a,), (r2, a2). The required equation is (a a) sin (a - aa ) sin- ) sin (a - 0) 0 r rl r2 4. If, from a point 0 taken arbitrarily in the plane of a triangle, straight lines be drawn to its summits A, B, C, and then, through the same point 0, perpendiculars be drawn to the three lines OA, OB, OC; to prove that the intersection of these perpendiculars with the sides BC, CA, AB, respectively, will lie all three in one straight line. If 0 being taken as pole, and the coordinates of A, B, C, be respectively (a,) r,), (a2, r2), (a,, r); then, putting for brevity a8 -a2 =, a1 - a= = s,2 a2 - al = En, we shall obtain for the equation to the line containing the three intersections, 2 sin 2s, sin 2si 2 Os -. sinl.sin 2.sin S, + -.cos (a, - ) +. cos (a2 - 0) r 7'1 7r sin 2s, + -.cos (a - 0) = 0. Bobillier: Gergonne, Annales de Mathematiques, tom. xvIIi. p. 185. SECTION IV. Rectilinear Loci. 1. To find the locus of a point equidistant from two given straight lines. Let the equations to the given lines be a cos a + y sin a = 8, and x cos a + y sin a' = '. Then xS, y1, being the coordinates of the point, its distances from the two lines will be + (x. cosa + y, sin a - 8), and + (x1 cos a + y1 sin a- 8'). 16 STRAIGHT LINE. Since these distances are equal, we must have xi cos a + y, sin a - 8 = + (x cos a' + y3 sin ' - 8'); this result shews that the required locus is two straight lines at right angles to each other and passing through the point of intersection of the two given lines. 2. A CB, D E, are two right lines of given lengths, C is the middle point of AB, and the line through E and B meets that through A and D in a point P; supposing AB to move parallel to itself, while DE remains fixed, to find the locus of P. Let xEO, parallel to BA, be taken as the axis of x, and yA 0, parallel to DE, as that of y. Let AC a = BC, CE=b, DE = c. / Then the equation to AP, which passes through the points (0, b) and (a, c), is O X ay - ex = b (a - x)....................(1) The equation to EP, which passes through the points (a, 0) and (2a, b), is ay = b (x - a).................... (2). At the intersection P, of (1) and (2), we have, adding together the two equations, 2ay = cx, the equation to the locus of P, which passing through O. is therefore a straight line 3. A straight line PP' is drawn parallel to the base of a triangle AOA'. From P, P', are erected perpendiculars upon OA, OA', respectively; to find the locus p\ of the intersection of these perpendiculars. Take Ox, parallel to A'A, as the o axis of x, and Oy, at right angles to A'A, as that of y. Let LAOy= s, zA'Oy='; AA' RECTILINEAR LOCI. 17 then the equation to OA will be x = y tan a........................(1). The equation to PP' will be of the form c y............................ (2). The coordinates of P will therefore be c tan s, c, and the equation to the perpendicular through P will be y - c = - tan s.(x - c tan )...............(3). Similarly, putting - s' in place of s, we shall have for the equation to the perpendicular through FP y - c = tan ' (x + c tan s')............... (4). At the intersection of (3) and (4), x (tan ~ + tan a') = c (tan2 s - tan2 a'), x = c (tan a - tan a'), and therefore, from (3) or (4), eliminating c, we shall readily get, for the equation to the locus of the intersection of the two perpendiculars x = tan (a - ); the required locus is therefore a straight line through 0. Puissant: Recueil de diverses propositions de G6ometrie, p. 124, troisigme edition. Lardner: Algebraic Geometry, vol. I. p. 29. 4. AB, CD, are two given parallel straight lines; through C and D are drawn two straight lines PCF, PDG, cutting AB in F, G, so that the ratio of AF to BG is invariable. To find the locus of the point P. Let AB, produced indefinitely, be taken as the axis of x, and a line through A, at right angles to AB, as that of y. Let (h, k), (h', k), be the coordinates of C, D, respectively; (x, y) those of P. Let AB = c. Also let n.AF= n'.BG.................. (1), n, n', being constant quantities. C 18 STRAIGHT LINE. The equation to PF, which passes through the two points C, P, will be (k - y) '- (h - ) y' = kx - hy............(2). Similarly, the equation to PG, which passes through the two points D, P, will be (k - y) x'- (' - x) y' = kx - h............(3). Putting y' = 0 in (2) and in (3), we get, from (2), AF = -' = kx - y k-y kx - Wky and, from (3), AG = x = -k - y, whence BG = c -- k.x k -y Hence, from (1), n (kx - hy) = n'c (k - y) - n' (x - h'y), (n + n') kx - (nh + n'h' - nc) y = n'ck, the equation to the locus of P, which is therefore a straight line. Leybourn's Mathematical Repository, New Series, vol. I. p. 45. 5. A, B, are two given points, CD, CE, two given straight lines. From any point F in CD / is drawn the straight line FBG, cutting CE in G; from G is drawn GP parallel to BA, and meeting/ FA in P. To find the locus of P. / Let CD, CE, produced inde- A finitely, be taken as axes of x, y,. respectively. Let a, b, be the co- ordinates of A; a', b', those of B. Also let CF = m. Then the equation to GBF is 'y x - m b a - Am and thence, putting x = 0, and y = CG, we have C -b'm a - m RECTILINEAR LOCI. 19 Hence the equation to GP, which passes through G in a direction parallel to BA, is 'mrn ' -b Y +, -=, a - m a -a or (a- m) {(a - a) y + bx} = b' {x (a -m) - (a )... a (1). Again, the equation to FP, which passes through F and A, is yx-m(2) = ~- ---.......................... (2). b a - mI Eliminating y between (1) and (2), we have { (a' - n) - ' (a - mn)}.{x (a - m) - m (' - a)} = 0, and therefore x (a' - m) - m (a' - a) = 0 and therefore, by (1), (a' - a) y + bx = 0, the equation to the required locus, which is therefore a straight line passing through C. Leybourn: Mathematical Repository, New Series, vol. I. p. 143. 6. The sides containing a given angle are in a given ratio and the vertex is fixed; supposing the extremity of one of the sides to move in a given straight line, to find the locus of the extremity of the other. Let OA, OB be the two sides; let z A OB = a, A Ox =, L BOx = O', Ox being a fixed straight line; OA = r, OB= r r = nr. Then, since A moves in a straight line, r cos (d - /) = 8, where / and 8 are constants. Hence r cos (0 - /) = n8, r cos (0' - a - ) = nS, the equation to the locus of B, which is therefore a straight line at a distance n8 from 0, the inclination of this distance to Ox being a + B. c2 20 STRAIGHT LINE. 7. From a fixed point A a straight line AB is drawn to a variable point B in a given straight line Ox, and upon AB is described an equilateral triangle ABP. To find the locus of P. Let yA 0, drawn through A at right angles to Ox, be taken as the axis of y, Ox being that of x, and let AO = c. Then the required locus will be a straight line represented by the equation equation 3 - y = c. 8. To find the locus of a point, the distances of which from two lines — +- = 1 + =1, a b a b' referred to rectangular coordinates, are always in the ratio of n':n. The required locus is a straight line defined by the equation -Y x Y - 4. - 1 + a b,a b' n. = n. Lhuilier: Elemens d'Analyse Ge'omntrique et d'Analyse Alg#brique, p. 122. Puissant: Recueil de diverses propositions de Geometrie, p. 190, troisieme edition. Garier: Gdometrie Analytique, p. 438, deuxieme edition. 9. To find the locus of a point, the algebraical sum of the distances of which from the sides of a given polygon is constant. If the equations to the sides of the polygon be xco + in, xcosa + ysina a' =, x cos a" + y sin a" = 8"......I RECTILINEAR LOCI. 21 the equation to the required locus will be x (cos a + cos a' + cos a" +...) + y (cos / + cos I + cos/" +...) = c + 8 + 8' + " +......, where c is the constant quantity. Timmermans: Gergonne, Annales de Mathematiques, tom. xvIII. p. 217. 10. Having given the positions of two points C, C', to find the locus of a point P under the condition that (PC')2 - (PC)' = a constant quantity. Let a, b, be the coordinates of C; a', b', of C'; and x, y, of P. Then, the axes being supposed to be rectangular and c' denoting the value of the constant quantity, the required locus will be a straight line at right angles to CC', defined by the equation (a - a').(2x - a - a') + (b - ').(2y - b - b) = c'. Lhuilier: Ellmens d'Analyse Geometrique et d'Analyse Algebrique, p. 121. 11. A parallel to the base of a triangle being drawn, and its points of intersection with the sides being connected with the extremities of the base, to find the locus of the intersection of the connecting lines. The required locus is a right line bisecting the base and passing through the vertex. Lame: Examen des dijffrentes methodes employees pour rdsoudre les problemes de Geometrie, p. 47. Puissant: Recueil de diverses propositions de Geometrie; p. 122, troisieme edition. Lardner: Algebraic Geometry, vol. I. p. 28. Garnier: Geomtrie Analytique, p. 436, deuxigme edition. 12. Ox, Oy, are two fixed indefinite straight lines; from a point P are drawn PA, PB, cutting Ox Oy, respectively in 22 STRAIGHT LINE. A, B: supposing the angles PAx, PBy, and the ratio of PA to PB to be given, to find the locus of P. Let Ox, Oy, be taken as axes of x, y. Let L PAx = a L PBy = 3, and let PA be to PB as m to n. Then the equation to the locus of P will be mx sin a = ny sin 3. L'Hospital: Traite Analytique des Sections Coniques, p. 249. 13. A straight line D CIH passing through two given points D, C, cuts a given straight line AB in 1, in such a manner that AH n BH- n' n and n' being constants. In the line AB, are taken any two points F, G, such that AF AH B BH' To find the locus of P, the intersection of FCG GD, produced indefinitely. Let HB, HCD, produced indefinitely, be taken as axes of x, y, respectively; let HC = k, HD = k'. Then the required A. G.I1' ] locus will be a straight line, parallel to AB, the equation of which is (n n\,. eybourn:n - New vol.. p. Leybourn: Mathematical Repository, New Series, vol. I. p. 45. RECTILINEAR LOCI. 23 14. AB, AC, are two straight lines given in position; a straight line DE meets them in D, E, respectively, so that AD + AE = a constant length; also DE is divided in a point P, so that DP bears a constant ratio to EP. To find the locus of P. Taking AD, AE, produced indefinitely, as axes of x, y, respectively, let the constant value of AD + AE be denoted by c, and let a.DP = 8.P; then the required locus will be a straight line defined by the equation x y c a ~ at+, ' Leybourn: Mathematical Repository, New Series, vol. II. p. 7. 15. A CB is a given triangle, 1, K) L, given points in the side AB; through H is drawn any straight line EHF, cutting B ACG BC, in E, F, respectively; l: through E and F are drawn right lines EL, FK, intersecting in P. / \ To find the locus of P. -c A j-7 Let HL=a, KB=b, HK=c, AL-d, AB=d A GB=m C=n; and let CE, produced indefinitely, and an indefinite line through C, parallel to AB, be taken respectively as axes of x and y. Then the locus of P will be a straight line defined by the equation ncdx = (ab + cd).ny. Leybourn: Mathematical Repository, New Series, vol. II. p. 89. 16. The loci of two points A, B, are two parallel straight lines. To find the locus of a third point P, such that PA, PB, are inclined at given angles to the loci of A, B, respectively, and bear to each other a constant ratio. 24 STRAIGHT LINE. Let a, 3, be the two given angles, and c the distance between the two parallel lines; also let PA: PB:: m: n. Then, the locus of A being taken as the axis of x, and that of y being at right angles to it, the equation to the required locus is me sin c Y m sin a + n sin,8 L'Hospital: Traite Analytique des Sections Coniques, p. 251. SECTION V. Transversals. Explicit Parameters; The word transversal is thus defined by Camot, in his Essai sur la Theorie des Transversales: "J'appelle transversale une ligne droite ou courbe qui traverse d'une manibre quelconque un systbme d' autres lignes, soit droites, soit courbes; ou meme un systeme de plans ou de surfaces courbes." 1. BA C is a triangle, and B', C', are points so taken in the sides AB, A C, respectively, or these sides produced, that AB' + AC'= AB + AC. To find the locus of the in- tersection of BC' and B'C, produced indefinitely. Let AB, A C, produced in definitely, be taken as axes of x, y, respectively. Let AB=a, A C=b, AB'=a', AC'=b'. Then the equation to A' B' s B' C will be x+y- = 1...................(1), a and that to BC' will be x Y -+ = I........................(2). TRANSVERSALS. EXPLICIT PARAMETERS. 25 Also, by hypothesis, a + b'- =a + b........................(3). From (1) and (2), there is a - a + -0b aa' bb' and therefore, by (3), + y = o....................... (4). aa 6bb' But, from (1) and (2), x y 1 y z 1 -+ - =-, and + — =aa ab a bb ab b and therefore, by addition, attention being paid to (4), x +y = a +, the equation to the required locus. 2. If A, B, 0, be two fixed lines and a fixed point; then, drawing any two lines Oba, Ob'a', through 0, and joining the points a, b', and a, b, the locus of the point C, in which ab' and a'b intersect, is a fixed line passing through the point of intersection of A and B. Let the intersection K of A, B, be the origin of coordinates, A and B being the axes of x and y. Leta = a, K = a', Kb = b, Kb'= b'; and let (X, /) be the coordinates of 0. Then the equation to ab' is + = 1, a y and to ab is -+ 1. a b 26 STRAIGHT LINE. At their intersection C, x(a-u- ) +aY (b- )=0............... (1). But, since ab6 a'b', both pass through O, +, + = I, a b + a, and therefore whence, by (1), xa a) b ).+Y= the equation to the locus of C, which is therefore through K. 3. If two lines AB, CD, intersect in 0, an AC, BD, meet, when produced, in E, and AD, BCC, in F; to determine the condition Athat the line EF may be parallel to AB. Let OA, O C be taken as the axes of x, y, respectively. Let OA = a, OB = b, OC = c, OD = d. Then the equation to A Cis a straight line id if the lines x y -+-=1; a c and the equation to BD is x Y b d hence, at the point E - ( —) y =a +......(.............(1). Again, the equation to AD is Y= a d and the equation to BC is x c TRANSVERSALS. EXPLICIT PARAMETERS. 27 hence, at the point Fl ( ) Y = a + b.....................(2). In order that EF may be parallel to AB, the values of y in (1) and (2) must be equal: hence b a a b c d c d' ac d) = c d( ' a = b. Thus the required condition is that AB be bisected in the point 0. F: Cambridge Mathematical Journal, vol. I. p. 87. 4. To prove that the perpendiculars from the three angles of a triangle upon the opposite sides all meet in one point. Let (a, )(, /3), (da, 3'), be the three angular points. The equation to the side through (a, 3'), (a", i/3), is (a - a') y - (/3" - 3' ) x = a3' - ad'". The equation to the perpendicular upon it through the angle (a, /3) is (a" - a) (x - a) + (3" - /') (y - /) = o, or (a - a) x + (/3" - 3) y= a (a" - a) + / (/3" - /)...(l). The equation to the perpendicular through (a', /') upon the opposite side is, by symmetry, (a - a") x + (/3 - ") y = a (a - d') + /3 (3 - /3)...(2). At the intersection of (1) and (2), we have, adding together their equations and changing signs, (a - a) x + (//' 3- )) " (/' - /3). This equation coinciding in form with what we know by symmetry to be the equation to the third perpendicular, we see that the intersection of any two of the perpendiculars lies in the third. Puissant: Recueil de diverses propositions de Geometrie, p. 120, troisieme edition, 28 STRAIGHT LINE. 5. To prove that lines drawn from the three angles of a triangle to bisect the opposite sides, all meet in a single point. Let the coordinates of the three angles be (a, /), (a', /'), (a", 8"). The equation to the line through (a, 8/), and through the middle point between (a', 3'), and (a", /"), of which the coordinates are - (a' + a"), 4 (/' + /'), is y - / x - a /3' + " - 2/3 a' + a" - 2a' or (a'+ d'-) y- (/' + a-"- 2a) xy = a'B- at/'+ad'3-a/3"...(1). The equation to the corresponding line through (a', /') is (a" + a — 2a') y - (" + 3- 2/3') = d"'' - a'/"'+ a- '- a'/3... (2). At the intersection of (1) and (2), we have, adding and changing signs, (a + a' - 2a") y- (i3 + 2 - a2/") =- a" - a"7 + a'/' - a",' a result which shews that the lines through (a, /3), (a', '), cut each other in the line through (a', /3"). 6. In every quadrilateral, the three lines which join the middle points of the opposite sides and the middle points of the diagonals, pass through the same point. Let ABA'B' be any quadrilateral, the diagonals of which intersect in O. Let OAx, OBy, be taken as the axes of coordinates. i/ - Let OA=a, OB=b, OA'=a', OB'= b'. Then the coordinates of the middle points of AB, and A'B', are respectively (2a) (-and Hence the equation to the line joining these middle points will be ( ) (a ' + a) = i(a'b - ab').........(). TRANSVERSALS. EXPLICIT PARAMETERS. 29 Again, the coordinates of the middle points of AB', A'B, are respectively (2) and (: and therefore the equation to the line joining these points will be (a' + a) y + (b' + b) x = (ab - a'b').........(2). Subtracting and adding the equations (1) and (2), we shall get, for the coordinates of the intersection of the lines they represent, x = (a - a), y= (b- '). Again, the intercepts on the axes of x and y of the line passing through the middle points of AA', BB', are (a-a'), - (b-Y). These intercepts are accordingly twice the magnitudes of the coordinates of the intersection of (1) and (2); the intersection of (1) and (2) lies therefore in the line joining the middle points of the diagonals of the quadrilateral. Gergonne: Annales de Mathematiques, tom. I. p. 353. 7. In every frustum of a triangle, with parallel bases, the intersection of the two diagonals is in a right line with the middle points of the bases of the frustum and the summit of the triangle. Let the angle of the summit of the triangle be taken as the angle of the coordinates. Then the equations to the bases of any frustum will be The equations to the two diagonals will be - + 1........................(3), a Xb +=1.........................(). 30 STRAIGHT LINE. At the intersection of (3) and (4), ( 1 )-bl- ) 0 x y or -Y........................ (5), a b which is the equation to a line which passes through the origin and through the intersection of every pair of diagonals. Combining (1) and (5) we have x= a, y=lb; combining (2) and (5) we have x = a, y = Xb. These results shew that the line (5) bisects the bases of every frustum. Fregier: Ger.qonne, Annales de Mathematiques, tom. VII. p. 167. 8. Let AB, A'B', A"B", be any three parallel lines. Let C C', C", be the points of concourse respectively of A'A", B'B"; of A"A, B"B; and of AA', BB'. Then the three points C', C", will lie in a right line. Take the axis of x parallel to the three lines AB, A'B' A"B". Then the equations to the extremities of these three lines will be, A... a A'... 7 a A"... ( = a \y y y= J -^ (x = a+ C cX a (= + C x a1 + cf B... =,... y, B"... Hence we get for the equations of AA', BB', AA'..... = — a ( - a), a-,a a- a ' BB'...... y - b = (x- a - c): - a-a +c-c' TRANSVERSALS. EXPLICIT PARAMETERS. 31 whence we have for the equations to the point C", where these two lines meet, ac - ac =- Yc - c-c C C By symmetry we have also, ac - ac X -- ------ C0 C b"d' - b'c Y C, -C ac -ac and Ca ci c Le - Let the axis of y be so chosen as to pass through C and C': we shall then have ac -a ac - ac = —, - f = O C - C C -- or ac = ac", ac = a'c, whence also dc = ac', ac - ac and consequently = 0: c-c the point C" will therefore lie also in the axis of y. Durrande: Geerqonne, Annales de Mathematiques, tom. vii. p. 183. 9. If, from any point 0 in a given line of indefinite length, equal distances OPt, PP, PP,... be measured off, and other equal distances OQ, Q1Q2 Q2Q,... in the opposite direction; and if P,, P,, P.,... be joined by straight lines with a point E, and Q1, Q2, Q,... with a point F, E and F lying in a line parallel to the given line; then the intersections of lines of the former with lines of the latter group will all lie in 32 STRAIGHT LINE. straight lines forming two groups of lines converging to two points in the line passing through E, F. Let OPP,... be taken as the axis of x, and a line OA r Ii ei" -.es * F d -P _ through 0, perpendicular to the two parallel lines, as the axis of y. Let OP, = a, OQ, = b, OA = c, AE = m, AF= n. Then, the intercepts of the line through E, P., are, as may be readily seen from similar triangles, ra. c ra ---- ra - m hence the equation to EP, is -+ r —)-(1)r.................... (. a a c Similarly, putting s for r, - b for a, - n for rn, we have for the equation to FQs, - + s-b =.....................(2). At the intersection of (1) and (2) we have, adding their equations, /I 1\. m ny xa - + (r+s — = r + s...... (3), \a bl \ a c and, subtracting, /1 ' / m n y X + + r+ - -- =r.-s...... (4). \a 1 a b_ c The equations (3) and (4) shew that the intersections of the two groups of lines always take place in one of a single pair of lines so long as r + s or r- s is constant, there being as many such pairs as there are values of r + s and r - s. TRANSVERSALS. EXPLICIT PARAMETERS. 33 Suppose y c: then, from (3), mn n a b -1 i a b n n a b and, from (4), x= - -. a b These values of x, as not involving r or s, shew that the lines of intersection (3) and (4) all pass through two points in EFP at distances mn n m n _ + _ ab a b 1 1' 1 1-' a b a b from the point A. F: Cambridge Mathematical Journal, vol. I. p. 88. 10. To prove that the middle points of the diagonals of the three simple quadrilaterals, forming a complete quadrilateral, lie all in one straight line. The figure, of which A, B. C, D, A', B' are the angular points, is a complete quadrilateral, consisting of the three simple,'' c.-, -. quadrilaterals, A CBDA, of which the diagonals are AB, CD; A'DB'CA', of which the diagonals are CD, A'B'; AB'BA'A, of which the diagonals are A'B', AB. Let CA= a, CA' = a', CB = b, CB' = b'. Let M, M, be the middle points respectively of the diagonals AB A'B'. D 34 STRAIGHT LINE. Through these two points draw a straight line cutting CD in N. Let the lines OA', CB', be taken as axes of x, y, respectively. The equation to AB' is Y - $=1; a b and the equation to A'B is -+ -1. a b The equations to the point D are therefore, b-b' a —a' x = aa., y = b. ab - a'b' ab - ab The equation to CD will therefore be ad (b - b') y = bb(a - a') x. The equations to M are x = a, y= b; and those to M' are = ad, y = Ub'. Hence the equation to MMl' is 2 (a - ) y - 2 (b - ') = ab' - ab. Combining this equation with that to CD, we shall have for the equations to N, b- b' a -a = aa. b - a'b'' Y = ab - a'b The coordinates of N are therefore respectively half of those of D. The point N is accordingly the middle point of CD. Thus the middle points of the three diagonals are in a single straight line. Rochat: Gergonne, Annales de Mathematiques, tom. I. p. 314. 11. Having given the position of a point relatively to a known angle and in the same plane with it, to find in this plane two other points such that if we draw through them, in an TRANSVERSALS. EXPLICIT PARAMETERS. 35 arbitrary direction, two parallel straight lines cutting the two sides of the angle, the given point shall always lie in the direction of one of the diagonals of the trapezium intercepted between the parallels and the two sides of the given angle. Let the two sides of the angle be taken as axes of coordinates. Then the equations to a pair of parallels will be, (x' y') and (x"' y") being the two required points P' P", y y' = c (x - x'), and y - y = a (x - x"). The abscissa of the intersection of the former of these lines with the axis of x will be ax - y a and the ordinate of the intersection of the latter with the axis of y will be y" -a. Hence, (A, k) being the given point C, which by hypothesis lies in the line joining these two intersections, we have ah k -,, + 1 ax - y y - ax or' x.( - * ),2 a- {jy ( - h) + x" - kx'a + ya (y"- ) = 0. Since a is indeterminate, this equation resolves itself into the three following x" (x' - ) = 0, y (y" - ) = 0, y" (x -;) + x"y' -- x' = 0. The problem is therefore indeterminate, since there are only three equations between the four coordinates x', y ', x' y". The first two equations can be satisfied by one of the four systems of values f X' = oI \ /xk i - It ty - j 9 y-) 0Y, \y- =. It is easily seen that of these systems only the first and the last can satisfy the third equation. 36 STRAIGHT LINE. The first, which shews that the point P is in the axis of x, and the point P" in the axis of y, reduces the third equation to x' (y" - k) - hyk = 0; from which it appears that the two points P, P", are in a right line with the point C. Thus the required points will be the intersections of the two sides of the given angle with any line whatever drawn through the given point. The last system, which indicates that the required points lie in lines drawn parallel to the two axes through the point C, reduces the third equation to y - kx' = 0, or y - yx = 0, which shews that the points P', P", are in a right line with the origin. Thus we see that the points sought will be the intersections of a straight line, drawn arbitrarily through the summit of the given angle, with parallels to its two sides passing through the given point. J P / /eO o B' - - -o-K Schumacher: Gergonne, Annales de Mathematiques, tom. I. p. 193. Garier: GSom4trie Analytique, p. 53. 12. The equations to two lines being x+ -, 1 +Y-1 a ~ a b' to find the equation to a line passing through their point of intersection and through the origin of coordinates, a, b, a' b', being subject to the relation a + b = a' + b. TRANSVERSALS. EXPLICIT PARAMETERS. 37 The required equation is x _Y aa' b' 13. To find the equations to the diagonals of the parallelogram, the sides of which are defined by the equations x=a, x=a', y=b, y=b', and to determine the coordinates of the point of their intersection. The equations to the two diagonals are (a' - a) y - a'b = (b' - b)x - b'a, and (a - a') y - ab = (b' - b) x - a'b'; and the coordinates of their intersection ( (a+ a'), 1(b + '). 14. To find the condition that the three lines +Y, Y X X + Y -+ =1 -+ -1 F w+b =1 -a b a a b may all pass through a single point. The condition is expressed by the following equation, 1 1 1 1 1 1 $-~~,- + ~+ 0. a'b" d'b' db ab" ab' a'b Lamie: Gergonne, Annales de Mathematiques, tom. VII. p. 230. 15. If p1, p2, p8, denote the perpendicular distances of any point in a side of any quadrilateral from the three other sides; and a, b, assigned lengths; to prove that the equation Ax + Y=pt a b will represent one of a family of lines which all pass through a single point. 38 STRAIGHT LINE. 16. To find the equation to a straight line passing through the intersection of the two lines x cos + y sin= 8 xos' +ysin ' = 8" and cutting at right angles the line x cos a + y sin s" = "'. The required equation is x cos y sin - 8 x cos ' + y sin s' - 8' cos (s - ) Cos (" - ') O'Brien: Plane Coordinate Geometry, p. 31. 17. If, on the three sides of a triangle, taken in turn as diagonals, be constructed parallelograms, the sides of which are parallel to two given straight lines, to prove that the other three diagonals of these parallelograms will pass through a single point. Sturm, Vecten, Querret: Gergonne, Annales de Mathematiques, tom. xv. p. 103. 18. If, through the extremities of the base AB of any triangle ABC, be drawn straight lines AP, BQ, of arbitrary length, and respectively parallel to the sides BC, A C; and if, through the points P and Q, be drawn, parallel to BQ, AP, respectively, straight lines meeting in a point D; to prove that the three lines A Q, BP, DC, will pass through a single point. Gergonne: Annales de Mathe'ratiques, tom. xv. p. 88. 19. If any two straight lines be drawn, one of which bisects the other, and the straight lines joining their extremities be produced to intersect, to prove that the line joining the two points of intersection shall be parallel to the line bisected. 20. If OAA'A", OBB'B", be two straight lines, harmonically* divided, the former in the points A, A', and the latter in the * A straight line ABCD is said to be harmonically divided in the points B, C, if the whole AD: either extreme AB:: the other extreme CD: the mean BC. TRANSVERSALS. EXPLICIT PARAMETERS. 39 points B, B', to prove that the straight lines AB, A'B', A"B", will, produced indefinitely if necessary, either pass through a single point, or be parallel to each other. De La Hire: Sectiones Conicce, lib. I. prop. 18. 21. From three points A, B, D, in a straight line ABCD, straight lines are drawn through a single point P; through C is drawn a straight line ECF, parallel to AP, cutting PB produced in E, and PD in F. To prove that AD.BC: AB. CD CE: CF. De La Hire: Sectiones Conicce, Appendix, prop. 3. 22. From an angle C of a parallelogram ABCD, a perpendicular CE is drawn to the diagonal BD; to prove that perpendiculars drawn to AB, AD, at the points B, D, will intersect each other in the line CE. Leybourn: Mathematical Repository, New Series, vol. III. p. 179. 23. APB, CQDj are two parallel straight lines. Also AP:PB:: DQ: QC; to prove that the right lines PQ, AD, BC, meet in a single point. 24. Euclid I. 47. In the figure AL is perpendicular to BC; B er to prove algebraically that DB, EC, intersect AL in the same point. 40 STRAIGHT LINE. 25. If through the three angular points of any given triangle, lines be drawn parallel to the opposite sides, to prove that the straight lines joining their points of intersection and the opposite angles of the original triangle will intersect each other in a single point. SECTION VI. Transversals. Implicit Parameters. 1. To prove that the three lines bisecting the angles of a triangle all pass through the same point. Let the equations to the three sides be u = 0, v = 0 w = 0O where u, v, w, are of the form x cos s + y sin e - 8. Then, the origin being anywhere within the area of the triangle, the equations to the bisecting lines will be u - =0 v- 0 - = 0; these equations shew that the bisecting lines pass through a single point.* 2. If lines be drawn, bisecting the angles of a triangle and the exterior angles formed by producing the sides, these lines will intersect in only four points besides the angles of the triangle. Taking x cos a + y sin s - 8 = 0 as the general type of the equation to a straight line, we may represent the equations to the sides of the triangle by u = 0 v = 0 w = O and the equations to the six bisecting lines by the equations v + w = O...(1), w + u = 0...(2), + v = 0...(3), v - w =...(1)', w - u = 0...(2)', u - = 0...(3)'. * See problem (5) of this section. TRANSVERSALS. IMPLICIT PARAMETERS. 41 The pairs of lines (1) and (1)'; (2) and (2)'; (3) and (3)'; meet respectively at the angles of the triangle. Also the triple systems of lines {(1), (2), (3)'`} {(2), (3), (1)'}, {(3), (1), (2)'}, {(1)', (2)', (3)'}, meet each in a point. Now, generally, the number of intersections of six lines is equal to 15: hence, in the present problem, deducting from this number the number 3, for intersections at the angles of the triangle, and the number 2, for each of the triple systems which pass through only one point instead of three, we have, for the greatest number of intersections, 15 - 3 - 8 = 4. 3. From the angles A, B, C, of any triangle are drawn three straight lines AA', BB', CC', bisecting the angles of the triangle: through A, B, C, are also drawn three straight lines, respectively perpendicular to AA', BB', CC', to meet the sides BC, CA, AB, produced, respectively in G, H, K. To prove that GC,, K, lie in a right line. Taking x cos e + y sin s = 8 as the general type of the equation to a straight line, the equations to the several lines of the figure will be BC......... u=O, CA......... v = O, AB......... w =, AA'... v - w = O, BB'... w - u = O, CC'....u - v = AG... v + w = O, BH.... w + u = 0O CK......u + v = O. From the equations to BC and A G, it is evident that G will lie in a line, of which the equation is u + + w = 0. Symmetry shews that this line must contain also H and K. Vecten: Gergonne, Annales de Mathematiques, tom. x. p. 202. 4. To find the equation to a straight line which passes through a proposed point and through the intersection of the two lines u ==0 v = 0O. 42 STRAIGHT LINE. The general equation to a line passing through the intersection of these two lines must be u = Xv........................... (1), X being an arbitrary constant. This is evident when we consider that the values of x and y, which satisfy the equations u = 0, v = 0, must also satisfy the required equation; or, which is the same thing, that whenever u = 0 and v = 0, simultaneously, the required equation must be identically satisfied. Let u vu, be the values of u, v, when the coordinates of the proposed point are substituted for x, y. Then, since the equation (1) must be satisfied by these values of x and y, we have = Xv........................... (2). From (1) and (2) we have u V U1 VI the equation to the required line. Suppose for illustration the two proposed lines to be X Y1 X )l -+ = 1,,+ = 1, a a y and the proposed point to be a, /. Then the required line will have for its equation x Y x y -+ - 1 — BY,-1 a b a b cc =a a + - f,+ _- I a b a b 5. To prove that any three lines, which intersect in one point, may be represented by the three equations v- W - 0 O - U u-V = 0. Let the equations to any two of the lines be p=0, q=0: that to the third will be p + q = 0. TRANSVERSALS. IMPLICIT PARAMETERS. 43 Hence any three lines whatever, which cut each other in one point, may be represented by the equations p = 0, q = O, -p - Xq = 0; or, putting p = v - w, Xq = w - U, by the three equations v - w =O - u = -= - 0. 6. Three straight lines A, B, C, pass through a single point; X, Y. Z, are three other straight lines such that Y, Z, intersect in A; Z, X, in B; and X, Y, in C. To prove that the lines A, B, C, X, Y Z, may be respectively represented by the correlative system of equations v-w= O w-u=0 O u -v= 0 v + w=\ W+ u =, x u + v =X. The first three equations, as we know by problem (5), will always represent A, B, C. Again, since Y, Z intersect in A, the equations to Y, Z, must be of the forms p= 0, a + v - w =; or, putting a2p = X - u -v, of the forms U + v =_ X w + = X. Again, it is obvious that a line of which the equation is V + W = X intersects Y in B, and Z in C: this line must therefore necessarily be X. Hence the system of equations is A B C - =O w-u=O - =O X r z Iv+w=X w+u=X I u+v=X V + W ==X W +U =X U + V =X 7. HKL, PQR, are two triangles; to prove that if the straight lines HP, KQ, LR, meet in one point 0, the intersections of KL, QR; LH, RP; HK, PQ; lie in a straight line. 44 STRAIGHT LINE. Since OPH, 0 QK, ORL, all meet in a point, their equations may be expressed respectively in the forms - = 0, W - = 0, U -v==0. Again, since KL, LH; LH, HK; HK, KL; intersect respectively in ORL, OPH, O0QK, the equations to KL, LH, HK, are X = v + w, = w X=w + = u +v. For a like reason, the equations to QR, RP, PQ, are respectively X'=v+w, X'=w + u, X'=u+v. Hence KL, QR; LH, RP; HK, PQ; all intersect in a straight line line X = X'. Gergonne: Annales de Mathematiques, tom. VII. p. 187. Vincent: Cours de Geometrie, p. 210. Greathead: Cambridge Mathematical Journal, vol. I. p. 171. Frost: ib. vol. Iv. p. 113. 8. From the angles A, B, C, of a triangle, lines are drawn, through a point 0, to meet the opposite sides in E, F1 G, respectively: FG, GE, EF, are produced to meet BC7, CA, AB, respectively, in P, Q, R. To prove that P, Q, R, lie in one line. P? q --- Let the equations to BC, CA, AB, be = 0, v = 0, w=0. The equations to CG, BF, are respectively U V U1 V1 and UI W1 where uO, v1 wt are the values of u V: w, at the point 0. TRANSVERSALS. IMPLICIT PARAMETERS. 45 The equations to the points G, F, are therefore, respectively, u V 1 1 1 U W and.1 W V1 Hence the line GFP is defined by the equation v w u V1 W1 U1 Hence the equations to the point P are w - -0 P therefore lies in the line u v w BE w. W -+- + - = 0. "u VI WI Symmetry shews the same to be true of Q and R. Hence P, Q, R, lie in one line defined by this last equation. We subjoin also another solution of this problem. Since AE, BF, CG, meet in a point, their equations may be written respectively in the forms AE v-w =...................... (1), BF w - u= 0......................(2), CG u - v = 0......................(3). Since AB, AC, intersect in (1); BC, BA, in (2); CA, CB, in (3); the equations to CA, AB, BC, will be respectively CA w+ u=2X....................(4), AB u+ v=2X................... (5), BC v + w = 2.....................(6). 46 STRAIGHT LINE. Since FG passes through the intersections of CA, BF, and of AB, CG; GE, EF, passing through analogous points; the equations to FG, GE, EF, will be respectively FG = x........................ (7), G E v = x.........................(8), EF w = X.........................(9). Hence P, Q, R, the intersections of (6), (7); (4), (8); (5), (9); all lie in a line u + v + w = 3X. F: Cambridge Mathematical Journal, vol. I. p. 87. 9. To prove that, the circumstances of the preceding problem remaining unchanged, the lines joining B, Q; C, R; A, E; meet in one point, as also those joining C, R; A, P; B, F; and those joining A, P; B, Q; C, G. The equations to GF, BC, AB, CA, being GF u = X, BC v + w=2X, AB u+ v = 2X, CA w+ u=2X, it is plain that the line denoted by the equation 2u + v + w = 4X passes through the intersection of GF, BC; and of AB, CA; the line is therefore the line joining A, P. The equations to AP, BQ, CG, are therefore, regard being paid to symmetry, AP 2u + v + w = 4X, BQ 2v+w + = 4X, CG -v=0. Since any one of these equations results from the other two, these three lines must all pass through one point. The same thing must, by virtue of symmetry, hold good respecting the intersections of the similarly related lines. TRANSVERSALS. IMPLICIT PARAMETERS. 47 10. Let there be any complete quadrilateral; let its sides be produced indefinitely; let its three diagonals be also indefinitely produced: the diagonals will intersect, two and two, in three points. Through each of these points let there be drawn straight lines to the two extremities of the diagonal in which it does not lie; we shall thus have six straight lines, each of which will determine two points on two sides of the quadrilateral; so that we shall have in all twelve of these points, distributed, three and three, on the four sides of this quadrilateral. It is required to prove that these twelve points will lie, two and two, on twelve new straight lines, meeting, four and four, in the three points of intersection of the diagonals of the proposed quadrilateral. Let AA'A"B'BB" be the proposed quadrilateral, of which the diagonals are AB, A'B', A"B", intersecting, AB and A'B' in C", AB and A"B" in C', A'B' and A"B" in C. Let the point C be joined to the points A, B, by two straight lines, of which the former cuts the sides BA", BA', in n, q, and the latter the sides AA", AB', in m, p. Since the construction would evidently be the same for the point C', relatively to the diagonal A'B', and for the point C", relatively to the diagonal A"B"; it will be sufficient to prove, first, that the lines np, mq, meet in the point C', and, secondly, that the lines mnn, qp, meet in the point C". Since the three lines AB', A'B, A"C', meet in one point, we may take for their equations AB'.........u - v = 0, A'B............ v - w =, A"C'........... w - = 0. 48 STRAIGHT LINE. Again, since A"B, AA", AB, intersect, two and two, in ABf A'B, A"C', their equations will be A"B............u + v = 2X AA".......... v + w = 2X, AB............w + u = 2X. Since A'B' passes through the intersection of AA", A'B, and of A"B, AB', its equation will be A'B'...........v = X. Since An passes through the intersection of AB, AA", and of A'B' A"C', its equation will be An............ - u + 2v + w = 2X. Since Bm passes through the intersection of A'B', A"C', and of AB, A"B, its equation will be Bm............u + 2v- w = 2X. Since mn passes through the intersection of Bm, AA", and of An, A"B, its equation will be mn............u + 4v + w = 6X. Since pq passes through the intersection of AB', Bn, and of A'B, An, its equation will be pq............- u + 4v - w = 2X. At the intersection of mn, pq, we have, from their equations, w+u=2X their intersection therefore coincides with C", the intersection of AB, A'B'. Again, since np passes through the intersection of A"B, An, and of AB', Bin, its equation will be np............3u - w = 2X. Since mq passes through the intersection of Bmn AA", and of A'B, An, its equation will be mq............- u + 3w = 2X. TRANSVERSALS. IMPLICIT PARAMETERS. 49 At the intersection of np and nq, we have, from their equations, w - U = 0 ) w+u=2X their intersection therefore coincides with C', the intersection of AB, A" C'. COR. If r, t, be the intersections of mn, pq, with A"B", and s, v, those of np, mn, with A'B'; rs vt, will intersect in B, and rv, st, in A. Availing ourselves of the preceding equations, we shall easily obtain for the equations to rs, vt, rv, st, rs............ 3u + 4v- w= 6X, t............ u + 4v - 3w = 2X7 rv............- u + 4v + 3w = 6X, st..........- 3u 4v + w = 2X. From these equations the truth of the proposition asserted in the corollary is easily seen. The truth of the corollary is also evident from the consideration that the four points r, s, t, v, have the same relation to the quadrilateral mnpq, which the four points m, n, p, q, have to the quadrilateral A"A'B"B'. Legrand, Rochat, Penjon: Gergonne, Annales de Mathmatiques, tom. II. p. 369. The reader is referred also to a memoir by Vecten, in the 15th volume of the Annales, p. 146, where he will find other properties of the complete quadrilateral. 11. Let AB, A'B', A"B", be any three parallel lines. Let C, C', C, be the points of concourse respectively of A'A", B'B", of A"A, B"B, and of AA', BB'. Then the three points C, C', C", will lie in a right line. The lines AB, A'B', A"B", A'A", B'B", AA', BB', may be represented by the system of equations AB......... u = c A'B'..........u = c' AB".......... = c"; A'A".........v = O B'B"........w = O; AA'.........u - v = c BB'.........u - w = c'. 50 STRAIGHT LINE. The equations to the points A, A", will then be A...ac,} A". {U = % } v =c-c v =c Hence the equation to the line A"A will be A"A.........(c' - c)u + (c - "')v = c" (c - c). Again, the equations to the points B, B", will be B............. B" W, - eC - C=0 The equation to the line B"B will therefore be B"B......... (c' - c) + ( - c")w = c (c - c). From the above equations it is plain that the intersections of A'A", B'B"; of A"A, B"B; and of AA', BB'; will all three lie in a line of which the equation is v = W. Durrande: Gergonne, Annales de Mathenatiques, tom vII. p. 183. 12. Let ASB, A'S'B', A"S"B", be three angles of which the sides are respectively parallel. Let S'A', S"B", meet in M) and S'B', S"A" in N: let S"A", SB, meet in M', and S"B", SA, in N': let SA, S'B', meet in M", and SB, S'A', in N". Then the three lines MN, M'N', M"N", will meet in one point. Let the equations to the sides of the three angles be AS.........u = a, BS.......v = 3; A'S'..... = a, B'S'......... v=; A"S".........u = a', B"S".........v = '. Then the equations to the points M, N, will be M X...... a} N i........ The equation to MN will be, therefore, MN........ (1' - ) + (a - a") v = al' - a". TRANSVERSALS. IMPLICIT PARAMETERS. 51 By similarity, for the equations to M'N', M"N" we have M N......... (o3 - 3) ( - a) a " - M"T".......... ( - 3') u + (a - a') v = at - ao'f'. Adding together these last three equations, we obtain an identical equation; the three lines MN, M'N', M"N", therefore all meet in one point. Durrande: Gerqonne, Annales de Mathmnattiques, tom. VII. p. 183. 13. If three straight lines, drawn from the summits of a triangle, meet in one point, their respective parallels, drawn from the middle points of the opposite sides, also meet in one point. The equations to the three former lines may be written in the forms v - w == p v............ OA w - u = v - X............ OB, u - v= X -............OC; and the equations to the three latter, being respectively parallel to the three former, in the forms v -w =...(4), w - u = m...(5), u - v = n......(6). In these six equations u, v, w, are supposed to denote linear functions of x and y without constant terms; 1, m, n, X, /A, v being all constants. Let,, w,, be the values of v, w, at the middle point of BC; w,, u,,, those of w, u, at the middle point of CA; and u,,,, v,,,, those of u, v, at the middle point of AB. Also let (u', v', w), (u", v", w"), (u"', v"', w"'), be the values of (u, v, w), at the points A, B, C, respectively. Then it is plain, from the geometrical conditions, that 2v, = v" + v"', 2w, = w + w"'; 2w-, = w' + w' 2u, = u"' + U'; 2%, = u/ + u", 2v,,. = v' + v. But, from (4), (5), (6), we have 2 (1 + + +n) = 2 (v, - w) + 2 (- ) + 2 (,,- v,,,); E2 52 STRAIGHT LINE. hence 2 (1 + m + n) = v"-w" + v"'-w"' -w - "-'+w -u' + ' - v' + /" - v =v -u +w -v' + -W. But, from the equations to OA, OB, 0C, v - w = -v, w - = v-X, u"' - = X - hence 2 (l+ +n) = /- X+v- / +X-v and therefore 1 + m + n = 0. This result shews that the lines (4), (5), (6), pass through one point (('). CoR. It is easily seen that the equation to the line joining 0, 0', if we put I = '-v, m = v-' n= ' - ', is (X'- X) (( - v) + (u' -, () (v - X) + (' - ) (X - ) = (' - ) ( - w) + (,a' - b^) (w - u) + (V - v) (u - v) = ( - ) (0' -v') + (M' -L) (V- (V' - V) (X - '). Fregier: Gergonne, Annales de Mathematiques, tom. VII. P. 170. 14. To prove that the sides A, A', A", A"',... of any polygon, and the sides X, X', X", X"',... of any polygon inscribed within it, may be represented by the system of equations A A A" A"' A =0 u =0 '0 =...... =0 X....... a+ ' + u"+......+ u() = 0 X......au + au' + U' + u" +......"+ (" = 0.......au + '+ uau" + +.....+ u( =.0.................... + +................................+ 0.................................................................... X)...... au + au' + au" + au"' +......+ au(") = O. TRANSVERSALS. IMPLICIT PARAMETERS. 53 Let the sides A, A', A"7...... of the polygon be represented by the equations A= 0 A'= 0 A"= 0......) and the sides X, X', X"...... by the equations X=0, X'=0, X"= 0....... Since A, X, X'(), pass through a single point, X(") = XA + /.X, X and / being constants. In like manner X = XA' + utX', X' = X2A +4 XI X" = A"' + /kX,........................e....-.............................. X(8-l) = XA(") + X(n). Multiply these n + 1 equations in order by 1, e, 1A7 P,,27................ h1.2-; then, representing X("' /1X, X'XI, *......... 4.........- X(n-1)> by ) yY Y',......... YOl) and XA, IX1A', AuqX 2A"......... * / *........ * 1 * XnA( )) by ) ' u......... A'(n we have, putting a for P/d1l'2....../P,, y() =u + Y, Y =u' + Y, Y' =u" + Y", Y.. U"' + Y"'......................... Y("-l) = u~) + aY(,o. 54 STRAIGHT LINE, From these equations we have (l-a) Y("' = u+ u' + U +......+ (1- a) -Y au + u + u+ +......+. uz~ (1- a) Y' = au + a u+ u" +......+ u("..=..U+..............,.......,..,............... o,, o..o o.o o.ooo,.oo............................ ooo (1 - a) y(a-l) = au + aCu + au" +......+ au") Since u, u,' u"...... are constant multiples of A, A', A"...... and the right-hand members of the last n equations are constant multiples of y(n)" Y, Y...... and therefore of X(n, X X'...... the truth of the proposition is established. Ex. Having given the equations to the sides of any quadrilateral, to find the equations to the sides of an inscribed quadrilateral the opposite sides of which intersect in two given points. Let the equations to the sides of the original quadrilateral be k = 0, =0, r= 0, n = 0. Then) since u, u', u" u "' must be some constant multiples of k, 1, mn n, respectively, the equations to the sides of the inscribed quadrilateral will be of the forms ak + /i3l+ /m + 8n = 0.................(1), (k + /3) + ym + n = 0..................(2) a(k + / + rym) + an = 0..................(3), k + 13 + 7r + an = 0..................(4). Let the values of k, 1, m, n, be k1, 1, m,, n1, and k2, 12, m, n2, at the two given points. Then we shall have, for the determination of the constants ak + 1 + 7yn + 8n, = 0..................(5), a (k, + 12) + y + n = 0..................(6), a (k1 +, + y1m,) + n = 0................(7),, 31 +, +,, = 0..................(8). TRANSVERSALS. IMPLICIT PARAMETERS..55 From (5) and (7) we have /1l + /m, = 0, a.k + Sn, = 0; and, from (6) and (8), k2 + 1 3 = 0, rm, + 8n, = 0. Substituting the values of a, 3, ry, 8, given by these four equations, in (1), (2), (3), (4), we shall obtain, for the required equations to the sides of the inscribed quadrilateral, nk - kin lmr - mJl 2 71 1 2 k112 m21% + = O. km - ml n % - kn 12m1 nLk2 = mn - n km kl - l2k + =0. mn2 k211 Cambridge and Dublin Mathematical Journalfor May 1850. 15. If the lines u = 0, v = 0 be parallel, to shew that, k being an arbitrary constant, u +k v = is parallel to each of them. 16. If, through the three angles of a triangle, straight lines be drawn at right angles to the base, to prove that, the general type of the equation to a straight line being (Xz - ) y - (y2 - y) x = X2Y1 - XY2 the equations to the three sides of the triangle and to the three perpendiculars may be represented by the system of equations uM = w 01 = 0, W = 0, v —w= 0, w —u = 0, u - = 0. 17. A straight line passes through the intersection of two lines u = v0 = 0; 56 STRAIGHT LINE. and through that of two lines ' = O v' = 0. To prove that it is impossible to express generally the equation to this fifth line in terms of u, v u', v'. 18. If lines be drawn through any two of the points A, B, C, &c., and other lines through any two of the points a, b, c, &c.; to prove that the intersections of AB with ab, of AC with ac, &c. will lie in one straight line, provided that the lines through the intersections of any two of the first series of lines and the corresponding intersections of the second series all pass through the same point. 19. The four angles of a quadrilateral ABCD are bisected by four straight lines: the bisectors of A, B, meet in a; of B, C, in /; of C, D, in ry; and of D, A, in 8. To prove that the directions of 88, ay, pass through the intersections of the directions of AD, BC, and of AB, CD, respectively. 20. In any quadrilateral ABB'A', let the diagonals AB', A'B, be drawn, intersecting in B", and the sides AA', BB', produced to meet in A": also let A"B" be drawn, cutting AB and A to meet in A": also let A"B" be drawn, cutting AB and AB' in C' and C respectively. To prove that, a being the intersection of the diagonals of the quadrilateral A C'CA', and / that of the intersection of the diagonals of the quadrilateral BC'CB', the three points a, B", /3, will lie in one straight line. RECTILINEAR AREAS. 57 21. A is any point whatever in the plane of a triangle BCD: from A are drawn to B,, D, the \ three straight lines AB, A C, AD. On AB, A C AD, are taken any points mn, n, p, respectively; and \ Bn, Cm, C p, Dn, Dm, Bp, are joined: through the points of / intersection D', B', C' are drawn the transversals Ad, Ab, Ac. To prove that r (1). The transversals Dd, Bb, Cc, cut each other in a single point A'. (2). The four transversals AA', BB', CC', DD', pass all through a single point. This is a particular case of a theorem in Carnot's Essai sur la Theorie des Transversales. Cambridge and Dublin Mathematical Journal, for Feb. 1850. SECTION VII. Rectilinear Areas. 1. To find the area of the figure included by the lines defined by the equations x-y =, x+y=0, x- y =a, x+y=b. The construction of the four lines shews that they form a rectangle, the first and third being two parallel sides. Now the distance between the first and third, which is the same thing as the distance of the origin from the third, is equal to. V/2 Similarly, the distance between the second and fourth is equal to Hence the area of the figure is equal to ab, V22 58 STRAIGHT LINE. 2. From the origin of coordinates, 0, perpendiculars 0 C, 0 C', are let fall upon two straight lines x y x+1 -a+ = 1,+ = y, respectively: the points C, C', are joined: to find the area of the triangle COC'. It is plain that B~ sin z CO C' = sin L C OA'. cos / COA B - sin COA cos z C' OA', = cos A' sin A - cos A sin A', 0 A - a' - a/3 (a2 + /'2) * (a2 + 13'2) -(a2 + 12) (a2 + /3'P) ' _ ^73 - ^/ (a2 + /2). (a" +/") ' OB c a Also CO-= OA. AB = (a ) and, similarly, C'O = '2 ( +1 /Ia'()8 - a~ ) Hence area CO C' -= I f) (aa" 2 ) 3. To find the area CC' C", included between the three lines AB, A'B', A"B", the equations to which are respectively x Y x x x y -+ = 1 -5, + = -+ 1. (x - a Z - 1 a b -o RECTILINEAR AREAS. From the geometry we see that CC'C" = AA'- AC'CA', A C"A' + A'CA"- A"C'A. Now area A C"A' = I AA' x ordinate of C" 1 1 a ab ab' a'b bb' -= (C'a -a) - a. In the same way it may be proved that area A' CA"= (ad - a b'),-b - b b" bb" and area A" C'A (a" - a) " - ). 2 s / a~b - ab" Hence area CC' C" is equal to a- a)' (a - a (a-) a ca ca a a a'.b' b b" b' b b J This area may similarly be shewn to be equal also to f(' - b)2 (- - b')' ( - b")2 b +b' b" ~ b" _a a a a a I a a a a a a J 4. To find the area of a triangle, the angular points of which are given. Let A1A2A8 (fig. 1) be the triangle, the coordinates of A., A2, A8, being respectively (X? Y1), (X2) Y2) ()3, Y-)Then the area of the triangle is equal to AA,2H2HI + A2AAH2f - A1A fI3HH = (X2 -x1) (Y2 + Y1) + (- -x,) (: / -KY) - coordinates of the Fig. 1.. I -A, I Ir o0 ff, - 7 -.TZ ~s (j3- C1) (y3 + ), 60 STRAIGHT LINE. whence, P denoting the required area, 2P = xy, - xy2 + xy, - X2y3 + X1Y3 - X3,y Fig. 2. y 2 -A' 14 2 3 Had the figure been such as fig. (2), we might have shewn in a similar way that 2P - xY2 - xy + x2y3 - x3y2 + x3y1 - xy3. Thus we see that the area of any triangle, the angular coordinates of which are (x,, y), (x, y2), (x3, y3), is equal to ~+ -{X2Y - X1Y2 + 3y - xY32 + x1Y3 - 3}, the + or - sign taken accordingly as the point A, and the indefinite point x are on the opposite or on the same side of the indefinite line through A1, As. Puissant: Recueil de diverses propositions de Geometrie p. 115, troisieme edition. 5. To find the area of a polygon the angular coordinates of which are (x,a y), (x, y2), (x3, y3)......(x, y.). fig. 1. fig. 2. fig. 3. i0~. ~-_ - 0 Taking a quadrilateral, we have, for the area of the triangle A,12AA in fig. (1) and (2), the expression -1(X2Y - X2 + 2Y - X2Y3 + x1Y3 - XY) and, in fig. (3), the expression I(x1Y2 - X21 + X2y - X3y2 + x3Y1 - xy3) RECTILINEAR AREAS. 61 Again, the area of the triangle A1A3A4 is equal, in fig. (1) and (3), to the expression 1 (X3Y, - XY3 + X - X34 + x y4 - X4Y); and, in fig. (2), to the expression (XY3 - X31 + XY4 - X43 + 4Y1 - x1Y4). Hence the quadrilateral area A1A,2AA4 is equal, in fig. (1) and (2), to the expression (x2Yl - x1Y2 + x3Y2 - X2y3 + x4y3 - x3y4 + 1Y4 - x4Y1) and, in fig. (3), to this expression affected by the negative sign. Thus we see that the area of a quadrilateral A1A2A3A4 is equal to ~+ (x, - xy2 2 + X Y2 - x2Y3 + x4S - XY4 + xY4 -41) the + or the - sign being taken accordingly as the point A2 and the indefinite point x are on the opposite or on the same side of the indefinite line through A1, A,. Proceeding in the same way, we shall evidently obtain, for the area of the proposed polygon, the expression + (xy - xY2 + xy- x2y, + X4y3 - X3y+ 3 4 +......+ xIn - xYl) De Stainville: Gergonne, Annales de Math4matiques, tor. I. p. 190. Puissant: Recueil de diverses propositions de Geom4trie, p. 116, troisieme edition. 6. To find the area of the quadrilateral the equations to the sides of which, taken in order, are = 0.............................(1), y = x............................ (2), (y - a) /3 = x - a................... (3), y = x - c......................... (4). At the intersections of (1) and (2), (2) and (3), (3) and (4), (4) and (1), respectively, we shall have cV3 X a + --- - 01 a C J4 ~ \Y1 = 0J \z = ad t3, + c ) a= _ OV. lY3 = a + 2/3 _ 1l 62 STRAIGHT LINE. But, P denoting the required area, 2P = x2y, - xy2 + X32y - X2y + X43 - xy4 + xY4 - x4Y: hence, putting for the coordinates in this expression their proper values, we have 2P=a a+ /3 - a a+ 3_1) +c (a+ / 2ac + P= ac + 2 (V3 + 1). 7. To find the area included between the lines x x = = 0 = 0, + 1. x=O, y= 2 3 Required area = 3. 8. To find the area of the triangle formed by the three lines x = a, y = mx, y = - mx. Required area = ma2. 9. To find the area included between the four lines x=a, x=a' y=b, y =b'. Required area = (a' - a).(b' - b). 10. To find the area of the triangle included by the three lines y =,,y=aa, y = ax. Required area = ib2 (- Id - 11. To find the area of the triangle formed by the line x+y= 1, a perpendicular upon it from the origin, and the axis of x. Required area = }. 12. To find the area included between the three lines x= 0, ax+y= 3, 2a + y = 23. Required area =2a RECTILINEAR AREAS. 63 13. To find the area of the triangle included between the three lines x + 2y = 5, 2x + y = 7, y= x + 1. Required area = i. 14. To find the area included between the lines of which the equations are xz y x = i = 1_ a l' a '' a =b The required area is equal to 2a af'' - b 2 (a/' + b').(a/3 + ba) 15. To find the area included between the lines of which the equations are x=O0 y=O, x=a, y=b. The area is equal to ab. 16. To find the area of the quadrilateral included between the four lines _ Xthe four lines = a, y = ax, y = x. Required area = I (a'2 a2).(a' ' a). 17. To find the area included between the lines of which the equations are X y X =0O, - = + - =+ =1 +. a b a ' The area is equal to (ab db'). 18. To find the area of a parallelogram, the equations to the sides of which are lx + my ~ 8 = 0, 'x + n'y + 8' = 0, o being the angle between the coordinate axes. The required area is equal to 488' sin co lm', I'm 64 STRAIGHT LINE. 19. To find the area of the quadrilateral included between the four lines y = ax, y = a'x, y = mx + c, y = nx + c'. Required area = (a - ). -m'(- n') - ( -m).(-),,- ( 2 - v ) * ()(a -m ) (am).(-An) 20. To find the area of the quadrilateral included between the four lines x + y = 4, -3x +2y= 5, 12x +7y=-35, 4-5y = 25. Required area = 30*8. 21. If x,, Y1; x2, y2; x, y;...... (xl, yj); denote the coordinates of the angular points of a polygon taken in order, the axes being inclined to each other at any angle o, to prove that the area of the polygon is equal to + i-sin ) {2Y1 - x1Y2 + X3- X-23+Z 3 -X3Y4 +..+ x1y,-Xy1}-. 22. To find the equations to straight lines passing through a given point and including a given triangular area between given axes of coordinates. Let a, b, be the coordinates of the given point, a the angle between the axes, and c2 the given area. Then the equations to the lines, which satisfy the conditions of the problem, will be + 2c2 (x - a) (y - b) + sin a {a (y - b) - b (x - a) }2 = 0. Puissant: Recueil de diverses propositions de Geometrie, p. 139, troisieme edition. 23. To bisect a trapezium A OBC, the angles A, 0, of which are right angles, by a straight line at right angles to OA. Let OA = a, CA = h BO = h' and, the lines OA, OB, being taken as axes of x, y, respectively, let x be the abscissa of the bisecting line. Then X a - h' th2 + h'2\ h-hA -- -V 2.) Puissant: Recueil de diverses propositions de Geometrie, p. 160, troisieme edition. RECTILINEAR AREAS. 65 24. A, B, are two given points; P is a point so placed that the sum of the squares of the distances PA, PB, bears to the area of the triangle APB the ratio of f to g, and that the angle APB has a given value. Supposing the middle point 0 between A and B to be the origin of rectangular coordinates, OA produced indefinitely being the axis of x, to find the coordinates of P. Let QA a = OB, L APB = a. Then x = + "a -. (f 2 sin2 a-16g2)4, x sin a - 4g cos a (f 4ag sin a fY sin a - 4g cos a L' Hospital: Traite Analytique des Sections Coniques, p. 362. 25. To find the lengths of the sides AA', A'A", A"A, of a triangle and its area, the equations to its sides AA', A'A", A"A, being respectively x cos a"+ y sin a"= ", x cos a + y sin a = cos a'+ y sin a'=. The required lengths are A'A = s" sin (a - a') + 8 sin (a' - a") + S' sin (a" - a) sin (a - a").sin (a' - a") sin(a' - a") + 8' sin (a' - a) + 8" sin (a - a') sin (a' - a).sin (a" - a) ' sin (a" - a) + 3" sin (a - a') + 8 sin (a' - a") A"= sin (a" - a').sin (a - a') and twice the area of the triangle is equal to { sin (a' - a") + 8' sin (a" - a) + '" sin (a - a') '2 sin (a - a'). sin (a' - a"). sin (a" - a) Lhuilier: Eldmens d'Analyse Geometrique et d'Analyse Algebrique, p. 116. F ( 66 ) CIRCLE. SECTION I. Referred to two Perpendicular Diameters. Tangents. 1. To find the relation among the quantities a, b, c, that the line y -+ - 1 a b may be a tangent to the circle x2 + y2 = 2. If a be the inclination of the radius of any point in the circumference to the axis of x, the equation to the corresponding tangent is z cos a + y sin a = c. In order that this equation may coincide with that to the given straight line, we must have cos a 1 sina 1 c a c b' Squaring and then adding these equations, we get, for the required relation, 1 1 1 2. a+ c 2. Tangents are drawn to a circle x2 + " = 2 at two points (X': y'), (x", y"): to find the distance of a point (h, k) from a line passing through the centre of the circle and the intersection of the two tangents. The equations to the two tangents are xx + yy = c2 xx" + yy" = c. PERPENDICULAR DIAMETERS. TANGENTS. 67 Subtracting one of these equations from the other, we get for the equation to the line, which passes through the centre and the intersection of the two tangents, ( ) y' - y") + y ('- Y") = 0. Let 8 denote the distance of (A, k) from this line; then {2 h(x'-x") + ~ (y'- 'Y)2 (x, - x,) + (y - y') 2 3. To find the equations to straight lines touching the circle x2 + y2 = 10 at points the common abscissa of which is unity. The required equations are x + 3y = 10. 4. To find the equation to a circle, the centre of which is at the origin of rectangular coordinates, and which is touched by the line y = 2x + 3. The required equation is x2 + y2 =. 5. To find the intercepts on the axes of coordinates of the tangent to a circle drawn parallel to a given straight line. If c be the radius of the circle, and 1, m, the direction-cosines of the given line, the required intercepts are equal to c, C. Puissant: Recueil de diverses propositions de Ge'ometrie, p. 148, troisieme edition. 6. If, from a given point S, a perpendicular SY be drawn to the tangent PY at any point P of a circle of which the centre is C, and, in the line MP at right angles to CS and produced if necessary, a point Q be always taken such that MQ = SY to find the locus of Q. F2 68 CIRCLE. Let C be the origin of coordinates, CS, produced indefinitely, being the axis of x; then, denoting CS by a, and the radius by r, the equation to the locus of Q will be ax - ry = r2. SECTION II. Referred to two Perpendicular Diameters. Chords. 1. If a chord be drawn parallel to a diameter of a circle, and from any point in the diameter lines be drawn to its extremities; the sum of their squares will be equal to the sum of the squares of the segments of the diameter. Let the chord QR be drawn parallel to the diameter AB; let Q, R, be joined by straight lines to any point P in AB. Draw QM, RN, at right \ angles to AB. -t A 7 Let QM= y = RN, CP= a, r = the radius of \ f the circle, CM = x = CN. Then pQ' + R2 = y2 + (x - a)' + y2 + (x + a)2 = 2 (X2 + 2 + a2) = 2 (r + a2) = (r - a)2 + (r + a)2 = PA' + PB2. 2. To prove analytically that the angle in a semicircle is a right angle. Let (a, b), be the coordinates of one end of the diameter of the semicircle; (- a, - b) will be those of the other end. The equations to the two chords will be x-a y-b 1' m t ',7~ ' PERPENDICULAR DIAMETERS. CHORDS. 69 At the intersection of these two lines xd~~~~' - ~~a~ a2 -b'. (1). aF = y,*....................(1. it' mm.n But, c being the radius of the circle, x2 + y' = c, a2 + b' = c, and therefore 2 _- a2 + y2 b2 = 0..................(2). From (1) and (2) we see that II' + mm' = 0, the condition for the perpendicularity of the two chords. 3. C is a fixed point in the diameter BA, produced, of a circle; from C any line CLK is drawn cutting the circle in two points L, K; AK, BL, are joined, inter- secting in P; to find the locus of P. P Let OCx be taken as the axis B o _ C of x, and Oy, at right angles to it, as that of y. Let a denote the radius of the circle. Then the equations to AK, BL, will be, respectively, of the forms y = m (x-a), y = n (x + a); both of these equations being comprehended in the following one, (y - mx + na). (y - nx - n) = 0, or y2 + mnx - (m + n) xy + a (m - n) y - mna=...(1). The equation to the circle is 2 + y =........................ (2). Eliminating y2 between (1) and (2), we get, relatively to the points of intersection of AK, BL, with the circle, y2 - ny - (m + n) xy + a (m - n) y = 0; whence, dividing by y, we have (1 - mn) y - (en-,+ ) x + (m - *n) a = O...... (3), 70 CIRCLE. as the equation to the chord KL. Let 00 = c; then, since KL, produced, passes through C, we have, from (3), (m - n) a = (mn +n) c...............(4). At the point P, in which AK, BL, intersect, we have m (x - a)= n (x+ a).................. (5). From (4) and (5) there is a (x + a) - (x - a)} = c {(x + a) + (x - a)}, whence x = - c which is the equation to the locus of P, being therefore a straight line at right angles to the axis of x, passing through the point at which a tangent from C touches the circle. 4. To find the length of a chord of the circle x2 + 2 = c' the equation to the chord being -+ =1. a 6 The required length = 2 (c-2 a 2+ 2 5. If 2a, 2a', be the inclinations of two radii of a circle x2 + y = C2 to the axis of x, to find the equation to the chord joining the extremities of the radii. The required equation is x cos (C + a') + y sin (a + a') = c cos (a - a'). 6. From any point Q in the circumference of a semicircle two chords are drawn to the extremities A, B, of its diameter. From any point M in this diameter a perpendicular to it is drawn, which cuts A Q in 1H BQ in K, and the curve in P. To prove that IMH. MK = (MP)2. PERPENDICULAR DIAMETERS. 71 7. If H represent the harmonic mean between the abscissae, and K that between the ordinates, of the points, in which a circle x2 + y/2 = C2 is cut by a chord x + my = 8, where 1, m, are the direction-cosines; to prove that HI + Km = 8 - c28' 8. To determine, in the circumference of a given circle, a point P, such that, if it be joined to two given points A, B, the lines AP, BP, may intersect the circumference in the extremities A', B' of a chord A'B parallel to AB. The equation to the circle being x2 + yr = c2 and (a, /), (a', /'), being the coordinates of A, B, respectively, the required point P will be determined by the intersection of the given circle with the straight line defined by the equation ax + 3y - C2 a'x + y - c' a2 + /2 c2 a '2 + 2 2- c2 This is a particular instance of the more general problem, "To inscribe in a given conic section a triangle the sides of which are subjected to pass through three given points." Lamd: Examen des diferentes methodes employees pour resoudre les problemes de Geometrie, p. 74. SECTION II1. Referred to two Perpendicular Diameters. Points. 1. Having given two points A and B, to find the position of a point P in the circumference of a given circle, such that, if AP, BP, be joined, and P be joined to the centre C of the circle, sin CPA m sin CPB n 72 CIRCLE. Let Cx, parallel to AB, be taken as the axis of x, and Cy, at right angles to Cx, as that of y. Let (a, I8), (a', 8/), be the coordinates of A, B, respectively, and x, y, those of the required point P. Let c be the radius of the circle. Then the equation for determining x or y will be an equation of the sixth degree resulting from the elimination of y or x between the two equations x2 + y = c2, and n. ('y - 3_x) {(a - x))2 (f - y)} - m.(- cy + /) {(a - _ )2 + ( - y)2}. Descartes: Lettres, Lettre 65, tom. III. p. 357. Puissant: Recueil de diverses propositions de Geometrie, p. 168, troisinme edition. 2. The distances of a given point C from two points P, Q, in the circumference of a circle x2 + y2 = r2 are each equal to nr: to find the equations to the two lines PC, QC. If (a, b) be the coordinates of C, the equations to PC, QC, are included in the equation {a2 + b2 + (n2 - 1) 2}2 {(X - a)2 + (y- b)2} = 4nr. {a (x - a) + b (y - b)} SECTION IV. Referred to any Rectangular Axes. 1. To determine the relation among the constants of the two equations Ix + my = 8, (x - a)2 + (y - b) = C in order that they may touch each other REFERRED TO ANY RECTANGULAR AXES. 73 The coordinates of the centre of the circle are a, b, and therefore the square of the distance of the centre of the circle from the straight line is equal to (al + bm- 8)2 12 + m2 But this distance, the line being a tangent, must be equal to the radius: hence (al + bm - 8)2 = (12 + m2) C2 which is the required relation. COR. If 1, m, be the direction-cosines of the straight line, 12 + m2 = 1 and the relation is reduced to + (al + bm - 8) = c. 2. To find the equation to a circle having for a diameter the line joining the points of intersection of the line y = ax and the circle y = 2cx - 2 At the intersections of the line and circle ax2 = -2cx - X2 and therefore the coordinates of these two points are. =0\ Xy = o2 2c ] and 2ac Let x', y', be the coordinates of the centre of the new circle, and c' its radius. Then - c = - 2 X 2 C / 1- - ac ac a = 2y2 = 1 +~ C2 and c'2 =x' +/ y" I a+ 74 CIRCLE. Hence the equation to the new circle is C \2 / ac 2 c2 x l-+a2) + - l a2 la-2 a2 1 + (2l or (1 + a2). (2 + y2) = 2c (x + ay). 3. A straight line cuts two given circles in such a manner that the parts of it intercepted by them are of given lengths: to find the point in which this line cuts the line joining the centres of the two circles. Let the line joining the centres of the circles be taken as the axis of x, and one of their centres as the origin of coordinates. Their equations will then be of the forms 2 y = C2..................... (1), (x + a)2 + y2= C2..................... (2). The chords intercepted by the two given circles being equal and known, the perpendiculars from the centres of the two circles upon the intersecting line must also be given: let them be represented respectively by p, p'. Let the equation to the intersecting line be m n Then, p denoting the distance of the centre of (1) from (3), 1 P = ( + and, p denoting the distance of (- a, 0), the centre of (2), from (3), a -+1 -p a Hence m+ p ~ap + P -P REFERRED TO ANY RECTANGULAR AXES. 75 which determines the distance of the required point from the origin. Lacroix': App1ication de I'Alebre a la G6ometrie, No. 109. Puissant: Recueil de diverses propositions de Geomdtrie, p. 152, troisieme edition. 4. Diameters of two circles, the centres of which are A, B, make up a diameter of a greater circle,, the centre of which is C. To compare the lengths of the two chords EF E'', of the greater circle, which touch t~ -~ - both the smaller ones. \ Let a, b, be the radii of the two smaller circles, a + b being therefore the radius of a the larger. Let EF = c, E'F' c'. Let the equation to EF be x y - -t+ = 1. m n Then, p denoting the distance of C from EF, 4= (a + b)2 p2 = (a + )2- m+,...............(1). Again, the distances of the centres of the circles A, B, from EFl being a, b, respectively, we have n (b - m) a (i2 2+ 2) (M "+ n )8' 76 - n (a + m) (m2 + W2)z From these two equations we may, by a few obvious algebraical processes, ascertain that 2 (a2 + b2\2 a = - b ' _ (a2 + b2)2 4ab Hence, substituting these values of m2 and n2 in (1), and reducing, we shall get 2 a + ab + b2 C2 = 16ab. (a + b)2 76 CIRCLE. Again, c' denoting the chord E'F, we have:c'" = 2a.2b, or c2 = 16ab. Hence the ratio between the lengths of the chords EF, E'F', is equal to (a + ab + b2)j a + ab ah b 5. If three fixed circles be cut by any other, to prove that the three chords will form a triangle, the loci of the angular points of which are three fixed straight lines meeting in a point. Let the coordinates of the centres of the three fixed circles be (a, b), (a', '), (a", b"), and their respective radii c, c', c". Let (p, q) be the centre of the variable circle, and r its radius. The equations to the first of the fixed circles will be (x - a)2 + (y - b)2 = c2 and to the variable circle (X -P) + ( - )2= r2 Subtracting the latter equation from the former, we have 2x (p - a) + 2y (q - b) + r2 - C2 = p2 - a2 + q2 - b2, which is the equation to the chord of intersection. The equation to the chord of the second of the fixed circles will be, similarly, 2x ( - a') + 2y (q - b') + r - c = p - a2 + -. At the intersection of these two chords, there is, one equation being subtracted from the other, 2x(a - a) + 2y (b- V) + c2 -_2 = a2 a2 + b2 - b2) which is the equation to the rectilinear locus of one of the angular points of the triangle of chords. The equations to the other two loci are, by similar reasoning, 2x (a' - a) + 2y (b - ") + c'2- c"2 = a'2 a"2 + b' - b"2 2x ( - a) + 2y(b" - b) + c" - c2 -= a a2 - b1 - b2 REFERRED TO ANY RECTANGULAR AXES. 77 If we add together the equations to these three loci, we shall evidently obtain an identical equation. They must therefore pass through a single point. 6. To find the equation to a circle the diameter of which is the common chord of the two circles X2 + y2 = 2, (x -a)2~ + y = c2. The required equation is x2 - ax + y2 = 2 -2 a2 7. To find the equation to the straight line which passes through the centres of the two circles x2 + 2x + y2 = 0, y2 + 2y + x2 = 0. The required equation is x + y +1 =0. 8. To find the equations to the circles which touch the three lines x= a, y2 y=2b, y=2b'. The required equations are (x - a + b' T )2 (y - b - b)2 = (b - b)2. 9. To find the equation to a circle which passes through three points the coordinates of which are (1, 2), (1, 3), (2, 5). The required equation is - 9x + - _ 5y + 14 = 0. 10. To find the equation to a circle, which passes through the three points (0, 0), (- 8a, 0), (0, 6a). The equation to the circle will be (x + 4a)2 + (y - 3a)2 = 25a2. 11. To find the centre and radius of the circle x2 + y2 - 6x + 4y + 4 = 0. The coordinates of its centre are (3, - 2), and its radius is 3. 78 CIRCLE. 12. To find the equation to the common chord of the two circles (x - 1)2 + (y - 2)2 = 6, (x - 2)2 + (y 3)2 = 8. The required equation is x +y =3. 13. To find the equation to a circle having for diameter the distance of a given point from the origin of coordinates. If a, /, be the coordinates of the given point, the required equation will be x2 + y2 = ax + y. 14. To find the length of the common chord of the two circles (x - a) + (y -/3)= 72, (x - 3)2 + (y a)2 = y2. The required length = {42 - 2 (a - /)2}1. 15. To prove that all circles represented by the equation (x - ) (Y - 2)2 - + y + k{ a)2 + (y - /)2 - 2} = 0, where a, /f, 7a, a, 3,, y,, are constant, and k any quantity whatever, pass through two fixed points; and to find the equation to the straight line which passes through the two points. The required equation to the straight line will be 2 (a - a,) x + 2 (3 - /3) y + a2 + /3, _ a-,32 = 7,2 - 72 16. To find the equation to a straight line passing through the origin of coordinates and touching the circle of which the equation is x2 + y2 - 3x + 4y = 0. The required equation is 3x = 4y. 17. To find the equations to the lines which touch both the circles A - y2 = C2, (x - a)2 + y2 = c2 The required equations to the four common tangents are (c + c') x + a {a- (c ~ c'')23y = ac. Puissant: Recueil de diverses propositions de CGometrie, p. 150, troisieme edition. REFERRED TO ANY RECTANGULAR AXES. 79 18. The abscissae of the centres of two circles are al, a2, and the lengths of tangents drawn to them from the origin are 1, 12; to find the position of a point in the axis of x such that tangents drawn from it to the two circles shall be equal to each other. The distance of the required point from the origin is equal to 72 _ 72 12 112 2 (2 - a) 19. To find the equation to a circle which touches the three lines x=O, y=O, -+ b 1, 0a in the positive quadrant. The required equation, if we put a2 + b2 = c2, is x2+ yv - (a b - )(x + y) 4 (a + b - c) =0. 20. To find the equation to a circle which touches the two lines y xy a b ' a ' and of which the centre lies in the line ax = by. The required equation is ( + + ( -( + y2) - ( + (+ ) + + X- 0. 21. A fixed circle is cut by a series of circles, all of which pass through two given points; to shew that the straight lines which join the points of intersection of the fixed circle with each of the series, all converge to one point, and to determine the position of this point. If the equation to the fixed circle be x2 + 2 = r and (a, b), (a, 3), be the two given points, the coordinates of the 80 CIRCLE. required point will be (a - a) r2 - a (a2 + b) + a (a2 + /2) a2 + l32 - 2 _ 2 and (/ - b)r' - / (a2 + 6') + b (a2 + /32) a2 + 32 _ a2 _ b2 Leybourn: MathematicalRepository, New Series, vol. I. part II. p. 39. 22. If P be the point of intersection (not the origin) of the circles 2x 2 X' + y2 - 22 + Y2 =b and FP the point of intersection (not the origin) of the circles 92 y2 2x ~ 2 2y X + Y = a'; +Y#= b'; to prove that P =( + b2).(a,' + b'2). {(a - a) + (b - 23. Two parallel tangents are drawn to a given circle A, and two circles B, C, touch each other and A externally, touching also the two tangents respectively; to find the locus of the point of contact of B and C. Let a denote the radius of A. Then, the centre of A being taken as the origin of coordinates, and the diameter of A, which is per- pendicular to the two parallel tangents, as the axis of x, the equation to the required locus will be (x2 + y2)2 = a2 (X2 + 22). SECTION V. Referred to two Tangents as Axes of Coordinates. 1. To find the equation to a circle referred to two tangents as axes of coordinates. REFERRED TO TWO TANGENTS AS AXES OF COORDINATES. 81 Let c denote the inclination of the axes, c the radius of the circle, and a, a, the equal coordinates of its centre. Then the equation to the circle will be (x - a)2 + (y - a)2 + 2 (x - a).(y- a). cos c = c2: but from the geometry it is plain that c is equal to a sin o; hence the equation becomes x2'+ 2xycos o + y' - 2a (1 + cos w).(x + y) + 2a2(1 4 cos ) =a2sin w, or (x + y)2 - 2a (1 cos co) (x + y) + (1 cos co )2 2xy (1 - cos w), or ( + y - 2a cos2- =4 sin'2.Xy \ 2 / 2 or x- y + 2 sin.(xy)l = 2a cos2 2 2' 2. To find the equation to a circle, of which the radius is c, and which is referred to two rectangular tangents as axes. The required equation is x2 + y - 2c (x + y)+ c2= o. 3. To find the equations to two circles which touch the rectangular axes of x and y, and pass through a given point (a, b); to find also the equation to their common chord. The equations to the two circles are + y2 - 2 {a + ~ + (2ab)}.(x + y) + la + b + (2ab)}2 = 0, and the equation to the common chord is x+y= a -b, 4. Supposing the inclination between the coordinate axes to be, to prove that 3 1 x + y - a = (xy)' is the equation to a circle the radius of which is equal to a-. G 82 CIRCLE. 5. To find the equation to a circle touching a given circle and the two sides of a given angle. The two sides of the given angle o being taken as axes of coordinates, let c be the radius and A, k, the coordinates of the centre of the given circle. Then the equation to the required circle will be x +y -la cos = 4 sin 2.x, a being determined by the equation (h - a)2 + (k - a)2 + 2 (h - a) (k - a) cos o = (c + a sin 0)2. There being four values of a, it is plain that there will be four circles satisfying the conditions of the problem. Puissant: Recueil de diverses propositions de Geometrie, p. 159, troisieme edition. SECTION VI. Referred to any Oblique Axes. 1. To determine the radius of the circle represented by the equation x2 + y2 + 2xy cos w = ax + by. The general equation to the circle referred to oblique axes including an angle co is (X - h)2 + (y - k)2 + 2 (x - ) (y - k) cos o = c2 where c is the radius. Equating the coefficients of x, y, and the constant terms in these two equations, we have 2 (h + k cos o) = a.....................(1), 2 (k + h cos w) = b................... (2), c2 = h + k2 + 2hk cos.................. (3). Multiplying (1) by h, (2) by k, and adding the resulting equations, we see that 2 (h2 + k2 + 2hk cos co) = ah + bk, POLAR COORDINATES. 83 and therefore, by (3), 2c2 = ah + bk.......................(4). From (1) and (2) we readily find that a - b cos co b - a cos co h= k= 2 sin2 o 2 sin co and therefore, by (4), a2 + b - 2ab cos ro 4c = — 2 sin' ct which determines the magnitude of the radius. 2. To determine the inclination of the coordinate axes in order that the equation x" - xy + y2 - ax - ay = 0 may represent a circle; and to find the magnitude of its radius. The inclination of the axes = w, and the radius = a. 3. The axes Ox, Oy, cut a circle in points A, A', B, B', respectively; to compare the values of x, y, at the intersection of the chords AB', A'B. Let OA = OA' = a, OB = 3, OB' = P; then x A' - a' -a SECTION VII. Polar Coordinates. 1. From a point O, without a circle, two tangents OF, O G, are drawn to the curve; the chord FG is joined. From 0 any straight line OPQR is drawn, cutting the circle in the points P, R, and the chord in Q. To prove that the line OR is harmonically divided in the points P and Q. G2 84 CIRCLE. Join 00, C being the centre of the circle, cutting FG in M. Let a = the radius of the circle, OC = c. Let LROM= 0, and let OP or OR be denoted by r. Then, by the polar equation to the 9a circle, 0 being the pole and 00 the 0 > <C direction of the prime radius vector, there is (r cos 0 - c)2 + r2 sin2 0 = a2, or r.- 2c cos 0.r + e - 2 = 0, or, since c' - a =f2, f denoting OF, r2 - 2c cos.r +2 = 0. Hence it appears, by the theory of equations, that OP.OR = f, and OP+ OR = 2c cos 0, 1 1 2 cos0 whence -OP OR f+ But, joining CF, we see by the similar triangles OMF, OFC, that 2 OQ cosS = OM=-. 1 1 2 Hence -O + 0 OP OR OQ De La Hire: Sectiones Conicce, lib. I. prop. 21. 2. If on any three chords, drawn through the same point in the circumference of a circle, as diameters, three circles be described; the points of intersection of these circles lie in a single straight line. Let the point of the original circle through which the three chords are drawn be taken as the pole and the diameter through it as the prime radius vector. Let c represent the diameter of the original circle, and c of one of the three new ones. Let a represent the angle between c and c'. POLAR COORDINATES. 85 Then the polar equation to the circle, of which c' is the diameter, will be r = c cos (0 - a): but it is plain that c' = c cos a; hence r = c cos a cos ( - a) = I-ccos0 + cos ( - 2a)}..............(1). Similarly, the equations to the other two new circles will be r= ccos a' cos ( - a') = c cos o0 + cos (0 -2a')}...(2) r = c cos a" oss ( - a") = 2c {cos 0 + cos (0 - 2a")}...(3). At the intersection of (1) and (2), 0 = a + a', r = ccos a cos a; this intersection therefore lies in a straight line represented by the equation r cos (a + a' + a" - 0) = c cos a cos a' cos a". By symmetry it is plain that this line will contain also the intersection of (2), (3), and that of (3), (1). Greathead: Cambridge Mathematical Journal, vol. I. p. 168. 3. To determine the magnitude and position of the circle of which the equation is r2 - 2 (cos 0 + /3 sin 0)r = 5. The coordinates of the centre are 0 = T7r r = 2, and the radius is equal to 3. 4. To prove analytically that if, from a point 0 without a circle, two straight lines OPQ, 04, be drawn, one of which cuts the circle in P, Q, and the other touches it in A, the square of the line OA which touches the circle is equal to the rectangle contained by the straight line OPQ which cuts the circle and the part of it OP without the circle. 5. Through a given point to draw two straight lines at right angles to each other, such that the parts of them intercepted by a given circle may bear towards each other a given ratio. '86 CIRCLE. Let c, c', be the lengths of the two intercepted chords, and b the inclination of the former to the diameter passing through the given point; let r denote the radius of the circle, and a the distance of the given point from its centre. Let c = tGc', M being a constant. Then (8r2 - 4a2 /8rc - 4a'2/ and {-r (1 -) + }a'} Puissant: Recueil de diverses propositions de Geometrie, p. 145, troisieme 4dition. 6. To find the position of the centre of a circle which passes through a given point, touches one given straight line, and intercepts from another given straight line a chord of given length. Let the summit of the angle included between the straight lines be taken as the pole, the straight line which touches the circle being the prime radius vector. Letf= the length of the given chord; let (/3, p) be the coordinates of the given point, a the radius vector of the point of contact, c the perpendicular distance of the centre from the prime radius vector, a the angle between the two straight lines. Then a and c are to be determined from the two equations 2cp sin 3 = p2 + a2 - 2ap cos,, p2 (4a + f2) sin2 / = {2ap sin (,3 - a) + (p2 + a2) sin a}2, the latter being a biquadratic in a. Bossut: Geometrie, p. 345. Puissant: Recueil de diverses propositions de Geometrie, p. 173, troisieme edition. POLAR EQUATIONS TO TANGENTS AND CHORDS. 87 SECTION VIII. Polar Equations to Tangents and Chords. 1. If a constant moveable angle has its summit fixed at a point in the circumference of a circle, the chord, which it subtends in the circle, is always a tangent to another circle, concentric with the original one. If r = c be the equation to the circle; X, A, the angular coordinates of the chord in any position, and s the constant angle; then, the perpendicular distance of the origin from the chord being ce os 2X and 2 + being the inclination of this ~2 u2 distance to the prime radius vector, the equation to the chord will be - / X + =A cos cros - + - 2 2 * But from the geometry it is easily seen that X - = 2 (7r - s). Hence the equation to the chord becomes c cos (rr-) =r cos ( - X-+ which is the equation to a series of straight lines, varying in position with the value of X+ +_ the perpendicular distances of all of them from the origin being c cos (r - s). The chord therefore always touches a concentric circle, the radius of which is c cos (7 - m). Bobillier: Gerqonne, Annales de Mathematiques, tom. XVIII. p. 190. 2. To find the equation to the chord of a given circle, supposing it to subtend a right angle at the centre, and having given the position of one of its extremities. 88 CIRCLE. Let a be the angular coordinate of the given extremity, and c be the radius of the circle. Then the equation required is c = 2. r,cos (o - a ~. (. 4) SECTION IX. Poles and Polars. 1. If any number of circles touch each other in one point, all their polars, which correspond to a common pole, will pass through a single point. Let the point of contact of the circles be taken as the origin of coordinates, the common diameter through the origin as the axis of x, and the tangent as the axis of y. The equations to any two of the circles will be x2 + y2 = 2r'x x2 + y1 = 2r"x. Let (x, y,) be the common pole. Then the equations to the polars of these two circles will be (, - r) x + yy = r'x, and (x, - r") x + yy = rx. At the intersection of these two polars, (r -') =- (r" - r') x,, 2 = - XI7 y = Y-. Since the values of x and y are independent of the radii of the circles, the truth of the proposition is obvious. 2. To find the locus of the pole, when the polar of a given circle always passes through a given point. POLES AND POLARS. 89 Let the equation to the circle be x2 + y _ r2' the axis of x being so chosen as to pass through the given point. Then (x,, y,), being the pole, the equation to the polar will be x + yy y= But x = a, y = 0, simultaneously, a being some constant quantity. Hence r2 I = -a a the equation to the locus of the pole, which is therefore a line perpendicular to the line which passes through the centre of the circle and the given point. 3. A fixed circle is cut by a series of circles touching each other at one point; to find the locus of the pole of the fixed circle, the polar of which passes through the two points in which the fixed circle is cut by any one of the series of circles. The equation to the fixed circle being (X - a)2 + (y -_ )2 = 72, and that to any one of the series of circles x2 + y2 = 2rx, the equation to the required locus will be (y2 - a2 + P2) (y- /3) + 2y2,3 + 2a,/ (x - a) = 0. 4. The equation to the polar of a circle, denoted by the equation x2 + y2 = c2 x y being -a+b to find the coordinates of the corresponding pole P, and, AB being the portion of the polar intercepted by the circle, to find the inclinations of AP, BP, to the axis of x. 90 CIRCLE. C2 C2 The coordinates of P are -, - and, t denoting the tangent a b of the inclination of AP or BP to the axis of x, a2 2 c+ 2c2 62 -2 t ab + 2 0. SECTION X. Radical Axes, Centres of Similitude, &c. 1. If three circles, described in one plane, intersect, two and two, their three chords of intersection pass through a single point. Let the equations to the three circles be (-a)2 + ( -b)= C2..................(1), (x - a')2 + (y - )2 = 2................. (2), (x- a")2 + (y - b")2 = c"2................. (3). Subtracting (2) from (1), we have, for the equation to the chord of intersection of the first two circles, 2 (a'- a) x + 2 (b'- b) y+ - a' + 2 - 'a" = c - c'...(4). Similarly, for the intersections of (2), (3), and of (3), (1), we have 2(a" - a') x + 2(b" - b') y + a"- + b'2 - b"= 2 - c. (5) and 2 (a - a") x + 2 (b - b")y +a2 a2 + b"2 -b "2 - c2...(6). The three lines, denoted by (4), (5), (6), as their equations indicate, pass through a single point. The lines (4), (5), (6), have been called by Gaultier the radical axes of the pairs of circles (1, 2), (2, 3), (3, 1), respectively, and the point of intersection of (4), (5), (6), has been called by him the radical centre of the three circles. The terms radical axes are applied to the lines denoted by (4), (5), (6), and radical centre to their point of intersection, even when the circles do not cut each other. Gaultier-de-Tours: Journal de l'Ecole Polytechnique, Cah. xvi. Carnot: Geometrie de Position, p. 347. Durrande: Gergonne, Annales de Mathematiques, tom. XI. p. 13. RADICAL AXES, CENTRES OF SIMILITUDE, ETC. 91 2. The tangents drawn to two circles, from any point whatever in their radical axis, and terminated at their points of contact, are equal to each other. Let the equations to the two circles be (x - a)2 + (y - b')= c' 2 (x -a)2 + (y- b)2 = C2. The equation to the radical axis will be 2(a - a') x + 2(b" - ')y +a'2 - a + b'2 - b"2 = c' Suppose the origin to be so chosen as to lie in the radical axis; then a'2 - a"^ + b'2 _ b"t2 = c'2 - c'2 or a"G + b'"- c' - = ca" + b"2 _ c"2 an algebraic expression of the truth of the proposition. Durrande: Gergonne, Annales de atathematiques tom. VIII. p. 322. Plucker: Gergonne, Annales de Math4matiques, tom. XVIII. p. 33. 3. To find the equations to the internal and external polars of similitude of any two circles. The term centre of similitude was introduced by Monge. The external centre of similitude of two circles is the point of intersection of the two common external tangents, and the internal centre is the point of intersection of the two common internal tangents. Any straight line containing three of the centres of similitude of a system of three circles is called an axis of similitude; an external axis, when it contains the three external centres of similitude, and an internal axis, when it contains one external and two internal centres of similitude. The internal polar of similitude is the polar of the internal centre of similitude, and the external polar of similitude is the polar of the external centre of similitude. Let the equations to the two circles be ( - a) ( -- b)2 = c..................(1), (x - a)2 + ( b')2 '2.................. (2). 92 CIRCLE.e The equation to a tangent to (1), at a point (x, y), is (X- a). (x - a) + (Y- b). (y - b) = c...... (3); and that to (2), at a point (x', y'), is (X- a'). (x' - a') + (Y- b'). (y' - ') = c'2......). If the lines (3) and (4) are coincident, then, the coefficients of X, Y, having the same ratios in both equations, we must have (x'- a') = x- a.....................(5), and (y- b) = y - b.....................(6): hence, from (3) and (4), (a' - a). ( - a) + (b' - b).(y - b) = c - Xc'... (7). But, from (1), (5), (6), (X'- a')2 + (y_ - ^')2 and therefore, (x', y') being a point in (2), c' = o, orX=+. The equation (7) is therefore reduced to the form (a - a) (x - a) + (b' - b) (y - ) = c ( c'), and is equivalent to the two (a'- a) (x - a) + (b' - b) (y -b) = c (c - c')... (8), and (a - a) (x - a) + (b' - ) (y -b) = c (c + c')... (9). Either of these equations, combined with (1), will give two pairs of values for the coordinates of contact: they will accordingly represent two chords of contact, perpendicular to the line joining the centres of the circles, viz. the two polars of similitude relatively to the circle (1). Salmon: Treatise on Conic Sections, p. 100. 4. To find the coordinates of the centres of similitude of two circles. The centres of similitude evidently lie in the line joining the centres of the two circles, the equation to which, if we adopt the RADICAL AXES, CENTRES OF SIMILITUDE, ETC. 93 circumstances and notation of the preceding problem, will be Y-b X-a b -b a -a' that to a common tangent to the circles at a point (x, y), being (X- a) ( - a) + (Y- b) (y - b= c From these two equations we have (X - a) ((a' - a) (x - a) + (b - b) (y - b)} = (a - a), and therefore, if we suppose x, y, to be a point in the former of the polars of similitude, a -a ac - ca b'c - c'b X - a = c ---- = = X-a=- C - c - C C - C From these values of X, Y. we see that X - a, Y- b, have the same signs as X- a', Y- b', and that accordingly the point X, Y, lies in the prolongation of the line joining the centres of the circles. Hence these values of X, Y, belong to the external centre of similitude, and the equation (8) to the external polar of similitude. If we suppose the point x, y, to be in the latter of the polars of similitude, we shall get for the coordinates of the internal centre of similitude a'c +c 'a b'c c'b X= Cc'? C +- c' the equation (9) belonging accordingly to the internal polar of similitude. Salmon: Treatise on Conic Sections, p. 101. For further information respecting centres and polars of similitude, the student is referred to memoirs by Durrande; Gergonne, Annales de Mathmatiques, tom. II. p. 1; tom. XVII. p. 285; and Plucker, ib. tom. XVIII. p. 29. 5. To find the locus of the radical centre of three circles, the radii of which are r + p, r' + p, r" + p, where p is a variable quantity, the centres of the three circles being fixed. 94 CIRCLE. Let (a, b), (a', b'), (a", b"), be the central coordinates of the three circles: then the required locus will be a straight line defined by the equation {r (a' - a) + r' (a - a) + r" (a - a')} + {r (' - b") + r' ( - b) + r (b - ')} y = 0, the position of the radical centre of the three circles, when p = 0, being taken as origin of coordinates. Plucker: Gergonne, Annales de Mathematiques, tom. xvIII. p. 35. 6. To prove that the locus of the radical centre of the three circles (x - a)2 + (y - 6)2 = (r + p)2, (x - a')2 + (Y - b) = (r' + p)2, (X - a)2 + (y - b")2 = (r" + p)2, where p is a variable quantity, is at right angles to their external axis of similitude. Plucker: Gergonne, Annales de Mathematiques, tom. XVIII. p. 36. 7. The four tangents, which are common to two circles which do not intersect, and are terminated at their points of respective contact, have their middle points on the radical axis of the two circles. Gergonne: Annales de Math4mnatiques, tom. VIII. p. 323. 8. Let two circles be touched respectively by a single straight line AA' in A and A', and by a single circle BB'C in B and B': if the straight line and the circle touch in the same manner the two circles, the point C of the concourse of AB and A'B', will lie on the circumference of the circle BB'C and on the radical axis of the two other circles. Durrande: Gergonne, Annales de Mathematiques, tom. VIII. p. 324. 9. The external centres of similitude of three circles, taken successively two and two, all lie in one straight line; and each RADICAL AXES, CENTRES OF SIMILITUDE, ETC. 95 of them is situated in a right line with two of the internal centres of similitude. Fuss: Nova Acta Academice Petropolitance, tom. XIV. p. 140. Durrande: Gergonne, Annales de Mathe6matiques, tom. XI. p. 8. Puissant: Recueil de diverses propositions de Geometrie, p. 165, troisieme edition. Fuss says, in relation to the former portion of this problem: " I1 y a deja plusieurs annees qu' un jeune Francois, employe alors au Corps Imperial des Cadets de Terre, me parla d' un Theoreme de G4ometrie qui, dans le tems qu' il etoit encore a Paris a 1'Ecole Royale militaire, avoit eu quelque celebrite et qu' on avoit pretendu tenir de feu Mr. d'Alembert. Je lui en donnai une demonstration, dont j' ai retrouve depuis peu le brouillon en fouillant mes papiers. En relisant cette demonstration j' ai vu que la belle propriete qui en fait le sujet, peut conduire a d' autres non moins remarquables. En rassemblant mes idees sur cette matiere il en est resulte le petit Memoire que j'ai l'honneur de presenter ici l' Acad6mie pour la collection des Memoires traduits en Russe, qu'elle se propose de publier, ou bien pour les Actes m6mes, si elle le juge digne de cet honneur. II y fera sans doute plaisir a plus d'un amateur de la G6ometrie, et peut-etre meme a quelque Geometre de profession." 10. The three circles (x - a) + (y - b) = r, (x - a')2 + (y - 6)2 = r'2 (x - a)2 + (y- b")2 = r"2, are touched by a fourth circle: to prove that the point, in which the third circle is touched, lies in the line of which the equation is (a - a"). (x - a") + (b - b"). (y - b") + (r - r") r" (a - a ") + (b - b")2 _ (r - r")2 (a' - a"). (x - a") + (V - b"). (y - ") + (r' - r") r" (a - d) + 'V - b) _ (r _- r)2 For full information on the subject of the contact of a fourth circle with any three proposed circles, the reader is referred 96 CIRCLE. to Euler,1 Fuss,2 Schubert,3 Gergonne, Puissant,5 Montucla,6 Hearn,7 Salmon.8 SECTION XI. Inscribed and Circumscribed Polygons. 1. ABCP and A'B'C'P' are two concentric circles, ABC and A'B'C' any two equilateral triangles inscribed in them. If P, P', be any two points in the circumferences of these circles, to shew that (A'P)2 + (B'P)2 + (C'P)2 = (AP') + (BP') + (CP'). Let the equations to the two circles be x2 + 2 = r2, 2 + y2 = r2 Let (x, y) be the coordinates of P, and (x', y,'), (x2 'y,2') (x', y3'), those of A', B', C', respectively. Then (A'P)2 + (B'P)2 + (C'P)2 - (X'- )2 + (y1-_y)2+ (X- X)2+ (y2-y)2 + (x'-X)2+ (y3-)2 1= Y12 + y 12 + 2 + 32 + 2 + Y + + 3 (;'j + y2) - 2x (x' + x2' + x,3) - 2y (Yi + y2 + Y3) = 3 (r'2 + r2) - 2x (x1' + 2' + ') - 2y (y' + y' + y' ). But, r'V3 being the side of the triangle in the circle A'B' C'P', 3r'2 = (X21 - X2')2 + (Y1' - y).:2r22 - 2 (Xl'x1 + Y1 Y2 ), 2 (XlX2' + Y1Y2') = - r'2 = (x1 + y1). Similarly 2 (x2'x' + y2'y) = - (x2 + Y2 ), and 2 (xx1' + y') = - ('2 + Y2). Nova Acta Academie Petropolitance, tom. vi. p. 95. 2 Ibid. tom. vi. p. 102. 3 Ibid. tom. x. p. 77. 4 Annales de Mathematiques, tom. vii. p. 289. 5 Recueil de diverses propositions de Geometrie, p. 176, troisieme edition. 6 Histoire des Mathematiques, tom. I. p. 263. 7 Researches on Curves of the Second Order, p. 22. 8 Treatise on Conic Sections, p. 104, &c. INSCRIBED AND CIRCUMSCRIBED POLYGONS. 97 Adding together these three equations, we have (X1' + X2 + X3)2 + (y1 + Y;' + y,')2 = 0, and therefore X1 + X2 + X3 = 0) +,z + Y3' = 0. Hence (A'P)2 + (B'P)2 + (C'P)2 = 3 (r2 + r2). Also, by symmetry, (AP')2 + (BP')2 + (CP')2 = 3 (r2 + r'). 2. Any quadrilateral being described about a circle, to prove that the line, which joins the middle points of the diagonals, passes through the centre of the circle. Let the equation to the circle be x2 + y2 = c': this equation is equivalent to the two following, C +y /(- 1) = ca, x-y (- 1) c where a is arbitrary: whence we have - / 1\ ( c / - X= C (a+ a) The equation ( a+ - / a - = 2c.......... (1) represents therefore a tangent to the circle, which we will suppose to coincide with a side of the quadrilateral. The equation X (a ) + (a'- ) = 2c............2) \ a/ V(-1) V a / a may be taken to represent the next side. At the intersection x1,2, y,12, of these two lines, as may be easily shewn, 1 + aa' 1- aa' a+ aL Y C(-1) a +' X,,, =C ~ ~ ~ ~ ~ 98 CIRCLE. Similarly, at the intersection of the other two sides, 1 + a"a"' 1 - a"a" 3'4 = o d-'+a" ' Y3,4=C (-1) a",a"' If X, Y, denote the coordinates of the middle point of the diagonal through these two intersections, we have ( + aa I' + a 2 a 3+a' a+a"a"' /1 - aa' 1 - a"a"' Y=24V(-1) -f+ i 2; whence, performing obvious simplifications, Y aM a~+ a-'"(1 + a... aa') XY ( a ( +- a'a") + a'(1 - a"a") a+ (1 a"a) a"'(l aa') This result, being symmetrical in regard to a, a', a", a"', shews that in the case of the other diagonal, we shall have the same ratio for the coordinates X', Y', of its middle point. Hence y' Y X which shews that the two middle points are in a line through the centre of the circle. Durrande: Gergonne, Annales de Mathematiques, tom. XIV. p. 309. 3. To prove that the equation to a circle, circumscribed about a triangle, the equations to the sides of which are u = x cosa + y sina -p, v = x cos + y sin / - q, w = x cos y + y sin y -r, is vw sin A + wu sin B + uv sin C = 0, A, B, C, being the angles between the sides (v, w), (w, u), (u, v), respectively. Salmon: Treatise on Conic Sections, p. 92. 4. If two triangles are the one inscribed in a circle and the other circumscribed about it; so that the summits of the INSCRIBED AND CIRCUMSCRIBED POLYGONS. 99 inscribed are the points of contact of the circumscribed triangle; to prove that, (1), the points of concourse of the directions of the opposite sides of the two triangles lie in a single straight line, (2), that the lines joining their opposite summits meet all three in a single point, and, (3), that this point is the pole of the aforesaid straight line. Durrande: Gergonne, Annales de Mlcathtmatiques, tom. XIV. p. 45. Salmon: Treatise on Conic Sections, p. 94. 5. The feet of the perpendiculars let fall, upon the directions of the sides of any triangle, from any point in the circumference of the circumscribed circle, all lie in one straight line. Servois, Gergonne: Gergonne, Annales de Math4matiques, tom. IV. p. 251. Durrande: Ibid. tom. VII. p. 253. Querret: Ibid. tom. XIV. p. 285. Salmon: Treatise on Conic Sections, p. 94. 6. The base OA of a triangle POA is given, and the triangle is such that a circle of given radius may always be inscribed in it: to find the locus of P. If OP= r, LPOA = 0, OA = a c = the radius of the inscribed circle, the equation to the locus of P will be sin 0. (a sin 0 - 2c) r = 2c (a sin 0 - c cos 0 - c). MIanderlier: Quetelet, Corr. Matl. et Phys., tom. II. p. 320. 7. To find the radius of a circle inscribed in a triangle the equations to the sides of which are x cosa + y sin a = 8, x cos a + y sin a = ', x cos a" + y sin a" = ". The radius is equal to I V I sin (a' -a"I) ~ ' sin (af'-") + al sin (a — 4 sin. sin. sin 2 2 2 Lhuilier: Elmnens d'Analyse Geometrique et d'Analyse Alqgbrique, p. 119. H2 100 CIRCLE. 8. The equations to three straight lines are u = 0 U, = 0, ut2 = 07 x sin 0 - y cos 0 - c = 0, being taken as the type of the equation to a straight line: to prove that the equations of the four circles, to each of which these lines are tangents, are si 0 - 0 1. 0 -0 2 - _ 0 u Bsin l-lu's~sinn -- u sin2 1 + u 2 + U2 sin -1- = 2 2 2 2? sin 2 — + U1cos 202 + u 2 Cos 1 = 0, 2 2 2 u cos 2- - + u, sin 22 + cos - = O, _0 2- 0 1 0- 0, 01-0 u cos + u cos 2 + u sin 2 =0. 2 2 2 Salmon: Treatise on Conic Sections, p. 94. 9. Having given the equations to the three sides of a triangle, to find the equations to the lines joining the angles of the triangle with the points in which the escribed circles touch the opposite sides. The notation of the preceding problem being retained, the equations to the three lines will be 0-02 20-0,O01- 0 202- 01 u cos 2 2= u, cO u cos 8 = U Cos" - 2 2 2 2 2 2 0 0 20 u cos 2- 1 cos 2 These equations shew that the three lines pass through a single point. 10. If any number of quadrilaterals, inscribed in a given circle, have a common side, to prove that the lines which, in these quadrilaterals, join the intersection of the diagonals and the point of concourse of the sides adjacent to the common side, CIRCULAR LOCI. 101 will all meet in a single point of the perpendicular on the middle point of the common side. Durrande: Gergonne, Annales de Mathematiques, tom. XIII. p. 308. Puissant: Recueil de diverses propositions de Geometrie, p. 137, troisieme edition. SECTION XII. Circular Loci. 1. The distances of a point P from two fixed points A, A', are r, r'. To find the locus of P, supposing that nr2 + n'r'2 is equal to a constant quantity c, where n, n', are given quantities. Let (a, b), (a', b'), be the coordinates of A, A', respectively, (x, y) being those of P. Then n {(x - a)2 + (y- b)2} + n' {(x - a')2 + (y - )'} =c'2 (n + n') (x + y2) - 2x (na + n'a') - 2 (nb + n'b') + n (a' + b2) + n' (a'2 + b12) = 2, na+n'a', nb +n'b' C2-n (a2+ b2) -n'(a2+b'2) _ -21 + '1 x +yb -2.y= - I 1o T 'I I - V /I T I I / na + na' ( / nb - n'b' x - ____- 4 - --— _ C2 nn =,+ n (,, + r,.' {(a-a,)2 + (b b ')2}, n. n (n n)2 (a a) (6 the equation to the required locus, which is therefore a circle, its radius being equal to n [(n + n') - nn' {(a - a')2 + (b - V')23, n + n' and central coordinates to na + n'a' nb + nb' n + n n + n' Lhuilier: Eldmens d'Analyse Gomnetrique et d'Analyse Alyebrique, p. 131. 102 CIRCLE. 2. To find the locus of a point such that the sum of the squares of its distances from any number of proposed points may have a given value. Let there be n proposed points (al, (), (a,, 83), (a, 3)),...... (an, I3B), and let (x, y) be the coordinates of the moveable point. Also let c2 be the constant value of the sum of the squares. Then, the axes being rectangular, we have ( - a)2 + (y - /1)2 + (X - 2)2 + (y - /,) + (X - a)2 + (y - P3) + (X - a,, + (y - )2 =, or 2 +2_2 2 '" +y _ - (al + a,2 + a3 +... +a) X - - (1 + /,2 +/S+. + + +) Y n n t2 - _(a,13r2 + a2 + C2 + e... + a2)_- (812 +3 22 32 +... + 32)1 the equation to the required locus, which is therefore a circle the centre of which has for its coordinates 1 I -(a,+2+ a... + a. 31+32 + a1 +... + a,... Apollonii Pergcei Locorum Planorum Libri II. Bestituti a Roberto Simson, p. 159. Montucla: Histoire des Mathemnatiques, tom. I. p.284, 2e edit. Puissant: Recueil ce diverses propositions de Geometrie, p. 194, troisieme edition. Garnier: Gceometrie Analytique, p. 183, deuxiene edition. 3. There are 2n given straight lines, which make with another given straight line angles a, 3, 7...... A point P is taken such that the sum of the squares of the perpendiculars drawn from it upon these 2n lines is constant. To prove that the necessary and sufficient condition, that the locus of P may be a circle, is n+ cos 2 (a - 3) + cos 2 (8 - y) + cos 2 (a - ) +... 0. CIRCULAR LOCI. 103 Take the fixed line of reference as the axis of y, and any line at right angles to it as that of x. Then the equations to the 2n lines will be of the forms x cosa + y sin a = a, x cos / + y sin / = b, x cos 7y + y sin 7y = c,..............,............ The condition of the problem, A denoting the constant value of the sum of the squares, gives us (x cos a + y sin a - a)' + (x cos, + y sin 3 - b)2 + (x cos 7y + y sin r - c)2 +......= A. In order that this equation may represent a circle, it is sufficient and necessary that the coefficients of x2 and y2 be equal, and that the coefficient of xy be zero: hence cos 2a + cos 2, + cos 2y +...... =0, and sin 2a + sin 2/3 + sin 2 +...... = 0. These two equations are equivalent to the single equation (cos 2a + cos 23+ cos 2y +...)2 + (sin 2a + sin 2/8+sin 2y +...)2=0O or, squaring the two members, simplifying and dividing by 2, n + cos 2 (a - 3) + cos 2 (, - y) + cos 2 (a - ) +... = 0. Lhuilier: El'emens d'Analyse Geometrique et d'Ancalyse Algebrique, p. 142. 4. To find the locus of the intersection of a perpendicular from a fixed point upon a line passing through another fixed point. Let the former fixed point be taken as the origin of polar coordinates: then, p denoting the perpendicular and q its inclination to the prime radius vector, the polar equation to the line mentioned in the problem will be p = r cos (0 - ). 104 CIRCLE. Let r', 0', be the given point: then p = r cos (0' - b). This result shews that p, f, are the polar coordinates of any point in a circle of which r' is the diameter, this circle being therefore the required locus. 5. To determine the geometrical locus of the point which divides into two parts, in a constant ratio, the straight line drawn from a given point to any point in the circumference of a given circle. Let the circle be referred to two diameters as axes of x and y, the axis of x being so chosen as to pass through the given point. The equation to the circle will be, c denoting its radius, x2 + y2 =c2......................(1). Let a be the abscissa of the given point C, and x', y', the coordinates of the dividing point P. Let CP' meet the circle in P, and let CP' = n.CP, n being, by the condition of the problem, constant. Then it is easily seen, by similar triangles, x, y, being the coordinates of P, that a - x = n (a - x), or nx = x + (n- 1) a, and ny = y'. But, from (1), (nx)2 + (ny)2 = (nc)2: hence we have, for the equation to the required locus, {(' + (n - 1) a) + 2= (nc)2, which is therefore a circle, the radius of which is nc, its centre lying in the axis of x at a distance (1 - n) a from the origin. COR. Suppose the equation to the locus of P to be given, and let it be (x' + h)2+ y2 c2 CIRCULAR LOCI. 105 Then, for the determination of the position of C and of the magnitude of n, we have (n - 1) a, c' = nc, and therefore n = - a = c C -C REMARK. Thus, two circles being given, we may find the position of such a point and also determine such a ratio, that, if we draw through the point any straight line to cut the circumferences of the two circles, the ratio between the parts of this straight line, comprised between this point and the circumferences, may be equal to the ratio. The proposition which constitutes the subject of this remark belongs to a class of propositions, to which (according to Simson*) the ancients gave the name of porisms. While a locus is determined by certain independent conditions; on the other hand, when a locus is proposed, it determines certain relations among the conditions by which it would have been itself determined; so that two or more of these conditions are simultaneously determined by the proposed locus. Thus, the locus of the point P', which is the object of the present problem, being proposed, the position of the point C and the ratio of OP to CP' are conjointly determined. Playfairt has taken a different view of the nature of a porism; he has arrived at the conclusion that the ancients gave the name of porisms to those propositions which affirm the possibility of finding conditions such as to render a particular problem indeterminate or susceptible of an infinite number of solutions. Simson's definition of a porism is the following: " Porisma est Propositio in qua proponitur demonstrare rem aliquam, vel plures datas esse, cui, vel quibus, ut et cuilibet ex rebus innumeris, non.quidem datis, sed quse ad ea quae data sunt eandem habent relationem, convenire ostendendum est affectionem quan * De Porismatibus Tractatus; Opera qucedam reliquca. 1776, Glasgow. t Transactions of the Royal Society of Edinburgh, vol. iii. 106 CIRCLE. dam communem in Propositione descriptam;;" or, in Playfair's translation of these words, "a Porism is a proposition, in which it is proposed to demonstrate, that one or more things are given, between which and every one of innumerable other things, not given but assumed according to a given law, a certain relation, described in the proposition, is to be shewn to take place." Lhuilier: Eldmens d'Analyse Geometrique et d'Analyse Algeirique, p. 39. Lame: Examen des differentes Meathodes employees pour resoudre les Problmes de Geometrie, p. 22. 6. To find on the circumference of a given circle a point such that the sum of the squares of its distances from two given points shall be equal to a given area. Let the equation to the given circle be + Y =....................... (1); and let (a, b), (a', b'), be the coordinates of the given points. Then, m2 representing the given area, we have, by the condition of the problem, (x - a) + (y - b), + (x - a)2 + (y - 6)2 = M, or x"+y2-(a+ a')x-(b+b')y=(m2-a2-_b2- a'2- b2)...(2). From (1) and (2) we have (a + a')x + (b + V')y = - (2c2- m2 + a2 + b a + )...(3). Thus the required point will be an intersection of the circle (1) with the chord (2). COR. Suppose that a+a' = 0 b + b'=0 and 2c2 - n2 + a2 + 62 + a'2 + b'2 = O, or da + b2 + c2 = a2 = a'2 + 2 + c. Then the equation (2) becomes an identical equation, and the problem becomes indeterminate; any point whatever in (1) satisfying the conditions of the problem; the required point being thus replaced by a circular locus of appropriate points. CIRCULAR LOCI. 107 This problem affords an exemplification of Playfair's interpretation of the ancient signification of the word porism. Playfair: Transactions of the Royal Society of Edinburgh, vol. III. Lhuilier: Elemens d'Analyse Geometrique et d'Analyse Algebrique, p. 204. 7. A and B being two fixed points and P a point such that AP = p.BP, to find the locus of P. The locus is a circle, the centre of which is in the line AB at a distance from A equal to p/a 2 — 1' and of which the radius is equal to pfa Newton: Arithmetica Universalis, prob. xxvI. L'Hospital: Traite Analytique des Sections Coniques, p. 252. Lhuilier: Elenens d'Analyse Geometrique et d'Analyse Agebrique, p. 129. 8. To find the locus of a point such that the square of its distance from a point (a, b) is equal to the rectangle contained by its distance from a line x cos a + y sin a = 8, and a given line 2c. The required locus is a circle defined by the equation {x-(a+c cosa)}2+ {y- (b+c sin a)}2 = 2c {8+ (a cosa+b sina) + c}. Lhuilier: Elemens d'Analyse Geometrique et d'Analyse Algebrique, p. 131. 9. To find the locus of the intersection of two straight lines, which pass through two given points, and are inclined to each other at a given angle. 108 CIRCLE. Let the axes be so chosen that the one point coincides with the origin and that (a, 0) are the coordinates of the other point. Then, c denoting the tangent of the constant angle included between the two lines, the locus will be a circle defined by the equation a x2 y = (Cx + y) Puissant: Recueil de diverses propositions de Geometrie, p. 199, troisieme edition. 10. AB, AC, are two straight lines given in position; DE is a straight line of given length terminating in them at D, E, respectively; from D, E, are drawn perpendiculars to AB, AC, respectively, intersecting each other in P. To find the locus of P. Let AB, A C, produced indefinitely, be taken as axes of x, y. Let DE = c, L BA O= c. Then the required locus will be a circle defined by the equation x' + y2 + 2xy cos co = sin- Leybourn: Mathematical Repository, New Series, vol. I. p. 115. 11. Given one side of a triangle and the opposite angle, to find the locus of the intersection of the bisectors of the other two angles. The middle point of the given side being taken as the origin, and a perpendicular to it through this point as the axis of y, the locus required will be a circle defined by the equation x, + y2 + 6aycot, = 9a', 2a being the given side and / the given angle. Lardner: Algebraic Geometry, p. 110. 12. Straight lines are drawn from the extremities of a given diameter of a circle to the extremities of a chord which always subtends a given angle at the centre: to find the locus of the intersection of the straight lines. CIRCULAR LOCI. 109 Let a = the given angle, a = the radius of the circle. Let the centre of the circle be taken as origin of rectangular coordinates, and the given diameter as axis of x. Then the required locus will be a circle represented by the equation a x2 + y2 + 2ay tan = a2, the + or - sign being taken accordingly as the diameter and chord are joined towards the same or opposite parts. 13. To find the locus of a point, such that, if lines be drawn from it through the three summits of a triangle, and through each summit be drawn a perpendicular to the line passing through it, the three perpendiculars shall always pass through some one point. Let one summit of the triangle be taken as the origin of rectangular coordinates; and let the coordinates of the other two summits be (a, b) and (a', b'). Let c, c', be the lengths of the two sides which meet in the origin, and a the angle between them. Then the required locus will be a circle, defined by the equation cc' (w2 + y2) sin o + c'2 (bx - ay) - c2 (b'x - a'y) = 0, being therefore the circumscribed circle of the triangle. Stein: Gergonne, Annales de Mathematiques, tom. xv. p. 73. 14. A straight line is drawn through a given point C within a circle, to cut it in points P, P. If a point p be taken in this straight line, such that (C)2 = CP. CP to find the locus of p. If a be the distance of C from the centre, and r be the radius of the given circle, the locus of p will be a circle of which C is the centre, and of which the radius is equal to (r - a2)4. 110 CIRCLE. 15. A straight line, drawn through a point 0, cuts two given circles, the centres of which are C, C', in P, Q, and P', Q', respectively. To find the locus of 0, having given that OP.OQ = OF.P'. Q'. The equations to the two circles being of the forms (x- a)2 + Y = C2 '( a - ')2 + y2 c12 the locus of 0 will be a circle, defined by the equation (x - a)2 + y - c2 = { (x - af)' + y -_ c}. Con. 1. The centre of this circle will lie in CC' or C'C produced accordingly as JL is < or > 1. COR. 2. If i/ = 1, the locus of 0 becomes a straight line. 16. From a point are drawn perpendiculars upon the sides of a regular polygon. To find the locus of the point, supposing the sum of the squares of the perpendiculars to be equal to a given area. Let the centre of the polygon be taken as origin of rectangular coordinates; let n be the number of the sides of the polygon, c the distance of its centre from each side, and S the given area. Then the required locus is a circle represented by the equation 2 ( 2 Lhuilier: Elemens d'Analyse Geomntrique et d'Analyse Algebrique, p. 135. Garnier: Geometrie Analytique, p. 187. 17. The equations to the four sides of a quadrilateral taken successively are u = x cos a + y sin a - 8 = 0 u' = x cos a' + ysin a' -8' = 0, u"0 = Xcos a"+ ysina"- U=c a sin a a"- "=, = 0: the rectangle between the distances of a point P from the first and third of these lines is to that between its distances from the second and fourth, as n to n'. To find the conditions that the locus of P may be a circle. CIRCULAR LOCI. 111 The conditions are n = n' and a + a" = a + a"', or a + a" = 27r - (a' + a) Lhuilier: Elemens d' Analyse Geomntrique et d'Analyse Algebrique, p. 144. 18. To find the locus of a point in the plane of a triangle, such that, if perpendiculars be drawn from it upon the directions of its sides, the area of the triangle, formed by straight lines joining the feet of the three perpendiculars, may be constant. If the equations to the three sides of the triangle be xcosa+y sin a =, xcos a'+y sin a'= ', xcos a" y sina"= 8", and k" be the constant area, the required locus will be two circles denoted by the double equation 2k = (x +y2) ina sin( sin' si(-a-a' ini (a-) i(d'-ca) sin a + sin a sin a + 8 sin (a' - a") x cos (a + a" - a) + y sin (a' + a" - a)} + 8' sin (a" - a) {x cos (a" + a - a') + y sin (a" + a - a')} + 8" sin (a - a') x cos (a + a' - a") + y sin (a + a' - a")} 8 {ssin n (a (a- a) sin (a" - a)} Querret: Gergonne, Annales de Mathkmatiqzes, tom. XIV. p. 280. Sturm: Ibid., tom. xIv. p. 286. ( 112 ) PARABOLA. SECTION I. Referred to the Axis and its Tangent. Ordinates. 1. A parabola, of which the equation is 2 = 1X, is cut in four points by the circle ( - a)2 + ( - ))2 = V: to find the product of the distances of the four points of intersection from the axis of the parabola. Eliminating x between the equations to the two curves, we get (y2 -_ la) + 12 (y - fi)2 = 122V or y4 +......+ 12 (a'2 + /S - 72) = o. Whence, by the theory of equations, the required product is equal to 1 (U2 + i3 _ 2) or, a being the distance between the centre of the circle and the vertex of the parabola, to 12 (a2- y2). 2. Two ordinates of a parabola meet its axis in points equidistant from the focus. If the vertex be joined with the point, where one of the ordinates meets the parabola, to find the equation to the locus of the point where this line intersects the other ordinate. The equation to the parabola being y2 = 4mx, let the abscissa3 of the two points be x and x'. THE AXIS AND ITS TANGENT. ORDINATES. 113 Then, by the hypothesis, x + x'= 2........................ (1). The equation to the line joining the former point and the vertex is Y y, =.x,.21 I Cx 2 = 2 (,.....................(2), and the equation to the ordinate of the other point is X, = X..............................(3). At the intersection of (2) and (3), 2 x + x' = 4m + x, and therefore, by (1), (2m - x) y,2 = 4mx,, the equation connecting the coordinates of every point of the curve formed by the intersection of the line through the vertex and the other ordinate, that is, the equation to the required locus. 3. The rectangle contained between two ordinates y,, y,, of a parabola y2 = 4mx, is equal to n2, the distance between these ordinates being equal to m: to find the magnitudes of y, and y,,. y, = i (/5 - 2)1, y,, = n (V5 + 2)1. 4. To prove that the area of a triangle inscribed in a parabola is equal to 8m (y Y... ) (y y)' where y', y", y"', are the ordinates of the vertices of the triangle, y2 = 4mnx being the equation to the curve. 5. The abscissa and double ordinate of a segment of a common parabola are a and b, and the diameters of its circumscribed and inscribed circles D and d; to prove that D + d = a + b. 1 114 PARABOLA. 6. If a parabola intersects a circle in four points, to prove that the ordinates of the points of intersection, which lie on one side of the axis of the parabola, are together equal to the sum of the ordinates of the points of intersection, which lie on the other side of the axis. De La Hire: Sectiones Conicce, lib. v. prop. 29. SECTION II. Referred to the Axis and its Tangent. Tangents. 1. Two tangents are drawn to a parabola; the one touches it at a point (a, b), the other at a point (a', b'). To find the point in which the two tangents intersect. If 4m denote the latus-rectum,_ the equations to the tangents will be by = 2 (x + a), by = 2m (x + a'). At the intersection of these two lines, eliminating x, and observing that b2, b'2, are respectively equal to 4ma, 4ma', we have (b - b') y = 2m (a - a') = (b2- b_ 2) whence y = (b + b'). Also, eliminating y, we see that ab' - a'b b - b' Thus the coordinates of the required point are ab' - a'b a b+' V and. b-b' 2 2. In the parabola, of which the equation is y2 = 4mx, two tangents are drawn at points of which the abscissa are in the ratio of 1:/f; to find the equation to the locus of their intersection. THE AXIS AND ITS TANGENT. TANGENTS. 115 If (x', y'), (x", y"), be the two points of contact, the equations to the two tangents will be yy' = 2mx + -y'2, yy" = 2mx + 1 Y2. But y2 = 4mx', y" = 4 = 4Px" mx', and therefore y"= ~ y I.........................(1). At the intersection of the two tangents, we have, from their equations, Y(y" - y') = 1y - y2) y = (y + y') = -(1 +~P) y by (1). Hence from the former of the equations to the tangents, we see that 2mx=y (y-y )=.I y )mx 92 - ( '2 +/X; the appropriate equation to the locus being, by virtue of (1), y= (,a + j 4) mX or y2 =- (4 - L 4)2 mx, accordingly as the points of contact are on the same or on opposite sides of the axis of the parabola. 3. To find the area of the triangle included between the tangents to parabola 2 =2 4=x y' = 4mx, y, = 4tkx, at points the common abscissa of which is a and the portion of the ordinate intercepted between the two curves, The required area is equal to 2 ( Umt ' ).a 116 PARABOLA. 4. To find the magnitude of the ordinate of such a point in a parabola, that the intercepts, on the axes of coordinates, of a tangent drawn to the curve at this point, may be equal to each other. The required magnitude = half the latus-rectum. SECTION III. Referred to the Axis and its Tangent. Magical Equation to the Tangent. 1. A straight line, inclined at an angle b to the axis of x, touches both the curves y2 = 4mx, x2 + 2 = 2: to find the value of 0. Put tan P = a: then the equation to the line, since it touches the former curve, will be m y = ax + -, and, since it touches the latter, y=ax + c (1 + a2)4 Since these two equations represent the same line, they must be identical; hence m c(1 +02)-=-M whence c sec = m cot c, or c sin b = m cos2 cq; an equation which determines the value of f. 2. Two straight lines, which are always tangents to a given parabola, are such that the sum of the cotangents of their inclinations to its axis is constant; to find the equation to the locus of their intersection. MAGICAL EQUATION TO THE TANGENT. 117 The equation to either tangent is rn = ax +a I1 y I x whence 1 - + =0. a m a m Hence, a', a", denoting the two values of a, 1 +1 y a a m but, by the condition of the problem, 1 1 7- ~ - = ^ a a,3 denoting some constant quantity. Hence y = m/3, the equation to the required locus, which is therefore a diameter of the parabola. 3. If al, a2, be the trigonometrical tangents of the inclinations of any two tangents of a parabola to its axis, to find the equation to the tangent at the extremity of the diameter which passes through their intersection. The equation to the parabola being y2 = 4mx, the equations to the first two tangents will be m = ax + -, and Y a +and m a2 and y - a\x +-. At their point of intersection, yl being its ordinate, l (a -a,) = M(at -; and therefore Y -A (a, + a2 which is the equation to the diameter which is the equation to the diameter 118 PARABOLA. The equation to the required tangent will therefore be 2mn y -P- (x + x) = y ( 2m = - x + -2Y1, y1 2ala2 al + a,2 or y= x + m. al +- a2 2ala2 4. From the vertex of a parabola a straight line is drawn, inclined at an angle Ir to the tangent at any point; to find the equation to the locus of their intersection. Let the equation to the straight line be y = x........................... (1); then, by the hypothesis, the equation to the tangent being a +........................ (2), a' — c 1. + a we shall have 1 = =- a; 1 + a a 1 - a hence the equation (1) becomes 1+a 1....................... ) Obtaining a from (3), in terms of x and y, and substituting its value in (2), we shall get Y-x y+x y +x y-x or (y2 + x).(y - x) = m (y + X)2, which is the equation to the required locus. 5. To prove that the three altitudes of any triangle circumscribed about a parabola all pass.through a single point in the directrix. The equations to the three sides of the triangle will be of the forms y=a + =a..... (1), y=+-......(), y +......(3). a a a....() MAGICAL EQUATION TO THE TANGENT. 119 The equation to one altitude, viz. that which passes through the intersection of (1) and (2) at right angles to (3), will be, as may easily be ascertained, aac (X + a"Y) = m (+ + a"a)...........(4). By similarity it is clear that the equation to the altitude, which is at right angles to (1), is a'a" (x + ay) = m (1 + c"a + c+ a')............ (5) At the intersection of (4) and (5) x = - n, y =, (1 + ac'" + a"a + an'). These results, being symmetrical in relation to, a', a", shew that the three altitudes all pass through a single point. The value of x shews that the point lies in the directrix, which is therefore the locus of the intersection of the altitudes of a variable tangential triangle. Steiner: Gergonne, Annales de Mathematiques, tom. XIX. p. 59. 6. To find the equation to the tangent of a parabola y' = 4mx, which passes through the point in which the directrix cuts the axis. The required equation is y = x + m. 7. To find the locus of the intersection of two straight lines, which always touch a parabola y = 4mx; the product of the trigonometrical tangents of their inclinations to the axis of x being a constant quantity k. The locus is a straight line of which the equation is m X =. Rochat: Gergonne, Annales de Mathematiques, tom. Ii. p. 229. 8. To find the distances of the vertex and focus of a parabola from the tangent, in terms of the inclination of the tangent to the axis of x. 120 PARABOLA. If ) denote the inclination, the required distances are respectively equal to m 2 Co m mcos2'b and - sin a sin 9. If, from the focus of a parabola, lines be drawn to meet the tangents at a constant angle, to prove that the locus of the points of intersection will be that tangent to the parabola, the inclination of which to the axis is equal to the given angle. 10. Two tangents to a parabola make angles 0, 0', with its axis. To find the locus of their intersection, having given that sin O.sin O' is invariable in magnitude. Put sin 0.sin 0' = a; then, y2 = 4mnx being the equation to the parabola, the required locus will be a circle defined by the equation (n (X - _)' + y2 = - _ 11. Two tangents to a parabola make angles 0, 0', with its axis. To find the locus of their intersection, supposing cot 0 - cot O' to be invariable. The equation to the parabola being y2 = 4mx, the required locus will be another parabola, the equation to which, a denoting cot 0 - cot 0', is y2 = 4mx + a2m2. 12. To prove that the area of the triangle formed by three tangents to a parabola, the cotangents of the inclinations of which to the axis are successive terms of an arithmetical progression, of which the common difference is X, is equal to X'm2, 4m being the latus-rectum. 13. From two points in the diameter of a parabola, two pairs of tangents are drawn to the curve; the trigonometrical tangents of the inclinations of the one pair to the axis are a,, a, and, of the other pair, a., a: to prove that 1 1 1 - + = -- + - a, a, a, a, THE AXIS AND ITS TANGENT. NORMALS. 121 14. If any hexagon be described about a parabola, to prove that its three diagonals will all pass through a single point. (This problem is a particular case of the same proposition in regard to any conic section). Lubbock: Philosophical Magazine, August 1838. Ellis: Cambridge Mathematical Journal, vol. I. p. 204. 15. If x, y1; x2, y2; x3, y;...... be the coordinates of the angles of any re-entering polygon of 2n sides, circumscribing a parabola, to prove that X1X3X5...... X2_n-1 = X2X4X...... X2n and Y1 - Y2 + Y3 -..... - Y2 = 0. Ellis: Cambridge Mathematical Journal, vol. II. p. 48. 16. To prove that the continued product of the abscissae of the points of intersection of any number of tangents to a parabola, is equal to the continued product of the abscissa of the points of contact, provided that no three points of intersection lie in the same straight line. Ellis: Cambridge Mathematical Journal, vol. II. p. 48. SECTION IV. Referred to the Axis and its Tangent. Normals. 1. In a parabola y,2 l Cx, the ordinates of three points, such that the normals pass through the same point, are yl, Y, y3: to prove that Y1 + 2 + = 0 and to find the equation to the circle, passing through these three points. Let (x, y) be the coordinates of any one of the three points; a, 3, being the common point of the normals. Then ' 2 (a- X) P-y=-~.2 Eliminating x between this equation and y' = lx, we have 2y3 = 1(2a - ) y + 12.................. (1). 122 PARABOLA. The form of this cubic shews that 1 + Y2 + Y3 = 0. Let the equation to the circle passing through the three points be x + y + Xx + p/y + v=0. From this equation and that to the parabola, we have, for the points of intersection, y4 + l(1 + X) y2 + 1y + 2 = 0............(2). Multiplying this equation by 2 and attending to (1), we get 1 (2a + 2) y + 1 ( + 2) y 2 2) y + 212v = 0. This equation must be satisfied by Y1 Y2) y, as roots: hence we must have 2a + I + 2X = O, / + 2p/ = 0, v = 0. These three equations reduce the equation to the circle to the form 2 + y2 - (a + l) x - My = 0. This equation shews that the circle passes through the vertex of the parabola. 2. To find the locus of the intersections of the normals at any two points of a parabola, on opposite sides of its axis, the ordinates of which are as 1 to 2. The equations to such a pair of normals are. (1, Y ~ y ^ Y - 4m /.................. 2y'= (................. (2). Multiplying the equation (1) by 8 and adding the result to the equation (2), we have, at the intersection of these two lines, 9y - 6y' - 3y x 3my 2m - x THE AXIS AND ITS TANGENT. NORMALS. 123 Substituting this expression for y' in (1), and reducing, we shall get for the equation to the required locus F= (- (x - 2m)3, Y 27m ( ) ' which is the equation to the evolute of the parabola. 3. If SL be drawn from the focus S of a parabola perpendicular to the normal at any point P, to find the locus of the point L. If tan-1a be the angle made by the tangent at P with the axis, the coordinates of P will be m 2m Hence the equation to the normal at P will be 2rn 1 m\ Y - - --.................(1). Also, the equation to SL will be y = a(x - m).................,......(2). From (1) we have y + ax = m + 2a, and, from (2), ax - y = am. Hence, at L, 1 a) m +a Y=a; and therefore, from (2), y1 = m ( -m), which is the equation to the locus of L. The locus is therefore a parabola the latus-rectum of which is one-fourth of that of the original parabola and of which the vertex is S. 4. To find the equation to the normal of a parabola, which is inclined at any proposed angle to the axis of the curve. 124 PARABOLA. If a = the tangent of the normal's inclination to the axis of the parabola, the equation required will be ax = y + (a + 2) am. 5. To prove that, y2 = 4mx being the equation to a parabola, three normals may be drawn to the curve from any point within the area of the curve defined by the equation 27my' = 4 (x - 2m)3, and, from any point without the area, only one. 6. Two normals to a parabola are always at right angles to each other: to find the locus of their intersection. The equation to the parabola being y' = 4mx, the required locus will be another parabola defined by the equation y2 = m (x - 3m). Bobillier: Gergonne, Annales de Math4matiques, tom. xvI. p. 281. 7. To draw, from a given point, normals to a parabola. Let (a, b) be the coordinates of the given point, and let y2 = 4mx be the equation to the parabola. There will be generally three normals passing through the intersections of the parabola with the circle represented by the equation x2- (2m + a) x + y2 - Fby = 0. James Bernoulli: Analysis et Constructio Problematis Hugeniani, Opera, tom. II. p. 700. 8. The axis Ax of a parabola is divided into a number of successive portions AM,, M~MV iM,2M,1...... equal respectively to 1, 31, 51,...... where 1 denotes the latus-rectum. Circles are described with AMi, M~M2, 2f.,...... as diameters. To prove that two radii of every circle, except the first, are normals to the parabola. De la Hire: Sectiones Conicce, lib. VII. prop. 43. THE AXIS AND ITS TANGENT. CHORDS. 125 SECTION V. Referred to the Axis and its Tangent. Chords. 1. If chords be drawn to a parabola, all passing through the point where the axis meets the directrix, to find the equation to the locus of their middle points. Let (x,1 y), (x,, y), be the two extremities of any one of the chords. Then, y denoting the ordinate of its middle point, Y = 2(Y1 + Y2)....................... (1). Now y1, y2, are the roots of a quadratic resulting from the elimination of x between the equations y = a (x + ).......................(2), and y2 = 4mx; that is, of the equation 4m y2 - y +4m2 = 0: 4mn hence Y1+ 4 = and therefore, by (1), 2m y =.(....................... (3). Eliminating a between (2) and (3), we get for the equation to the locus, y2 = 2m ( + n), which belongs to a parabola similar to the original one; the distances between their vertices being m. 2. If a circle, described upon a chord of a parabola as a diameter, meets the directrix, to prove that it also touches it; and to shew that all the chords, for which this is possible, intersect in a single point. Let (x', y'), (x", y"), be the two ends of the diameter. Then the equation to the circle will be {x - (xX + X)} + {y- (y' + Y)}' = I{(x" - ') + (y-y')'}, or x2 +Y - (I' + X') - (y' + y") y + x'x" + yy" = 0. 126 PARABOLA. Let the equations to the chord and parabola be y = ax + /, y2 = 4mx. At their intersection, a2' - 2 (2m - a/3) x + /2 = 0, and ay" - 4my + 4m/ = 0: hence x + x" 2 = 2 (2m - a), XIX 2 4m, 4mn/3 Y +Y Y - a The equation to the circle is therefore 2 2 4m /32 4m/3 + y2 + (a /-2rn) x- y+ + = 0. W Ca d a If the circle meets the directrix, x = -, and therefore 4m 4n2 (/3N 2 Y2_- y + - - + + = o............(1). In order that y may be possible, we must have + = 0................ (2), which shews that the equation to the chord is y = a (x - m); the chords therefore all pass through the focus. The roots of (1), under the condition (2), are equal, and therefore the circle touches when it meets the directrix. 3. To find the equations to all the common chords of the two curves y2 = 2cx - x2, y2 = 4mx. The required equations are = 2 (c-2m), 8 + -(.2 4. To inscribe in a given parabola a chord Pp of given length, passing through a given point Q. THE AXIS AND ITS TANGENT. FOCAL PROPERTIES. 127 Let Pp intersect the axis Ax of the parabola in 0, and let QN be the ordinate of Q. Let AN= a, QN= b, Pp = c, ON= z, 1 = the latus-rectum. Then the value of z will be defined by the biquadratic equation 14 - 4-Wz3 + b'l (4a + 1) z2 + 4 (4a1 - c) = 0. Newton: Arithmetica Universalis, prob. XIv. SECTION VI. Referred to the Axis and its Tangent. Focal Properties. 1. P' P", being any two points in a parabola, and 0 the point of concourse of the tangents at these points, and S the focus: to prove that (P 0)2 _ (p /0)2 P'OS - P"0S p's PI's Let the equations to the tangents at P', P", be qa y=ax + i7 y = a"x +,* Combining these two equations we obtain, for the coordinates of the point 0, m m,, 1 ' ) Y1 = Ia' (a' + a") Also, combining each of the equations to the tangents with that to the parabola, viz. y2 = 4mx, we have, for the coordinates of P', P", respectively, and, 2m 2m Hence (P' 0)2 = (xI - x') + (y - y') = 2 (.+ l) (l 1)2 =m [l+^)^ ~^)f 128 PARABOLA. and (P'S)2 = (xI _ m)2 + y'" - m (i+ ) We have therefore (P'O)2 /( 1)2 PIS' _ at at Symmetry shews that we must have also (P -0)2= M (- 1)2 pit o a" a') Hence (P' 0)2 (P 0)2 P'S P"S~ Massabieau, Guillaume, Gobert, Berard: Gergonne, Annales de Matlhmatigues, tom. Iv. p. 183. 2. To find the locus of the intersection of a tangent at one extremity of a focal chord of a parabola with the ordinate at the other produced. Let x', y, be the coordinates of the point of contact of the tangent; and let (x, y) be the point of intersection of the tangent and ordinate. Then yy = 2m (x + x'), yY2 = 4m' (x + x')2, y = m (x+ x')........................(1). ButI 1 1 1 But.... m SP +P' m + x m m + x ' 1 x whence = m + x m m + x)' m2 and = -. Hence, from (1), Y = m (x + - ) mxy2 = (X2 + n2)2, the equation to the required locus. THE AXIS AND ITS TANGENT. FOCAL PROPERTIES. 129 3. To prove that a series of circles, of which the centres are in a parabola, and which pass through the focus, all touch the directrix. 4. PT, PG, are the tangent and normal of a parabola, at a point P, terminating at T, G, in the axis: SQ, SR, are perpendiculars upon PT, PG, respectively, from the focus S: to find the area of the rectangle SQPB, and to determine the position of P when the rectangle is a square. The equation to the parabola being y2 = 4mx, the area of SQPR, if x, y, are the coordinates of P, is equal to }y (x + m); and, when the rectangle is a square, the point P is at an extremity of the latus-rectum. 5. If from the focus of a parabola as centre and with the focal distance of a point in the parabola as radius, a circle be described, to prove that the intersections of the tangent and normal to the parabola at the point, with the axis, will lie in the circumference of the circle. 6. If SL be drawn, from the focus S of a parabola, at right angles to the normal at any point P: to prove that the abscissa of the point L is equal to SP, and that, AM being the abscissa of P, (SL)2= AM.SP. 7. In the preceding problem, to find the locus of the point L. The equation to the parabola being y' = 4mx, the required locus is another parabola defined by the equation y2 = m x - m) K 130 PARABOLA. SECTION VII. Referred to a Tangent and its diameter as Axes. 1. From a point 0 are drawn two lines to touch a parabola in the points P and Q: another line, parallel to PQ, touches the parabola in R and intersects OP, 0 Q, in S, T, respectively: if V be the intersection of the lines joining PT, QS, crosswise, to prove that 0, R, V, are in the same straight line. Let R T be taken as the axis of y, and the diameter through R as the axis of x. Since PQ is parallel to SR T, we may take (x', y') and (x', - y'), as the coordinates of Q, P, respectively. The equations to Q O, PO, will accordingly be of the forms 2m 2 (x + W) and y = (x + x). These two lines therefore intersect in the axis of x. Moreover the coordinates of T are 1 -= 0 2mx' Y1= y = The equation to PT is therefore x' + 3y'x = xy Similarly, putting - y' for y', we see that the equation to QS is xy - yx = - xy. These two lines PT, QS, therefore also intersect in the axis of x. Thus 0, R, V, all lie in the axis of x. 2. From any point 0, in the arc of a parabola, a straight line OR is drawn, parallel to the axis of the curve, to meet a chord Qgq produced if necessary, in R. To prove that RQ.Rq c R O. REFERRED TO A TANGENT AND ITS DIAMETER AS AXES. 131 Let a tangent PT, and its diameter Px, be taken as axes of coordinates, PT being parallel to q Q. Produce OR to meet PT in E. The equation to the curve is of the form y'2 = lx. Let A, k, be the coordinates of R: then, if we put h for x, and k - y for y, the values of y in the resulting quadratic (y - k)2 = 11, or y2 - 2ky + k2 - lh = O, will be equal to R Q, Rq: hence RQ.Rq = 1 - Th, = (OE- RE), = l.RO, c R O. De la Hire: Sectiones Conicce lib. v. prop. I. 3. If two tangents be drawn to a parabola, to prove that a third tangent, parallel to the chord joining the points of contact, will bisect the parts of the other tangents which are included between their point of intersection and their points of contact. K2 132 PARABOLA. 4. BC, CD, are two consecutive arcs of a parabola, the sagittme of which, bisecting their chords and being parallel to the axis, are equal; to prove that the chord BD is parallel to the tangent at C. 5. From a point B, without a parabola, two straight lines are drawn, one touching the curve in P, and the other cutting it in Q. A straight line HEF is drawn, parallel to BP, cutting in the point H the diameter through B, and the curve in the points E, F. HQ is joined. To prove that (BP)2: HE.EF:: BQ: HQ. De la Hire: Sectiones Conicce, lib. III. prop. 28. 6. Qq is a chord of a parabola: from any point R in this chord is drawn REF, parallel to the axis of the parabola, meeting the curve in E, and the tangent, drawn through the point Q, in F: to prove that RQ: R:: EF: ER. De la Hire: Sectiones Conicce, lib. v. prop. 25. 7. Through a point P in a parabola, a chord is drawn to cut in Q, q, another parabola, equal to the former, the axes of the two curves lying in a single straight line: to prove that, Qq remaining always parallel to itself, the rectangle PQ.Pq will remain constant for all positions of P. De la Hire: Sectiones Conicce, lib. vi. prop. 13. SECTION VIII. Referred to two Tangents as Axes. If the parabola be referred to two tangents as axes of coordinates, its equation will assume the symmetrical form (a+ b=+1, REFERRED TO TWO TANGENTS AS AXES OF COORDINATES. 133 a and b being the lengths of the tangents from their intersection to the points of contact. Puissant: Recueil de diverses propositions de Geometrie, p. 105, troisieme edition. Gregory: Cambridge Mathematical Journal, vol. II. p. 14. 1. To prove that two parabolas cannot touch each other in more than one point. If possible, let the tangents at two points of contact be taken as axes of coordinates, and let a, b, be the distances of these points of contact from the origin. Then the equation to either parabola will be ()~ ( = + 1; and therefore the two parabolas coincide. Ilapa/3oXr 7rrapa/3oX\7 ov':c ea7TrTeTat KaTra 7rXelova a-/xelta } ev. 'AIIOAAQNIOT IIEPrAIOT KwvlK6,V tro T~Traprov, nporartais Kn. 2. The equation to a curve referred to oblique axes, which are inclined to one another at an angle 0, is x% + y = c: to prove that, when the axes are changed into a system of rectangular ones, of which the axis of x bisects 0, the equation will become 2a sin2 y' 2= co 0 (x- a cos 0) Let P be any point whatever. Let OM, PM, be the coordinates of P in regard to the original axes Ox, Oy; and OM', P ', in regard 0 \M to the new axes Ox', Oy'. Draw M M'K, parallel to xO, meeting PM in K. 134 PARABOLA. Then x= Q - M'K,sin - 0, cos -0 sin0 a sin 0 _ sin 0 Y sin 0s XIx ' l sin 1+ COSi 2 cos0 + 2 sin 10' Now x + y = at and therefore x + y + 2 (xy)> = a: hence ( y'-" = '2 2 a cos - 0 cos sin I 2C 0 - 0,2 2a sin20 y12 -2 (o f a cos2 0). 3. If three parabolas be described, having their principal diameters iconios in the continuations of the lines bisecting the three angles of any plane triangle, and if each parabola touch one side of the triangle and the prolongation of the other two: then, if A denote the area of the triangle; a, b, c, its sides; s half the sum of the sides; and r~t 'rTT9 r3 the semiparameters of the three parabolas, it is required to prove that (6 + c) (c +' a) (a - b) /1223 abc '\s,1 Hi Let HKT be the triangle. If S be the focus of the parabola in the A\ _l / diagram, TS will bisect the angle T \T s A = PTK. If P, Q, b e the points of contact, it is plain that PT = QT. f Let TP, TQ, be taken as axes of i coordinates. T hen the equation to REFERRED TO TWO TANGENTS AS AXES OF COORDINATES. 135 the parabola, if TP = m, will be xs + y = + m....................... (1). Let TH= a, TK= b. Then += 1........................... (2) is the equation to a tangent to the parabola. But the tangent at a point x,, y1, is + 1.................. (3). (mxg)- (My )4 Comparing (2) and (3), on the supposition of their representing the same line, we see that (mx1)4 = + a, (my,1) = b: and therefore, by virtue of (1), putting x, y,, for x, y, we get a + b = m = PT. But, by known properties of the parabola, PT= 2SPcos A, and PT sin'2 ~A = 2AS cos ~A; hence 7r = 2AS= (a + b) sin 4 sin2 LB Similarly r = (b + c) sin21C and 73 = (c + a). s i1C cos C, ' Hence W12TT _ +(b+c)(c+a)(a+b) s (s-a)(s-b)(s-c))} Hence 7r7rrabc = abc ' s Lady's and Gentleman's Diary, 1844. 4. From P, the point of concourse of two tangents to a parabola, viz. PQ, PQ', a straight line PABC is drawn, meeting 136 PARABOLA. the curve in A, C, and the chord Q Q' in B. To shew that PA, PB, PC, are in harmonical progression. Let PQ = a PQ' = b. Then the equation to the parabola, referred to these tangents as axes, will be +( =-+..................... () The equation to Q Q' will be a b + = 1.(2). The equation to PABC we may take to be = 3........................... (3). Let xA, x,, x,, be the abscissa of A, B, C, respectively: then, from (1) and (3), ( =b 1 - (a) ( }; 1 (h1N f (/3\~)2 ( 21\ whence ) + 1 {(( -) (; x \a/ \b/ X}/ 3 a 7b and, from (2) and (3), - = + I x a b 2 1 1 Hence we see that -=-+-. X2 X1 X3 But it is plain that x,1 x2 x3, are proportional to PA, PB, PC; hence 2 1 1 PB PA + P De la Hire: Sectiones Conicce, lib. II. prop. 21. 5. To determine the magnitude of the latus-rectum of the parabola represented by the equation xi ~ y = + a, the axes being rectangular. The latus-rectum = a /2. REFERRED TO TWO TANGENTS AS AXES OF COORDINATES. 137 6. TP, TQ, are two tangents to a parabola: any other tangent cuts these two in points p, q, respectively: to prove that T2p T 1 TP TQ 7. PT, QT, are two equal tangents to a parabola, P and Q being the points of contact. If PT, Q T be cut by a third tangent in E, F, respectively, to prove that PE=FT, and QF= ET. 8. If there are three tangents to a parabola, the triangle formed by their intersections is half of that the angular points of which are the points of contact. Gregory: Cambridge Mathematical Journal, vol. II. p. 16. 9. Three straight lines MN L, NL LM, produced if necessary, touch a parabola in the points P, Q, R, respectively; to prove, that. RL:LM::LQ: QN::VMN:NP. \ De la Hire: Sectiones Conicce, lib. III. prop. 20. 10. Two tangents OA, OB, are drawn to a parabola: in the line AB, joining the points of contact A, B, is taken any point whatever E: from E are drawn the straight lines EH, EK, parallel to BO, AO, respectively, and cutting AO, BO, respectively in ET, K: to prove that HK will touch the parabola in some point P, and that PE is a diameter of the curve. De la Hire: Sectiones Conicce, lib. III. prop. 21. 11. Two straight lines OA, OB, touch a parabola in A, B. The chord AB is joined. M is the middle point of AB; OM is joined. A third tangent to the parabola cuts OA, OB, respectively in H, K. 138 PARABOLA. If LMOA = X, z MOB = /., OA = a, OB = b, OHT= a, OK= /3, to prove that the equation to the directrix, OA, OB, being chosen as the axes of x, y, respectively, will be x cos X + y cos = cot (X + i). (a sin X + 83 sin i). COR. It may easily be shewn that the directrix always passes through a fixed point of which, o denoting the angle between the axes, the coordinates are equal to COS o COS co sin2 (P - a os cos), si (a - p cos c). 12. Two tangents to a parabola, the lengths of which are a, b, intersect in 0 at an angle co, and a circle is inscribed between the tangents and the curve; to find the distance of the centre of the circle from 0. The required distance = ab (a + b) sec o + 2 (ab)q.tan 2 c 13. Parabolas are described touching two given straight lines at right angles to each other; to find the locus of the vertices of the parabolas, supposing the chords of contact to be all parallel to a fixed line. Let the two given straight lines be taken as axes of coordinates; then, 1, m, being the direction-cosines of the fixed line, the locus of the vertices will be a straight line represented by the equation x y 3 - m3 SECTION IX. Referred to any Rectangular Axes whatever. Reduction. The most general form of the equation to a parabola is ax2 + by2 + 2cxy + 2a'x + 2b'y + c' = 0, the parameters a, b, c, being subject to the relation c2 = ab. ANY RECTANGULAR AXES WHATEVER. REDUCTION. 139 The object of the general problem of reduction is to ascertain the positions of the vertex of the parabola and its focus. This object may be effected by first turning the axes of coordinates through such an angle that the coefficient of xy in the transformed equation may be zero, and then changing the origin into such a position that the final equation may consist of only two terms, one term containing the first power of one of the coordinates, and the other term containing the second power of the other. The equation will then be reduced to the more simple form belonging to the axis and its tangent as axes of coordinates. 1. To find the position of the vertex and the magnitude of the latus-rectum of the parabola x q- y~ = +_ a. Squaring the equation, we have x + 2xey + y = a, and thence (x + y- a)2 = 4xy, or x2- 2xy + y2 - 2ax - 2ay + a = 0. If we turn the axes through an angle 0, the equation will become (x' cos 0 - y' sin 0)2 - 2 (x' cos 0 - y' sin 0).(x' sin 0 + y' cos 0) + (XI sin 0 + y' cos 0)2 - 2a (x' cos 0 - y' sin 0) - 2a (x' sin 0 + y'cos0) + a' = 0. Assume the coefficient of x'y' to be equal to zero; then -2 sin 0.cos 0-2 (cos2 0 - sin' 0) + 2 sin 0.cos = 0, and therefore cos 20 = 0, 20 = t7r, 0 = T7r. Hence, substituting for cos 0 and sin 0 their common value in the equation to the curve, we get Ix'2 + 'y'2 - 2 (i2 - y22) + '12 + _yI2 -a ('/2 - y'/2) - a (x'/2 + y'12) + a= 0, and therefore 2y'2 - 2/2 ax' + a2 = 0. 140 PARABOLA. Again, changing the origin of coordinates to a point a, 13, we have 2 (y + ) - 2/2.a (x" + a) + ' = 0. Equating to zero the coefficient of y" and those terms which involve neither x" nor y", we obtain 4/ = 0, / = 0, and 2/32 - 2V2.aa + a2 = 0, whence 2V2 azla' The equation then becomes y"2 = /2.ax". Thus the latus-rectum of the parabola is equal to a/2, and the coordinates of the vertex, referred to the original axes of coordinates, are la, Pa. The diagram will elucidate the analytical transformations. L XOx' = 7rT, OA = 2 OH = a cos |:r = ~la, AH = a sin i r = a. aa AS a= 2'2 POLAR EQUATION. FOCUS THE POLE. 141 2. To find the magnitude of the latus-rectum of the parabola of which the equation is (y - mx) = cx. The required magnitude =. (1 + m'2) 3. To find the position and dimensions of the curve 2+2xy + y2 - 3x + y + 1 = 0. The latus-rectum =-, the coordinates of the vertex are 5, -, and the inclination of the axis of figure to the axis of x is 45~. 4. To find the position of the parabola represented by the equation (x + a) + (y + a) = 2a, and the magnitude of its latus-rectum. Its vertex is at the origin of coordinates, its axis bisects the angle between the axes of x and y, and its latus-rectum is equal to 4aV/2. 5. The equation to a parabola being (x + y/3)2 _ 8x (1 + /3) + 8y (1 - /3) + 48 = 0, to find its latus-rectum and the inclination of its axis to the axis of x. The latus-rectum = 4, and the required inclination is equal to Wrr. SECTION X. Polar Equation. Focus the Pole. 1. If, with the focus S of a parabola as centre, a circle be described, passing through the vertex, to prove that the rectangle under the intercepts PQ, qp, of any focal chord PQSqp, included between the circle and the parabola, is constant. 142 PARABOLA. The polar equation to the parabola being r = m sec' 1 0, and that to the circle r = m, the difference between these radii vectores for any value of 0 is equal to m tan' 0. The corresponding difference for a value 7r + 0 of the angular coordinate is equal to n cot2 1 0. The quantities mtan2 0 mcot2 0, are the values of the intercepts PQ, pq: the area of the rectangle included by them is equal to the constant quantity n2. 2. To find the locus of the vertex of a series of parabolas which have a given focus and touch a given line. Let EF be the given line, and S the given focus. Draw SY at right angles to EF. Let P be the point at which one of the parabolas of which A is the vert is touched y of the parabolas, of which A is the vertex, is touched by EF. Let EF cut the axis of the parabola PA in T. Let LASY=, zASP=, YS=c, AS= n. Then, by the nature of parabolas, c = ST. cos 0 = SP cos 0. POLAR EQUATION. FOCUS THE POLE. 143 But SP 2 1 + cos,~~~hence C 2m cos 0 hence c = 1 + cos ( 2m cos 0 1 + cos 20 m cos ' or C Cos 0 _= m, which shews that the locus of A is a circle of which SY is the diameter. This proposition is easily proved geometrically. We know, by a property of the parabola, that L SA Y is a right angle; hence the locus of A is a circle on SY as diameter. Lardner: Algebraic Geometry, p. 129. 3. If r, r', be two radii vectores of a parabola at right angles to each other, and 1 the semi-latus-rectum, to prove that (r ) -+ t1r ) = 2' 4. In the focal distance SP of any point P in a parabola, Sp is taken equal to the distance PN of P from the axis; to find the equation to the locus of p. The equation to the parabola being 2m r =1 + cos 0 that to the required locus will be r 2mtan 0. 5. If, from any point P in a parabola, a perpendicular PK be drawn to the directrix, to prove that, S being the focus, SP, SK, and the latus-rectum, are in geometrical progression. 144 PARABOLA. 6. If A be the vertex, S the focus, and PSp a focal chord of a parabola, to prove that the rectilineal triangle PAp varies as (Pp). 7. From any point in the directrix of a parabola, two tangents are drawn to meet the curve: to determine the length of the latus-rectum of the parabola, the lengths of the two tangents being given. If a, b, be the lengths of the two tangents, the length of the latus-rectum will be equal to 4a2b2 (a2 + b2) 8. A chord QSQ' is drawn through the focus S of a parabola, parallel to the tangent at a point P; to prove that, L denoting the latus-rectum, SQ.SQ' = L.SP. 9. If P be a point in the radius vector produced of a parabola r (1 + cos0) = 2m, such that its distance from the focus is equal to the corresponding focal chord; to find the equation to the locus of P, and to shew that in this curve, if two radii vectores be taken at right angles to each other, the sum of their reciprocals is equal to the reciprocal of the latusrectum of the given parabola. The equation to the required locus is 4m r = sin 0 10. If PT, pt, be the tangents, and PG, pg, the normals at P, p, the extremities of any focal chord PSp in a parabola, T7 t, being the intersections of the tangents and G, g, of the normals, with the axis, to prove that PT.PG _ t ASP pt —.= tan pt.pg 2 POLAR EQUATION. VERTEX THE POLE. 145 11. Let S be the focus of a parabola P1P]P3...P,, of which the vertex is A, axis AM, and P-Pn latus-rectum L. Draw any line SP, and make n angles PSP2 / PSP,,...... PSP,, round S, all Pa equal to each other. Draw SQ A ---- so that the angle MSQ = n times the angle MSP,. It is required to prove that SP. SP82. SP3..... SP-L = L.SQ. p Herschel: Leybourn's Mathematical Repository, New Series, vol. IV, p. 67, SECTION XI. Polar Equation. Vertex the Pole. 1. Two given parabolas have a common vertex A and a common axis: P, P', are points in the two curves, such that AP, AP', are at right angles and equal to each other: to find the magnitude and positions of AP, AP', Let c, c', be the two latera recta; let AP be inclined to the axis at an angle 0, and let r denote its magnitude. Then, by the polar equation to the curve AP, rsin 0 = c cos 0...................(1). Similarly, by the polar equation to the curve AP', r cos2 = csin 0..................... (2). From (1) and (2) there is tan = ( which determines the position of AP, AP'. Also, from (1) and (2), r = c cos 0 + c' sin 0 Cc3 + c'e) 12 (C + C)' which determines the magnitude of AP or AP', L 146 PARABOLA. 2. Having given r, r' two radii vectores of a parabola at right angles to each other, the vertex being the pole and the axis the prime radius vector, to find the latus-rectum. If I denote the latus-rectum, 12 =(rr' ) r3 + r'3 SECTION XII. Polar Equation. Pole a point in the Axis. 1. A point C is taken in the axis of a parabola, at a distance from the vertex A greater than half the latus-rectum; to determine the least distance CP of this point from the curve. Let CA = c, and 4m = the latus-rectum. Then, C being taken as the pole, the polar equation to the curve, if we put LPCA = 0, CP= r, will be r? sin2 8 = 4m (c - r cos 0), whence r cos2 0 - 4mr cos 0 = r2 - 4mc, (r cos 0 - 2m)2 = r2 - 4n (c - m). From this result it is evident that {4m (c - m)}~ is the least value of r. The corresponding value of 0 is given by the equation 2m / m i cos = - =r \c- m the value of cos 0 being always possible, because c is greater than 2m, and therefore c- m than m. If the ordinate drawn from P meets the axis in M, then, from the value - of cos 0, we see that CM is equal to half the r latus-rectum. Apollonius: Conicorum Liber Quintus, prop. 8. 2. In the axis of a parabola is taken a point C, at a distance from its vertex A equal to half the latus-rectum. P is any POLAR EQUATION. POLE ANYWHERE. 147 point in the curve; CP is joined. PM is an ordinate to the axis. To prove that CA is the least value of CP and that (CP)' - (CA)2 = (AM)2 Apollonius: Conicorum, lib. v. prop. 4. SECTION XIII. Polar Equation. Pole anywhere. 1. Through a given point within a parabola, a chord is drawn; to find the position of the chord that the rectangle of its two segments may be equal to a given square. Let h, k, be the coordinates of the given point. The equation to the parabola, referred to its axis and tangent, being y = lx, the polar equation, (h, k) being the pole, will be (r sin 0 + k)2 = I (r cos 0 + A), or r2 sin2 0 + (2k sin 0 - I cos 0) r + k2 - lh = 0. If r,, r, be the roots of this equation, 'F- lh r1.r2= sin ' The point (h, k) lying within the parabola, k2 - 1h is negative: hence, c denoting the side of the given square, we shall have, equating c2 to the area of the rectangle between the two segments of the chord, Ih - k' sin2 0 ' lh - k2 or sin" = 2 This equation gives two values of 0, or two positions of the chord; provided that c2 is not less than Ih k2, when the problem is impossible. L2 148 PARABOLA. COR. If the given point were without the parabola, we should have- sin = - De la Hire: Sectiones Conicce, lib. v. prop. 39. 2. Through a given point 0, within a parabolic area, is drawn a chord POQ: to find the inclination of POQ to the axis in order that the rectangle PO.OQ may be the least possible. The chord must be at right angles to the axis. De la Hire: Sectiones Conicce, lib. VII. prop. 28. 3. If, from a point T without a parabola, two tangents be drawn to the curve, meeting it in P and Q, and the axis in E and F respectively; to prove that PT ET QT- FT' SECTION XIV. Linear Equation. 1. To prove that, in a parabola, the semi-sum of the focal radii vectores which terminate at the extremities of any arc of the curve, is always equal to the radius vector which terminates at the end of the diameter drawn through the middle point of the chord, together with the part of this diameter intercepted between the arc and the chord. Let x, x', be the abscissae of the ends of the are, and x the abscissa of the end of the diameter. Let r, r' r, be the corresponding radii vectores. Let also 4m be the latus-rectum. Then, by the linear equation to the parabola, we have r = m + x, r = m + x, and therefore i (r + r') = m + - (x + x'). But -i (x + x') is equal to the abscissa of the middle point of the chord, that is, to x, + c, where c represents the intercepted POLAR EQUATION TO THE TANGENT. 149 part of the diameter: hence - (r + r') = m + x, + c = r, + c. 2. The abscissa of two points in a parabola, reckoned along the axis, are x, 3x, and the corresponding focal distances r, 2r: to find the position of the former of these points. The required point is an extremity of the latus-rectum. SECTION XV. Polar Equation to the Tangent. 1. If PT, QT7 be tangents at the points P, Q, of a parabola, the focus of which is S, to prove that SP.SQ = ST2. A being the vertex of the parabola, let z ASP= a, and AS = m. Then the polar equation to PT will be 2m — = cos0 + c os (- a...............).). r Similarly, if L ASQ == 3, the polar equation to QT will be 2m — 2 = os 0 + cos ( - 3)................. (2). r At the intersection T, of (1) and (2), cos (O - a) = cos ( - /), whence 0- a = - ( - ), 0 = (a + i), and therefore 2m a +/3 a- f a o 8 ST o = c os + cos = cscos ST=. cos -a.cos I/ * This equation to the tangent of a parabola is given by Mr. Davies in the Philosophical Magazine for 1842, p. 191. 150 PARABOLA. Whence, by the polar equation to the parabola, ST 2= m m cos 2 _ cos2' -2 = SP.SQ. 2. To prove that a circle, described about the triangle formed by three tangents to a parabola, will pass through the focus. The polar equation to a parabola being 2mrz 1 + cos 0 the polar equations to three tangents will be 2m - = cos 0 + cos ( - a)................. (1), -- = cos 0 + cos (0 - a')................ (2), 2m -= cos + os ( - a")............(3), a, a', a", being the angular coordinates of the three points of contact. If,, r, be the coordinates of the intersection of (1) and (2), -= (a + a'), - = cos. cos-. ir, 2 2 Similarly, (0, r), (0,, r,), being the coordinates of the intersections of (2), (3); (3), (1); we shall have nfn fa a 0= =(a + a) -= cos cos —; and 0, = (a" + a), - COS cos -. r, 2 2 It is evident from these results that the three points (0, r), (0,, r,), (0,, r,,), all lie in a circle of which the equation is m / a + a'+ ca" a a a" rcos 0 - 2- = cos 2 cos a 2 POLAR EQUATION TO THE TANGENT. 151 The equation to the circle shews that it passes through the focus of the parabola. Poncelet: Gergonne, Annales de Mathmatiques, tom. vIII. p. 9. Davies, Goodwin: Philosophical Magazine, vol. XXI., 1842, pp. 190, 219. Greathead: Cambridge Mathematical Journal, vol. I. p. 170. Ellis: Cambridge Mathematical Journal, vol. I. p. 206. Haydon: Cambridge Mathematical Journal, vol. IV. p. 192. 3. To determine the magnitude of the latus-rectum of a parabola the equation to which is a + () = 1, w being the angle between the coordinate axes. The form of the equation shews that the axes of x and y touch the parabola, a, b, being the P distances of the points of contact from the origin. a Let PT1 QT, touch the parabola in P, Q, respectively, PT being de- \ - noted by a, and QT by b; then LPTQ= o. Join SP, SQ, ST, S being the focus; let A be the vertex. Put / PSA = a, z QSA =/3. Then the equations to PT, QT, are, respectively, 2m 2m — = cs + os (- a), = cos 0 + cos (O - /), r r and therefore at T, the intersection of these two lines, 0 = ~ (a+/3) = AST, whence z PST = } (a - 8) = QST. But L SPT= i (7r - a), L SQT= T (wr + O). Hence L PTS = 1 (?r + 3), z QTS= (r - a), and consequently w = L PTQ == - ~ (a - )...........(1). 152 PARABOLA, From the triangle STP, a - P sinm 2 a= ST. r - a sinl 2 nm sin c) cos la.cos 1 ' cos - a' m sin o a o a a = cos. cos.(2). — C S' 2.COS2.................. (2). Similarly, from the triangle STQ, m sin 2 6 =Cos - Cos - ~~~i.~~~ (3). - 2.cos 2........ -(3. From (2) and (3), we see that a mb. 3 ma cos -. sin C, cos - 0 = -2 Sin ow: 2 a 2 but, from (1), cos a cos 1/3 + sina sin = - cos Wc) cos a cos2a - + 2 cos o cos - a cos }/3 + cos2 o = sin2 -a sin2 18 = 1 - cos2 a - cos2 3 + cos2 a cos'2/ cos2 a + 2 cos t) cos } a cos }/3 + cos2 /i3 = sin2 CD; hence, putting for cos la, cos ~/, their values, we have m (a2 + 2ab cos a) + b') = (sin co), (ab)" and therefore 4a&b' sin' e) latus-rectum = 4 3sin (ad + 2ab cos co + b2) 4. If a straight line be drawn from the focus of a parabola, intersecting at an angle / the tangent at any point of the curve; to find the equation to the locus of the intersection. The equation to the parabola being 2m 1 + cos 0 POLES AND POLARS. 153 the required locus will be a straight line of which the equation is m r sin/3 m sin (O + 1i)' and which is therefore a tangent to the curve at a point of which the angular coordinate is 7r - 23. 5. A right angle moves in such a manner that its sides always touch respectively two confocal parabolas, the axes of which lie in the same line; to find the locus of the summit of the angle. The equations to the parabolas being 2m 2m' - = 1 + cos 0 = 1 + cos r r the required locus will be a right line, perpendicular to the axis of the parabolas, defined by the equation r cos 0 = m + '. Bobillier: Gergonne, Annales de Mathematiques, tom. XIX. p. 323. SECTION XVI. Poles and Polars. 1. If, from any point of the exterior of two similar parabolas, having the same axis, tangents be drawn to the interior one, to shew that they will touch it at the extremities of diameters, the distance between which is constant. Let the equation to the interior parabola be y2 = 4mx........................ (1), and to the exterior y2 = 4m (x + c).......................(2). The equation to the polar in the parabola (1), corresponding to a pole (x,, y), is yy1 = 2qm ( + ox) or 2yy1 = + 4 mx 4m.................... (3). 154 PARABOLA. At the intersections of (1) and (3), we have y2 - 2y1y + 4mx, = 0. Let y', y", be the ordinates of these intersections: then y" + y' = 2y1, yy' = 4mx1 (y" -y') = 4y 1 - 16m................ (4). Suppose the point (x,, y,), to lie in (2): then, from (2) and (4), (y" - y')2 = 16mc, y" - y = 4 (mc)l, which is the distance between the two diameters. 2. To find the locus of a pole, relatively to a given parabola, the portion of the corresponding polar intercepted by the curve being constant. Let the equation to the parabola be y2 = 4m x........................... (1). Then, (h, k) being the pole, the equation to the polar will be ky = 2m (x + h)......................(2). Let (x,, y1) be the middle point of the intercepted portion of the polar: then, putting in (1) x = xw + r cos0, y = y + r sin0, we shall have r sin'0 + 2r (y, sin0 - 2m cos0) + y2 - 4mx = 0...(3). Also (2) becomes k (r sin0 + y) = 2m (r cos 0+ + + ); but, (x1, y,) being a point in (2), kyl = 2m (1 + A)................... (4), and therefore k sin0 = 2m cosO..................... (5). At the intersection of (3) and (5), 0 is common to both, and the values of r in (3) are c and - c, 2c denoting the length of the intercepted portion of the polar. Hence, by the theory of equations, y, sinO = 2m cosO....................(6), c2 sin 0 a = 4mx - y*.................. (7). POLES AND POLARS. 155 From (5) and (6) we see that,= k........................... (8). From (4) and (7) c2 sin2O + y,2 = 2ky1 - 4mh; whence, by (8), c sin2O = k - 4mh or c2 sin20 = (ke - 4mh) (cos2O + sin'O), and therefore, by (5), 4m2c = (k2 - 4mh) (k2 + 4m2), the equation to the required locus. 3. If the normal at one extremity of the latus-rectum of a parabola be the polar, to prove that the corresponding pole will lie in the diameter through the other extremity of the latusrectum, and to determine its exact position in this line. If y' = 4nmx be the equation to the parabola, the coordinates of the pole will be x=- 3m, y =-2m. 4. To find the locus of the intersection of a perpendicular drawn from the vertex of a parabola to meet a polar, the pole of which lies in a similar external parabola having an axis coincident with that of the inner parabola. If the equation to the inner parabola be y2 = 4zmx, then, a denoting the distance between the vertices of the two parabolas, the equation to the required locus will be (m + x) y2 = (a - wx x'. 5. Any number of parabolas are described having the same vertex and axis, and any straight line is drawn without the parabolas at right angles to the common axis. If any points whatever in this line be taken as poles, to prove that all the polars belonging to all the parabolas will intersect in a single point. 156 PARABOLA. 6. The polar belonging to a pole P, relatively to a given parabola, intersects the curve in points E and F: the triangle EPFbeing supposed to have a constant area, to find the locus of P. The equation to the parabola being y, = Il and c2 denoting the constant area, the required locus will be a parabola defined by the equation 2 = x + (1 C21)3. SECTION XVII. Intersection of Parabolas. 1. If two parabolas have their axes perpendicular to each other; to prove that, if they cut each other in four points, those four points will lie in a circle. The equations to the two parabolas, supposing the axes of coordinates to be parallel to their two axes, will be of the forms y2 + i3y + ax + 7y = 0 x2 + 3'x + a'y + y' = 0. At the intersections of these parabolas, we shall have 2 + y2 + (a + ') + ( + a') y + 7 + 7 = 0. This equation shews that all their points of intersection lie in a circle. 2. Three parabolas, of which the axes are parallel to each other, intersect: to prove that the three chords joining their points of intersection pass through a single point. 3. If a parabola, of which the equation is y2 = lx, be intersected in four points by a parabola, the diameters of which are parallel to the axis of y, to prove that the sums of the ordinates of intersection on opposite sides of the axis of abscissa, will be equal to each other. De la Hire: Sectiones Conicce lib. v. prop. 30. INTERSECTION OF PARABOLAS. 157 4. Ax is the axis of a parabola AE, Ay being the tangent at A. Also Ay is the axis of a para- F bola AF, Ax being a tangent to it at A. To prove that, E, F, being the latera-recta of the parabolas A E, N AF, respectively, E: PM:: PN: F, P being the point of intersection of the two curves, and PM, PN, at right M A angles to Ax, Ay, respectively. This is one of two methods of finding two mean proportionals between two given lines E and F. discovered by Menechmus, a geometrician of the School of Plato. His two solutions of the problem of the two means afford the earliest instances of the application of geometrical loci and conic sections to the solution of problems which cannot be solved by means of the rule and compass, called by the ancients solid problems. The problem of the two means excited much interest among the ancient philosophers by reason of its intimate connection with the celebrated Delian problem of the duplication of the cube. Many solutions were accordingly discovered by various Greek geometricians; by Eudoxus, Plato, Hero, Philo, Apollonius, Diocles, Pappus, Sporus, Menechmus, Architas, Eratosthenes, and Nicomedes. The solution of Diodes, as depending upon the construction of his cissoid, has been rendered more complete by Newton, who has given in his Universal Arithmetic a method of describing this curve by continuous motion. For an exposition of the various solutions of this problem the reader is referred to the Commentaries of Eutochius on the Second Book of Arch~`imedes, Hepi 2calpas Ical KvlXvSpov; to the Mathematicce Collectiones of Pappus, lib. III., where he will find an account of several of the solutions; and to Montucla's Histoire des Math4matiques, tom. I. p. 186. AzXios, ryap Xotl^/oactv, e'XprEev 5o Ar6XXwov, araXXa7yao-Oat0 Tov Xotpovo, el Trv O/%I/Ov 8&7rXaoCtdoaovotv, KCV/3Lq~V eXovra aXco'Xa. o 8 e7Trico/oJ-uvaav 7rpoo'0vreve s T rporepp 158 PARABOLA. f) rpo cv/3ov to-ov' aJXX' A Tv iOVo Cvf3ewv oavvOjKfrl, TO TOV fic6Vov xcr'rla X 'oilwe rye'ryove &e avT'l Cv/3ov, &oKc[. TOD Xot/pov 8oe /L/ 7ravo-a/jpeov, eXpraev 0o 06os pj 7r7rrorflcevaat avrov9 TO 7rpo-TaXOev. o 6 fev yap 7rpoeTa-e 7wrXactra'at To7V cv/,ov' TOVTe7T&o, f3Eoiobv tcaTaT'UcevaatL IKV/3LKV TOV TrpoTrEpoV 8trrXao-ov Ol 8e, CKVR3o e7rl KV'C0 e7rEO7av. iXOov oiv 7rpo s HXaTr va C 7LTOVVTe JL~00EooV, 7rcog av TOYv cvf3ov 8t7rXaoTtcaLev ev' 6 pe 'rpo? avT'sov (qcro-v, eKcev V/JLV oVEtleteV 6 O eco &? adp.eXov(' )/eOpefjTp'ta. 6 &e TOV fCV/30ov 87rXaotao-u'o", evpe60('ralTaL (qa'tv, etl VO evOLei v 8Vo peoat avadXoryov evpeOeLev' cKa TOVOTO TO 7rpo/3'X/cLa TOS pcaO?7'Tal? 7rpoe/3A\XXev Ol'TtVE9 Kcal 7rept TOVTOV ryeryp acrc'v (S9 86eSVwrTaC e'caaTro' o tv oveVTt 7epLE'cgeTLeTa /beXpt 70TV vVv, d\\X' ov8' 6 ryeoupeTp'9 7rep6 Tro3Tv eTreo-wpjpva'o' ao o /Jo v yap 80oe0ELcv eV0 eLtCV, pLo-'ov avdaXoyov EVpe, pe6V EOeTO Tr7V '7ro'8 tv 'IQANNOY TOT rPAMMATIKOY, EI TA "YSTEPA 'ANAATTIKA 'APISTOTEAOYT, THOMNHMA. Venice, 1534, p. 24. Kopt,'o^pevos, ibv a'r' Atyv7'rrov w'nepl Kapiav Aq\iXwv Ttvves a7rl7vT7]l-av, 8eo0,evoit IaTrovos 0 7 yecoe-TpLKOVc Xvo-at Xp?7O7Jov avTos, arTOW7OV v7rO TOV) Oeov 7rpo,3e,/3Xq)/evov. )v 8e XpqroSb, AqVX\oL'0 IcaKa Toos a\\XXos "EXXo-t 7ravXav T'&v 'rapovTrwV Kcaicv 'eo'eOat 87r'Xaca'do'ao' TOY v eA Xt /3ci)wpo6v' orTe 8e \T\v 8.votav eKcelvob 'v/i/3daXXe6v 8vvdIevo, Kcal 7rep TT)V TOV /ojOV KcaTaCtKevrv yeXota 7rao-'ovTe (ecao yap TCV Teo-o-apwv 7rXeVpS)v 8t7rXaoaaopje'v's, ')XaOov Tr' av'eTt TO7TOV 'Tepeov oVKTa7rXa-crov arepyao-apevo., S caretpiav avavXorylac, q Tw fL/xce ICE 7rdXartov vrapeXeTa,) FIXaT)wva T?''? a7roplas e7r~caxovvTo /3o'OOv. o6 8, 'TO Ai7VT'rilov jvfurOelLs, 7rpoo'7rai^'Etv 'bf TOVv Oebv 'EXX/ctrV, o3Xyiwpovaol vra7 elas, olov efvfl3prtovTa T)V al/aOlav fl/tWov, Kca, icexevovTa ryeco/JeTpacp a6TGrreoaOaL f/ib rapepyov. ov yaprTo ofavDXov, ov8' da/fL3Xv &8avo'lac 6poo'W, acp(oA 8e T~a' rypapLLad?^oC')/Levrs Sepyov eva, Vca Ovolv Ueowv avAoov Xy'ttv pov &87rXacTa'Ta& or-x ia Kcv3Lecov 'row/aTOo, eK jrado? o6poois avSo:/xevov c&aarTdoew. ' TOVTo Lev ovv Eivo!ov avTols TOv Kvi'ov, iq To'v KvL'crvbov 'EXwc6va, avvTeXere',v. IIAOYTAPXOT HIIpl Tro6 SQKPATOY AAIMONIOY. PARABOLIC LOCI. 159 SECTION XVIII. Parabolic Loci. 1. A triangle is constructed on a given base, its vertex lying in a given line parallel to the base: to find the locus of the intersection of perpendiculars from the extremities of the base on the opposite sides. Let AB be the base, and C the vertex. The perpendiculars if: c h if 1 A n L B. AHt BK, CL, from A, B, C, on the opposite sides of the triangle, will meet in a single point P. Let Ox, Oy, be the axes of coordinates, 0 being the middle point of AB. Then, putting AO = a= BO, CL =b, OL = x, PL =y, we have Y = tan PAL=cot CBL= a- a + x b whence a - x2 = by, the equation to the locus of P, which is therefore a parabola, the latus-rectum of which is equal to the distance between AB and the locus of C, and which passes through A and B. 2. Two straight lines SPs, TPt, are drawn through two points S, T, the former of which is fixed and the latter arbitrary, 160 PARABOLA. of an indefinite straight line xAx'. PM is at right angles to xx'. Supposing that the angles SPT, STP, are always equal, and that, A being a given point, AT is always equal to AM, to find the locus of P, in which the two straight lines intersect. Let Ax be taken as the axis of x, and a line through A, at right angles to it, as that of y. Then, denoting AS by m, the equations to Tt, Ss, will be respectively of the forms y = ax + /3........................ (1), y = ' (x - m)........................(2). Since LSTP = LSPT, we have ax -cx a= 1 + aa....................... (3); 1 +T cxa and, since AT = AM, we have /s + Ma' af//3-+a.....................(4). From (3) and (4) there is m=/S+ ' p + Ma' l + aa' ' a,/ = m........................... (5). From (4) we have 2a/3 = a'(/3 - ma), and therefore, by (5), 2m = a'(, - ma)..................... (6). From (2) and (6), 2m (x - m) = y ( - ma)............... (7). From (1) and (7), y2 2m (x - m) y2 + 2x (x - m) Y (x + m) y (x + m) Substituting these expressions for a and, in (5), and simplifying the result, we shall get {y2 + (x - m)2}. (y2 - 4mx) = 0. Rejecting the former factor, the equating of which to zero would give impossible values to the coordinates, we have Y2 = 4mx, PARABOLIC LOCI. 161 the equation to a parabola of which A is the vertex and S the focus. Raymond: Gergonne, Annales de Mathmzatiques, tom. III. p. 143. 3. To find the locus of the centre of a circle inscribed in a sector of a given circle, one of the bounding radii of the sector remaining fixed. Let a be the radius of the circle of which OBQ is a sector, OB being the fixed radius. Let Q p denote the radius of the inscribed circle: r, 0, the polar coordinates of P, the centre of p the inscribed circle, LPOB being equal to 0 and OP to r: then, manifestly, o 0 r + p = a, and p=r sin: a hence r = 1 + sin9' the equation to a parabola of which 0 is the pole and focus, and of which the latus-rectum is equal to twice OB. 4. Two lines DE, DN, indefinitely produced, are given in position: MEN is a given angle, the summit E of which always _N/ X C M D A B E P moves along DE, while the side EN always passes through a given point C: moreover EM is a third proportional to NC, CE. To find the locus of M. Draw CA parallel to ND; and draw CB so that L CBE = CEM. From A draw A Q, meeting the indefinite line DN M 162 PARABOLA. in Q, so that /BAQ = /ABC. Draw MP parallel to QA, meeting the indefinite line DE in P. Let AD =, AB= b, BC= c, AP=x, PM= y, AE=z. Then, by the construction, LEPM = DAQ = L CBE: also LPEM+ LCEM = LCEP= LCBE + LBCE = zCEM + LBCE, and therefore LPEM= LBCE. The two triangles PEM, BCE, are therefore equiangular. By the condition of the problem, (CE) = EMNC................ (1) But, by the similar triangles PEM, BCE, CE BC c EM=f EP = x- @ a o e (2).z Also, by the similar triangles CAE, NDE, CE AE z N G ~ = AD ~....................... (3)a z - - z From (1), (2), (3), 7 = C...........................(4). Again, by the similar triangles PEM, BCE, PM l BE PE B C y z-b (5). or - =........................ (5). x - z c Eliminating z between (4) and (5), we shall readily get, putting a + c = f, for the equation to the required locus, x2 - y = 0. a a " Le Comte Roger de Vintimille a propose ce Probleme avec quelques autres dans le Journal de Parme du mois d' Avril de l'annee 1693, ce qui a donne occasion au Pere Saquerius de faire imprimer un petit livre a Milan, dans lequel il avoue qu'il PARABOLIC LOCI. 163 n' a pa resoudre celui-ci, quoiqu' il fasse assez paroitre par la solution des autres qu' il est fort verse dans la Geometrie." L'Hospital: TraitN Analytique ces Sections Coniques, p. 254. 5. To find the equation to the locus of a point the distance of which from the origin of coordinates is equal to its distance from the line (y - c) cos 0 + x sinO = 0, and to refer it to axes inclined to the original ones at an angle 7r- 0. The required locus is a parabola the equation to which, referred to the new axes, is y = c cos0 (c cos0 - 2x) 6. In a plane triangle ABC, if tanA. tan 1B = 2, and AB be fixed; to find the locus of C. If AB= c, BC= r, LABC = 0 the locus of C will be a parabola defined by the equation c 1 + cos O 7. From any point Q, in a fixed ordinate of a parabola at right angles to its axis, a diameter is drawn to meet the curve in P. A being the vertex of the parabola, and AB the abscissa of Q, BA is produced to C~ so that AC = AB. To find the locus of the intersection of A Q, P, produced indefinitely. The required locus is the parabola itself. 8. A is the origin, B a point in the axis of y, BQ a line parallel to the axis of x: in A Q (produced if necessary), P is taken such that its ordinate is equal to BQ: to find the locus of P. If AB = c, the locus will be a parabola defined by the equation y2= Cx. M 2 164 PARABOLA. 9. If, from any point P of a circle, PC be drawn to the centre C, and a chord PQ be drawn parallel to the diameter A CB and bisected in R, to find the locus of the intersection of CP and AR. If c be the radius of the circle; then, ABC being taken as the axis of x and a diameter, at right angles to it, as the axis of y, the required locus will be a parabola of which the equation is y2 = 2cx + c2. 10. OA is a fixed straight line of length a, POP' a straight line through 0 making an angle 0 with OA. To find the locus of P or P', having given that the product of the triangular areas A OP; A OP', is equal to a4, and their quotient to cot' 2 0. The locus of P or P will be a parabola represented by the equation a 1 + cosS 11. Having given one side of a triangle, and the sum of the tangents of the adjacent angles, to find the locus of the vertex. Taking the origin of rectangular coordinates at the middle point of the given side, the given side being chosen as the axis of x, we shall have, for the equation to the required locus, m (a2 - x2) = 2ay, 2a denoting the given side and m the given sum. Lardner: Algebraic Geometry, p. 116. 12. To find the locus of the centre of a circle, which passes through a given point and touches a given straight line. Let the given straight line be taken as the axis of x, and a line perpendicular to it, through the given point, as that of y. Let c be the ordinate of the given point. Then the required locus will be a parabola defined by the equation x2 + c = 2cy. Puissant: Recueil de diverses propositions de Geometrie, p. 196, troisieme edition. Garnier: Geometrie Analytique, p. 180. PARABOLIC LOCI. 165 13. To find the locus of the centre of a circle which touches both a given circle and a given straight line. Let the given straight line be chosen as the axis of x, and a perpendicular to it, through the centre of the given circle, as that of y. Then, y being the radius of the given circle, and / the ordinate of its centre, the required locus will be a parabola defined by the equation x = (3 ) 2y ~ - p). Puissant: Recueil de diverses propositions de Geometrie, p. 197, troisieme Edition. Garnier: Geometrie Analytique, p. 439, deuxieme edition. 14. If SY be the perpendicular drawn from the focus of a given parabola upon the tangent at any point P; to find the locus of the centre of the circle circumscribing the triangle SYP. The equation to the given parabola being y' = 4mx, the required locus will be a parabola defined by the equation y' = m (2x - m). 15. A straight line DCE, passing through two given points D, C, cuts a given straight D line BA, produced, in E. Two points F, G, are taken in EA, such that AF E BG DE /' 2 — B and the straight lines CF, D G, are produced to meet in P. To find the locus of P. p Let EAB, ECD, produced indefinitely, be taken as axes of x, y, respectively. Let EC = c, ED = k', EA = a, EB = a'. Then the required locus will be a parabola defined by the equation (a'- a'k) (kk'+ y2) + kk'(k - k') x + {a'k2 - ak'2 (a'- a) kk'} y = 0. Leybourn: Mathematical Repository, New Series, vol. I. p. 45. 166 PARABOLA. 16. To find the locus of the focus of a parabola, which has a given vertex, and which touches a given right line. Let A be the given vertex: from A draw AB at right angles to the given right line, meeting it in B. Let BA, produced indefinitely, be the axis of x, and Ay, at right angles to AB, be the axis of y. Then, a denoting the distance AB, the required locus will be a parabola represented by the equation y' = ax. Lardner: Algebraic Geometry, p. 130. 17. Two straight lines OA, OB, are divided each of them into n equal parts. The ends of the rth and (r + 1)th divisions of OA, reckoning from A, are joined respectively to the ends of the rth and (r + I)th divisions of OB, reckoning from 0. If H be the middle point of AB, OHx be the axis of x, and Oy, parallel to BA, that of y; to prove that, if OH = p, AH = q the locus of the intersection of the two joining lines will be a parabola defined by the equation n2py2- 2n'q2x + (n2 - 1)pq2 = 0. Brianchon: Journal de 1' Ecole Polytechngiue, 19e cahier, tome XII. Puissant: Recueil de diverses propositions de Geometrie, p. 201, troisieme edition. Traite de Topographie, No. 182, 2e edition. Garnier: Goome6trie Analytique p. 443, deuxiemne edition. SECTION XIX. Parabolic Envelops. 1. To find the locus of the ultimate intersections of perpendiculars drawn to the normals of a parabola at the points where they cut the axis. The abscissa of the intersection of a normal to a parabola y = 4mx, drawn at a point (x, y) of the curve, with the axis, PARABOLIC ENVELOPS. 167 is x + 2m: the equation to the perpendicular to this normal will therefore be 2m, y, (= - x - 2m), or yy' = 2m (x' - 2m)- y. Differentiating this equation with regard to the parameter y, we get y = -Y: hence the equation to the required locus is - y2 = 2m (x'- 2m) - y"'2 or y'2 = 4m (2n - x'); which is the equation to a parabola similar in form to the original one, its concavity being opposite and its axis coincident, and its vertex being at a distance 2m from the origin. 2. A straight line PQ moves with its extremities P, Q, in the sides OA, OB, of a triangle A OB, so that the rectangle of O P A x the segments OP, O Q is always equal to that of the segments AP, BQ: to determine the curve to which PQ is always a tangent. Let OA = a, OB = b, OP= rm, OQ = n. Then, OAx, OBy, being chosen as the axes of coordinates, the equation to PQ will be + 1........................ (1), tm n in, n, being subject to the condition inn = (a - n) (b ), m1 n oa =....................... (2). 168 PARABOLA, Differentiating (1) and (2), we have dm + dn= 0, dm dn a + ED = O~ a b whence, X being an arbitrary multiplier, X x X y a wme' b n /m n\ x y whence X + — + \a b nm n or, by (1) and (2), X = 1. We have, therefore, m2 = ax, n = by, and therefore, by (1) or (2), a -) r which is the equation to a parabola to which OA, OB, are tangents at the points A, B. Servois: Gergonne, Annales de Mathematiques, tom. IV. p. 156, 159. 3. Circles are described on successive double ordinates to the axis of a parabola as diameters: to shew that their envelop is an equal parabola, and that, of this system of circles, those of which the diameters are less than the latus-rectum, do not admit of an envelop. 4. If P be any point in a parabola, the equation of which is y2 = la and M, N, be the points where the perpendiculars from P meet the axes of x, y, respectively; to find the locus of the continued intersection of lines MN. The required locus is a parabola represented by the equation y2 = - 41x. 5. To find the locus of the consecutive intersections of straight lines drawn through points KI L, which move in MISCELLANEOUS PROBLEMS. 169 parallel straight lines so as to make OK always equal to OL, O being a fixed point in the locus of K. Let 0 be the origin of rectangular coordinates, the locus of K being the axis of x. Let B be the intersection of the axis of y with the locus of L. Let OB = b. Then the required locus will be a parabola defined by the equation X = b.(b - 2y) 6. A straight line cuts off from a parabola a segment equal to a given area: to find the equation to the curve to which this line in any position is a tangent. Let c represent the given area; then, the equation to the parabola being y = 4mx, that to the required envelop will be y= 4m x -- }. SECTION XX. Miscellaneous Problems. 1. The equation to a parabola, referred to rectangular axes, being y' + 4ay cot a = 4ax, to find its equation when referred to oblique axes inclined to each other at an angle a, the axis of x remaining the same. The required equation is. 4a sin2 a 2. P is any point in a parabola of which the vertex is A, and focus S; T is the point where the directrix intersects the axis; TP is joined and produced to cut the latus-rectum in N; SPQ is drawn to meet NQ, which is parallel to ST, in Q: to find the locus of Q. 170 PARABOLA. The equation to the parabola being yf = 4nmx, the required locus will be a circle represented by the equation x2 + y2 = (2x + 3m). 3. To find the locus of the intersection of the tangent at one extremity of a focal chord of a parabola with the ordinate at the other extremity produced. The equation to the parabola being y2 = 4mx, that to the required locus will be mXY2 = (x2 + y2)2. 4. A circle is inscribed in a parabola, the distance of the centre of the circle from the vertex of the parabola being to the radius of the circle as n2 + 1 to n2 - 1. To find the positions of the points of contact. The equation to the parabola being y'2 = x, the abscissa of the points of contact is In (n + 1)2 5. Having given a diameter of a parabola, and a tangent through the vertex of the diameter, to find the locus of the vertex of the curve. The given diameter being taken as the axis of x and a perpendicular to it through its vertex, as that of y, the required locus is a straight line of which the equation is y = 2x tan a, where a denotes the angle between the given tangent and diameter. Lardner: Algebraic Geometry, p. 132. 6. Under the conditions of the preceding problem, to find the locus of the focus. MISCELLANEOUS PROBLEMS. 171 The axes of coordinates being the same as in the preceding problem, the required locus is a straight line represented by the equation y = x tan 2a. Lardner: Algebraic Geometry, p. 132. 7. The tangent to a parabola, at a point P, meets the latusrectum produced in L, and the directrix in I; to find the locus of the centre of the circle inscribed in the triangle SML, S being the focus. The equation to the parabola being r = m sec2 0, that to the required locus is r = n -m.sec. sec2' -0 '2 tE ( 172 ) ELLIPSE. SECTION I. Referred to its Axes. Ordinates. 1. IF a, /, are the coordinates of the centre of a circle to radius y, which cuts in four points an ellipse a'y2 + b2 =2 = b..................... (1); to find the value of the product of the four ordinates of intersection. The equation to the circle will be (x _ a)2 + (y -/3)2 = V Multiplying this equation by b2 and subtracting the result from the equation to the ellipse, we have (ad - 6a) y + 22b2/y - b282 + 2b2aX - b2a = 62 (a' - y)...(2). Multiplying (1) by 4b62 we have 4a'62bay2 + 4b4a2xz = 4a2,b4a2, and therefore, by (2), 4a2bO&y2 + (b2 (a2 + a2 + /32 - 72) - 2b3y - (a2 b62) y2}2 = a4ac2, or (2 - b2)2y4 +......+ b4 {(a + a +,32 - 72)2 - 4a2a}) = 0. Hence, by the theory of equations, y1, y,2 y2, yy4 being the four ordinates of intersection, we have YYY4 = (a2 + a2 + 32 - 7'2)2 4a2a2 Y1921Y34 = b - (a - b 2 -2 (a2 _ b)2e 2. If ordinates y1, y2, divide the major axis of an ellipse into two parts x,, x,', and x2 x'2t respectively, to shew that REFERRED TO ITS AXES. ORDINATES. 173 I denoting the latus-rectum, X1.X 2a' and Y2 = Xl'X Y2 X2.X2 'Eav eV KVXblvSpov rTO/tU evOetaL aO'bv 'r TOv 8aLLeTpov TeTray7uevcov' eCoTa& Ta a7r avTrov yrerpayoyva 7rpo? KfEv Ta 'repLeXopJeva Xwopa v't7o 7rv a7roxaaP3avoevjLv, vr' avTv 7trpo, Trot0 7rpaCoT r79 7rXay/a9 TO ' e8Sovq TrXevpaS, &q Tro eov97 v 6pOlba 7rXevpa vrpo ' TrV 7TrXarytav vrpto? eavTa 6 co T'ra 7repLeXopjIeva Xopia v7ro Tr5v, c, e'prfTat, a7roXa/pj3avop/jevv evO) eeLv. SEPHNOY ANTINIEUS bIAOZOOT IlTpl KvXtivpov Tor's;. HpoTrarLs Ll. 3. A perpendicular AH is drawn to the semi-axis-major CA of an ellipse, through its extremity A, equal to half the latusrectum; an ordinate MP to the axis cuts the straight line CH in Q. To prove that (PM)2 = 2 area QAEH. Apollonius: Conicorum, lib. v. prop. 1. 4. An ellipse and a circle having the same centre, the axis major of the ellipse is equal to three times the radius of the circle: to determine the points of intersection of the two curves, supposing the foci to cut the distance between the extremities of the diameter and axis in a given ratio; and to determine that ratio when the ellipse and circle just touch each other. Let AE= c, AS= -, A being one vertex of the axis, m S the nearer focus, E the nearest point of the circle. Then 3c = + 3m -- 1 (4m2 - 6m + 1)>; and, when the ellipse and circle just touch each other, m = (3 + V5). Leybourn's Mathematical Repository, New Series, vol. IV. p. 130. 174 ELLIPSE. 5. About the centre of an ellipse a2 b2 _ + - 1, is described a circle, with a radius equal to the ordinate: to find the locus of the intersection of this circle with the ordinate. The required locus is an ellipse defined by the equation a2 + b2 2 Yl a2b *x + = 1. Lardner: Algebraic Geometry, p. 151. SECTION II. Referred to its Axes. Tangents. 1. If (a, /), (a, /3'), be the coordinates of two points in a diameter of an ellipse, and be subject to the condition aa' /313' a2 + 6 = to find the equations to the tangents at the extremities of this diameter. If x, y, be the point of contact, X2?Y2 -.( =1 i................... (1). But, the points (a, /), and (a', /3'), lying in the same line with the point (x, y), x a a' Y G and therefore = y aa /3/3' ( //aa l e hence, from (1), - + -2 = 7 and consequently, by the condition of the problem, = +(aa) similarly y = + (/33'). REFERRED TO ITS AXES. TANGENTS, 175 Hence the equations to the tangents are (2 +.y'=+l, 2. To find the area of the parallelogram formed by tangents at the extremities of diameters 2r, 2r', of an ellipse, the angle between these diameters being 0. Let the semi-axes of the ellipse be a, b, and (x, y), (x', y'), the extremities of 2r, 2r', respectively. Let 4 denote the angle between the tangents at their extremities. Then xx — + = cos (,-~~,-^,7, and therefore sin2 = 1 - cos2 S = (xY — 2 - C= C4 ( a4b4 / x b4 (XY - Iry)2 2 8, ~', being the perpendiculars from the origin upon the two tangents. But, u denoting the area required, it is easily seen that 88' u = 4 q; sin <, hence, X, X', denoting the inclinations of r, r', to the major axis, 4a'2b' 4a2b2 xy' - xy rr' (sin X' cos X - cos X' sin X) 4a2b2 rr' sin (X' - X) 4a2b2 rr sin 0' 3. In the tangent at any point P of an ellipse is taken a point T, such that the angle PCT, subtended by PT at the 176 ELLIPSE. centre of the ellipse, is always a right angle: to find the locus of the point T. The coordinates of P being a cos 0, b sin 0, the equation to PT will be - cos 0 + Y sin 0 = 1....................(1). a b Also, A being the extremity of the axis major, 1b sin 0 tan z POA = a c a cos 0 Hence the equation to CT, which is perpendicular to CP, is sin 0 ax cos - y........................ At the intersection T of the lines (1) and (2), the equation to the former of which may be written x - cos 0 + Y sin 0 = (cos' 0 + sin2 0, a b (/, we have. y y. 2y + a2) a b (a)2 bS (a b)2 which is the equation to the required locus. 4. To find the equation to the tangent of the ellipse x2 + 2y = 3, at the extremity of the latus-rectum. The required equation is x y 1 /6 + 3 5. If h, c, represent the intercepts of the axes of coordinates, made by any tangent of an ellipse a2 y~ a+ W = 12 a2 b2 to prove that.. +, = 1 REFERRED TO ITS AXES. 177 6. A tangent at any point P of an ellipse meets the semiaxis-major, produced, in T, and the semi-axis-minor, produced, in t; from P are drawn PM, Pnm, perpendicular to CT, Ct, respectively. To compare the areas of the triangles PMT, Pint. The coordinates of P being x, y, triangle PMT: triangle Pmnt:: a4y4: b4x4 7. The circumstances of the preceding problem remaining the same, to find the locus of a point P' in Tt, such that P'T n Pt - m The equation to the required locus is /ma (nb, +) ( + ( n). 8. If, from a point without an ellipse, two tangents be drawn to the curve, and two straight lines to the foci; to prove that the angle contained by one tangent and the line drawn to one focus, will be equal to the angle contained by the other tangent and the line drawn to the other focus. Leybourn: Mathematical Repository, New Series, vol. II. p. 40. 9. To find the locus of the intersection of an ordinate of the ellipse 2 Y2 with the perpendicular drawn from the centre upon the tangent at the extremity of this ordinate. The required locus will be another ellipse represented by the equation 2 2y2 -i d —4 = 1. a+ a Lentheric, Vallbs, Bobillier: Gergonne, Annales de Mathe matiques, tom. XVII. p 377. N 178 ELLIPSE. 10. AA', BB', are the diagonals of a rhombus ABAB', which intersect in C. If a, /, be the semi-axes of any ellipse described with C as a centre and axes in directions AA', BB', so as to touch the sides of the rhombus; then a, /, are coordinates of a point in the ellipse having AA', BB', for its axes. 11. In two ellipses, concentric and similarly placed, are taken two points P, Q, the abscissae of which are as their major axes; P', Q' are two other such points. To prove that, 0, 0', p, O', being the angles which the tangents at P, P', Q, Q', respectively, make with the major axes, tan 0 tan b tan O' tanl ' SECTION III. Referred to its Axes. Magical Equation to the Tan gent. The equation to an ellipse being a2 b2 the straight line defined by the equation y = ax + (a'2' + b), will be a tangent to the ellipse, varying in position with the value of a. This equation may be seen in Leroy's Geometrie and in Waud's Algebraical Geometry: it seems however, together with analogous equations to the tangents of parabolas and hyperbolas, to have been first employed by Mr. A. Smith for the demonstration of various properties of conic sections. Owing to its great utility, particularly in problems of tangency which do not involve the consideration of the point of contact, it has been called the Magical Equation to the tangent. Smith: Cambridge Mathematical Journal, vol. I. p. 9. MAGICAL EQUATION TO THE TANGENT. 179 If I, m, be taken to represent the direction-cosines of the tangent, its equation may be expressed in the more symmetrical form I form Ix + my = (l2a2 + mi2b2). 1. Two straight lines, such that the product of the trigonometrical tangents of their inclinations to the axis of x is constant, touch an ellipse a2 2+ to find the equation to the locus of their intersections. The equation which connects the coordinates of any point in a tangent to an ellipse with a, the trigonometrical tangent of the inclination of the tangent to the axis of x, is (y - x)2 = a' + b,2 or (x2 - a2) a - 2xya + y -. 2 =. If therefore x, y, be the coordinates of the intersection of the two tangents, the two values of a in this quadratic will be the trigonometrical tangents of the inclinations of the two tangents to the axis of x. Hence, m denoting the constant product, we have for the equation to the required locus - == m (x'2- a2) which represents an ellipse or an hyperbola accordingly as m is negative or positive. Rochat: Gergonne, Annales de IVmathematlques tom. II. p. 225. Garnier: Gometrie Analytique, p. 275, deuxieme ecition. 2. To find the locus of the intersection of a pair of tangents to an ellipse, at right angles to each other. The equation to one tangent being x cos a + y sin a = + (a' cos2 a + b2 sin2 a), that to the other will be - x sin a + y cos a = + (a2 sin2 a + b1 cos2 a): adding together the squares of these two equations, we get x2 + y2 = a2 + b2, N2 180 ELLIPSE. as the equation to the required locus, which is therefore a circle concentric with the ellipse, having (a2 + b2)' for its radius. Lame: Examen des diffrentes methodes employees pour resoudre les problmes de Geometrie, p. 77. Puissant: Recueil de diverses propositions de Geometrie, p. 214, troisieme e'dtion. 3. If p, p', are the distances of the foci of an ellipse from a tangent, to prove that pp' = b2. The equation to a tangent being Ix + my = (12a2 + m2b2)^, where i, m, are its direction-cosines; and the coordinates of the foci being {(a2 - b2), 0}, and {- (a2 - b2)&, 0}, we have p = - (2a2 + m2b2) + l(a2 - 2), ' = - (12 + m2b2) - (a2 - b); whence pp' = l"a2 + m'2b2 - P (a2 - b2) = (12 + m2) b2 = b2. 4. To find the locus of the middle points of chords in a circle which touch a concentric ellipse. Let the equation to the ellipse be _ -1 a2 y The equation to the tangent will be, 1, m, being its directioncosines, l + my = (a2 + m2b2).................(1). Let the equation to the circle be + ye = c2......................... (2). At the intersections of (1) and (2), x-2 - 2(a' + 2b).X + ab)^ + 2 + m%'2 2 - m = 0. MAGICAL EQUATION TO THE TANGENT. 181 Hence, x, denoting the abscissa of the middle point of the chord, and x', x", the abscissae of its extremities, x = (X' + x") = l(l1a + m2b2). Similarly y = (y' + y") = m (12La + m2b2). Hence x,2 + y,2 = 12c +- m2b2. But also mx, = ly,, and therefore, by virtue of the equation 12 + m2 1, we have 12 (x,2 + y2) =,2, m2 (X,7 + y,2) = y,2. Hence (x 2 + y,2)2 = a2X2 + b2y2, which is the equation to the required locus. 5. To find the locus of the summit of a moveable right angle, one side of which touches one, and the other side the other, of two confocal ellipses. Let the equations to the ellipses be - -+ C _ 12 +, a2+6{ ~ a,2 6~2 where a, b, a', 6', will be subject to the condition a2 - b2 = a2 - b2..................... (1). The equations to the two tangents will be respectively Ix + my = (12a2 + Mn22), - m + y = (m2'2 + 2b'2)". At the intersection of these two tangents, we have, squaring and adding their equations, (12 + n) (2 + y2) = 12 (a2 + b'2) + in2 (a 2 + b2) = (12 + 2) (a2 + b'2), by (l), or = (12 + m2) (a'2 + 62): thus the locus of the summit of the angle is a concentric circle represented by either of the equations a + b'2 = x2 + y2 = a'2 + b2. Bobillier: Gergonne, Annales de Mathematiques, tom. XIX. p. 317. 182 ELLIPSE. 6. To find the equation to a tangent to an ellipse 3x2 + y2 = 3, inclined at an angle of 4 r to the axis of x. The required equation is y = x +2. 7. To find the distance of the centre of an ellipse from a tangent inclined to the major axis at an angle b. Distance = a (1 - e2 cos2 <)&. 8. To find the distance of the focus of an ellipse from a tangent inclined at an angle q to the major axis. Distance= a {e sin b + (1 - e2 cos' ")i}. 9. The tangent of an ellipse is inclined to the major axis at an angle b: to find the area of the triangle included between this tangent and the two axes. Required area = - (a2 tan Q + b2 cot q). 10. A tangent to an ellipse X2 y2 a2 + 2= 1l is inclined to the axis of x at an angle sb; to find the product of the distances of the ends of the axis major from this tangent. The required product is equal to b2 cos2 b. 11. To find the equation to the locus of the intersection of a tangent to an ellipse with a perpendicular upon it from a given point. The coordinates of the given point being a, /, and the equation to the ellipse being xt2 2 ct2+ MAGICAL EQUATION TO THE TANGENT. 183 the required locus will be defined by the equation {x (x - a) + y (y - )}2 = a2 (x - )2 + b2 (y - /9)2. If the given point be the centre, the equation becomes (x"2 + y2)2 = a'22' + b2y2, a result given by Vallbs: Gergonne, Annales de Mathematiques. tom. XVII. p. 379. 12. To find the equation to the curve, from any point of which if two tangents be drawn to a given ellipse, the angle contained between them shall be constant. If - denote the value of the tangent of the constant angle, n the required equation to the curve will be m (x2 + y2- a2 -_ 2)2 = 4n (a2y2 + b2X2- b)_. 13. If the two sides of a moveable right angle are always tangents to a given ellipse, its summit will describe a circle concentric with the ellipse, the radius of which is equal to the chord joining extremities of the major and minor axes. Rochat: Gergonne, Annales de MJatmheatiques, tom. II. p. 228. 14. If two straight lines, always touching a given ellipse, move in such a way that the product of the trigonometrical tangents of the angles which they form with one of the axes be constant and negative, the point of intersection of the two tangents will describe a second ellipse. If we conceive two tangents to this second ellipse, moveable like the first, and subjected to the same conditions, the intersection of these last will describe a third ellipse from which, following the same course, we may deduce a fourth, and so on successively. Then (1). The areas of these ellipses will form a geometrical progression having 2 for its common ratio. (2). The tangents, the intersection of which shall describe any one of these ellipses, will be always parallel to two supplementary chords of the ellipse which shall immediately precede in the order of their successive generation. Rochat: Gergonne, Annales de Mathelmatiques, tom. II. p. 228. 184 ELLIPSE. 15. The semi-axes of an ellipse being a, b; to find the area of a parallelogram which circumscribes it, the inclinations of its sides to the major axis being given; and to determine the least of all such parallelograms. If tan-l m and tan- m' be the given inclinations, the required area will be equal to 4 a2b2 + 9bm - m99 ~4 {ab + (a2 2n~' + b)'2}' which is evidently the least possible when a2 mm' + b2 = 0, or when the parallelogram is a conjugate parallelogram. Durrande; Gergonne, Annales de Mathe'matiques, tom. xII. p. 142. 16. If parallelograms, which circumscribe an ellipse, have their areas constantly equal to k times that on the major and minor axes; to find the equation to the loci of their angular points. The equation to the ellipse being X2 Y2 a2 2+ 1 the angular points lie in two ellipses represented by the equation + 2 = 2k {1 + (k2 1)4}. SECTION IV. Referred to its Axes. Normals. 1. A normal at a point P of an ellipse, cuts the major axis in G. To prove that GP is the shortest line which can be drawn from G to the curve. The equation to the ellipse being x2 y2 a+ b It REFERRED TO ITS AXES. NORMALS. 185 let h be the abscissa of P; then, C being the centre, CG = e2h. Let r be the distance of G from any other point P' in the curve; then, x, y, being the coordinates of P', 2 = (x - eAh)2 + y2 = (x - e2)2 + (1 - e2) (a2 - x2) = e2 ( - )2 + (1 - e2) a2 - e2 (1 - e2) = e2 (x- h)2 + (1 -e2) (a2 e2h2), and therefore r is the least possible when x = h, or when P' coincides with P. De la Hire: Sectiones Conicce, lib. VII. prop. 13. 2. P is a point in an ellipse, D a point in the major axis, such that PD is equal to the minor semi-axis: PQ is a normal at P meeting in Q a perpendicular to the major axis through D. To find the locus of Q. Let (x', y') be the coordinates of P: then, the equation to the ellipse being x" y'2 a2 -1, that to the normal at P will be a2x 2y -.................. (1) Again, since PD is equal to b, we have, x denoting the abscissa of D, (x' _ x2 + y,2 = b2 and therefore, by the equation to the ellipse, (' - ) 2 + W (2- x'2) = 62 (y +- x ) = 6 x, (X' - X)2 = 62 X t b x - x = +- x a ax. (2). a + b................. 186 ELLIPSE. Substituting this value of x' in (1), we have, x, y, denoting the coordinates of Q, b= b -a2 + a(a b) = b(b + a), ^............................. (3). Substituting the expressions for x', y', given in (2) and (3), in the equation X+i y1, 2 + 2"~ = 1, we have, for the equation to the required locus, x2 + y2 = (a + b)2, which represents either of two circles, concentric with the ellipse, and having a -b, a + b, as radii. 3. To find the equation to the normal at a point of an ellipse, defined by the equation 2x2 + 3y2 = 47 the abscissa of the point being 1. The required equation is Y = ~ (3 - 1). 4. To find the equation to the normal of an ellipse a2 b2 _ + Y = 1 in terms of a, the inclination of the normal to the axis of x. Putting cos a = 1, sin a = m, we shall obtain, for the required equation, x y a - b_ 1 m (12a2 + m2b2)' 5. To find the length of that portion of the normal to an ellipse, which is included between the curve and the major axis, in terms of the inclination of the normal to this axis. REFERRED TO ITS AXES. NORMALS. 187 If 0 denote the inclination of the normal, and 2a, 2b, the major and minor axes of the ellipse, the length of the normal will be equal to b2 (a' cos20 + b^ sin2 0) 6. To find the tangent of the angle between the normal at any point (x, y) of an ellipse X'2 a + x' y' a2 62 and the straight line drawn from this point to the centre of the ellipse. The required tangent is equal to tI 1A 7. To find the length of the longer normal, drawn from a point in the minor axis of an ellipse at a given distance from the centre, and intercepted between that point and the curve. If c denote the given distance, a the semi-axis major, and e the eccentricity, the length of the normal will be equal to 8. If a right angle moves so that its sides are constantly normals to an ellipse, to find the locus of the summit of the angle. The equation to the ellipse being a2 His a + 2 = 1 that to the required locus will be (ao + b ). (X2 + y2) (aey2 + b2x2)2 = (a2 - b2)2. (ay2 _ 62x2)2. Bobillier: Gergonne, Annales de Mathematiques, tom. XVII. p.277. 188 ELLIPSE. SECTION V. Referred to its Axes. Chords. 1. Chords are drawn from the extremities of the transverse axis of an ellipse to any point in the curve, and perpendiculars, drawn to these chords from the same point, are produced to cut the same axis: to find the part of the axis which is intercepted between the perpendiculars. The principal axes of the ellipse being taken as axes of coordinates, the equation to one of the chords will be, if x, y, be the coordinates of the point, 'y= _ Y (' _ a), x - a and that to its perpendicular y 'y- y (x- x). y Putting y' = 0 in the last equation, we have, for the intercept of the perpendicular on the axis of x, I Y 2 X' - =X —. x- a b2 = x - (a + x). Similarly, the intercept of the other perpendicular, - a being substituted for a, is equal to b62 x = - - (- a + x). Hence the part of the axis intercepted between the two perpendiculars 2b2 a = the latus-rectum. Ivory: Leybourn's Mathematical Repository, New Series, vol. II. p. 175. REFERRED TO ITS AXES. CHORDS. 189 2. If, from any point in the circumference of an ellipse, two chords be drawn so as to touch the circle described about the minor axis as diameter, and if the chord joining the points of contact be produced to meet the major and minor axes in H, K, respectively, to prove that, a, b, being the semi-axes of the ellipse, b2 a2 (CHU)2 + (CK)2~ b Let (h, k) be the coordinates of any point in the ellipse from which the two chords are drawn. Then the equation to the chord of contact will be hx + cy = b2, b2 b2 and therefore CH = - CK= - h ' k but, by the equation to the ellipse, h2 k1 a2 a=l; b2 a2 a2 hence (CH)2 + (CK)2 b2 3. To find the equation to the locus of the middle points of all chords of a given length in an ellipse. Let M be the middle point of any chord KL. Let x', y', be the coordinates of L, and x, y, those of M. Also let MK=r=ML. Then, 0 denoting the inclination of KL to the axis of x, X' = + r coS0 y' =y-r sin0; and therefore, by the equation Xt2 y12 a2 + b2 1 we have (a + r cos0)2 (y - r sin0)2 a2 b2 190 ELLIPSE. But, if KL = 2c, the two values of r in this equation must be each equal to c in magnitude and must have opposite signs: hence x cos0 ysino 2 62 a b2 2 y2 2 /cos2 sin 0\ and = A+ b2 + c 2 a2 + b2 1, cos2 0 sin2 0 ax t, - + ~ or - + 2 + c. a 6+ cos2 + sin2 From these two equations we get x2 y2,2 ay2 + b2x2 a2 b2 a y2 + b which is the equation to the required locus. 4. P is any point in an ellipse, A' its axis major, MP an ordinate to the point P: to any point Q in the curve are drawn the chords A Q, A' Q, meeting MP in R, S, respectively. To prove that MR.MS = MP2. 5. If, from a point P in the tangent at the extremity B of the axis minor of an ellipse, a tangent PQ be drawn, to prove that PQ will be equal to the chord B Q if BP: AC:: 3: 1. 6. If the equation to the chord of an ellipse be x y ca /3 to find the length of the chord. If 21 be the length of the chord, 12 = (a + /2). '~ a? + b2X2 - a_2s (a2/82 + 2a2') 7. PR; QR, are two normals of an ellipse, cutting each other at right angles in R: CRH is a semi-diameter. To prove that the tangent at H is parallel to the chord PQ. REFERRED TO ITS AXES. CHORDS. 191 8. A chord of an ellipse, inclined at an angle tan-'c to the axis major, passes through an extremity of the axis minor: if x, y, be the coordinates of the second point of the chord's intersection with the ellipse, to prove that, a, b, being the semi-axes, (tan-' )- + 2 tan-1 ( = r. 9. To find the locus of the middle point of a chord of an ellipse, the equation of which is _ y2 2 a2 4- =; the chord being supposed to touch a concentric circle of radius r. The equation to the required locus is,(2 + r2 / 2 + Y 10. If, through the extremity of either axis 2a of an ellipse, any two chords be drawn at right angles to each other, to prove that the chord joining the points in which the two chords cut the curve, will always pass through a point in the axis at a distance from its extremity equal to 2 a2 -b a. a a2+ b 2 11. To find the length of a chord of an ellipse, which is a tangent at any proposed point of a confocal ellipse. Let x, y, be the coordinates of the point of contact, a, b, the semi-axes of the interior, and a, /3, those of the exterior ellipse. Then, c denoting the required length, X2 y c -2=/.,a,=( _ =c a4 ~ 4 12. PQ, PR, are chords of an ellipse, drawn through any proposed point P in the curve, at right angles to each other: 192 ELLIPSE. to find the equation to the chord QR, to prove that QR always passes through a fixed point I, and, P being afterwards regarded as variable in position, to find the locus of the point L If m - -, denote the tangents of the inclinations of PQ, PR, to the axis of x; then, a cos, b sin0, being the coordinates of P, the equation to the chord QR will be ab( - m). (ay cos 0 x + sin)+ (a2 + b2) (ay sin- bx cos0) + ab (a - b2) = 0: the point, through which this chord always passes, while m varies, is the intersection of the two lines ay cosO + bx sinO = 0, (a2 + b2). (ay sin0 - bx cos ) + ab (a2 - b2) = 0, and, 0 being now considered variable, the locus of this point is an ellipse defined by the equation x' y2 (a2 - ^b22 + -2 _2_ a" b2 a2 + b) SECTION VI. Referred to its Axes. Focal Properties. 1. PT, QT, are tangents to an ellipse at P, Q; S is one of the foci, C the centre, R the point where CT cuts the curve; T RN, parallel to ST, meets the major axis in N: to shew that -SP.SQ = RN2. REFERRED TO ITS AXES. FOCAL PROPERTIES. 193 Let x', ", be the abscissae of the two points of contact. Then SP. Q = (a - ex').(a - ex"). The values of the abscissae if xI, y, be the coordinates of T, will be obtained from the two equations a2 y2 " a2 6 2, Ws + = +- = 1, Eliminating y, we have (-a2) = VY (I 2) 1-2 (c]2 y + -2xx++1 0: a' \a 2 a2 b2 hence SP.SQ = a - ae. (x' + x") + ex'x" 1 Y2Y 2 2x b2 a - ae. -2 + e2a2. 2 2 a2 b62 a2 2 _Y 2 + (1 - ae)2 a2 2 = =2 /1c 2 S t 2 RN~. a2 2 a2 b 6 We have replaced ST, x1, y2, by RN, x"') y"', where x"'. y"', are the coordinates of R, by virtue of the similarity of the triangles CST, CNR. 2. A tangent at any point P of an ellipse meets the major axis produced in T, and perpendiculars upon it from the centre and focus in Y, Z: to prove that TY (TZ\2 PY kPZ' 3. If P be any point in an ellipse, the major axis of which is AA', minor semi-axis BC, and foci S, S': to shew that (AP+ SP). (AP- SP) + (A'P + S'P). (A'P- S'P) = 2(BB)2. 0 194 ELLIPSE. 4. If 0 be the angle between the focal distances of any point of an ellipse, and f the angle between the lines joining the point with the extremities of the major axis, to prove that, e denoting the eccentricity, 0 and a will be connected by the equation cot b. (cot b - e cot 0) = 1. 5. To find the locus of the centre of a circle inscribed in the triangle SPH, S and H being the two foci and P any point of a given ellipse. The equation to the ellipse being (1 - e2) x2 + y2 = a2 (1 -e2) the required locus will be a concentric ellipse defined by the equation (1 + e) y2 + (1 _e) x2 = ae2 (1 - e). Lardner: Alqebraic Geometry, p. 125. 6. To find the locus of the centre of a circle which, the diagram remaining the same as in the last problem, touches SP, HP produced, and HS produced. The required locus is a straight line defined by the equation x = a, and is therefore a tangent to the ellipse through one of its vertices. Lardner: Algebraic Geometry, p. 126. 7. To find the locus of the centre of a circle which, the diagram remaining the same as in the two preceding problems, touches SH, PS produced, and PH produced. The locus is an ellipse defined by the equation (1 - e) y2 + (1 + e) x2 = ae2 (1 + e). Lardner: Algebraic Geometry, p. 127. 8. If a circle be described through the two foci of an ellipse, and any point in the conjugate axis produced; to prove that the right line joining that point and one of the points where the circle cuts the ellipse, will be a tangent to the ellipse. Leybourn: Mathematical Repository, New Series, vol. I. p. 187. REFERRED TO ITS AXES. CONJUGATE DIAMETERS. 195 9. If SP, HP, be any two focal distances of an ellipse, the vertex of which is A, e the eccentricity, and a, b, the semi-axes, and if a straight line AQL be drawn cutting SP in Q. and bisecting HP in L: to find the locus of Q. The required locus is an ellipse represented by the equation Ix - le (I + de a' " 2 2 + Y v/ 3- = 1. {x (1 - +e)a}2 (1 - e). SECTION VII. Referred to its Axes. Conjugate Diameters. 1. CP, CD, are conjugate semi-diameters of an ellipse; to prove that the sum of the squares of the distances of P, D, from a fixed diameter is invariable. Let a cosO, b sin, be the coordinates of P: those of D will then be -a sin0, b cos0. Let 2, p2, represent the respective distances of these points from a fixed diameter represented by the equation lx + my = 0, where 1, m, are its direction-cosines. Then p = la cos 0 +mb sin 0, p = - la sinO + mb cos, and therefore p2 + p2 = 12a2 + m2b2. 2. CP and CD are conjugate semi-diameters of an ellipse. If normals at P and D intersect in K, to prove that KC is perpendicular to PD. Let (x, y), (x,, y1), be the coordinates of P, D, respectively: then the equations to the normals will be I, a b 2 (S- ) ~)-=(y -Y ) Y) * 2x ~ b 02 196 ELLIPSE. At the intersection of these two lines (a2 - b) (y1 - y) = - a2 1 X I 2 - a2) (x, - ) = - b2 ( yX and therefore Y -Y a y x - x b2 xx y but, by the nature of conjugate diameters, a b 1= - Y=, X: hence _ = t - y which shews that CK is perpendicular to PD. 3. If C be the centre of an ellipse and, in the normal to any point P, PQ be taken equal to the semi-diameter CD which is conjugate to the semi-diameter CP; to find the locus of Q. Let (x, y) be the coordinates of P, and (x', y') those of Q. Then, by the equation to the normal, we have b2 x( - )= (y'-y) This equation shews that, X being some arbitrary quantity, x' -x = X.............(1). But, the coordinates of D being we have, by the b a a, we have, by the condition of the problem, (x' - )2 + (Y )2 '2 =+ y2 and therefore by (1) 2 = 1, X = + 1. and therefore, by (1), X2 == 1, X = + 1. REFERRED TO ITS AXES. CONJUGATE DIAMETERS. 197 Hence, by (1), we have ' =(a+ b) -, y' = (a + ) Y a =(b ~+ a). From these equations, combined with the equation to the ellipse, we have, for the equation to the required locus, x12 + y2 = (a + b). This shews that the locus is either of two circles concentric with the ellipse, of which the radii are a - b and a + b. 4. If a series of chords of an ellipse pass through a fixed point, to prove that the chords of the corresponding conjugate arcs have the same property. Let the coordinates of the extremities of any one of the series of chords be (a cos0, b sin0) and (a cos ', b sin0'). The equation to the chord will therefore be x (sin0 - sin0') (cos - cos0') = sin ( - 0'), a b and therefore, (h, k) being the fixed point through which every chord of the series passes, - (sin - sin') - (cosO9 - cos ') = sin ( - 0')...(1). Again, the coordinates of the extremities of the conjugate chord are (- a sinO, b cos0), (- a sin ', b cos '): its equation will therefore be - (o 0- cos 0') + Y (sin -- sin ') = sin (0 - '), or, by virtue of (1), ( + ( (cos0 - cos ') + (sin -0 sinG') = 0. This equation shews that the conjugate chord passes always through a fixed point ak bic x=- b Y = ba 198 ELLIPSE. 5. CP and CD are conjugate semi-diameters of an ellipse, and PF is a perpendicular let fall upon CD: to determine the locus of the point F. Let x, y, be the coordinates of P. Then the equation to the normal PF will be a2 b2 W (X' x) = _'* ({x'-x) = (y'- y)..................(l) x and the equation to CD will be xx+ yy' +........................(2). At the intersection of (1) and (2), we have x' (x - ) + y' ( - y) = 0, or xZ' + y'= xx' + y', and therefore, by (2), X,2 + yt2 xx' x' b2 b2 aL7 X a (x'2 + y'2) or -2) XI a (a2 - b2) x' y_ b (x2" + y'2) Similarly b = (^2 a2) y But the equation to the ellipse is a2 62 -+92=1: a b2 P hence ('2 + y') a2 2 (a_ - b2)2. + 1, which is the equation to the required locus. 6. The equation to an ellipse being 2x'- + 3y2 = 4, to ascertain the equation to a diameter conjugate to one represented by the equation y = 2x. The required equation is y = - ix. REFERRED TO ITS AXES. CONJUGATE DIAMETERS. 199 7. CP,'CD, are two conjugate semi-diameters of an ellipse, of which S is a focus: to prove that the distance of P from a diameter, drawn parallel to the line joining S and D, is equal to the minor semi-axis. 8. To find the value of the sum of the squares of two normals of an ellipse drawn from the extremities of two conjugate diameters to the major axis. The required value is equal to I. (a' + b2). 9. If CP, CD, be conjugate diameters of an ellipse, and ordinates through P, D, meet another ellipse, described on the same major axis, in Q, E, respectively: to prove that CQ, CE, are conjugate diameters of the second ellipse. 10. PSP', QHQ', are chords drawn through the foci, H, of an ellipse, parallel to any pair of conjugate diameters: if SP = r, SP' = r', SQ = p, SQ = p't to prove that, 2b, 21, denoting the axis minor and latus-rectum respectively, rr' + pp' = b2 + 12. 11. Two conjugate semi-diameters a', b' of an ellipse, are inclined at angles 0, b, respectively to the semi-axis major: to prove that a'2 sin 20 + b2 sin 20 = 0 12. C is the centre, S the focus, and P any point of an ellipse: to prove that, Q being the intersection of PS with the diameter conjugate to CP, PQ = the semi-axis major. 13. If (a,, b6) and (a2, b2) be the coordinates of the extremities of any two diameters of an ellipse, to find the angle between the two conjugate diameters. 200 ELLIPSE. The required angle is equal to tan-' a2. a4bb2 + b 4aa which, if the given diameters be conjugate, reduces itself to t {(a - b2).albl tan. 'h2 + ba2 ~.a 14. If, from the extremities P, D, of two conjugate semidiameters CP, CD, of an ellipse, straight lines be drawn, parallel to the semi-axis minor CB, to meet, in points P, D', the circumference of a circle which has the axis major for its diameter: to prove that the angle P' CD' is a right angle. 15. If D represent any diameter of an ellipse and P the parameter of D, to find when D + P is the least and when the greatest possible. It is the least when D is the axis major and the greatest when it is the axis minor. De la Hire: Sectiones Conicce, lib. VII. prop. 38. 16. If CP, CD, be any conjugate semi-diameters of an ellipse APBDA', A and A' being the ends of the major and B an end of the minor axis, and if BP, BD, be joined, and also AD, A'P, these latter two intersecting in 0, to prove that BDOP is a parallelogram, the greatest area of which is equal to ab.(V2 - 1). 17. If a, b, be the semi-axes of an ellipse, and a', b, conjugate semi-diameters, to prove that, a' being inclined to a at an angle a, and b' to b at an angle /, a'2- 6b2 cos(a + 3) a2 - b2 cos (a -/3) 18. To find the locus of the middle points of the chords joining the extremities of conjugate diameters in an ellipse. AXES PARALLEL TO THE AXES OF THE CURVE. 201 The equation to the ellipse being x2 y2' the required locus will be an ellipse defined by the equation a2 y2 'I a + bY - 2. 19. If tangents be drawn from different points of an ellipse, and of lengths equal to n times the conjugate semi-diameters at those points; to find the locus of their extremities. The equation to the ellipse being X'2 Y/2 a Y b2= 1 the required locus will be a concentric ellipse defined by the equation 2 2y a+|. =l+n2. SECTION VIII. Referred to Axes parallel to the Axes of the Curve. 1. Two ellipses lie in the same plane, are similar, and have their major axes parallel. Each is cut in four points by lines drawn from its centre to touch the other. To prove that the eight points of intersection lie on two parallel lines. Let the equations to the two ellipses be a2 b2.(1), (2 ~ 2 a —2 + Y = 1........................, (a,+(I, )........ 42.w.e*(2) Draw a tangent from (a, /8), the centre of the latter, to touch the former. 202 ELLIPSE. Then, (x, y), being the point in which (1) is touched, the equation (3 ) as + - = 1........................(3) ac b2 will be satisfied when a and 3 are put for x' and y'; hence ax,V. 1 (4) 2- + 4 1........................ (2). a b Combining (4) with (3), we get b2 (/3- ay') = -, (Cay' - /x) = - / and thence, by (1), (ay' -_ ') =) a2 (y' -/)2 + b2 (x' -a)2, or, dropping accents, (ay - /3)2 = a2 (y - /3) + 2 (X - a)2, the equation to the system of two tangents to (1) passing through (a, 3). Combining this equation with (2) multiplied by a2b', we get ay - Sx = + mab, the equations to two parallel lines on which the four points of intersection lie. The distance of either from the centre of either of the ellipses is obviously mab (a2 + /32)' In order to determine the corresponding system for the tangents drawn from the centre of the former to touch the latter of the two ellipses, write ma, mb, for a, b, respectively, and for m; when mab (a2 + /32)4 remains unchanged. The two systems therefore coincide. 2. To find the equation to an ellipse, placed with its axes parallel to the rectangular axes of coordinates, so as to touch the lines x = 2a, x = 2a 2b', y = 2b'. POLAR EQUATION. CENTRE THE POLE. 203 The required equation is (x- a-a') (y - b - b') (a-a')2 (b - ')2 SECTION IX. Polar Equation. Centre the Pole. 1. CP, CD, are two conjugate semi-diameters of an ellipse, of which CP is inclined to the major axis at an angle a: to find the magnitude of CD. The polar equation to the ellipse, the centre being the pole, gives CP2 (a2 sin2 a + bV cos' a) = a2b2. Hence CD2 = ad + b2 - COP = a + - b- 2 a2 sin2 a + b2 cos aa a4 sin a + b4 cos2 a a2 sin2 a + b2 cos2 a 2. If CP, CD, be two semi-diameters of an ellipse at right angles to each other, to prove that the distance of the centre from the chord PQ is equal to ab (a + b2) Let the equation to the ellipse be 1 cos2 0 sin2 0 2 2 b2 r -a2 + b2 and that to the chord PQ r cos (0 - a) = 8. At the intersection of these two lines os (0 a) = 82 (cos 0+ sin2 0. \ ' + /' 204 ELLIPSE. Let X, ~Ir + X, be the angular coordinates of P, D, respectively: then cos2 (X- a) = 82 (s si + X) ~Cos\ a + *~. / \ sin2 X\ cos Xb2 and in2 (X - c) = 2 i- + coS X \ a2 + b Adding these two equations, we get 1 =82 ( ) a2b2 ad + b2'~ 3. If A be the elliptic area contained by two semi-diameters including an angle a, and B that contained between two semidiameters at right angles to the former, to prove that fa b\ 2A 2B + - cot a = cot - + cot \b a/ ab ab The polar equation to the ellipse, the centre being the pole, is r cos2 a sin2 \0 r Va" + b~ =1' hence A = fr r 'd = cosi'2 2 2 Cos2 Sin2 B = tanen d tan 0 tan' + tans 0 *l tan 8' a< where 0 - 0'= a: or a, -, /tan0\ 0 tan ) A = ab {tan-1 (a tan 0) - tan-1 (a tan O')} 2 - b ' a2 2A b 1 + tan tan t' whence cot t —=- n -tanaZt a tan 0 tangO POLAR EQUATION. CENTRE THE POLE. 205 Putting 0 + ~-r for 0, and 0' + ~-r for 6', we have + tan 0 tan O' cot =. ah a *tan 0 - tan O' 2A 2B b hence cot -- + + cot(0- 0') ab ab a \ 6V / / (a b\ = ~ + - cot a. \b a 4. If two semi-diameters of an ellipse be drawn at right angles to each other, to prove that, r, r', denoting their lengths, 1 1 1 1 ' -- a2 + 2 Abonne: Gergonne, Annales de Mathdmatiques, tom. XVIII. p. 369. 5. To prove that of all diameters of an ellipse, the axis major is the greatest and the axis minor the least, and that a diameter nearer to the axis major is greater than one more remote. Apollonius: Conicorum, lib. v. prop. 11. 6. If CP make an angle 0 with the axis major of an ellipse, C being its centre and P a point in its circumference, and be produced to meet a tangent at the extremity of the axis minor, so that the part produced is equal to the semi-axis minor; to prove that 0 is always less than -; and, when 0 = -1 that 8. 7r e = - sin. 5 5 7. If (0, r), (0', r'), be the coordinates of any two points in an ellipse, the centre being the pole and the major semi-axis the prime radius vector, to prove that, 2a, 26, being the axes, 1 1 - 1 = (a' - 2) (sin2 0 - sin2 0'). 8. The two axes of an ellipse are 6 and 4, and, from its centre, two straight lines, including a right angle, are drawn 206 ELLIPSE. to the curve. Supposing these lines to bear to each other the ratio of 3 to 4, to find their actual lengths. The required lengths are 15 10 2 13 an 13 9. To find the locus of the extremity of a straight line drawn from the centre of an ellipse, such that the rectangle contained by it and the diameter perpendicular to it, may be equal to half the rectangle contained by the axes. The equation to the required locus, the centre of the ellipse being the pole, will be 2 sin2 0 cos'2 0 a~b2= a2 + b2 SECTION X. Polar Equation. Focus the Pole. 1. In an ellipse HP is bisected in L: AL is joined, cutting SP in Q. To find the locus of Q. Since S = HC, and HL = PL, it follows that CL is parallel to SP. Hence SQ -AS CL- A 1 -Put SQ=p, zLASQ=0: then p = (1 - e) CL = I (1- e).SP - ( _ ) a (1- e2) - ~ 1 ~- ecosO ~ (1 - e)2.(1 + e).a 1 + ecos which shews that the locus of Q is an ellipse, similar to the original one. The semi-axis major = (1 - e)2(1 + e)= i (1 - e)a. 1 — e POLAR EQUATION. FOCUS THE POLE. 207 1 2. The eccentricity of an ellipse is -, and, through the focus, a straight line is drawn, inclined at an angle of 45~ to the axis major, meeting the ellipse in two points: to shew that the tangents at these points meet at an angle the tangent of which is 2. The intersection of PT, QT, the two tangents, will take T-.1 ~place in the directrix TH. Join ST, which will, by a known theorem, be at right angles to PQ. Then TTC A TA C AS(1L 1+.e 1-F e a ES ==AH+ AS= A8(1 a (1 -e) *+e= - J L A J \e I}2I e /HS = ST - =a~ cos 45 c a(1 - e2) 1 - e cos 45 = a(1-e'2) _a. SQ = 1 + e cos 45~ 3 If L PTS = a, and L Q TS= 13, we shall therefore have tan a = SP = 1, tan = -ST =- 3 and therefore tan (a + /3) = 1+ = 2. 1-3 3. S and H are the two foci of an ellipse: from any point P in the curve a normal PG is drawn to cut SH in G. To prove that the projection of PG upon HP or SP is constant. 208 ELLIPSE. From G draw GL at right angles to PH, and let L PHS = 0, e = the eccentricity, 2a = the axis major, CG = x, C being the centre of the ellipse. Then PL = HP - HL = HP - (ae + e2x) cos 0 = HP- e.HP.cos 0 = P (1 - e cos O) = a (1 - e) = the semi-latus-rectum. The same proposition is true also in relation to the hyperbola and parabola. Pagbs: Liouville, Journal de Mathematiques, tom. II. p. 437. 4. If Pp, Qq, be chords of an ellipse drawn through one of the foci at right angles to each other, to prove that 1 1 2 - e2 Pp Qq 2a (1-e2) 5. If P be any point in the periphery of an ellipse, S and H being the two foci, to prove that, the angles PSH, PHS, being denoted respectively by 0, 0', 0 0' 1-e tan -.tan 2 2 1 +e 6. If an ellipse and parabola have a common vertex and common focus, to prove that, Sp, SP, being respectively focal distances inclined at a common angle 0 to the axis, P2 -_ 1-.e tan 7. A chord is drawn through the focus of an ellipse making an angle 0 with the major axis, and tangents are drawn at the extremities of this chord: to find the angle contained between these two tangents. POLAR EQUATION. FOCUS THE POLE. 209 If (b be the required angle, then 2e sin 0 tan = 2-. 8. In an ellipse, of which S is a focus, SA (the least distance) = D, LASP = 0, and the eccentricity = e; to prove that, 1 -e if / =1 +, and powers of f3 above the first be neglected, D SP = Cos2- -.(l + / tan0 2) * 9. To find the locus of the middle points of all focal chords of an ellipse. If C be the centre and S the focus of the original ellipse, the required locus will be a similar ellipse having SC as its major axis. 10. Five radii vectores of an ellipse being represented by a, b, c, d? e, and its parameter by p, to prove that sin a + sin3 + sin 7 + sin S + sins s sin a sin/ sin sin si ns a b + c d e where a = L (, c) + L (, e), /3 = (c, d) + t (e, a), y=z (de) +(a, b), 8= (e, a) + t (b, c), = (a, b) + z (c, d). This problem was proposed for solution by Gergonne, to whom it had been communicated by Mr. Talbot, in the 17th volume of the Annales de Mathdmatiques, p. 283, and solved, in a more general form, by Lentheric, p. 366. 210 ELLIPSE. SECTION XI. Polar Equation. End of the Axis Major the Pole. 1. A circle is described upon AA', the major axis of an ellipse. P is any point in this circle: AP, A'P, are joined, cutting the ellipse in points Q and Q' respectively: to shew that AP A'P a 2 +b2 AQ+AQ'' b 2 A Q + alat= x,2 The polar equation to the ellipse, A being the pole, and AA' the direction of the prime radius vector, is 2 cos 0 A Q- __a __ AQ= 8Cos2 a sin2O a2 + But, by the equation to the circle, AP = 2a cos 0: AP (cos2 0 sin2 0 a2 hence AQ - b )a Similarly, 17r - 0 taking the place of 0, AP / sin2 0 cos20 a2 A Q a2' + b AP A'P a2 + b2 Hence AQ+ A'Q b A —Q +A-' - b' 2. Through A, the common vertex of two similar ellipses ABB', ADD', the greater axes of which are coincident in direction, are drawn chords ABD, AB'D': B, B', and D, D', are joined: to prove that BB' is parallel to DD'. 3. If CP be any semi-diameter of an ellipse, and A Q O be drawn from the extremity A of the major axis parallel to PC, and meeting the curve in Q and the minor axis produced in 0: to prove that AO.AQ 2CP2. Herschel: Leybourn's Mathematical Repository, New Series, vol. IV. p. 152. POLAR EQUATION. END OF THE AXIS MINOR THE POLE. 211 SECTION XII. Polar Equation. End of the Axis Minor the Pole. 1. To determine the longest straight line which can be drawn from the extremity of the minor axis of an ellipse to the curve. The equation to an ellipse, an end of the axis minor being the pole, and the axis minor the prime radius vector, is 2b cos 0 1 - e' + e2 cos2 0 Our object is to find when r is the greatest, or, putting cos 0 1 - e2 + e' cos' 0, when u is the greatest. Now e'tu cos2 - cos 0 = - (1 - e2), (eu cos 0 2 = 4e-( -e2) u. From this it appears that u can never be greater than 1- ei) and, since cos cannot be greater than 1, can 2e(1- e2)1' be so great only when 1 2e2u or u e2 1 1 or 2e(1 - e'2) =2e" e ( -e2, If e2 <, the value of u will be greatest when eu cos 0 - 2e P2 212 ELLIPSE. is least, or, since cos 0 is always smaller than -1 when cos 0 is greatest, that is, when 0 = 0. Thus, if e' > or = -, we have, r being greatest, ( 1 -- e'2) ) cos 0 e2) and, when e2< ~, 0 = 0, r= 2b. SECTION XIII. Polar Equation. Point in the Axis the Pole. 1. 0 is a point, in the semi-axis major CA of an ellipse, at a distance from the vertex A equal to half the latus-rectum. P is any point in the curve. PO is joined and PM is drawn at right angles to CA. To prove that OA is less than OP, and that (OP) - (OA)2 Oc (AM)2. If OP = r, L POA = 0, e = the eccentricity, and CA = a, the polar equation to the ellipse will be e2r cos2 0- 2ae2 (1 - e2) r cos 0 = r2 - (1 - e2) (1 -e4) a2. This equation may be written in the following form, (er cos 0 - ae (1 - e2)}2 = r2 - ( - )2; and therefore, observing that r cos 0 = 0 a ( - e2) = OA, we see that e.(AM)2 = (OP)2 - (OA)2. This result shews that OP is greater than OA. Apollonius: Conicorumn lib. v. prop. 6. 2. In the semi-axis major CA of an ellipse, a point 0 is taken, at a distance from the vertex A greater than the ratio of (BC)2 to AC. 0 is joined to any point P in the curve, and PM is drawn perpendicularly to CA. To determine the least value of OP, and the corresponding value of OM. POLAR EQUATION. POLE ANYWHERE. 213 Let CO = c; then, when PO is the least possible, PO = (1 - ) - e MO=c.1 -. e Apollonius: Conicorum, lib. v. prop. 10. 3. In the semi-axis minor BCB' of an ellipse, a point 0 is taken at a distance from B equal to the ratio of (AC)2 to BC; to prove that OB is the greatest distance of the point 0 from the curve. Apollonius: Conicorun, lib. v. prop. 16, 17, 18. 4. A point 0 is taken in the axis minor BCB' of an ellipse, at a distance from B greater than the semi-axis minor BC, but less than the ratio of (A C)2 to BC. 0 is joined to any point P in the curve: to determine the magnitude and position of the greatest value of OP. Let / POB = 0, and OP = r, CO = c. Then the required magnitude and position are given by the equations = (a+ c) cos 0 e2 1 / e (a2'e2 + c2)' Apollonius Conicorumn lib. v. prop. 20. SECTION XIV. Polar Equation. Pole anywhere. 1. To prove that two chords of an ellipse, which are not diameters, cannot bisect each other in their point of intersection. Taking the point of intersection of the chords as the pole, we may write the equation to the ellipse in the form (p cos 0 + h)2 (p sin 0 + k)" 1 a I +~ 214 ELLIPSE. Suppose that, disregarding signs, the two values of p given by this equation are equal, for a certain value of 0; then, the signs of the equal values of p being opposite, we have h cos0 ksin0 + =.................. (1). Suppose the same thing to be true for some other chord through the pole, 0 replacing 0: then hcos k sin 0 2 3-+- o- 0.................. (2). From (1) and (2) we have sin b cos 0 = sin 0 cos, sin (9 - 0) = 0, = 0. This result shews that the latter chord must coincide with the former, and therefore establishes the proposition. 'Eav ev eXXe1elfL KVIsCXOV 7 repLpepeca S o evOelab 7TreOLVwoLV aXXqrXa /!a 8BaL To70 fevvTpov ovo-at~L ov TrePLVOVorv dX)XXa 8i1xa. 'An[OAAQNIOYr IEPrAIOT Kwvtwv 70V TO zeT-7 povo IpTraOtLs '. 2. Through a given point 0, is drawn a straight line cutting an ellipse in two points P, Q: to determine the position of the line in order that the rectangle OP. 0 Q may be a maximum or minimum. Let 0 = the inclination of the line to the axis major: then the rectangle will be a maximum if 0 = 0, and a minimum if 0=2r7r. De la Hire: Sectiones Conicoe, lib. vII. prop. 31. 3. If there be two similar concentric ellipses, similarly placed, the ratio between either axis of the one to the corresponding axis in the other being V2 to 1; to prove that, if from any point in the periphery of the inner ellipse, any right line be drawn to terminate in the periphery of the outer ellipse, the REFERRED TO CONJUGATE DIAMETERS. 215 rectangle under the segments of that line, is equal to the square of that semi-diameter of the inner ellipse, which is parallel to the right line. Ivory: Leybourn's Mathematical Repository, New Series, vol. II. p. 169. SECTION XV. Referred to Conjugate Diameters. 1. CP, CD, are two conjugate semi-diameters of an ellipse. If PD be joined, and, through any point Q of the curve, be drawn a line parallel to PD cutting CP, CD, respectively in P', D', to prove that (QP')2 + (QD') = (PD)2. Draw QV parallel to DC, to cut CP in V. Let CP= a, CD = b6 and accordingly, A' (-/-_ _ X being some arbitrary quantity,.. CP' = Xa, CD' = b. Also let CV=x, QV=y, PD=c, QP'=r. Then, the triangles QVP', DCP, being similar, a b x = Xa - -.r y = r: c c but, by the equation to the ellipse, a2 ~2 hence 1 (Xa - - r + = 1, and therefore r" - Xcr + 1 (X2 - 1) C2 = 0. The two roots r,e r, of this equation, which is symmetrically related to the axes of x and y, represent the lines QP', QD': but r, 2 + r,2 = (r, + r,,) - 2rr, = X2 - (2 - 1) c2 = c2: the proposition is therefore established. 216 ELLIPSE. In the case of an hyperbola, we might prove in a like way that r, - r,, = c. Encontre: Geryonne, Annales de Mathermatiques, tom. rv. p. 294. 2. If, in a given ellipse, be inscribed any parallelogram HKLM, the sides of which are parallel to conjugate diameters; s/' and if an ellipse be inscribed in this parallelogram, so as to touch / its sides in their middle points; to prove that this new ellipse will also touch the sides of a parallelogram H'K'L'M', the diagonals 'L', K'M', of which I: are conjugate diameters of the given ellipse parallel to the sides HK, KL, of the parallelogram HKLM. Let HK = 2a HM = 2b, HL' = 2a', KJ' = 2b'. Then, since the ellipse inscribed within HKLMA touches the sides in their middle points, it follows that the lines joining their middle points are conjugate diameters: hence the equation to the inscribed ellipse, CH', CK', being taken as axes of coordinates, is - + = 1...................... ( ). a 2 b2( Again, the equation to H'K' is 1..........................(2). a b But, since H, the coordinates of which are a, b, lies in the given ellipse, a2 b2 a +2 1-........................ (3). The relation (3) expresses the condition of tangency between the lines (1) and (2). Similarly the other sides of H'K'L'M' may be shewn to be tangents to the inscribed ellipse. Rochat: Gerconne, Annales de McattemMatiques, tom. ITI. p. 27. REFERRED TO CONJUGATE DIAMETERS. 217 3. To find the area of the greatest triangle which can be inscribed in a given ellipse. Let PQR be the greatest triangle which can be inscribed in the ellipse. Then Q must be the j point in which a tangent parallel /D to PR touches the ellipse, for Q will v Q in this case be the point of greatest perpendicular distance from PR. \ Similar remarks hold good in relation to the points P, R, and the sides QR, PQ) respectively opposite to them. Let C be the centre of the ellipse: join P.C, QC, RC, and produce these lines to meet the sides of the triangle in p, q, r. Then, from what has been said, it is clear that RQ is a conjugate chord to the diameter PCp, PR to Q Cq, and PQ to RCr. Hence p, q, r, are the middle points of RQ, RP, PQ; and therefore, by a known property of triangles, Cr = CR, Cq = CQ, Cp = CP. Hence, by the equation to the ellipse referred CR, and its conjugate CD, area of APQR = Rr.Pr.sin L PrR CD = |CR. -C. (CB4 - C ).si) n L D CR. CR.CD sin L DCR = -V3.ab. CoR. The area of the ellipse is equal to 7rab: hence area of greatest triangle: area of ellipse:: 3/3: 47r. A different solution of this problem will be given subsequently in this volume. 4. If lines, drawn through any point in an ellipse to the extremities of any diameter PCP', meet the direction of its conjugate diameter DCD' in points M1, N, to prove that CM. CN = CD2. 218 ELLIPSE. 5. If a series of parallel chords be drawn cutting two similar and concentric ellipses; to prove that the product of the part Q Q' intercepted between the two curves, and the remaining part Q'q, will be constant. Cr 6. If a parallelogram be described about an ellipse, its sides being parallel to a pair of conjugate diameters, to prove that the portions of its diagonals, which are intercepted by the ellipse, form another pair of conjugate diameters. 7. From the ends of two conjugate diameters of an ellipse are drawn lines parallel to any tangent line: and, from the centre C, is drawn any line cutting these lines and the tangent in points p, q, r, respectively: to prove that (Cp)2 + (Cq)2 = (Cr). 8. CP, CD, being any two conjugate semi-diameters of an ellipse, PD is joined, a semi-diameter CP' is drawn parallel to DP, and PP' is joined: to find the area of the trapezium CP'PD. Required area = (A/2 + 1). 2a/2 9. If two points be taken on any semi-diameter of an ellipse, so that the rectangle of the segments between them and the centre may be equal to the square of the semi-diameter; and from these points straight lines be drawn to any point in the periphery of the ellipse, meeting it again in two other points: to prove that the straight line joining these two other points will be parallel to the tangent at the vertex of the diameter. Leybourn: Mathematical Repository, New Series, vol. III. p. 31. 10. To prove that the tangent to the interior of two similar ellipses, the axes of which are coincident, cuts off from the exterior curve a constant area. De la Hire: Sectiones Gonicce, lib. VI. prop. 17. REFERRED TO ANY TWO DIAMETERS. 219 11. A tangent to an ellipse, at a point P', intersects conjugate semi-diameters CP, CD, produced, in Hi K, respectively: CD' is conjugate to CP'. To prove that (CD') = HP'. KP'. De la Hire: Sectiones Conicce, lib. v. prop. 17. SECTION XVI. Referred to any two Diameters. 1. If, at the extremity of each two diameters (not conjugate) of an ellipse, a tangent be drawn meeting the other diameter, the line joining the points of intersection will be parallel to that joining the extremities of the diameters. Let the equation to the ellipse be X + xY y2 aW a3 b2 a, b, being the semi-diameters taken as coordinate axes. The equation to the tangent at any point (x, y) will be (X X) +a2 + a) y ) o, + 7) = ~ or z+ Y y' + =- 1 Put x = 0, y = y' = 0: then, x' denoting the distance of the intersection of the axis of x from the origin, bx', /3 a/3 ' a/ calQ3'b Similarly, y' denoting a like quantity in relation to the axis of y, a cY_ Hence a an equation which establishes o proposition an equation which establishes our proposition, 220 ELLIPSE. 2. To find the magnitude and position of the principal diameters of an ellipse defined by the equation ax2 y + + cxy =f................. (1), the axes of coordinates being supposed to include an angle a. Let r be the distance of any point of the ellipse from its centre. Then r 2 = 2 + y2 + 2xy cosa..................(2). At the extremities of the principal diameters, r is a maximum or minimum: hence, differentiating the two equations, and putting dr = 0, we obtain O = xdx + ycy + (xdy + ydx). cosa, o =- axx + bydy + (xdy + ydx). c. From these equations, X being an arbitrary multiplier, we get X (x + y cosa) = ax + cy, and X (y + x cosa) = by + cx. Multiplying the former of these two equations by x, and the latter by y, and adding, we see that, by virtue of (1) and (2), Xr2 = f Hence x (f- ar2) = y (cr' - fcosa), and y (f- br2) = x (cr2 -fcosa). Eliminating x and y, we get (f- ar') (f- br2) = (cr2 -fcosa)2. From this equation we shall have two values r'2 r"2, of r2. The quantities 2r', 2r", will be the lengths of the principal diameters. The equations to these diameters will be x (f- ar'2) = y (cr'2 -f cosa), x (f- ar"2) = y (cr"2 -f coa). Berard: Gejqonne, Annales de Mathkematiques, tom. III. p. 105. Bret: Gergonne, Annales de Mfathteqnatiques tom. V. p. 357. ANY RECTANGULAR AXES WHATEVER. REDUCTION. 221 SECTION XVII. Referred to any Rectangular Axes whatever. Reduction. The equation to an ellipse referred to any axes whatever is ax2 + by2 + 2cxy + 2a'x + 2b'y + c'= 0, a, b, c, being subject to the condition c2 Z/ab. If we substitute x' + a y' + /, for x, y, respectively, and, in the resulting equation, put equal to zero the coefficients of x', y', we shall reduce the equation to the form ax2 '2 + + 2 2c1x'y' + c' = 0, the corresponding values of a, /, being the coordinates of the centre of the ellipse. Again, putting, in the last equation, the axes being supposed to be rectangular, = x" cos - y" sin 0, y = x" sin 0 + y" cos 0, and then equating to zero the coefficient of x"y" in the resulting equation, we shall reduce it to the simple form a2x12 + b6y"2 + c2' = 0, the corresponding values of 0 determining the positions of the principal axes. 1. To construct the ellipse of which the equation is 5x2 + 2xy + 5y' - 12x - 12y = 0. Transform the origin to a point a, /: the equation will then become 5 (x'+a)2+2 (x'+a) (y'+/) + 5 (y'+/)2- 12(x'+a)- 12 (y'+1) =0. Equating to zero the coefficients of x' and y', we have 10a + 21 - 12 = O, or 5a + / - 6 = 0 and 2a + 10 - 12 =, or a - 5 - 6 = 0. 222 ELLIPSE. From these two equations we see that a=l, /3=1. The equation to the curve then becomes 5x'2 + 2x'y' + 5y2 - 12 = 0. Again, turning the axes through an angle 0, we shall have for the transformed equation 5 (x" cos0-y" sin0)2 + 2 (x" cos0-y" sin 0). (x" sin 0+y" cos0) + 5 (x" sin 0 + y" cos 0)2 - 12 = 0. Equating to zero the coefficient of x"y", we obtain - 10 sin0 cos0 + 2 (cos20 - sin"2) + 10 sin0 cos0 = 0, whence cos20 =0, 20 = r7, r = Tr. * /4. A/ The equation to the curve then becomes 6x"2 + 4y"2 = 12, 2112 y12 or = 1. 2 3 The coordinates of the centre of the ellipse, referred to the original axes, are therefore 1, 1; and the semi-axes are V2, /3, the former being inclined to the axis of x at an angle 7r. The diagram will serve to illustrate the geometrical signification of the transformations AC=V2, BC= V3, OE= 1 = CE, LxOx" = 7r. 2. To find the positions and magnitudes of the principal axes of the ellipse denoted by the equation 7x2 + 5y2 + 2 3.xy + 6x + 10 3. y + 7= 0. ANY RECTANGULAR AXES WHATEVER. REDUCTION. 223 The coordinates of its centre are (0, - /3); and its semi-axes are 4, 1, the latter of which is inclined to the axis of x at an angle 1Tr. 3. To find the positions and magnitudes of the principal axes of the ellipse represented by the equation 3x2 + 2xy + 3y2 - 16y + 23 = 0. The coordinates of its centre are (- 1, 3); and its semi-axes are }-, - the former of which is inclined to the axis of x at an angle T7r. 4. To find what the equation 9x2 + 3y2 - 6xy - 6x - 2y + 7 = 0 becomes, when the curve is referred to its principal axes. The required equation is '2 i + +2 1( - 5. The equation to an ellipse being x2 + y2 + xy + x + y = 1 to find the coordinates of its centre, and its equation referred to its axes. The coordinates of its centre are -,-, and the required equation is 3X2 + y2 =_ Garnier: Geometrie Analytique, p. 117. 6. To find the equation to the ellipse 2x2 + y2 - 2xy - 2x = 0 when referred to its axes. The required equation is (3 - V5) x2 + (3 + V5) y2 = 2. Garnier: Geornetrie Analytique, p. 129. 224 ELLIPSE, SECTION XVIII. Linear Equation. 1. If r, p, be the distances of the two foci of an ellipse from a point P in the curve; and r', p', from a point D in the curve; P and D being the extremities of conjugate diameters; to prove that rp + r'p' = a2 + b2. Let a cos 0 be the abscissa of P: then we know that - a sin 0 will be that of D: hence r = a ( - e cosO), p = a ( + e cosO), whence rp = a (1 - e2 cos'2). Similarly, - sin now occupying the place of cos O, rp' = a" (1 - e2 sin2 0). Hence rp + r'p' = a2 + a2 (1 - e2) = ac + b". Brassine: Liouville, Journal de Mathematiques, Avril, 1842. 2. If in an ellipse there be taken three abscissae in arithmetical progression along the major axis, the radii vectores, drawn from one focus to the extremities of the corresponding ordinates, will also be in arithmetical progression. 3. To prove that, P and D being the extremities of any two conjugate diameters in an ellipse, (SP- AC)2 + (SD - AC)2 = SC2. SECTION XIX. Intersections of Ellipses. 1. If two ellipses, so placed in a plane that two conjugate diameters of the one are parallel to two conjugate diameters of the other, intersect each other in four points; these four points will lie in a third ellipse, the equal conjugate diameters of which INTERSECTIONS OF ELLIPSES. 225 shall be parallel respectively to the said conjugate diameters in the two former ellipses. Let the axes of coordinates be taken parallel to the parallel conjugate diameters of the two proposed ellipses. Their equations will be ax2 + by2 + 2a'x + 2b'y + c = 0, ax' + L3y2 + 2ac'x + 2/y + ry = 0. Multiplying these two equations by r and p, and adding, we get (ra+ pa)x2 + (rb +p1) y2 + 2 (ra'+ pa') x+ 2 (rb'+p/3') y +rc+ pey7=0, the equation to an ellipse in which the two proposed ellipses intersect. Putting ra + pa = rb + p/3, r /3-a we have p a- b r if therefore this value be given to the ratio -, the coefficients p of x2 and y2 in the equation to this third ellipse will be equal, and consequently its two equal conjugate diameters will be parallel to the coordinate axes. Par un Abonn6: Ger.qonne, Annales de Mathematiques, tom. v. p. 88. 2. Two ellipses, of equal eccentricity and of which the major axes are parallel, can have only two points in common. To prove this, and to shew that, if three such ellipses intersect two and two in the points A and A', B and B', C and C', respectively, the lines AA',BB', CC', intersect in one point. The equations to two such ellipses, referred to axes of coordinates parallel to the principal axes, will be (X - a)2 (Y - P) ( 2 + - - 1.................. (xa b) 1 a /.())2, a2 - +..2 -.*.. ( 2)^ Q 226 ELLIPSE. and, e denoting the eccentricity of either, 2 b'12 1 e 2(3) 1 -.^ 4 4 ~~ ~~~~~..................... From (1), (2), (3), we have (1-e) (x-a)2 + (y-)2 =b2............ (4), (1 - e2) (x - a') + (y - /3')2 = 12........... (5). Subtracting (5) from (4), (l-e2{2 ('-a) x -(a'2-a()} + 2 (/s'-/f) y-(R/2-/32) = 6-2...(6), from which equation it appears that all points of intersection of (4) and (5) lie in a straight line; and, since a straight line can cut an ellipse in only two points, the former part of the proposition is established. Let the equation to the third ellipse be (1 - e2) (x - a")2 + (y - -/,)2 = b"2......... (7). The equations to the lines AA', BB', CC', will be (6), (8), (9), (1 - e2) {2 (a"- a') x - (2 - a"2)} + 2 (/" - /') y - ( 112 - 12) = 12 - 1/2... (8), (1- e) 2 (a - a") x - (a2 - -a2)} + 2 - ~) y - (3s - _B"2) = 12_ b32... (9). Adding together the equations (6), (8), (9), we obtain an identical equation. This shews that the three lines AA', BB', CC', intersect each other in a single point. 3. Two equal similar ellipses, which have the same centre, have their axes inclined at a given angle: to find the angle between the curves where they intersect. Let C be the common centre of the two ellipses, E their point of intersection, ET the tangent to one of the ellipses at E, meeting its major axis, produced, in T. Let CE be produced indefinitely to N. Let a = the inclination of the axes of the ellipses; then it is evident that ZECT= a or (r - a). Let ZLETC = 0. Then the angle between the curves at E, being evidently double the INTERSECTIONS OF ELLIPSES. 227 angle NET, is equal to 2 (~ a +0) = +2 ta n- tan )' o 2 2 0 = r-+2ta coto a/ 4. To find the coordinates of the points of intersection of the two ellipses x2 y2 x 2 2 W + =-1 = I + WL-I and to determine the tangent of their inclination to each other at these points. If 0 be the required angle, and x, y, the coordinates of the required points, ab x = + - ) =Y - (a2 + b) ' and tan 0 = a 2a'. 5. There are any number of ellipses having a common centre and their major axes in the same position. To prove that, if all the ellipses be twisted about their common centre through the same angle 0 in the same direction, the locus of the intersection of each ellipse with its original position, is two straight lines defined by the equations y = x tan 0 y - x cot 20. 6. If a quadrilateral be inscribed in two ellipses, which cut each other in four points, and of which the major axes are parallel or perpendicular to one another: to prove that any two of its opposite angles are together equal to two right angles. 7. Two concentric ellipses, which have their axes in the same directions, intersect, and four common tangents are drawn so as to form a rhombus, and the points of intersection of the ellipses are joined so as to form a rectangle: to prove that the product of the areas of the rhombus and rectangle is equal to half the continued product of the four axes. Q2 228 ELLIPSE. SECTION XX. Polar Equation to the Tangent. 1. If two straight lines be drawn at right angles to one another through the focus of an ellipse, and tangents to the ellipse be drawn at their extremities; to find the equation to the locus of the intersection of these tangents. The equation to one of the tangents being a (1 - e) - = cos(a - 0) + ecos 0 r or cos (a - 0) = a(1 - e cos0............ (1) r that to the corresponding tangent will be, -1 r + a being written for a, a(1 - e') - sin( - 0) = -e cos............(2). Squaring and then adding (1) and (2), we get, for the equation to the required locus, ( = i ( - )-e COS } r 1 a(1 e)+ 2el ' cos0 - 4ea - e)} cosO, = {a ( 1- - e) r Z=+ + e cos, a2(1 M- e) = + 1 e/2 cosO, r the equation to an ellipse, parabola, or hyperbola, as e2 is less than, equal to, or greater than -. 2. If tangents be drawn, from any point without an ellipse, to the curve: to prove that their lengths are inversely as the * This equation to the tangent of an ellipse is given by Mr. Davies, in the Philosophical Magazine for 1842, p. 192. POLAR EQUATION TO THE TANGENT. 229 sines of the angles which they make with the lines drawn from their points of contact to either focus. Let T be the point of concourse of the two tangents, P, P', their points of contact. Then, S being the focus, and CSA the semi-axis major, let PT=p, P'T=p', ST= r, LAST= O L PSA = LP'SA = a'. Then, the polar equations to the tangents give us, at the common point T, the relation e cosO + cos( - a) = - ) = e cosO + cos( - a'), whence 0 - a =-0 + a', 0 = (a + a'), and therefore L PST = i (a - a') = P'ST.. a-a' sin Hence = r sin 4. a-a sin - P 2 r sin q' p sinq' and therefore p - sin. p' sinm 3. To find the locus of the point E, in which the radius vector SP of an ellipse cuts the diameter conjugate to the semi-diameter CP. The equation to the ellipse being a(1 - e2) 1 + e cos0' the equations to SP and to the tangent at P will be respectively of the forms = a..............................(1) a( - e) = cos(0 - a) + e cosO........... (2). r 230 ELLIPSE. The equation to the diameter conjugate to CP, which passes through the point (ae, 7r), and is parallel to (2), is - (e + cosa). - = cos(0 - a) + e cos0.........(3). At the intersection of (1) and (3), there is - (e + cos ). - = 1 + e cos 0, r a(1 -e) or r +a a - 1 + e cos ' the equation to the required locus. 4. The tangent, at any point P of an ellipse, cuts the major axis, produced, in T: S is the focus. To prove that cos SPT = e cos STP. 5. From a point 0, two tangents OP, OP', are drawn to an ellipse, P being nearer to an end of the axis major than P' is: to prove that OP' is greater than OP. De la Hire: Sectiones Conicce, lib. VII. prop. 44. 6. To find the locus of the intersection of two tangents at points P, P', of an ellipse, such that, S being one of the foci, the sum of the inclinations of SP, SP', to the major axis may be invariable. Take SA as the prime radius vector, A being the extremity of the major axis nearest to S; then, putting L ASP L ASP' = 23, the locus of the intersection will be a straight line defined by the equation 0 = 3. POLAR EQUATION TO A CHORD. 231 SECTION XXI. Polar Equation to a Chord. 1. PSp is any focal chord of an ellipse, A the extremity of the axis major; AP, Ap, meet the directrix in two points Q, q: to prove that z QSq is a right angle. If a - 3, a + /3, be the angular coordinates of the extremities of any chord, the focus being the pole and the prime radius vector coinciding with the major axis, the equation to the chord will be - = sec3. cos(0 - a) + e cosO, c denoting a (1 - e). Let 2ry be the angular coordinate of P, then, 0 being that of A, we have a- = 0 a +3 = 2, and therefore a = = /3. Hence the equation to AP is - =secy. cos(0 - 7) + e cos..............(1). Again, the angular coordinates of A, p, being a-/3 = o, a +/ =-(r -2y), we have a = - (- - y) = 3. Hence the equation to Ap is - = _ cosecy. sin ( - vy) + e cos............(2). Again, the equation to the directrix is r cosO = -.........................(3). e At Q, the intersection of (1) and (3), we have e cosO = secy cos(O - y) + e cos0, and therefore cos ( - 7y) = 0, 0-7 =-1, o =-(r-7), or z. AS- = Q r - ey. 232 ELLIPSE, Again, at q, the intersection of (2) and (3), 0 = sin(O - 7), 0-y = 0 0 = Y7 or LASq = 7y. Hence zQSq = LASQ - ASq, — 2 lr~ 2. PSQ, P'SQ', are any two focal chords of an ellipse: to find the locus of the intersection of the chords PP', QQ'. The required locus is the directrix nearest to S. SECTION XXII. Polar Equation to the Normal. 1. To deduce the polar equation to the normal of an ellipse from that to the tangent, the focus being the pole. Let the polar equation to the normal be r (A cos O + B sin O) + C = 0. Let a be the angular coordinate of the point of contact of the corresponding tangent. Then, c denoting a (1 - e2) (A cosa + B sina) + C = 0 1 + e cosa and therefore A cos0 - osa )+B (sinO- c sin 0....( A y r cos- e 1 + e cosa Since this line is perpendicular to the tangent, which is defined by the equation -= e cosO + cos(O - a) = (e + cosa) cos0 + sina sin0, we must have the relation A sin a B e + cos a POLES AND POLARS. 233 and therefore (1) becomes / ( ccosa \ / (. cina \ sin a. r c osa0 - -1 o- = (e + cos). r sn- \ 1 + e cosa/ ' / \ 1 + e cos a ec sin a or r f(e cosa) sinO - sina cos} 1 -+ ec a cl 1 + e cosaa or c 1 + e cos a. te sin + sn( -a), r e sin a which is the required equation to the normal. 2. If the normal through any point P in an ellipse cuts the major axis in G, to prove that, S being the focus, SG = e.SP. SECTION XXIII. Poles and Polars. 1. There are two ellipses the axes of which are in the same straight lines. To find the locus of a pole of the exterior ellipse the corresponding polar of which is always a tangent to the interior. Let a, b, be the semi-axes of the interior, and a, /, of the exterior ellipse. Let the coordinates of the pole be x,, y,, and let x', y, be those of the point in which its polar touches the interior ellipse. Then, the axes of the ellipses being taken as axes of coordinates, the equation to the polar will be =......................( but, since it is a tangent to the interior ellipse, its equation will also be XXI+ -1 2 + = I _........... (2)........ Since the equations (1) and (2) must be identical, we have a2 b2 x =-x,, and y = y,: X2 '2 1 hence, by the equation - + y =, 234 ELLIPSE. a2 12 72 2 we have a" = 1, for the equation to the locus of the pole; which is therefore a a2 132 similarly situated ellipse with semi-axes -, a b 2. To find the locus of a pole in relation to a given ellipse, the part of the polar which is intercepted by the ellipse being of a constant length. If the equation to the ellipse be a2 Y2 -+ E = 1 a2 + "17 and 2c be the length of the intercepted portion of the polar, the equation to the required locus will be c"2.y2"~ /ay' b'X2 /x". 2 — 1)\ C2 ([2 + )2 2 + 2 b * 2 2 2 3. The polar corresponding to a pole P, relatively to a given ellipse, intersects the ellipse in points E and F: to find the locus of P, supposing the area of the triangle EPFto be constant. The equation to the ellipse being X2 y2 2 + = 1 and m2 denoting the constant area, the locus of P will be an ellipse defined by the equation an A62 yes 2 2 2 y2 aV2b + + 2 = + - )1 SECTION XXIV. Inscribed and Circumscribed Polqyons. 1. To find the locus of the foci of all ellipses, described with their major axes parallel to the base of an isosceles triangle, and touching its three sides. INSCRIBED AND CIRCUMSCRIBED POLYGONS. 235 Let HKT be the isosceles triangle. Let A C, BC, be semiT Ad2 B M I-I x axes of one of the inscribed ellipses, S a focus. Draw SM at right angles to BH. From any point Q, in 7TH, draw QN at right angles to TB. Let _ HTB = tan-' a =LKTB; BM= x, S =y, CN=x', QN= y', CA =a, CB = b, BT= c. Then, THbeing a tangent to the ellipse, y' + aa' = (ab2 + d)\2. When x' - b, we see that y' = ac: hence a (c - ) = (a&b + a2)'. But = (2 - b2), y =b; hence a2 = '2 + b2 x2 + y2; and therefore, for the equation to the required locus, we get a (c - Y) = (a2y2 + x2 + y2)4 or ac" - 2acy = X2 + y2, which is the equation to a circle. 2. To determine the ellipse of greatest area which can be inscribed in a given triangle. Let two of the sides of the triangle be taken as coordinate axes. Then the equations to the three sides will be of the forms x = 0O y = 0O ax + by = 1; and the equation to the ellipse will be (acx) + (3y)4 + (ax + by - I) = 0, where a and / are arbitrary constants. 236 ELLIPSE. Clearing the equation to the ellipse of radical indices we shall get (a - a)'x' + (13 - b)yy2 + 2 {(- a) (1 - b) - 2a/c} xy + 2 (a - a) x + 2 (, - ) y + 1 = 0. The coordinates of the centre of this ellipse will be = *(13-b) e- ji(a - a) a/3 + ba - ab a/ + ba - ab' and the equation to the ellipse, referred to the centre as origin, will be (a-) 22+ (/_b)2 y2 + 2 {(a - a)(/3 - b) - 2a} x'y' = a a/3+bac-ab I Let ao be the angle between the coordinate axes: then, A denoting the area of the ellipse, we have, the axes being determined as in prob. 2, Section xvI., A2 = 1 2 a/ sins co 4 (ab-a3-ba)' In order that A may be a maximum, we must have 3 log (ab - a/3 - ba) - log a - log / = minimum: hence, differentiating with regard to a, we get 3b 1 -- = 0 a3 + ba - ab a 2ba - a/3 + ab = 0. Similarly, by differentiating with regard to 3, 2a/3 - ba + ab = 0. From these two equations we have a= - a, = - b. Hence the equation to the ellipse becomes a2x2 + by2 + abxy - ax - by + 0 = 0, or (ax - ~)2 + (hy - )2' + abxy =, or (a~ - 1~ + ( -)1 + I ( - (by - 1 - ) =. INSCRIBED AND CIRCUAMCRIBED POLYGONS. 237 2Also A ' = r, ab sin'2 7r2 sin2 o Also -4 2* = 17oA Also 27 a:b3 108a2b2 v A 7r sin C) 6V3.ab ' Supposingf and g to represent the intercepts of the line ax + by = 1, on the coordinate axes, the equation to the ellipse is (x 2)( )2 xY. 1 or 2 _~ + y_ ~ _ _f =,4 [ y \2 _I 2 \ X or (x)2(Y) o f 37 ) ( 3 ) (f 3 3) 7 -and A -fg sin o COR. 1. From the two forms of the equations to the ellipse, it is evident that it touches the sides of the triangle at their middle points, and that its centre is coincident with the centre of gravity of the triangle. COR. 2. Let B denote the area of the triangle. Then B = lfg sin co. Hence A: B:: 7r: 31/3. Berard: Gergonne, Annales de Mathematiques, tom. IV. p. 284. 3. Among all quadrilaterals inscribed in an ellipse, to determine that which contains the greatest area. Let the equation to the ellipse be x2 2 a. + - 1~ and let the angular points of the quadrilateral be (xI, y1), (x2, y2), (X, y3)) (x4, y4). Then, u denoting the area of the quadrilateral, 2u = X2y1 - XY2 + - 3Y2 - X2y3 + X4Y3 - X3y4 + X1y4 - X4Y1 -2 + b2 1 2 + 2 2= 1 + 2 + = 12 a b a b a b a b 238 ELLIPSE. In order that u may be a maximum, we have, differentiating these last four equations, putting du = 07, and using the method of indeterminate multipliers, Xx Xlyl + Y - =Y2 0 b2 - X4 + x2 = 0~................................................... Hence we have a? (X4 - ~) b(1) a ( - (Y - Y )............... (2), - - = (y,-,)............... (2), W3 (x.-x (3) = 2 ( - X)(Y) 2............... From (1) and (3), we have 3............ (5), X3 X1 and from (2) and (4), Y_4 =.(6). 4 2 Also, from (1) and (2), a ( - x2- =- — ) (y1y4 Y2,). From the last three equations we see that 1 I \ _ (X X a2 (x1x4- x2x3) =- y 1Y2 ~ Q ), and therefore YiY.2 -.... (7). x -......................... The equations (5) and (6) shew that the diagonals of the quadrilateral are diameters of the ellipse, and (7) shews that they are conjugate diameters. Durrande: Gergonne, Annales de lllathernatiques, tom. XII. p. 223. INSCRIBED AND CIRCUMSCRIBED POLYGONS. 239 4. If any hexagon be circumscribed about an ellipse, the three diagonals, joining opposite angles, will all pass through one point. The equation to the tangent of an ellipse is XX' yy cxx''* ''' ~''= ' '''*'.(1), a2 b2 x' and y' being connected by the equation Xt'2 x.( 42). a2 b2 a2 + =........................ (2) Equation (2) is equivalent to the following system of equations - - 1 = a1, X' y' 1 -Y - 1=-, a b al al being an arbitrary quantity: hence,, / 1,-1 2-lra a) Ya-2\/_1 (a hence, the equation (1) becomes 2a (al+ + 2b /- 1 Put a = a' bV-1 = b'; and the equation to the tangent, which we will suppose to be one of the sides of the hexagon, assumes the form x (1 ) ~b' 1).(3). 2a' (a+1 + a —) = 1............(3) The equation to the next side of the hexagon will be, in like manner, of the form a + 2.............(4) At the intersection of (3) and (4), x12, y 2,, being hde coordinates of the corresponding angle of the hexagon, we shall get, 240 ELLIPSE, by the combination of the equations, 1,2-a cc + a v Y1s 2 a- +a The equation to the diagonal, through the angular points (X1, 2 1,2), (4,5, 1/4,5), will be x (Y4,5 - Yl,2) - (X4,5 - 1,2) = X1,2Y4,5 - 4,5Y1,2; or, substituting for the coordinates of the angular points their values bx {(a1 - a4) (1 + aa,) + (a, - a,) (1 + aa4)} - ay {(a - a4) (1 - aa5) + (a2 - a5) (1 - aa)} = 2a'b' (ala2 - a,)....... (5). By similarity, it is evident that the equation to the diagonal, through (x, 3, y (,),, Y5,6), will be b'x {(a, - a) (1 + aa6) + (a - a,) (1 + a2a5)}- a'y {(a )(- a -a,) + (a, - a6) (1 - 2a5)} = 2a'b' (a2a - ).......(6). At the intersection of these two diagonals, multiplying (5) by a- - a6, (6) by a4- a,, adding the latter of the resulting equations to the former, and dividing the final equation by a2 - a5, we shall thus get b'x ({(a - a) (1 + a, +aa) (a - a) (1 + aa) - ay (a - a) (1 - a) + (a4 - a) (1 - aa6)} = 2a'b' (aa4 - aa1).......(7). But, by symmetry, we know that (7) is the equation to the third diagonal: hence the diagonals (5), (6), intersect in (7). Cambridge Mathematical Journal, vol. IV. p. 165. 5. To find the axes of the least ellipse which can be circumscribed about a given rectangle. If 2f, 2g, be the sides of the rectangle, and 2a, 2b, the:-;-e; of the ellipse, a = f 2, b = g2.,!.:,"., i4cta Academice Petropolitance, pro anno 1780, pars.:' D. 3. INSCRIBED AND CIRCUMSCRIBED POLYGONS. 241 6. To find the ratio of the side of an equilateral hexagon inscribed in an ellipse, with two sides parallel to the major axis, to the side of an equilateral hexagon inscribed in a circle of which the axis major is a diameter. If e be the eccentricity of the ellipse, the required ratio is equal to 4- 2e' 4-ed 7. To determine the ellipse of least area which can be circumscribed about a given triangle. If two of the sides be taken as axes of coordinates, and f, g, be the intercepts of these axes made by the third side, the equation to the ellipse will be )3 +) ( 3) ( -3) a (- ). The centre of this ellipse coincides with the centre of gravity of the area of the triangle, and its tangents at the three angles are respectively parallel to the opposite sides. Moreover the area of the ellipse is to that of the triangle as 47r: 3 /3. Euler: Nova Acta Petropolitana, tom. Ix. p. 146. Gergonne: Annales de Mlatkematiques, tom. IV. p. 288. Gregory: Examples in the Differential and Integral Calculus, p. 123, 2d edition. 8. If one triangle be inscribed within another, so that their sides are parallel each to each, a single ellipse will be, at the same time, the greatest which can be inscribed in the greater triangle and the least which can be circumscribed about the smaller triangle. Gergonne: Annales de Maathematiquzes tom. IV. p. 291. 9. To prove that, in every parallelogram circumscribed about an ellipse, the diagonals intersect each other in the centre of the ellipse and are in the directions of two conjugate diameters. Ferriot: Gergonne, Annales de Mathematiques, tom. XVI. p. 373. R 242 ELLIPSE. 10. To find the ellipse of least area which can be circumscribed about a given quadrilateral. Let the diagonals be taken as axes of coordinates: then (a, 0), (- a', 0), (0, 3), (0, - P'), being the four angular points of the quadrilateral, the equation to the ellipse will be + 2cxy + 3+ - (a- a) + - = The equation to the ellipse will therefore be of the form ax2 + by2 + 2cxy + 2a'x + 2'y + c' = 0, where c is the only unknown parameter. The determination of c will depend upon the solution of the cubic equation c'c' - 4a'b'c2 + (3a'2b + 3ab2 - abc') c - 2aba'b' = 0. Euler: Nova Acta Academice Petropolitance, tom. IX. p. 132. 11. The straight lines which bisect the angles of a triangle, meet the opposite sides in the points P, Q, R: to prove that an ellipse may be inscribed in the triangle such as to touch the sides in P, Q. R. SECTION XXV. Elliptic Loci. 1. From P, a point in an ellipse, lines are drawn to A, A' the extremities of the major axis, and from A, A', lines are drawn perpendicular to AP, A'P: to find the locus of the intersection of these lines. The equation to the ellipse being X2 y2 a, bk = 1. and x,, y, being the coordinates of P, the equation to AP will be y = Y-c (x —a), XI - a ELLIPTIC LOCI. 243 and the equation to the perpendicular through A will accordingly be a x y = ( - ( _ a).....................(1). Similarly, - a replacing a, the equation to the other perpendicular will be - a + I(x + a).................. (2). At the intersection of (1) and (2), 2 ( 2 2 a -( as - 2 (X2 - Ca), 2 2 or + a b2 a' [2 the equation to the required locus, which is therefore a concentric and similarly placed ellipse of which the semi-axes are a, -. 2. If all the ordinates of a circle, referred originally to rectangular coordinates, be moved through a given angle, the abscissa and magnitude of each ordinate remaining the same; to find the locus of the extremities of the ordinates, the origin of coordinates being the centre. If c denote the radius of the circle, the equation to the required locus, referred to the original axis of x and to an axis of y, parallel to the new ordinates, is evidently x2 + y2 c6' which represents an ellipse referred to its two equal conjugate diameters as axes. Let a, b, denote the semi-axes of the ellipse, and 0, 0', the inclinations of the axes of x, y, respectively, to the direction of a. Then, cw denoting the angle between the two new axes, ab = c sin o, a' + b = 2c', 2 244 ELLIPSE. which equations are obviously satisfied by a = cV2 cosIc, b = c42 sin o. Also tan 0.tan (o + ) = - - = - tan" o, and therefore 0 =- o, 0'= o + 0 = -Co. We have thus ascertained the magnitudes and positions of the axes of the ellipse. 3. A straight line of given length 2c is made to move so that its ends lie always in two other given straight lines including an angle a: to find the locus of its middle point. Let the two given lines be taken as axes of coordinates. Let the intercepts of the moveable line on these axes, in any one of its positions, be a, b. Then 4c2 = a'2 + b2 - 2ab cos 2a. But, if x, y, be the coordinates of the middle point of the moveable line, x= 2a, y= b: hence c2 = x2 + y - 2xy cos 2a, which is the equation to the locus required. The locus is therefore an ellipse. Con. It may be shewn that the semi-axes of the ellipse are c cot a, c tan a, the former one bisecting the angle between the positive directions of the coordinate axes. 4. A, B, are two given points: to find the locus of the ly P C A_ 0 B3 D point P which is so situated that perpendiculars (drawn through it) to AP, BP, always cut off a given length on AB. ELLIPTIC LOCI. 245 Take 0, the middle point of AB, as origin, OBx as the axis of x, and Oy, perpendicular to it, as that of y. Let ACO= u BD=v, AB= 2a, CD = 2b. Then, from the right-angled triangle CPB, we see that 2 y = (u + + X).(a - X), or -= u + a + x................... (1); and, from the right-angled triangle APD, we see that y2 = (v + a - x).(a + x), or - v + a -......................(2). a + x From (1) and (2), we have Y2 - = u + v + 2a: a -x a+x but u + v + 2a = 2b: hence + Y' 2b a-x a+x or ay2 = b (a - -2), the equation to an ellipse the axes of which coincide with the axes of coordinates. 5. The focus of a parabola lies always in the circumference of a circle represented by the equation x + y2 = c2 the axis of the parabola being parallel to the axis of a, and its arc passing through the centre of the circle. To find the locus of the vertex of the parabola. The equation to the parabola will be of the form,ax + 2yy + y2 = 0. 246 ELLIPSE. Change the origin to a point h, k: its equation will then become (x' + A) + 2v (y' + ) + (y' + k)2 = 0. Assume the coefficient of y', and the terms which involve neither x' nor y', to be zero: then kh + 2vk + k2 = 0, v + k = 0, k2 and therefore L = - The equation to the curve becomes k2 y - -h ' Let x, y, be the coordinates of the focus: then, since he k, are the coordinates of the vertex, %x= ~h-4, ' =/ck; and therefore, the focus being a point in the circle, 16 2 C=x 2+= Y- + k" - ' 4h 4 4 e2 +e n to c the equation to the locus of the vertex, which accordingly lies always in the arc of one of two ellipses. De la Hire: Sectiones Conicce, lib. vIII. prop. 20. 6. A square CBPQ moves in the right angle x Oy so that the point C is always in Oy, and the point B always in Ox: to find the locus of P. If a represent the length of a side of the square, then Ox, Oy, being taken as coordinate axes, the locus of P will be an ellipse of which the equation is x2- 2xy + 2y2 = a. ELLIPTIC LOCI. 247 7. To determine the locus of the extremity of the straight line formed by augmenting the ordinate of the circle x' + y2 = c' by a length equal to the abscissa. The locus will be an ellipse defined by the equation 2x2 - 2xy + y = c2 Fuss: Nova Acta Academice Petropolitance, tom. xv. p. 72. 8. To find the locus of the intersection of two tangents to a circle, such that the product of the portions (measured from the centre) cut off by them from a given diameter, is always constant. If the given diameter be taken as the axis y; then, a denoting the radius of the circle and 1c2 a constant quantity, the locus will be an ellipse of which the equation is (a2 + k2) X2 + a2y2 = k2a2 9. In the given right lines AP, A Q, are taken variable points p, q, such that Ap:pP:: Qq: qA; to find the locus of the point of intersection of Pg, Qp. Let APx, AQy, be taken as axes of x and y: let AP = a, A Q = b. Then the required locus will be an ellipse represented by the equation b - x (a -) + ^ y ( - y) = (a - ) (b -y) a b COn. If we put x = 0, y = 0, successively, we get respectively (y - b)2 = 0, (x - a) = 0. Hence the ellipse touches Ax, Ay, at the points P and Q. 10. A straight line AB of given length, moves with its extremities A, B, always in contact with two straight lines Ox, Oy, respectively, inclined to each other at a given angle: to find the locus of an assigned point P in AB. 248 ELLIPSE. Let PA = a, PB = b, and x Oy = o: then, Ox, Oy, being taken as coordinate axes, the required locus will be anl ellipse defined by the equation a2 +, - 2y xy — x 2 - cos c = 1. a b2 ab Newton: Arithmetica Universalis, prob. 21. Bret: Gergonne, Annales de Mathematiques, tom. VI. p. 15. 11. To find the locus of the intersection of the lines represented by the equations x sin 0- y cos 0= (x2 + y:) cos2 0 sin2 0 1 and a2 + -2 =2 2 a bS x'2 + y' The required locus is an ellipse the equation of which is a'2 b2 12. From a point (a, 6) a normal is drawn to a parabola y = Ix; to find the locus of the intersection of the parabola and normal, I being a variable parameter. The required locus is an ellipse defined by the equation y (y -) +b 2x (x - a) = 0. L' Hospital: Traite Analytique des Sections Coniques, p. 259. 13. To find the locus of the middle point of a chord of an ellipse, the chord passing through a given point. The equation to the ellipse being a2 + = t1 and the coordinates of the given point being (a, 18), the locus required will be another ellipse defined by the equation (2x - )2.(2y - 3)2 a2 +32 a2 + a2 b2 ELLIPTIC LOCI. 249 14. A triangle ABC moves in such a manner that the angular points A, B, are always in two rectangular axes Ox, Oy, respectively: to find the locus of C. Let BC = a, CA = b, AB = c, and A CB = w. Then the equation to the required locus will be x + Y. 2. sin w = (cos w). 15. From the focus S of a given ellipse is drawn SQ always bisecting the angle PSC, C being the centre, and equal to the mean proportional between SC and SP: to find the locus of Q. The equation to the given ellipse being a2 y2 the required locus will be another ellipse defined by the equation (x - ae)2 + = ae 1 +e 1-e 16. A right angle BA C the sides AB, AC, of which are given, moves in such a manner that A, B, lie always in two perpendicular lines Ox, Oy, respectively: to find the locus of C. Take Ox, Oy, as axes of x, y, respectively. Let AB = a, A C = b. Then the locus of C will be an ellipse defined by the equation Ab' 2 + (a2 + b2) y2 - 2abxy = b4. Newton: Arithmetica Universalis prob. 22. 17. The circumstances remaining the same as in the preceding problem, except that LA OB, and z BA C, instead of being right angles, are each of them equal to a; to find the locus of C. The locus is an ellipse defined by the equation 2x2 + (a2+ b2 - 2ab cosa) y2 + 2b (b cosa - a) xy = b4. Newton: Arithmetica Universalts, prob. 22. 250 ELLIPSE, 18. The base and the sum of two sides of a triangle being constant, to find the locus of the centre of the escribed circle touching the base. Let the middle point of the base be taken as the origin of rectangular coordinates, the axis of y being perpendicular to the base. Let 2a represent the length of the base, 2b the sum of the two sides. Then the required locus will be an ellipse defined by the equation b + a 2 Y2 b - a(a-x 19. One extremity A of a straight line ABP and a given point B in it, move in straight lines intersecting at right angles in C: to prove that the locus of any other given point P in the line is an ellipse, and that, if the rectangle ACBD be completed, and PD joined, PD is a normal at P. 20. To find the locus of the points of quadrisection of a system of parallel chords in a circle. The equation to the circle being x2 + y' = aim and that to any one of the parallel chords lx + my = 8, where l=cosa, y=sina, the equation to the required locus will be (3m2 + 1) x2 - Glmxy + (312 + 1) y2 = 2. 21. Having given the point where a parabola intersects a given diameter, and also the parameter of that diameter, to find the locus of the vertex of the curve. Let p represent the given parameter; then, the given diameter and a perpendicular to it through its vertex being chosen as axes of x and y respectively, the required locus will be an ellipse denoted by the equation 4x2 + y2 + px = 0. Lardner: Alclebracic Geometry, p. 131. ELLIPTIC ENVELOPS. 251 22. Two semicircles are described on the segments of the base of a semicircle of which the radius is r: to find the locus of the centre of a circle touching these three semicircles. If the middle point of the base of the large semicircle be taken as the origin of rectangular coordinates, the base being the axis of x, the locus required will be the arc of an ellipse of which the equation is X2 (y + r)2 -4r + 162 1. 3 9 23. CP, CD, are conjugate semi-diameters of an ellipse: on the chord PD an equilateral triangle PDR is AI described: to find the equation to the locus of R. The equation to the proposed ellipse being ~2 ~,2 x" y. the required locus will also be an ellipse of which the equation is a2 y2 (a 3) (b + ba V3) SECTION XXVI. Elliptic Envelops. 1. CP, CD, are any two conjugate semi-diameters in an ellipse, of which the semi-axes are a and b: to find the locus of the consecutive intersections of all such chords as PD. If x', y', be the coordinates of P, we may put x = a cos y' = b sinO: we shall then have also, x", y" being the coordinates of D,! a b:x" ==-?/= - a si-n, = x = b cosO. b4 '!' a 252 ELLIPSE. Hence the equation to PD is, by the formula (" - X') y - (y" - y) x = ' - y", y (-a sinO - a cos ) - x (b cosO - b sin0) = - a sin. b sin - a cos. b cos, or (ay - bx) sin + (ay + bx) cosO = ab........(1). Differentiating with respect to 0, we have (ay - bx) cos - (ay + bx) sin 0 = 0.........(2). Squaring and adding (1) and (2), we get, for the equation to the required locus, 2a2y2 + 22x2 = a2b2. 2. If, from every point in the axis major of an ellipse, as centre, a circle be described, with radius equal to the ordinate at that point: to find the envelop of the circles. The equation to the ellipse being d' y2 + b" = 1, that to the envelop will be an arc of another ellipse represented by the equation a2 + b ~=2 3. Two straight lines are drawn from one extremity of the major axis of an ellipse, making with it angles the tangents of which are in a given ratio: to find the locus of the ultimate intersections of chords, each of which joins the two points where two corresponding lines cut the ellipse. The equation to the ellipse being b2 y- = D (2ax - 2), the envelop required will be another ellipse, the equation of which referred to the same axes will be, a denoting the given ratio 4ab2 =P (1T+a)2 a2 (2ax - x (If a) aGl MISCELLANEOUS PROBLEMS. 253 4. AEC, BFC, EF, are three straight lines given in position: from the given points A, B, straight lines are drawn, through any point in EF, meeting BC in Q, AC in P: PQ is joined. To prove that the curve, to which PQ is always a tangent, is an ellipse inscribed in the quadrilateral APQB, and touching the lines A C, BC, at the points E, F. Leybourn: Mathematical Repository, New Series, vol.III. p. 143. SECTION XXVIIL Miscellaneous Problems. 1. Of all systems of conjugate diameters in an ellipse, the principal diameters are those of which the sum is a minimum; and the equal conjugate diameters are those of which the sum is a maximum. Let a, b, be the principal semi-diameters of an ellipse, and a', b', any two conjugate semi-diameters, including an angle ry. Then we know that a'2 + b'" = a2 + b2 and a'b' siny7 = ab. From these two equations we have a' + = (a'+2 + - +sin /................. 1) and a -1'= (a + b' 2a6b ' ~~and a'-b '=~- sa i+ _.................(2, sin y - The equation (2) shews that sinry cannot be less than 2ab a' + b'' Thus siny is necessarily included between the limits 2ab 1 and + 2. a2 + b' Now a' + b' is greatest when sinr is least: in this case, from (1) and (2), a' + b' = {2 (a2 + b')}, 254 ELLIPSE. and a'- b' = 0; whence a = _ (a + - Also a' + b' is least when sin 7 is greatest: in this case, from (1) and (2), a' + b' = a + b, and a'- ' = a - b, whence a' = a b' = b. Durrande: Gergonne, Annales de Mathekmatijues, tom. xII. p. 168. 2. If a straight line be drawn from the focus of an ellipse, the eccentricity of which is e, so as to make a given angle 3 with the tangent; to shew that the locus of its intersection with the tangent will be a circle, which touches or lies entirely without the ellipse as cos / is < or > e. Let SY be the perpendicular from the focus S upon the i C s Ah Ttangent PT at P, which cuts the semi-axis major CA, produced, in T. Let Q be a point in PT, such that L SQP==,l. Let LPTS = b, SQ= r, QST= 0. Then r sin/3 =-SY= (a2 sin 2 + b'2 cos'2b)1 - ae sin b. Squaring and putting for b its value f - 0, we have, for the equation to the locus of Q, r2 sin2/3 + 2aer sin/ sin (/3 - 0) = a2 (1 - e), which is the polar equation to a circle. If any point of this circle lie in the periphery of the ellipse, (when, intersection being impossible, contact must take place), we have, equating the values of SY in the circle and the ellipse, a" (1- e) r = r2 sin2/, a2 e e2) = (2ar - r2) sin2/3, 2a — r (a - r)2 sin2/3 = a2 (e2 - cos2/3). 255 MISCELLANEOUS PROBLEMS. Hence, in order that contact may be possible, cos/3 must not be greater than e. If cos,/ be greater than e, the circle lies entirely without the ellipse. Bobillier: Gergonne, Annales de Mthkat atiques, tom. XVIII. p. 191. 3. A circle is described about an extremity of the minor axis of an ellipse as centre, with a radius equal to the distance of either directrix from the centre of the ellipse: to prove that this circle will touch the ellipse in two points, one point, or not at all, accordingly as the eccentricity is greater than, equal to, or less than. V2' Let the equation to the ellipse be + =.........................(1), or, e being the eccentricity, (1- e2) x' + y2= aY..................... (2). The equation to the circle will be x + (y - b) =, or (1 ) + -( - e2) (y - )2=........... Eliminating x2 between (2) and (3), we have, for all points where the circle and ellipse meet, the equation e2y2 + 2 (1 - e2) by + b- (1 - e2)2 = 0, which, being a perfect square of the equation ey + -(1 - =..............(4), shews that the curves touch wherever they meet. From (1) and (4), there is x2 (1- e2) a2 e4 256 ELLIPSE. This result shews that 1 - e must not be greater than e2, or that e must not be less than 2. If e = -2 x = 0, and there is contact at the opposite end of the minor axis; if e > 7, x has two possible values, and accordingly there are two points of contact. 4. To find the equations to the circles which touch the latera recta of an ellipse represented by the equation a + =1 in the major axis, and of which the radii are equal to the minor axis. The required equations are (x - ae ~- 6)2 + y' = b2, and (x + ae + b)'2 + y2 = b'2. 5. To find the equations of the circles the diameters of which are the distances between the extremities of the axes of an ellipse defined by the equation X2 Y2 -+ -. 4 9 The four circles are defined by the equation x2 + 2x + y2 + 3y = 3. 6. If one of the foci of an ellipse be the common focus of two parabolas the vertices of which are at the extremities of the major axis, to prove that these parabolas will intersect at right angles, at points the distance between which is equal to twice the axis minor. 7. If CD be the semi-diameter of an ellipse drawn parallel to a focal chord PSp, to prove that 2CD = AC.P2. MISCELLANEOUS PROBLEMS. 257 8. If a circle be described, touching the axis major of an ellipse in one of the foci, and passing through one extremity of the axis minor, to prove that the semi-axis major will be a mean proportional between the diameter of the circle and the semi-axis minor. 9. To find the relation between the central distance of any point of an ellipse and the perpendicular from the centre on the tangent at that point. If r = the central distance, andp = the perpendicular, aib2 a - P - a2 + b'- r2* 10. If two tangents be drawn at the extremities of the axis major of an ellipse, so as to intersect a tangent at any point in two points T, T', and a circle be described upon TT' as a diameter; to prove that this circle will pass through both foci of the ellipse. 11. A parallelogram circumscribes an ellipse, its sides being parallel to conjugate diameters. If 20 be one of the smaller angles of the parallelogram, when it is equilateral, to prove that its perimeter is equal to 4 /2. a sect. 12. A square is described touching an ellipse at the extremities of its minor axis: an ellipse upon the same axis major circumscribes the square. The same process is repeated in relation to the new ellipse, and the operation is continued until there are n ellipses: to prove that, if the eccentricity of the original ellipse be (- 1, the last ellipse will become a circle. n I 13. If, from the focus S of an ellipse, SQ be drawn, bisecting the angle PSC0 (C being the centre of the ellipse and SP any focal distance), to find the locus of Q, supposing SQ to be a mean proportional between SC and SP. s 258 ELLIPSE. The equation to the ellipse being 62 y2 1 a a (1 - e2) the required locus will be an ellipse defined by the equation (x -ea)2 Y =ea + e 1 —e 14. If P be any point in an ellipse, the vertex of which is A and nearer focus S, to prove that, if LPAS = 0, zLASP=, and e = the eccentricity, tan 0. tan l = 1 + e. 15. If 7e 0, be the angles which any diameter of an ellipse makes with its conjugate, and with the axis major, to shew that (a2 - 62) tan7 ~ (a2 tanO + b2 cot0) = 0. 16. To determine the least distance between the centre C of an ellipse and a point P in its periphery, and, PM being a perpendicular on the axis major, to obtain an expression for the difference between the squares of CP and of the least distance in terms of CM. The distance is least when P coincides with an extremity B of the semi-axis minor, and (CP)2 _ (CB)2 = e2. (GM)2. Apollonius: Conicorum, lib. v. prop. 11. 17. If PSQ, PHR, be chords drawn from any point P of an ellipse through the foci S and H, b and 0 the acute angles which the axis major makes respectively with the line QR produced and with the tangent drawn to the ellipse at the point P, to prove that tan = 1 +. tan0. 18. In two confocal ellipses are taken two points P, Q, such that, the ellipses being referred to the major and minor axes as MISCELLANEOUS PROBLEMS. 259 axes of x and y, their abscisse are as the major axes of the ellipses: P', Q', are two other such points. To prove that PQ = P'Q. 19. To shew that the equation ax 3y \2 (x' Y2 \ /a2. 32 \ 2 + - ). a + -)=(a2 b21 g)* +Ab2-) represents two straight lines, passing through the point (a; /3) and touching the ellipse x2 2 2 +-+ =1. a b 20. Having given the directions of two systems of conjugate diameters of an ellipse, to determine the directions and relative magnitudes of the two principal axes. Let 0 represent the angle between a principal axis and a given line drawn through the centre of the ellipse: let 2x denote the length of this axis, and 2y that of the other. Let a, /3, be the angles which the two conjugate diameters of one system make with the given line, and a', 3', the angles formed with it by the other system. Then the solution of the problem will be expressed by the following equations, tan 2 0 = os (a' + '). cos (a - 3) - cos (a + ). cos (a' - 3') sin (a' + /3') cos (a - 3) - sin (a + ). cos (a' - ') ' {- tan (0 + a) tan (0 + /)} = - = {- tan (0 + a') tan (0 + B')}. x Gergonne: Annales de Mathematiques, tom. XII. p. 374. 21. From any point C2 without a given ellipse, of which the centre is CG, is drawn a straight line cutting the curve in P,, P,. If a point P be taken in C2P, so that C2P is a mean proportional between CP, C2P, to find the locus of P. The equation to the given ellipse being x2 2 a2+ =b? 82 260 ELLIPSE. the locus of P will be an ellipse, of which C2 is the centre, represented by the equation (x aa)2 (y- ) f) 2 /2 a2 + a2 + 2 1 where a, A, are the coordinates of C,. 22. If, in the preceding problem, be drawn through CQ a straight line meeting the given ellipse in Q, and the locus of P in Q1i Q2; to prove that C Q will be a mean proportional between C Q1, C Q2. 23. If, under the circumstances of the two preceding problems, the given ellipse and the elliptic locus of P intersect in F and G, and if, about FCJ G C2 be described a third ellipse with axes parallel to those of the given ellipse, and if (a,, b ), (a,,, b,,), be the semi-axes of the locus of P and of this third ellipse respectively, to prove that a a, a, ab + ab, = 4a,,,,; and that the centres of the three ellipses lie in a straight line. 24. If SY, HZ, be perpendiculars from the foci of an ellipse upon the tangent at any point P, to prove that HY, SZ7 being joined, will intersect in the normal at the point P, and to find the locus of their intersection. The equation to the ellipse being X2 y2 a2+ b -,2 and e denoting its eccentricity, the equation to the required locus will be x Y 2 (1 + e2)2 a + b2 4 which is therefore a concentric and similarly placed ellipse, with axes (1 + e2) a and b. MISCELLANEOUS PROBLEMS. 261 25. P is a point in the circumference of an ellipse, S and H its two foci, and A the extremity of the axis major nearest to S. Supposing PH to be increased in length by a small quantity A, the position of P, the direction of PH, the magnitude and position of SP, remaining unchanged: to determine approximately the change in the eccentricity of the ellipse. If the axis major = 2a, SP= r, HP= r', L ASP= 0, and the eccentricity = e; then, approximately, hr 8e = 2ar' (e + cos0). 26. To find the envelop of the perpendiculars drawn through the extremities of all the diameters of a given ellipse. If the equation to the ellipse be x2 y= 2 as + - ~1, the equation to the required envelop will result from the elimination of t between the two equations a(t' + a _ b2)2 (t2 - 2) y2 a2 {(2b-2 a2) S + a(a2 - b2)}2 Roche: eroe d Math ue t6 Roche: Gergonne, Annales de Mathimatiques, tom. xIv. p. 207. ( 262 ) HYPERBOLA. SECTION I. Referred to its Axes. Ordinates. 1. A parabola and an hyperbola have a common axis, the vertex of the former coinciding with one of the foci of the latter: to shew that the product of the distances of the points of intersection from the axis is equal to "b"2, 1 being the latus-rectum of the parabola and b the minor axis of the hyperbola. The equation to the hyperbola being y= ' (x2 - a2), that to the parabola will be y = I (x - ae): at the intersections Y (2 y4 2ae y,2 ) or y4 +...... + b1 = 0: whence the required product, by the theory of equations, is equal to b212. 2. In two concentric and similarly placed hyperbolae, are taken two points, the abscissae of which are as the real axes of the curves. To find the locus of the middle point of the straight line joining the two points. The equations to the two hyperbolae being 2 b ~ AL aX 2 2 aW ab t h,2 I REFERRED TO ITS AXES. TANGENTS. 263 the required locus will be an hyperbola represented by the equation xz 2_ (a + -)2 (b + bT)2 4 SECTION IIo Referred to its Axes. Tangents. 1. The semi-axis CA of an equilateral hyperbola is intersected in T by a tangent to the curve at P: PM is a perpendicular from P upon CA produced: PC is joined. To prove that PC.PM = PT. CM. Let CA, produced indefinitely, be taken as the axis of x, and Cy, at right angles to it, as that of y. Then, putting CA = a CM= x% PM= y, and observing that a'2 CT is equal to -, we have x 2, 2v (PT)2 _ 2 + x- a-) (X a), Y + _ _ - (x2 + 2) (CPM)2 CP)2 or P. PM = PT. CM. De la Hire: tSectiones Conicc, lib, v, prop. 15. 2. If 8 be the perpendicular distance of the centre of the hyperbola x2 Y= from the tangent at y poin and if be the angles which froum thle tangent at any point, and if a, p3 be the angles which ~ 264 HYPERBOLA. makes with the positive directions of the coordinate axes, to prove that, X being an arbitrary quantity, a cos (- x) = 6 cos ( + ), 4 1_ 1 2 11 2 and -2 -= Y + X + X-) X 3. From the centre C of an hyperbola a perpendicular CY is drawn upon the tangent at the point P: to find the coordinates of P, when PY = CS. The coordinates required are x = (e2+e-1). a a y = (e2 - 1).. SECTION III. Referred to its Axes. Magical Equation to the Tangent. 1. To find the locus of the intersection of a tangent to an hyperbola with a perpendicular upon the tangent from the centre. The magical equation to the tangent of the hyperbola W2 =2 a2 b2 is y = ax + (a - b2)..................(1 ). The equation to a perpendicular upon this tangent from the centre is 1 y = -........................ (2). At the intersection of (1) and (2) we have, a being eliminated between these equations, (x2 + y2)2 = a2X2 - b2y which is the equation to the required locus. REFERRED TO ITS AXES. FOCAL PROPERTIES. 265 COR. If the hyperbola be equilateral, the equation becomes (xS + ')2 = a' (X' y-), the equation to the lemniscata. James Bernoulli: Opera, tom. I. p. 609. 2. To find the locus of the intersection of a tangent to an hyperbola with a perpendicular upon the tangent from a given point. The equation to the hyperbola being YX' x2 = 21 and the coordinates of the given point being a, /, the required locus will be defined by the equation { (x- a) + y (y - /)})2 = a2 (x - a) - a) (y - 3)2. 3. An ellipse and hyperbola having common axes, a pair of tangents is drawn to them, such that the sum of their inclinations to the transverse axis is equal to two right angles: to find the locus of their intersection. The equations to the ellipse and hpperbola being x2 ',2 2~ " y that to te = eqired alocus wb2l be that to the required locus will be 4xY4 = b. (a2- 2). SECTION IV. Referred to its Axes. Focal Properties. 1. If, about the exterior focus of an hyperbola, a circle be described with radius equal to the conjugate semi-axis, and tangents be drawn to it from any point in the hyperbola; to prove that the line, joining the points of contact, will touch the circle of which the transverse axis is a diameter. 266 HYPERBOLA. The equation to the circle, described about the exterior focus, is 2 ocus isx2 + y2 + 2aex + a2 = 0. Let (h, k) be any point in the hyperbola: then, tangents being drawn from this point to the circle, the equation to the chord of contact will be (h+ae) x + ky + a (eh + a) =............ (1). The equation to the hyperbola gives the equation k2= (e2_ 1) (h2- a2).................. (2). The distance of the centre of the hyperbola from (1) is equal ta (eh + a) {( + ae)2 k a (ek + a) from (2), {(h + ae)2 + (e2- 1) (2 - a2)} rom = a. This result establishes the proposition. 2. To find the locus of the centre of a circle inscribed in the triangle SPH, S and H being the foci and P any point of an hyperbola. The equation to the hyperbola being X2 y/2 s2 b2 = 1 the required locus is a straight line x = a, which is therefore a tangent at the vertex. Lardner: Algebraic Geometry, p. 128. 3. To find the locus of the centre of a circle, which, the diagram remaining the same as in the preceding problem, touches SP, HP produced, and HS produced. CONJUGATE DIAMETERS. CONJUGATE HYPERBOLA. 267 The required locus is an hyperbola defined by the equation (a + c)2 y~ - b"x = - bc2, 2c being equal to SH. Lardner: Algebraic Geometry, p. 128. 4. To find the locus of the centre of a circle which, the diagram remaining the same as in the two preceding problems, touches SH, PS produced, and PH produced. The equation to the required locus is x + a = 0, which is therefore a tangent at the vertex of the opposite branch of the hyperbola. Lardner: Algebraic Geometry, p. 129. SECTION V. Referred to its Axes. Conjugate Diameters. Conjugate Hyperbola. 1. At any point P of an hyperbola, a tangent is drawn, cutting the semi-axis CA in T, and the semi-axis BC, produced, in t: CD is drawn, parallel to tTP, to meet the conjugate hyperbola in D. To prove that PT.Pt = (CD). Let CA, CB, produced indefinitely, be the axes of coordinates. Let x, y, be the coordinates of P (x', y') those of D. The equation to the hyperbola being 0X2,2 as — =1, that to PTt will be _X = 1. a' b2 Putting x, = 0, we have oY, =-y b2 or Ct =- Y 268 HYPERBOLA. Hence, a being the inclination of PTt or DC to CA, PT.sina = y, Tt.sin = Ct = - Y b2 Pt.sina = y + - - and therefore PT.Pt.sin2a = y2 + b2 = -x2 Again, (CD)2. sin2 a = y'2- X2 Hence PTPt = (CD)2. De la Hire: Sectiones Coniccea lib. Iv. prop. 41. 2. An ellipse and a pair of conjugate hyperbolas are described upon the same axes, and, at the points where any line through the centre meets the ellipse and one of the hyperbolas, tangents are drawn: to find the locus of their intersection. Let A, k, be the coordinates of the intersection of the two tangents: then, (x, y), (x', y'), being the coordinates of the points of contact with the ellipse and the hyperbola respectively, fhix k 2, X2 Y2 a2 z — b2+ = 1 -............. (hx' k x" y12 and (= a b...............(2). But -Y, by the condition of the problem: hence, from x y (2), we have (a~ b) - -2 *........ (3). From (1) and (3), by addition and subtraction, we have h2x2 k2y2 x2 a4 + 4 - a'= and 2hkx = afy REFERRED TO ITS AXES. ASYMPTOTES. 269 Hence, eliminating x and y, we have 2 (1 + 4 )=-a2.................... (4), the equation to the required locus. The equation to the locus resulting from the intersection of the tangent of the ellipse with the corresponding tangent of the conjugate hyperbola is evidently k (1 + 4 ) = 2, being derived at once from (4) by interchanging (a, b) and (a, k). 3. To prove that, in a rectangular hyperbola, every diameter is equal to its conjugate diameter. 4. If CP, CD, be conjugate semi-diameters of an hyperbola, to shew that, K being the point of intersection of normals at P, D, KC is perpendicular to PD. 5. To find the area included by the normals to an hyperbola, which pass through the foci of the conjugate hyperbola. The equation to the hyperbola being 2 Y32 a2 b2 1 the required area is equal to 2 2 (a2 + b2).(a2 + 2b2.). a SECTION VI. Referred to its Axes. Asymptotes. 1. In the semi-axis CA of an hyperbola is taken a point M such that, S being the focus, CM: CA:: CA: CS; from P, any point in the curve, is drawn PR, parallel to one of the asymptotes, to meet a perpendicular to CA through Ml 270 HYPERBOLA. in the point R: to prove that, SP being joined, PR = PS. Let the hyperbola be referred to CA, produced indefinitely, as the axis of x% that of y being at right angles to CA through C. Let CIM= c. Then the equation to PR, c 3t A S x, y, being the coordinates of P, will be y - y = - (x - x). a Hence, putting c for x', and for y', putting y, = RM, we have y -y=b (c - x), and consequently (PR)2 = (x - c)2 + (y -,)' a=. + b' = (X - C),. = e2 (x - c)2, PR = e (x - c). But, by hypothesis, a = c.(a2 + b2) = cae, a = ce: hence PR = ex - a = SP. De la Hire: Sectiones Conicce, lib. VIII. prop. 18. 2. If, from a point in an asymptote to two conjugate hyperbolas, a tangent be drawn to each hyperbola, to prove that the points of contact will be in conjugate diameters. Let (x', y'), (x", y"), be the points of contact in the hyperbolas represented by the equations X2 y2 y2 '2 2The equations to the a2wi = The equations to the tangents will be xx yy a2 b2 yyl" XX"i 2 a2 -1. b2 a2 REFERRED TO ITS AXES. ASYMPTOTES. 271 At their intersection xx yy yy" xx" a2 b'22 b2 a2 But, at their intersection, x y::a b; hence X Y - Y ( h ca b b a....................(1). a b b a Now, by the equations to the curves, x'2 y"_ yt12 X,12 a2 b2~ 6 a2 hence, by (1), + -..=....................(2). a b b a From (1) and (2) we see that,i a I b I = b x = y', =a X These equations prove the truth of the proposition. 3. If, from a focus of an hyperbola, as centre, a circle be described with diameter equal to the imaginary axis; to prove that it will touch the asymptotes in points where the nearer directrix meets them. 4. A tangent EPOF to an hyperbola, at a point P, cuts the asymptotes in the points E, F, and the transverse semi-axis CA in 0: from F is drawn the straight line FKG, cutting A C at right angles in K, and the other asymptote CE in G; from 0, A, are drawn, at right angles to CA, OR, AI, cutting CE in R, I; from P, E, are drawn PM, EH, cutting CA, produced, at right angles in M, H: MP is produced to meet CE in Q. To prove that the three lines GK, AI, HE, are in continued proportion: and that the three lines OR, AI, MQ, are also in continued proportion. De la Hire: Sectiones Conicce, lib. IV. prop. 12. 272 HYPERBOLA. SECTION VII. Referred to its Transverse Axis and the Tangent at its Vertex. 1. If any chord AP, through the vertex of an hyperbola, be divided in Q so that AQ: QP:: AC'2: BC'2 and QM be drawn to the foot of the ordinate MP; to shew that Q 0, at right angles to QM7 cuts the transverse axis in the same ratio. Draw QN at right angles to the transverse axis, to cut it in N. P O A N AL Let AN=x,, QN=y, AM=x', PM=y', OA=x, L QOM=s. Then, from the figure, we see that y = tan ( + x).....................(1). Now, evidently, by the hypothesis, a a, a2, a2 and therefore tan = cot QlMN x -X~ = (a2 + b2) x' - a2x' 2 x yH oay a2y Hence also, from (1), l'x' 2y' = a2y, {a2X' + (a + b) X a2 b (x'2 + 2ax') = b"' {ax' + (a2 + b2) x}, 2Wa = (a2 + b2) x, a2 and therefore x = 2a This result establishes the proposition. This result establishes the proposition. REFERRED TO CONJUGATE DIAMETERS. ASYMPTOTES. 273 2. The straight line x + my = 8 is a tangent to the hyperbola y = + (2ax + X2): to find the value of 8 in terms of a, b, 1, m. The required value is given by the equation (~ + la)2 = a' 2- b2m'. SECTION VIII. Referred to Conjugate Diameters. Asymptotes. 1. CP is any semi-diameter of an hyperbola, of which C is the centre: a straight line QHK, parallel to PC, cuts the curve in Q and the asymptotes in H, K. To prove that QH.QK = (CP)'. Let CP, produced indefinitely, be taken as the axis of x, and CD, its conjugate, as that of y. Let CP = a, CD = b: then the equation to the curve will be a2 2- 1 Let k represent the constant ordinate of the line QHIK: then, 0 being the intersection of this line with CD, we see, from the equation to the curve, that (OQ)-a(1- ).................. (1). Again, at the intersection of the line y = k with the asymptotes, the equations of which are b Y = + -, al b ak and therefore OH = a =_ OK0 T 274 HYPERBOLA. Hence QH.QK= (OQ- OH).(OQ + OK) = (OQ)- (OH)2 =a (1i a) = a2 =(OP)2. De la Hire: Sectiones Conicce, lib. iv. prop. 4. 2. QQ', qq', are any two parallel chords of an hyperbola: a straight line FRf, parallel to the asymptote CI, cuts these chords, produced if necessary, in F, f, respectively, and the curve in R. To prove that FQ.FQ':fq.f':: RF: Rf. Let 'CP be a semi-diameter, conjugate to the two parallel chords: CD the semi-conjugate of CP. Let CP, CD, produced indefinitely, be chosen as axes of x, y, respectively. If CP = a' CD = b' the equation to the hyperbola is....................... (1). Let h, k, be the coordinates of F: then, writing h for x, and k - y' for y, we shall have a quadratic in y', viz. h2 (_ - y)2 a02 b,12 the two roots of which are FQ, FQ': thus FQ.FQ + b' - h2............... (2). a REFERRED TO CONJUGATE DIAMETERS. ASYMPTOTES. 275 Since Ff is parallel to C7, the equation to which is b' Y=-, X7 the equation to Ff must be of the form y = -, +c..................... (3). a But (A, k) is a point in Ff: hence k=, —, h + c. Consequently, from (2), FQ.FQ' = b + c2 - 2 ch..............(4). a At R, the intersection of (1) and (3), x = a2 (b2 + c2), and therefore x - h = -b (2 + c -2 ch) Let this value of x - h be denoted by L: then, by (4), 2b'c FQ.FQ' =. L. Similarly, it is evident that, the difference of the abscissie of RB f, being denoted by 1 2b'c f.f1 = — I.. a FQ.FQ' L Hence f f, = 7 fq:fq' 6 RF De la Hire: Sectiones Conicce, lib. IV. prop. 26. 3. From the extremity P of a diameter P'CP of an hyperbola, of which C is the centre, a tangent PE is drawn to meet one of the asymptotes in E: from E is drawn a straight T2 276 HYPERBOLA. line EQ, parallel to P'CP, to meet the curve in Q: from Q a line QN is drawn, parallel to EP, to meet P'CP, produced, in N. To prove that (GP)2 = PN.P'N. De la Hire: Sectiones Conicce, lib. IV. prop. 8. 4. From the extremity P of any diameter P'CP of an hyperbola, of which C is the centre, a tangent PE is drawn meeting one of its asymptotes in E: through E is drawn EQ, parallel to GP, and meeting the curve in Q: QE is produced to meet the other asymptote in F; through Q is drawn IQJ, parallel to EP, and cutting the asymptotes in I, J. To prove that QE. QF: QI. QJ:: PP: the parameter of the diameter PP'.* De la Hire: Sectiones Conicce, lib. IV. prop. 9. 5. CP is a semi-diameter of an hyperbola, CD its semiconjugate: through P is drawn, parallel to CD, a tangent PE, cutting one of the asymptotes in E: QQ' is a chord of the hyperbola, parallel to EP: from any point T in CP is drawn TO, parallel to CE, and cutting Q' Q produced if necessary, in 0: from Q is drawn QR, parallel to EC, cutting CP in R. To prove that O Q. 0 Q': -quadrilateral QR TO:: (PE)': triangle PEC. De la Hire: Sectiones Coniccea lib. IV. prop. 24. SECTION IX. Referred to Conjugate Diameters. Conjugate Hyperbola. 1. From any points Q, q, in the conjugate hyperbola, parallel straight lines QRR', qrr', are drawn to intersect the hyperbola in the points R, R', and r, r', respectively. * The name parameter was given by De la Hire to a third proportional to any diameter 2a' and its conjugate 2b'; the parameter being thus equal to 2b'2 --. Apollonius had called such a line the latus rectum of the diameter. a' CONJUGATE DIAMETERS. CONJUGATE HYPERBOLA. 277 To prove that QR. QR' = qr.qr'. Let the hyperbola be referred to CP, CD, produced indefinitely, as axes of coordinates, CP being a semi-diameter conjugate to BR', or rr', and CD being a semi-diameter conjugate to CP. Then the equation to the hyperbola is x2 y2 1 1........................(1), and to the conjugate hyperbola is y2 X2 bY1 a2................... (2). Let h, k, be the coordinates of Q: then, putting in (1), h for x, and k- y' for y, we shall have a'2 b"2 the two values of y' in this equation being QR, QR'. Hence QR. QR' = b'2 2 + L. But A, k, being a point in (2), k2 a2 b,2 a,2 = 1 hence QR. QR' = 2'2. Similarly qr.r' = 2b'2. De la Hire: Sectiones Conicce lib. IV. prop. 34. 2. To find the locus of the middle points of a system of parallel chords drawn between an hyperbola and the conjugate hyperbola. Let two conjugate diameters be taken as axes of coordinates, the axis of y being parallel to the chords. Let y1, y2 be the ordinates of the two ends of a chord of which x is the abscissa. Then, a, b, representing the mag 278 HYPERBOLA. nitudes of the conjugate semi-diameters, x2 /2 a- b2 = l - (1). _ = l........................(2). Let y be the ordinate of the middle point of the chord: then + = 2y........................(3). Adding (1) and (2), and paying attention to (3), we get b2 - Y1................. (4). Squaring (3) and (4), and then adding, we have y2 + y2 = 2y2+ 2...............(5). But, subtracting (1) from (2), we see that y12 + Y2- 2x( 6 = -... (6). From (5) and (6), there is 2y2 b2 2x'2 + a+ Ix y^_ 62 or as =y which is the equation to the required locus. 3. COP COP' are two semi-diameters of an hyperbola; CD, CD', are two semi-diameters of the conjugate hyperbola, respectively conjugate to CP, OCP: from P are drawn PE, PFE parallel respectively to P'C, CD', and cutting CD, or CD produced, in E, F. To prove that CE. C = (CD)2. De la Hire: Sectiones Conicce lib. Iv. prop. 38. ANY TWO DIAMETERS. CONJUGATE HYPERBOLA. 279 4. CP, CP', are two semi-diameters of an hyperbola; CD, CD', are two semi-diameters of the conjugate hyperbola, respectively conjugate to CP, CP': from P is drawn PT, parallel to DC, and cutting CP' in T; and from D is drawn DK, parallel to D'C, and cutting CP in K. To prove that triangle CPT = triangle CDK. De la Hire: Sectiones Conicce, lib. IV. prop. 39. SECTION X. Referred to any two Diameters. Conjugate Hyperbola. 1. The equation to an hyperbola being x2y 2 xy x~ y 2xy a& 32 mn to find the equation to the conjugate hyperbola? The equations to the two curves referred to their principal axes are 2 2 X 2 a2 "2 and 2-be= 1. If we change the directions of their axes of coordinates, by substituting Xx + s/y, X' + p'y,) for x, y, respectively, it is plain that the left-hand member of either of the equations will still remain the same as that of the other. Hence the equation to one being 2 y2 2xy 2 132 mn that to the other will be x' y 2mxy._ a 13' mn 280 HYPERBOLA. 2. CP, CP', are two semi-diameters not conjugate, the former of an hyperbola, the latter of its conjugate hyperbola. A tangent at P intersects CP', produced if necessary, in T, and a tangent at P' intersects CP, produced if necessary, in T'. To prove that the triangle PCT is equal to the triangle P' CT'. De la Hire: Sectiones Conicce, lib. Iv. prop. 36. SECTION XI. Referred to its Asymptotes. 1. A straight line touches an hyperbola in P and cuts its asymptotes in ], K; to prove that HK is the least possible when P coincides with the extremity of the transverse axis. The equation to the hyperbola, referred to its asymptotes as axes, is 2 X7y = C2. Let x, y, be the coordinates of the point P: the equation to the tangent HK will be xx + yy = 2c2. Hence, C being the centre of the hyperbola, 2c2 2c2 CH= - CK = x? y? and therefore, o denoting the angle between the asymptotes, (HK)2 = 14c ( + 1- 2cosco) 4c4 (x + ~y2- 2xy cosco) = 4c2 (x2 + y2 - 2C2 cosco). Let H', K', be the positions of H, K, when P is at the extremity A of the transverse axis: then, since CA bisects the angle between the asymptotes, the coordinates of A are c, c; REFERRED TO ITS ASYMPTOTES. 281 hence (H'K')2 = 4c2 (2c2 - 2c2 COS o), (HK)2 - (H'K')2 = 4c2 (x2 + y2 - 2c2) = -2 (4 + x2y2 - 2c22) 4dc = c (x2 _ c2)"2 This result shews that B'K' is less than HK; which establishes the proposition. De la Hire: Sectiones Conicce, lib. vII. prop. 33. 2. A straight line HPQK cuts the asymptotes Cx, Cy, of an hyperbola in the points H, K, and the curve in P, Q: to prove that PH = QK. The equation to the hyperbola is x c2........................... (1): let that to the straight line be = 1.......................... (2). At the intersections of (1) and (2), ac2 2 - ax + a- =0. Hence, (x1, y) being the coordinates of P, and (x2, y) those of Q, x + x, = a) (3). and Y1 + Y2 = Consequently, o being the angle between the asymptotes, (PH) = ( - x1) + y1 - 2 (a - x) y, cos o = (b - y2)' + 22 - 2 (b - Y2) x cos, by (3), = (QK), or PH = QK. 'Eav v7rep/3oXy evOela aov/imqrlT7r r ara 6vo o'Vleta' Eec/3akxXo&V evr/ ' eictlrepa avr7reoeetrat a-vvUT7rrTw7ToLs, Ica at a7roXa/,3av6o/uevat a7r avTqrm viro Tr 7ojTA7S Wrpos Ta'l avpuTrr7' roTso Lo'at aovrat. 'AHIOAA2NIOY HEPrAIOY KWVItKwv To' aEVTEpovU pO'rac's:j. 282 HYPERBOLA. 3. A tangent to the curve xy = c2 is intersected by a perpendicular upon it from the origin: prove that, if co be the angle between the coordinate axes, the equation to the locus of the intersection will be y (x cos o + y) + x (y cosco + a) = 2c (x cosco + y) (y cos c + x)}I. Let the equation to the tangent be y = ax + /: then the roots of the equation ax' + fx - c2 = 0 must be equal: hence 32 = - 4ac2, and the equation to the tangent will become y = ax + 2c(-ca)................... (1). Let the equation to the perpendicular upon it be y = a'x.......................... (2). Then, by the condition of perpendicularity, 1 + ca' + ( + a') cosco =0...............(3). From (2) and (3) there is x + oy + (ax + y) cosco = 0, x + y coso = - a (y + x cosco), and therefore, from (1), we get for the equation to the locus, y (y + x cosco) + x ( + y coso) = 2c{(( + y cos c). (y + cosco))}. 4. In every triangle inscribed in an equilateral hyperbola, the point of concourse of the perpendiculars, drawn from the angles to the opposite sides, lies in the curve. Let the coordinates of the three angular points of any inscribed triangle be (x, y), (x', y'), (x", y"). Then the equation to the side joining (x, y), (x', y'), will be ^Y -^ Y = t*- /(.i:, _ x t).................. (). y X - y, REFERRED TO ITS ASYMPTOTES. 283 But, by the equation to the hyperbola, supposing the asymptotes to be the axes of coordinates, xy = c2 xy' = c2 and therefore = -, 2 - X XXr The equation (1) therefore becomes C2 &- y = - -; (x1 - x). xx The equation to the perpendicular to this line through (x", y") will be I Y, - cy 2 (1 - x ).................. (2). Similarly, the equation to the perpendicular through the angular point (x, y) will be X'X" xx, Y, - c2 (X1- ).................(3). At the intersection of (2) and (3), we have, multiplying (2) by x", (3) by x, subtracting, and observing that x"y", xy, are both equal to c2, xx'x" Y1 --- Similarly, we must have x- yyy c- ~ By symmetry it is plain that these are the coordinates of a point through which the three perpendiculars pass. Moreover xy x y x"y" x1Y, c2 * C 2 - Cao hence (x, yl) are the coordinates of a point in the curve. Brianchon, Poncelet: Gergonne, Annales de Mathematiques, tom. XI. p. 205. 5. To prove that the straight lines drawn from any point in an equilateral hyperbola to the extremities of any diameter, are inclined at equal angles to the asymptotes. 284 HYPERBOLA. The equation to the hyperbola, referred to its asymptotes Cx, Cy, (which constitute a system of rectangular axes) will be xy = c2. /YP EC Let PCp be any diameter; (x, y) being the coordinates of P, (- x, - y) ofp, and (x', y') of any point P' in the curve. Let LP'Ex = 0, and P'Fx = qb. Then 1 1 y - 2Y X X c2 tano =, -c. X - X X -X XX Putting - x for x, and 0 for B, we have also 2 tan =,. XXFrom these results it appears that From these results it appears that 0 = 7T - 0 and therefore LP'EF = P'FE. COR. It may hence be easily inferred that, if from another point P", lines be drawn to P andp, LP'PP" = LP'pP". Newton: Arithmetica Universalis, prob. 35. Gergonne: Annales de Mathematiques, tom. II. pp. 32, 126. 6. If any hexagon be circumscribed about an hyperbola, the three diagonals, joining opposite angles, will all pass through one point. The equation to the hyperbola being 4xy = c( REFERRED TO ITS ASYMPTOTES. 285 let the equation to a tangent be a y then the roots of the equation '2 C2 --— x+ =-0 a1 41 which belong to the common point of the tangent and curve, must be equal: hence a1/ = '2. Thus the equation to the tangent is x aly -+ 4 — 1. a, c At the intersection of this tangent and another, represented by the equation x a2y -a+ C= we shall have 1 a2 These will be the coordinates of one of the angles of the circumscribed hexagon, the two tangents being two of the sides. We shall have analogous expressions for the coordinates of the other angles. The equation to the diagonal through the angles (xi,, Y1,2), (x4,5 4, 5)x Will be (4,5 -,2) - y (X4,5 - X1,2) = X1,2Y4,5 - x4,5Y2; or, if we substitute for the coordinates of the angular points, their values, c2x (a4 - a1) - (a2- a5)} + y {a4a1(a - a2) + asa2 (a4- a,)} = c2(a4a5 - aa2)......(). Similarly, the equation to the diagonal, through the angular points (x2,3, Y2, ), (x5,6, 6Y,6), will be c'x {(a, - (- )a2 - (a a + y {aa (a, - a) + a6a3 (a5 - a)} = c2(a16 - a21)......(2). 286 HYPERBOLA. At the intersection of these two diagonals, multiplying the equation (1) by ao - a%, and the equation (2) by a4 - a,, adding the latter of the resulting equations to the former, and dividing the final equation by a - as, we shall get cx {(a6 - a.) - (a4 - a,)} + y {at6a(a, - a4) + aa4 (a6 - as) = c"(a6l - a3a4).......(3). But, as is evident from symmetry, the equation (3) belongs to the third diagonal, namely, that which passes through the points (X3,4, Y3,4), (x6,l) Y6,)' Thus we see that the two diagonals (1) and (2) intersect in the third; which establishes the theorem. COR. Subtracting the sum of (1) and (3) from (2), we get for the value of y, at the point through which the three diagonals pass, Y {a12 (a4 + a5) - 2a3 (a. + a)+ ~ a3a4(a6 + a,) - a4a, (a1 + a2) + a5a6(a2 + a) - a6a (a3 + a4)) = c' (aa, - 2a3 + 3a4 - a4a a5a6 - a6a). The values of x will, as symmetry points out, be obtained by writing, in this equation, x for y, and 8 for a, throughout. Cambridcge Mathematical Journal, vol. Iv. p. 164. 7. To determine the hyperbola which has two given lines as asymptotes and which passes through a given point. Let h, c, be the coordinates of the given point, referred to the two given lines as axes of coordinates. The equation to the hyperbola is xy = h1k. ALVo;o0eit&'v evOetLv ryoviav 7repteXovacwv, Icai riaLeiov evro T9 rywviar rypafra4 &a to ft7/elov KCOVOV cTOV Trjv KcaXovt/LEV7LV V7Trep/0oX7V, Ca7re ao'V7rr7rTorovs etvat Tra SoOelo'a evOelas. 'AnOAAQNIOY IIEPrA10T KWVLKWu TO 8 vT-spov' po'rTacLv a'. Pappus: Mathematicce Collectiones a Comnmandino, lib. VII. prob. 15, prop. CCIV., p. 277. REFERRED TO ITS ASYMPTOTES. 287 8. To find the locus of the intersection of a pair of tangents to a rectangular hyperbola xy = c2, which are such that the product of the tangents of their inclinations to the axis of x is constant. If k denote the constant product, the required locus will be a straight line the equation of which is y = kx. 9. A straight line moves in the plane of a given angle, so as to form with the sides of the angle a triangle of given area: to find the equation to the locus of the centre of gravity of the triangle. The two sides of the angle being taken as the axes of coordinates, the equation to the curve will be 2k2 xy 9 sina where k denotes the given area, and a the given angle. Francoeur: Gerqonne, Annales de Mathematiques, tom. x. p. 79. Gergonne: Annales de Math4matiques, tom. x. p. 87. 10. If a concentric circle cut a rectangular hyperbola, referred to its asymptotes as axes, and PT, PT', be tangents to the circle and hyperbola respectively, meeting the axis of x in T, T', PM being the ordinate at the point of intersection: to prove that MT.MT' = MPP2. 11. To find the general form of the polar equation to the tangent of an equilateral hyperbola, an asymptote being the prime radius vector. The equation to the hyperbola being xy = c2, the polar equation to the tangent will be r a cos0 + sin = 2c where a is an arbitrary constant. 12. To prove that the secants, drawn from any one of the points of an hyperbola to two fixed points in the curve, always 288 HYPERBOLA. intercept, on either asymptote, a constant length equal to the length intercepted on the same asymptote between the secant through the two fixed points and the tangent at either of them. Sturm, Vecten, Querret: Gerqonne, Annales de Mathernatiques, tom. xv. p. 100. Quetelet: Corr. Math. et Phys., tom. II. p. 323. 13. To prove that every chord of an hyperbola divides into two equal parts the portion of either asymptote which is included between the tangents at its extremities. Sturm, Vecten, Querret: Gergonne, Annales de Mathematiques, tom. xv. p. 102. 14. If any two tangents be drawn to an hyperbola, and lines be drawn joining the points in which they intersect the asymptotes, to prove that these lines will be parallel to one another. 15. To find the equation to the conjugate diameter of any system of parallel chords in an hyperbola. If the equation to any one of the chords be x y — +- =1, m n the equation to their conjugate diameter will be x Y mn n 16. From any fixed point P in an hyperbola, are drawn PI, PK. parallel to the asymptotes; and, from another fixed point Q in the hyperbola, is drawn any straight line cutting PH, PK, in M, N, respectively, and the curve in R: to prove that RiMo RN. 17. If a pair of conjugate diameters of an ellipse be, when produced, asymptotes to an hyperbola, to prove that the points of the hyperbola at which tangents to the hyperbola will also be REFERRED TO ITS ASYMPTOTES. 289 tangents to the ellipse, lie in an ellipse concentric and similar in form to the given one. 18. A tangent at a point E of an hyperbola intersects one of the asymptotes in F: an indefinite line EN is drawn through E parallel to this asymptote: from F any line FGHI is drawn, cutting the curve in G, Ij and the line EN in H: to prove that the line FI is harmonically divided in the points G, H. De la Hire: Sectiones Conicce lib. II. prop. 22. 19. Cx, Cy, are the asymptotes of an hyperbola: HK is any chord: CP is a semi-diameter, conjugate to HK: from K, P, H, are drawn, parallel to yC, the three lines KN, PI, lHM intersecting Cx in N, 1l M. To prove that these three lines are continued proportionals. De la Hire: Sectiones Conicce, lib. IV. prop. 15. 20. If the abscissae of any number of points in an hyperbola, referred to its asymptotes, are in arithmetical, to shew that the ordinates are in harmonical progression. De la Hire: Sectiones Conicce, lib. IV. prop. 16. 21. From any two points P, P' in an hyperbola are drawn two straight lines, PM, P'M', parallel to the asymptote yC, and meeting the asymptote Cx in M, M': from the same two points are drawn also PN, P'N', parallel to xC7 and cutting Cy in N, N'. To prove that the mixtilineal area PNN'P' is equal to the mixtilineal area PMM'P'. De la Hire: Sectiones Conicce, lib. Iv. prop. 17. 22. A tangent is drawn at a point P of an hyperbola, cutting the asymptote Cy in E: from E is drawn any straight line EKH cutting one branch of the hyperbola in the points K, H: Kk, PM, Rh, are drawn, parallel to yC, to cut the other asymptote Cx in the points k, MI h. To prove that Hh + Kk = 2Pl. De la Hire: Sectiones Conicce, lib. IV. prop. 19. I 290 HYPERBOLA. 23. At a point P of an hyperbola is drawn a tangent P-l, cutting one of the asymptotes in M. From a point Q in the curve is drawn a line QNV parallel to PM1 cutting the same asymptote in N: from Q is drawn also QE, parallel to NC, cutting the semi-diameter CP in the point E. To prove that the triangle PCM1 is equal to the quadrilateral CEQN. De la Hire: Sectiones Conicce, lib. IV. prop. 20. 24. CP is any semi-diameter of an hyperbola, CD is its semiconjugate. At P is drawn, parallel to D C, a tangent PE, cutting one of the asymptotes in E: in CE, produced, is taken any point H, and from H is drawn, parallel to CD, a straight line HQ, Q being its first intersection with the curve: from Q is drawn the straight line QI, parallel to PC and meeting CE, produced if necessary, in I; from Q is drawn also QF parallel to EC, and meeting CD in F. To prove that the quadrilateral CIQF is equal to the triangle CEP. De la Hire: Sectiones Conicce, lib. IV. prop. 21. 25. If any right-angled triangle be inscribed in an equilateral hyperbola, the perpendicular let fall from the summit of the right angle upon the hypotenuse is a tangent to the curve. Brianchon, Poncelet: Gergonne, Annales de Mathematiques, tom. XI. p. 206. 26. If, on a chord of an hyperbola, considered as a diagonal, be constructed a parallelogram the sides of which are respectively parallel to the asymptotes of the curve, to prove that the other diagonal of this parallelogram, produced if necessary, will pass through the centre of the curve. Sturm, Vecten, Querret: Gergonne, Annales de Mathe'matzques, tom. xv. p. 102. 27. If, on the three sides of a triangle, taken in turn as diagonals, be constructed parallelograms, the sides of which are REFERRED TO ITS ASYMPTOTES. 291 respectively parallel to two given straight lines, to prove that the three other diagonals of these parallelograms will pass through the centre of an hyperbola which, being circumscribed about the triangle, has its asymptotes parallel to the two given straight lines. Sturm, Vecten, Querret Geqrqonne, Annales de Mathelmatiques, tom. xv. p. 103. 28. An hyperbola, denoted by the equation xy = c2 is intersected by a parabola, the equation of which is y = ax + f/: to prove that, y1, y2, being the lengths of the ordinates of the intersections of one branch, and y- the length of the ordinate of the intersection of the other branch, Y3 = y1 + Y2' De la Hire: Sectiones Conicae lib. v. prop. 32. 29. From a point P, in one of the branches of a rectangular hyperbola, xy = c2, a perpendicular PM is drawn to one of the asymptotes: PM is bisected in Q, and through Q is drawn, perpendicular to the other asymptote, an indefinite straight line: in this indefinite line is taken any point O; OP is joined. A circle is described, of which 0 is the centre and OP the radius. To prove that, if the circle cut the locus of xy = c2 in four points, the same branch will be intersected in another point P,, and the other branch in two points P2, P5, and that, y, Y2, y,, denoting the magnitudes of the ordinates of P,, P2, P, Y2 + Y3 = Yl. De la Hire: Sectiones Conice, lib. v. prop. 33. 30. From any point P in one branch of any hyperbola, is drawn PM, parallel to the asymptote yCy', and cutting the asymptote xCx' in M. PMis bisected in Q, and, in an indefinite straight line through Q, parallel to Cx, is taken some point O. An ellipse is described, passing through P and M, of which 0 is u2 292 HYPERBOLA. the centre, and OA', OB', two conjugate diameters parallel to Cx, Cy, respectively, such that OA': OB':: PM: PE, PE being at right angles to Cx. ^t / 3.^-T Ei/ To prove that, if the ellipse intersects the two branches in P, Pt, PF8 then, PM, P1M, PM,, being parallel to Cy, PIMI = Pi + PADe la Hire: Sectiones Conicce, lib. v. prop. 34. SECTION XII. Referred to any Rectangular Axes. 1. To prove that, in the equation to a rectangular hyperbola, referred to any rectangular axes whatever, the coefficients of x2 and y' are equal with opposite signs. The equation to a rectangular hyperbola referred to its principal axes is x' - y2 = a2. Turning the axes through an angle 0, we shall change this equation into (x cos0 - y sin 0) - (x sin 0 + y cos0)2 = a2, or (x2 - y2) cos20 - 2xy sin20 = a2. If we transfer the origin to any new place, the coefficients of x2 and y2 will not be affected. The proposition is therefore established. REFERRED TO ANY RECTANGULAR AXES. 293 2. If two points be taken in each of two rectangular axes so as to satisfy the condition that a rectangular hyperbola may pass through all the four, shew that the position of the hyperbola is indeterminate, and that its centre describes a circle which passes through the origin and bisects all the lines which join the points two and two. Let A, A', be two points taken in the axis of x, and B, B', two points in that of y. Let OA=a, OA'=a', OB= 3, OB'=/'. The equation to any conic section passing through these four points will be - bxy + y- - (a+)-i ( + ') = o (1) The hyperbola being rectangular, we must have aa' + / ' = 0....................... (2). Since b is independent of a, a', /, /', the position of the hyperbola is indeterminate. Suppose that the origin is changed to a point X, Y, the directions of the axes not being altered. Then (1) will become (X'~X)2 (a(+X) -b (+) (y'+ Y) + (y'+Y)2 -" wy+X Y___ (+ Y) x '+ (a + al)- ( +,') + 1 = 0. /9/9' aa' A1S' If X, Y, be the coordinates of the centre of the hyperbola, 2X + 2Y - + -' aC bY a =0, bX- 1 0, and therefore, eliminating b, and paying attention to equation (2), 2 (X2+ Y2) - (a + a') X- (/3 + /') Y= o, or(-4X f ( +)')~ / + ')=(+') — (~')^H..,.), or 4 + ( + 4 the equation to a circle passing through 0. The coordinates of the middle points of the lines AB, A'B, AB', A'B', are (2a,,/3), (ca', t/3), (la, 4i/), (1a', ') ): these coordinates, by virtue of (2), satisfy the equation (3). This shews that these lines are all bisected by the circle. 294 HYPERBOLA. 3. If uv = c' be the equation to an hyperbola, where ut = x cosa + y sina - A, v = x cos/ + y sin3 - k, and c is constant, to prove that uv =-, 0 2 = 0 _ _2 = m 2V, are the respective equations to the asymptotes, axes, and a pair of conjugate diameters, m being an arbitrary constant. SECTION XIII. Referred to any Rectangular Axes. Reduction. The equation to an hyperbola referred to any axes whatever is ax' + by2 + 2cxy + 2a'x + 2b'y + c' = 0, a, b, c, being subject to the condition c2 > ab. The process to be adopted for the purpose of finding the centre as well as the magnitudes and positions of the axes is the same as in the reduction of the general equation to the ellipse. 1. To determine the positions of the axes of the curve xy - x2 tana = ^b2 cota, and the ratio between them. Putting, in the equation, x = xcosO -y' sin, y = x sin + y' cosO, we have (x' cosO-y' sine) (x'sinO+y'cos) - (x'cosO-y'sinO)2. tan a = b2cot a. Equating to zero the coefficient of xy', we get cos20 + tana sin20 = 0, tan20 = - cota = tan(+~ 7r + a), 20 = ~+ r + a, which shews that the axes are inclined at angles {(rr + 2a), L (7-r + 2a), to the axis of x. REFERRED TO ANY RECTANGULAR AXES. REDUCTION. 295 Taking the former axis as that of x', the equation becomes (x' - y2) sin20 - tana {x' (1 + cos20) + y'2(1 - cos29)} = b2 cott, x'(cos - tan a sinac tana) -y'"(cosa+tana+sin tana) = -b2 cota, cos2 a x'2 (1- sina) - y"2( + sina) = b2 cOs2 a sin ' 2 2 X- _ Y = _-t2 cosec, + sina 1 -sina 7 --. ---.-i-. -- -,- -- i — -.-i-^ = ~rb' coseca. ' 12 12 b2 - iee a. (cos - a + sin ia)2 (cos I - sin l )2 =62 The ratio between the axes is therefore equal to 1 - tan -a r - 2a 2 = tan 1 + tan a 4 2. To prove that the equation 4xy - 3x2 = a2 will be changed into the equation X2 - 4y2 = a2 if the axes of coordinates be turned through an angle the tangent of which is 2. 3. The equation to an hyperbola being a2 a~ t t t t to find its equation when it is referred to its axes. The required equation is 2 ( 1 1) 2 1 1 = 2 4. The equation to an hyperbola being 4. The equation to an hyperbola being x - 2 + 2xy-x +y- 1=0 to find its equation when referred to its axes. The required equation is x - Y - 3 Garnier: Weomdtrie Analytique, p. 131. 296 HYPERBOLA. SECTION XIV. Polar Equation. Centre the Pole. 1. To prove that of all diameters of an hyperbola the transverse axis is the least. The polar equation to the hyperbola, the centre being the pole and the transverse semi-axis being the prime radius vector, is 1 cos 0 sin' 0 r-2 a' b19 whence sin ( The expression for 2 - 2 being essentially positive, it is evident 1 1 that 2 is less than -, or a2 less than r'7 or a than r, a conclusion which establishes the proposition. De la Hire: Sectiones Conicce, lib. II. prop. 36. 2. An hyperbola, of which C is the centre, is cut in two points P, P', by a given straight line: to find the tangent of the sum of the inclinations of CP, CP', to the transverse axis. The equations to the hyperbola and to the straight line being 2 /cos 0 sin2 0 and 8 = rcos(0- X), we have, at their intersections, 82 (os20 sinb2 ) = o2(0 ), and therefore (sin'X + 82 tan20 + sin2X.tan0 + cosX - - = 0 and consequently, 0,, 0,, being the inclinations of CP, CP' to the transverse axis, tan(0 + 0,,)= sin2.. cos 2X - + (1 POLAR EQUATION. FOCUS THE POLE. 297 COR. Suppose that -= c, e denoting the eccentricity of e the hyperbola: then we shall have 1 1 1 ai2 +2 = c2 X sin 2X and therefore tan(0, + 0) = ----- 62 cos 2X- Ca This result shews that the angle 0, + 0, remains invariable for all hyperbolas with axes coincident in direction, provided that their minor directors* are coincident. Booth: Application of a New Analytic Method to the Theory of Curves and Curved Surfaces, p. 7. 3. To prove that diameters of an hyperbola which are nearer the transverse axis are smaller than those which are more remote. De la Hire: Sectiones Conicce, lib. vii. prop. 1. 4. The distances r, s, of two points in an hyperbola and its conjugate respectively from the common centre are at right angles to each other: to prove that, a, b, being the semi-axes of the hyperbola, 1 1 1 1 r2 S2 - b' SECTION XV. Polar Equation. Focus the Pole. 1. If a straight line be drawn from the focus of an hyperbola, to prove that the part intercepted between the curve and the asymptote is equal to a sin a sina + sind 0 * A line drawn parallel to the transverse axis at a distance from the centre has been called the minor director, 298 HYPERBOLA. where 0 and a are the angles made by the straight line and asymptote respectively with the axis of the curve. Let p, P, be the points in which the p straight line, through the focus S, cuts / the asymptote and curve respectively. Then tan2 a = — =-1, ----- a e2= sec2a, e =sec. sin a Sp = ae sin( - a) SP= a (e2 - 1) 1 - e cos O Hence - ae sina a (e2 - 1) Hence Pp=. sin(0 - a) 1-ecosO, ( 1 sina ) = a tana (sin - a) a_ = {tsins ( - a) cosa - cost (a tang 1 sin a (cos a + cos) ~ sin i(O -c a) sin (0 + a) sin(0 - a) a tan a sin 0 cos a - sin a cosa sin(0 - a) ' sin(0 + a) sin-0 - sina = 2asa sin cos2a - cos20 sin - sina a sina = a sina sin 0 - sina + sin 0 2. To prove that r cos = a cos 2 is a focal polar equation 2 2 to an hyperbola, the eccentricity of which is 2 and the axes 2a 2a 3 \/3 POLAR EQUATION. POINT IN THE AXIS THE POLE. 299 SECTION XVI. Polar Equation. Point in the Axis the Pole. 1. In the semi-axis major CA of an hyperbola, produced indefinitely, a point 0 is taken at a distance from A greater than half the latus-rectum. P is a point in the curve, OP is joined. To determine the least value of OP. Let OP = r, zPOA = 0, CO = c. Then the polar equation to the curve will be r' sin2 0 = (e - 1).{(c - r cos 0)2 - a2}, or e'r Cos2 -- 2 (e' - 1) cr cos = r2 - (e2 - 1) (c - a2). This equation may be put under the form (e 0 e 1 c) = r2 e2-. (c2 - e'2a). (1 e e Now, since AO is greater than a (e - 1), c is greater than e2a, and therefore a fortiori than ea: hence, e2 -- 1 H e (c2 - e2a2) = an essentially positive quantity. Hence it follows that e-1 e (62 - eLa2) is the least value of r2. From (1) we have, for the determination of the corresponding value of 0, C e' which = eg geer e which, c being greater than e2a, gives possible values for 0. Apollonius: Conicorum, lib. v. prop. 9. 2. A point 0 is taken in the semi-axis major (produced) of an hyperbola, at a distance from its vertex A equal to half the latus-rectum. A point P is taken in the curve, PO is joined, and PML is drawn at right angles to AO. To prove that OA is the least value of OP, and that, e denoting the eccentricity, (OP)2 - (OA)2 = e2 (AM)2. Apollonius: Conicorunn, lib. v. prop. 5. 300 HYPERBOLA. SECTION XVII. Polar Equation. Pole Anywhere. 1. Through any given point 0 to draw a straight line cutting an hyperbola in two points Q, q, such that the rectangle contained by OQ, Oq, shall be equal to a given square. The equation to the hyperbola, referred to its principal axes, is x2 y2, Let (h, k), be the coordinates of 0. Then the polar equation to the hyperbola, 0 being the pole, and the prime radius vector being parallel to the axis of x, will be (rcosO + h)2 (rsin + k)2 a2 b2 Let r,, r2, be the two values of r in this quadratic: then h2 c2 - 1 a b2 rlr2 C= os2 sin Let d2 denote the area of the given square: then rl.r2 = + c2, the - or + sign being taken accordingly as 0 is supposed to lie between Q, q, or not: hence cos2O sinO _ /A2 2 \ a,2 Z2 -+ c2 a2 - 1) a a — ~ =_ ~ ~~? " b I. From this equation may be found generally two values of 0, and therefore the problem will generally admit of two solutions on each hypothesis respecting the position of 0. The possibility of the problem depends upon the possibility of the value of 0. De la Hire: Sectiones Conicce, lib. v. prop. 43. 2. Through a given point 0 is drawn a straight line cutting in the points P, Q, the opposite branches of an hyperbola: to find the inclination of PQ to the transverse axis in order that the rectangle between OP, OQ, may be the greatest possible. PQ must be parallel to the transverse axis. De la Hire: Sectiones Conicce, lib. VII. prop. 32. POLES AND POLARS. 301 3. Through a given point 0 is drawn a straight line cutting in the points P, Q, one branch of an hyperbola: to prove that the rectangle between OP, 0 Q, will be the least possible when PQ is parallel to the conjugate axis. De la Hire: Sectiones Conicce, lib. VII. prop. 34. SECTION XVIII. Poles and Polars. 1. With the asymptotes of an hyperbola, as conjugate diameters, ellipses are described, touching the hyperbola: to prove that, if a common polar of any two of the series of ellipses has its pole, relatively to one of the two ellipses, in the hyperbola, its pole, relatively to the other, will also be in the hyperbola. Let the equations to any pair of the ellipses be 2 + yb2 1........(1 a2 b.2) + - 2_ = 1 (2) + b 12' " '.................... (2). Let the pole, relatively to (1), be (h, k): then the polar will be hx ky a2 + -=....................... (3). Let (h', k') be the pole, relatively to (2), of the polar (3). Then the equation to the polar must be h'x k'y + = 1......................(4). Since (3) and (4) represent the same line, h h' k k' a2 ~ a'2) b bt2 } hk h'k' and therefore ab2 = a'b (5). ~~~~and therefore d 2 = ~1......................... (5). Since (1) and (2) touch the hyperbola, the equation of which is of the form xy = c2 302 HYPERBOLA. it may easily be shewn that a2b' = 4c2, a2b' = 4c2. Hence, by (5), hk = h'k'. But hk = c2, because (h, k) is a point in the curve: consequently h'k' c'2 and (h', k') is also a point in the curve. 2. If P be the pole, relatively to any polar of an equilateral hyperbola, of which C is the centre, to prove that q, q', being the inclinations of CP, and of the polar, to the real axis tan.tan ' = 1. Let the equation to the hyperbola be x2 _- y' = a' Then, (I, k) being the pole, the equation to the polar will be hx - y = a........................ (1). Also the equation to CP will be x..... h............................(2). From (1) and (2) it appears that k h tan = 7, tanb' = k; and therefore tan b.tanb>' = 1. 3. If a circle pass through the centre of an equilateral hyperbola and through any two points, 4, B, to prove that it will also pass through the intersection of two lines, one drawn through A, parallel to the polar of B, and the other through B, parallel to the polar of A. The equation to the hyperbola, referred to its asymptotes as axes, will be of the form xy = C2. 'Let h, k, be the coordinates of A, and h', k', those of B. Then the equation to the polar of A will be kx + hy = 2c2....................... (1), and that to the polar of B will be k'x + h'y = 2c2........................(2). POLES AND POLARS. 303 The equation to the line through A, parallel to the line (2), will be k' ( - A) + h'(y - ) = 0...................(3); and that to the line through B, parallel to (1), will be k(x- ') + h (y- ')= 0.................(4). The equation to any circle passing through the centre of the hyperbola will be x (x + a) + y (y + /) = 0.................(5). If this circle passes through the two points (A, k), (h', k'), we shall have h (h + a) + k(k + ) = 0.................(6), and h' (' + a) + k' ' + /) =...............(7). From (4) and (6), we have (- h') (h + c) -(y- ') (k + ) = 0, and, from (3) and (7), ( - A) (h' + a) - (y - ) (k' + /3) = o. From the last two equations, we get, by subtraction, (x + a) (A- ') - (y +) (k - ') = 0.........(8). But, from (3) and (4), by subtraction, (k- k') x + (h- ') y =............... (9). From (8) and (9), there is x (x + a) + y (y + /) = o, an equation identical with (5). Thus the intersection of (3) and (4) mhst lie in (5). Brianchon, Poncelet: Gergonne, Annales de Ml/athemnatiques, tom. XI. p. 208. 4. When three points, situated in the plane of an equilateral hyperbola, are such that each of them is the pole of the line joining the other two, the circle which passes through these three points will pass also through the centre of the hyperbola. Let the three points be (A, k), (h', k'), (h", k"). The equation to the polar of the first point will be kx + hy = 2c2. Since this line contains the two other points, we have kh' + hk'= 2c, k" + h" = 2c2, and therefore k (h" - h') + (k" - k') = 0............... (1) 304 HYPERBOLA. Similarly k' (h - h") + h' (k - k") = 0.............(2), and k" (h'- A) + h" (k' - ) = 0...............(3). Let the equation to the circle be x2 + y2 + ax + 8y + 7y = 0. Then, a, /, ry, will be subject to the relations h2 k + ah + /k + =0.............(4), h'2 + k'2 + ah' + /3' + y = 0.............(5), h"2r + k'2 + ah" + E/k" + y = 0.............(6). Subtracting (5) from (6), we get, by the aid of (1), - h (h" + h') + k(k" + ') - ka + k = 0......(7). Similarly, from (6), (4), and (2), there is - h' (h + t")+ k' (k + k") - a' + k' = 0......(8). Subtracting (8) from (7), we have h" (h' - a) - k" (k' - k) + a (h' - h) - / (k' - k) = 0, and therefore, by (3), h"2 + k"2 + ah" + /k" = 0............... (9). From (6) and (9) we see that y = 0. Hence the circle passes through the centre of the hyperbola. Brianchon, Poncelet: Gerqconne, Annales de Mathematiques, tom. XI. p. 210. 5. To prove that the diameter terminating at the point of contact of a tangent to an equilateral hyperbola, bisects two of the four angles, formed by the two diameters which pass through the points of intersection of this tangent with the polars of the two foci. Bobillier: Gergonne, Annales de Mathematiques, tom. XIX. p. 351. 6. To prove that the diameters terminating at the two extremities of a chord of an equilateral hyperbola, form respectively equal angles with those which pass through the intersections of this chord with the polars of the two foci. Bobillier: Gergonne, Annales de Mathematiques, tom. XIX. p. 351. HYPERBOLIC LOCI. 305 SECTION XIX. Hyperbolic Loci. 1. Each of a series of parabolas is described so as to pass through two given points, and have its axis parallel to a given fixed line: to find the locus of their foci. Let r, r', be the distances of the two given points from the focus or directrix of any one of the parabolas, and h, h', their distances from a fixed line drawn at right angles to their axes. Then, the directrix of each parabola, under the condition of the problem, being parallel to this fixed line, it is plain that r'- r = h' - h = c where c denotes some constant quantity. The locus of the focus is therefore an hyperbola of which the two fixed points are the foci. 2. From a given point (a, b) is drawn a normal to the hyperbola xy = c, of which the asymptotes are perpendicular to each other. To find the locus of the intersection of the normal with the curve, when the parameter c is regarded as variable. The equation to a normal at a point (x, y) of the curve is (x' - x) = y (y' - ). Let this normal contain the point (a, b): then x(a- x) = y(b - y), the equation to the required locus, which is therefore another rectangular hyperbola. CoR. If the origin be removed to a point (-a, lb), the equation becomes x_- = (a2 - ). L' Hospital: Traite Analytique des Sections Coniques, p. 263. 3. If tangents are drawn making given angles with the axes of all ellipses having the same given foci: to find the curve which is the locus of the points of contact. x 306 0HYPERBOLA. Let the equation to one of the ellipses be X'2 t2 1........................ ( The tangent at any point (x, y) will be a' b2 and, 8 being the distance of the origin from this tangent and a the inclination to the axis of x, we have x cosa y sina da 8 ' by 8 ~ x sina ycosa whence (2). ~~~whence~~ = -..................... (2). But, c being a constant quantity, - = c........................ (3). From (1) and (3), 2.(w2 - _2) 2+.(d -;') = c, and therefore, from (2), (x cosa + y sina) (x sina - y cosa) = csina cosa, or (x2 - y') sin2a - 2xy cos2a = c2 sin2a, which is the equation to an hyperbola concentric with the series of ellipses. 4. PQ is a chord of an ellipse at right angles to the axis major AB. The two lines AP, BQ, Q3~ produced indefinitely, intersect in R. To find the locus of R. Q - R The axes of the ellipse being taken as axes of coordinates, the equation to PR will be, y, b g te c as of P. x, y', being the coordinates of P. HYPERBOLIC LOCI. 307 The equation to BR will be y =-( + a). -. a +- x'' At the intersection of these two lines, we shall have y12 y= (x a2). -, X'2 or, by virtue of the equation to the ellipse, y2= b (X2 - a2). The required locus is therefore an hyperbola the axes of which coincide with those of the ellipse. 5. In two hyperbolas, concentric and similarly situated, are taken any two points the abscissae of which are as the real axes of the curves: to find the locus of the middle point of the line joining the two points. Let (x1, y,) be the point in one hyperbola, and (xI, y2) that in the other. Then, (a, b) and (a, /) being the semi-axes in the two curves, 2y 2 2( a2 132 a a From (1), (2), (3), we see that Y. = y2...... (4). b /-......................... Let x, y, be the coordinates of the middle point between (X1, Y1) and (x2, y2): then 2x = x1 + Xa, 2y = y1 + Y2? and therefore, by (3) and (4), x _ 2x a a+a' Y,_ 2y b b+ ' ~ x2 308 HYPERBOLA. Substituting these expressions for x_ and Y- in (1), we get a b xS 2 (a + a)2 - (b + 3)2 the equation to the required locus, being therefore an hyperbola concentric with the original ones and similarly situated. 6. To find the locus of T, the intersection of the tangents at two points P, Q, of a parabola, such that, S being the focus, SP, SQ, include a given angle s. Let A be the vertex of the parabola, and AS = m. Then, if LASP = a, the equation to PT will be 2m -= cosO + cos(O - a).................(1). Similarly, if LASQ = 3, the equation to QT will be 2m — = cos + cos(O- /3)................(2) At the intersection of (1) and (2), cos(O - a) = cos(O -,), 0- a =-(0 - /), + a = 2........................... (3). But, by the hypothesis, 8 - a =..........................(4). From (3) and (4), 0 - a = a2, and therefore, from (1), we get for the equation to the locus of T, 2m - = cos 0 + cos- s, r 2m sec I s r 2 1 sec.cos 0 the equation to an hyperbola of which sec ~ is the eccentricity. 7. To find the locus of the vertex of a plane triangle, having given the radius of the inscribed circle, and the difference between the angles at the base; the middle point of the base being a fixed point, and the base itself lying along a fixed line. HYPERBOLIC LOCI. 309 Let 0, the middle point of the base AB, be taken as the origin of coordinates, OA, produced indefinitely, as the axis of x, and Oy, at right angles c to Ox, as the axis of y. Let E be the centre of the inscribed circle. Draw CM, EH, perpen- B o 1- -- - -A dicularly to AB. Let L CAB = 2a, L CBA = 2 AB= 2c, EH= r, OM= x, CM= y, a - = s. Then it is clear from the geometry that y = (c - x) tan2a......................(1), y = (c + x) tan2/3...................... (2), 2c = r (cota + cot/)...................(3). Eliminating c and x between (1), (2), (3), we have y (tan2a + tan 2a) = r (cota + cot,/) tan2a.tan2/3, y sin(2a + 23,) = r (cota + cot/3).sin2a.sin2/3 = 4r.sin(a + /3).cosa.cos/3, ycos(a + /3) = 2rcosa cos/, = r {cos(a +) + cos(a - /)}, (y -r) cos( + ) = r cos(a -,) = rcosS.................. (4). Again, eliminating c between (1) and (2), we get y (tan2a - tan2/3) = 2x tan2a.tan2/3, y sin(2a - 2/3) 2x sin2a sin2,8 = x {cos (2a - 2) - cos(2a + 2/3)}, y sin2e = x {cos2s - 2 cos2 (a + /) + 1} = 2x {cos2 - cos2(a + /3)}. Hence, by (4), y sin2s = 2x {cos2 ( - r cos2 ytane.(y - r)2 = x {(y - r)2 _ r2} = xy - 2r), (y )2.tan = x (y - 2r), 310 HYPERBOLA. the equation to the required locus, which is therefore an hyperbola. Leybourn: Mathematical Repository, New Series, vol. II. p. 218. 8. From a given point E in the semi-axis major A C of an ellipse is drawn a straight line cutting the curve in any point P: PM is drawn to cut CA at right angles in M: MP is produced to a point P' such that MP'= EP. To find the locus of P'. Let CE = c: then, the equation to the ellipse being y2= (1 - e) (a2 _- x) the locus of P' will be an hyperbola represented by the equation e2y2 = (ex - c)2 + (1 -) (e2a2 - 2). P. Gregorius a Sancto Vincentio: Opus Geometricum Quadraturce Circuli et Sectionum Coni, p. 659. 9. Two straight lines, of given lengths, coincide with and move along two fixed axes, in such a manner that a circle may always be drawn through their extremities: to find the locus of the centre of the circle. Let a, b, be the lengths of the moveable lines on the axes of x, y, respectively, and let c be the angle between the axes. Then the required locus will be an hyperbola denoted by the equation a - x _- y=, 2 10. Two given straight lines OA, OB, intersecting in 0, are touched by a parabola in points H, K, respectively: the parallelogram OHPK is completed. Supposing the parabola to touch also a third given line, to find the locus of P. If the lines OA, OB, be taken as axes of coordinates, and the equation to the third given line be -+ =1, the equation to the required locus will be a - x y HYPERBOLIC LOCI. 311 which is therefore an hyperbola, the asymptotes of which are represented by the equations x= a y =. 11. Ox, Ox', are two indefinite straight position: from a given point A is drawn a straight line AEF cutting Ox, Ox', in E, Fl respectively: EF is divided into two parts EP, FP, which bear o to each other always a constant ratio. To find the locus of the point P. Join AO and produce it indefinitely to y: let Ox, Oy, be i chosen as axes of coordinates. Draw FG para OA = a; and let lines, given in llel to yO. Let 'F: JP:: n: m, OG: FG:: 1: n. and Then the required locus will be an hyperbola represented by the equation (ny.- ma- ) X = ly (y + a). Newton: Arithmetica Universalis, prob. 25. 12. The base OA of a triangle OAP is given: to find the locus of the vertex, the angle PAO being always double the angle POA. Let OA produced indefinitely be taken as the axis of x, and a perpendicular to OA through 0 as that of y. Let OA = c. The required locus will therefore be an hyperbola, the asymptotes of which include an angle of 120~, and which is represented by the equation y = 3x2 - 2cx. Pappus: Mlathematicce Collectiones a Commandino, lib. Iv. prop. 34. Newton: Arithmetica Universalis, prob. 36. 312 HYPERBOLA. 13. To find the locus of the vertex of a triangle, the base of which is given, the two adjacent angles differing by a given quantity. Let c = the base AB of the triangle, the angle at B exceeding that at A by a quantity a: let AB, produced indefinitely, be the axis of x, a perpendicular to it through A being that of y. Then the locus required will be an hyperbola defined by the equation (x2 - y2) sina + 2xy cosa = c (x sina + y cosa). Puissant: Recueil de diverses Propositions de Geometrie, p. 204, troisieme edition. 14. If normals be drawn to an ellipse '2 Y2 from a given point (a, 3) in its area, to prove that the points where they cut the curve will all lie in a rectangular hyperbola of which the equation is (a2 - '2) xy = a2ay - b2/x. 15. A circle always passes through the origin of coordinates and through the points in which the axes are intersected by a line x y - 4- 1, m n which always passes through a given point: to find the locus of the centre of the circle, m and n being variable, and the angle between the axes being given. If co denote the angle between the axes, and (a, /) be the given point, the locus will be an hyperbola represented by the equation 2 cos t) (x2 +y2) +2 (1 ++ cos )) xy- (a cos +8) x- (/cos w + a)y = 0. Leybourn's Mathematical Repository, No. xxIII. p. 74. 16. Two tangents to a parabola include a constant angle: to find the locus of their point of intersection. HYPERBOLIC LOCI. 313 The equation to the parabola being y2 = 4mx, and a denoting the constant angle, the required locus will be an hyperbola defined by the equation y2 = 4mx + tan'2.(x + m)2. De la Hire: Sectiones Conicce, lib. vmii. prop. 29. L'Hospital: Traite Analytique des Sections Coniques, p. 266. Puissant: Recueil de diverses propositions de Geometrie, p. 210, troisieme edition. 17. If any straight line, parallel to a given straight line, be drawn cutting an ellipse, and from the points of intersection normals be drawn, to find the locus of the intersection of these normals. The equation to the ellipse being a2 + 1 and the direction-cosines of the given line being proportional to a, 3, the required locus will be an arc of the hyperbola defined by the equation +x y\ a2ax + ~b2y a2a - b 232\ 2 Ka (a2-b,2 a_ b22)~2 b22S32 [ r (a2 - ~ a + b Goodwin: Cambridge Mathematical Journal, vol. IV. p. 190. 18. A series of circles pass through a given point 0, have their centres in a line OA, and meet another line AB. From the point, in which one of the circles again meets the line OA, is drawn a straight line, parallel to AB, to intersect, in points P and Q, straight lines drawn parallel to OA through the points in which the circle is cut by AB. To find the loci of the intersections P and Q. The line OA being taken as the axis of x, and 0 being the origin, let the equation to AB be x = ay + /3. 314 HYPERBOLA. Then the points P and Q lie generally in an hyperbola, of which the equation is (1 + 2a2) y2 - ax + 3a/3y _- 3x + /32 = 0. If the lines OA, AB, be at right angles to each other, a = 0, and the locus becomes a parabola. 19. A straight line D CE, passing D, C, cuts a given straight line BA, produced if necessary, in E. In AB are taken any two points F. G, such that AF n 1)/I nt X through two given points P /.DLv n E A F G B n and n' being constants. The straight lines FO, GD, are produced to meet in P. To find the locus of P. Let EB, ED, produced indefinitely, be taken as axes of x, y, respectively: let EC = k, ED = k' EA = a, EB = a'. Then the required locus will be an hyperbola defined by the equation / Ok nk' I I — y + k — = n'a + na'. Barrow: Geometrical Lectures, Lect. vI. Leybourn: Mathematical Repository, New Series, vol. I. p. 45. 20. Let AB be a given straight line, and Q a g without it: let PQR be drawn, from a point P, to meet AB in such a point R, that RQ AR PQ - BR', % To find the locus of P. 'iven point I B - Draw through Q the line Q 0 at right angles to the direction of AB, and cutting it in O: let OB, OQ, produced indefinitely, be taken as axes of x, y, respectively. Let OA = a, OB = a', O Q = c. Then the locus of P will be an hyperbola defined by the equation cxy + (y- c) ca' + a (y - c)} = 0. HYPERBOLIC LOCI. 315 COR. It may be easily shewn that the line BQ is a tangent to the hyperbola at the point Q. Leybourn: Mathematical Repository, New Series, vol.I. p. 146. 21. To find the locus of the centre of a circle which touches externally two given unequal circles. The locus will be an hyperbola the two foci of which are the centres of the two given circles. Puissant: Recueil de diverses Propositions de Geometrie, p. 207, troisieme edition. 22. To find the locus of a point, within a given angle, such that, perpendiculars being drawn from it upon the sides of the angle, the quadrilateral so formed may be of a given area. Let c' be the given area; co the given angle, its sides being taken as axes of coordinates. Then the required locus will be an hyperbola the equation of which is (x2 + y2) cos o + 2xy = 2c' cosecco. Puissant: Reczeil de diverses Propositions de GCometrie, p. 217, troisieme edition. 23. Having given one side of a triangle, and the difference between the tangents of the adjacent angles, to find the locus of the vertex. Take the middle point of the given side as the origin of rectangular coordinates, the side coinciding with the axis of x; then, m denoting the given difference, and 2a the given side, the required locus will be an hyperbola defined by the equation 2xy = m (a2 - x2). Lardner: Algebraic Geometry, p. 117. 24. Having given in position a right line AB, and a point 0 outside of it, a right line OP is drawn cutting AB in F, and from P is drawn a perpendicular PE upon AB: the magnitude of EF being invariable, to find the locus of P. 316 HYPERBOLA. Take Ox, parallel to AB, as the axis of x, and Oy, at right angles to AB, as that of y. Let EF= a, y OH= b. Then the locus of P is a rectangular hyperbola represented by the equation x y = ay + Abx. AI 1E Lardner: Algebraic Geometry, p. 134. o 25. Through a given point 0 a right line HKO is drawn, intersecting two right lines EC, FC, in H, K: a part OP of OH being supposed to be always equal to HK, to find the locus of P. c Let Ox, at right angles to the line CE, be the axis of x, and Oy, parallel to CE, that of y. Let / / denote the angle ECF. Then the / required locus will be an hyperbola defined by the equation xy + x2 cot - ay + bx = 0, where a, b, are the coordinates of C. Lardner: Algebraic Geometry, p. 135. 26. From the point of intersection of two given straight lines, is drawn a straight line of given length, bisecting the angle between them: to determine the locus of the middle point of a straight line, drawn through the extremity of this line, and terminating in the other two. Let the two given lines be taken as axes of coordinates, 2w denoting the angle between them: let c represent the given length of the bisecting line. Then the required locus will be an hyperbola, defined by the equation 1I 1 4cos w x y c 27. Two fixed axes Ox, Oy, intersect a variable circle in points (A, A'), (B, B'), respectively. To find the locus of P, the MISCELLANEOUS PROBLEMS. 317 intersection of the chords AB', A'B, supposing the two points B, B', to be fixed. If OB =, OB' = b', the equation to the required locus is x - y_ + (b + V) y b'. 28. A straight line BC moves between two fixed straight lines AB, A C so as to make the area ABC constant: to find the locus of the centre of a circle described about the triangle ABC. Let AB, A C, produced indefinitely, be taken as axes of x, y, respectively: let o be the angle between the axes. Then, the area ABC being constant, the rectangle between AB, A C will be constant: denoting this latter area by 4m2, we shall have for the equation to the required locus (x + y cos o). (y + x cos ) = m2. 29. A and B are two given points, Ax an indefinite fixed line: through B is drawn a straight line cutting Ax in T. and through A is drawn a straight line AP cutting BT in P. To find the locus of P, having given that AP is equal to PT. Let Ax be taken as the axis of x, and a line through A, at right angles to it, as that of y. Let 2a, 2b, be the coordinates of B. Then the required locus will be an hyperbola represented by the equation (x- a) y = bx. L'Hospital: Traite Analytique des Sections Coniques, p.271. SECTION XX. Miscellaneous Problems. 1. From a given point S, in the plane of a given circle, a straight line is drawn to intersect the circumference in P: from the point P an indefinite straight line PQ is drawn perpendicularly to SP: to prove that PQ is always a tangent to an ellipse or hyperbola the centre of which coincides with the centre C of the circle and of which S is a focus. 318 HYPERBOLA. Let C be taken as the origin of coordinates; CS, produced indefinitely, as the axis of x. Let SP = m, C = ae, a being the radius of the circle, and LPSx = 0. Then, x, y, being the coordinates of P, m cos0 + ae = x m sin = y and therefore m2 + 2aem cos 0 = a (1 - e2)...............(1). Also, the equation to PQ will be x cos0 + y sin0 = ae cos0 +............ (2). By virtue of (1), (2) becomes (x cos + y sin 0)2 = a2 cos20 + a(1 - e2) sin20. This equation shews that PQ is the tangent to an ellipse or hyperbola of which C is the centre and S a focus. This proposition affords a convenient method of drawing a tangent to an ellipse or hyperbola, which shall be parallel to a given line. From either focus S draw a line SP, at right angles to the given line, to intersect in P the circle described on the axis major. From P draw PQ perpendicularly to SP. Then PQ will be the required tangent. Maclaurin: Geometria Organica, sect. III. p. 102. Prony: Journal de ' Ecole Polytechnique, cahier x. p. 49. Gergonne: Annales de Mathematiques, tom. v. p. 49. 2. In every line of the second order, which has a centre, if we draw two tangents parallel to any single fixed line, and a third variable tangent; the product of the segments of the first two tangents comprised between their points of contact and the third tangent, will be a constant quantity. The points of contact of the two parallel tangents being the extremities of a diameter, we will take this diameter, which we will call 2a, for the axis of x, and its conjugate 2b for the axis of y. If x', y', be the coordinates of the point of contact of the third tangent, we shall have X12 /2 x. yt2 2 + 2 = 1 (........................ ); MISCELLANEOUS PROBLEMS. 319 and the equation to this tangent will be! - = 1 2........................(2). Let y, y,,, be the values of y, given by (2), when x is successively equated to a and - a. The quantities y, y,, are the segments of the two parallel tangents. Then we have b 2 ( b2 x') S2 / b V(a+ ) y = + (a - x'), y,, = + ay ay and therefore y.= a (a2 - x2) = + 2, by (1). Berard: Ger.qonne, Annales de Mathematiques, tom. v. p. 52. Brianchon: Ibicl. p. 53. 3. A straight line, remaining always parallel to itself, moves in the plane of two fixed straight lines. In every position of the moveable line, a point is taken in it such that the sum or the difference of the squares of its distances from the intersections of the line with the two fixed lines is equal to a given square; to determine the locus of the point. Let the two fixed straight lines be chosen as the axes of coordinates. Let x, y, be the coordinates of the particular point in the moveable line, and (x', y') of any other point in the line. Then, r being the distance between these two points, the equations to the line will be x' = ir + x y' =mr+ y, 1 and m being constants. Let r,, r0, be the values of r at the intersections of the moveable line with the axes of x, y, respectively: then O = lr,, + x 0 = mnr, y, x2 Y2 2 2 X2 2 and therefore r, ~+r = - ~ _. I d 1 2 - 320 HYPERBOLA. But rl + r, is equal to some constant c2: hence the equation to the required locus is x2 U2 -- 2 c 1F + n = 2 which belongs to an ellipse or an hyperbola. Bret: Gergonne, Annales de MIathdmatiques, tom. VI. p. 12. 4. The equation to an hyperbola, referred to rectangular axes, being ax2 + by2 + 2cxy = c', to find the position of its asymptotes, and the equation to the hyperbola referred to them as axes. But x = r cosO, y = r sin: then r2 (a cos20 + b sin20 + 2c cos. sin ) = c'. Now r = oo, provided that a cos20 + b sin20 + 2 cos0. sin0 = 0, and thus the equation to the two asymptotes is ax' + by2 + 2cxy = 0. Again, al, / representing the transverse and conjugate axes respectively, the equation to the hyperbola referred to its asymptotes is xy = (a 2 - b2). Differentiating r2 = x2 + y2 we have, for maximum values of r, xdx + ydy = 0, and, from the equation to the curve, axdx + bydy + c (xdy + ydx) = 0; and therefore, X being arbitrary, Xx = ax + cy........................ (1), y = by + c........................ (2). Multiplying (1) by x, (2) by y, adding, and attending to the equation to the curve, we have Xr2 = c' MISCELLANEOUS PROBLEMS. 321 But, from (1) and (2), (X - a) x = cy, (X - b) y = cx, and therefore (X - a) (X - b) = c2; hence (c - ar2) (c' - br2) = cA', (c" - ab) r4 + c' (a + 6) r' - c" = 0. Consequently, a2, 6b, being the roots of this equation,, a+b 2 c12 a1 + i =-c -2 a a 2 b2=- --- 1 l G -ab' ab' I2 2L 2V 2(a- 1 )2 + 4c2 and therefore (a, - b2) = c2 (, b) (c2 - ab)2 Thus the equation to the hyperbola referred to its asymptotes is t = (a-b) +4C21 C2 - ab 5. CP is any semi-diameter of an hyperbola, CD being its conjugate semi-diameter: PK is a tangent at P, equal to CD: from any point V in CP produced is drawn a straight line VEF, parallel to CD, cutting the curve in E and the straight line through C and K in F: to prove that (VF)2 - (VE)2 = (CD)2. Pascal: Essais pour les Coniques (Euvres, tom. IV. p. 5. 6. To prove that the radius of a circle, which touches an hyperbola and its asymptotes, is equal to that part of the latusrectum produced which is intercepted between the curve and an asymptote. 7. If an ellipse and hyperbola have the same foci, to find the equation to the locus of the intersection of two tangents to them at right angles to one another, The equations to the ellipse and hyperbola being x, y2 x, 2 y2 1 1 - + - - 1, and —. 1, the required locus will be a circle defined by the equation x + y2 = a, + b2. lr 322 HYPERBOLA. 8. P and Q being two points in an ellipse, if an hyperbola, the foci of which are P and Q, pass through one focus of the ellipse, to prove that it will pass also through the other focus. 9. To find the equations to the asymptotes of the hyperbola represented by the equation xy + ax + b + + 2 = 0. The required equations are x + b = 0, y + a = 0. 10. The equation to an hyperbola being ax2 + by2 + 2cxy + 2a'x + 2b'y + c' = 0, to shew that ax2 + by" + 2cxy = 0 is the equation to two straight lines through the origin parallel to its asymptotes. 11. Three hyperbolas have parallel asymptotes: to shew that the three straight lines joining the points of intersection of the hyperbolas, taken two and two, all meet in one point. 12. To prove that an ellipse and hyperbola, which have the same centre and foci, will cut each other at right angles; and that, if from any point in the circumference of the circle, which passes through their points of intersection, tangents be drawn to the two curves, they will be at right angles to each other. 13. CPLD is a parallelogram, the sides CP, CD, of which are conjugate semi-diameters of a rectangular hyperbola, inclined to one another at an angle of 60~: to find the equation to the ellipse which passes through the points C, P, L, D, and cuts the conjugate hyperbola at D at an angle of 15~. The equation to the rectangular hyperbola being x2 - y2 = 'a2 that to the required ellipse will be 1X2 + y2 + 2 (2- V3) y = 2 3 X 2,V3 MISCELLANEOUS PROBLEMS. 323 14. Any number of ellipses (or hyperbolas) concentric, similar, and similarly situated, are intersected by a line parallel to a directrix in P, P' P",.....: to prove that the extremities of the diameters respectively conjugate to the diameters through P P', P",...... are in a line perpendicular to the directrix. 15. If, in the preceding question, the curves be cut by any concentric hyperbola, the asymptotes of which have the same direction as their axes, in Q, Q' Q".......; to prove that the extremities of the diameters respectively conjugate to the diameters through, Q' Q"... are situated in another branch of the hyperbola. 16. A chord PP' is drawn in an ellipse, so that, AC being the semi-axis major, tanPCA + tanP' CA = 2 tana; to find the envelop of the chord. The required envelop is an hyperbola the equation of which is, the axes of coordinates coinciding with the axes of the ellipse, xy - x2 tana = 4 tanc 17. To prove that the tangent to the interior of two similar hyperbolas with coincident axes, cuts off from the exterior curve a constant area. De la Hire: Sectiones Conicce, lib. VI. prop. 17. 18. Supposing that, y', y", being any two values of y in the hyperbola xy = c2, such that y" = Xy', where X is constant, to prove that the area included between y', y", the axis of x, and the curve, is invariable in magnitude. Gregorius a Sancto Vincentio: Opus Geometricum Posthumumn ad Mesolabium, p. 252. ( 324 ) LINES OF THE SECOND ORDER. SECTION I. Referred to a Principal Diameter and its Tangent. Normals. 1. FROM a point P of a conic section a normal is drawn to cut a principal axis (the major axis in the ellipse, the transverse axis in the hyperbola) in a point G: to prove that, of all straight lines drawn from G to the curve, GP is the least. The equation to the conic section is y2 = (1 + e) {2emx- (1 - e) x2}. The equation to the normal, at any point (h, k), is (1 + e) {em - (1 - e) h}. (y'- k) = - (x' - ). At the point G, y' = 0, and therefore x = (1 + e) em + elz. Let r represent the distance of G from any point (x, y) in the curve: then r2= {- (1 + e) em - e'2}2 y' = 2 - 2ex {(1 +e) m + eh} + e2 {(1+ e) m + eh}2 + (1 +e) {2emx- (1 -e) X2} -e2 (x- h)2 + e2 (1 + e) m2 + 2e3 (1 + e) mrh - e2 (1 - e2) = (x - h) + (1 + e)2 m + (1 + e) {2emh - (1- e) h2} (X (x-ih)+ (1+ e)2 m2 + 2. Hence it appears that r is least when x = h, or when (, y) coincides with (h, k). De la Hire: Sectiones Conicce, lib. VII. prop. 13. PRINCIPAL DIAMETER AND ITS TANGENT. CHORDS. 325 2. AP is the arc of a conic section, of which the vertex is A; PG the normal, and PK a perpendicular to the chord AP, meet the axis in G and K. To shew that GK is equal to half the latus-rectum. 3. To find the locus of the middle point of the portion of the normal to a conic section which is included between the curve and the axis. The equation to the conic section being y2 = mx + nx2 that to the required locus will be (n + 2)'2 n - X2 mx + I-m2. (n + 1) = 0. Lardner: Algebraic Geometry, p. 149. SECTION II. Referred to a Principal Diameter and its Tangent. Chords. 1. From one extremity A of a principal axis of a conic section, a given straight line AP is drawn to cut the curve in P; to find the equation to the line joining P to the other extremity B of the axis. Let the axis AB and the tangent at A be taken as axes of x and y respectively. Then the equation to AP will be of the form y = ax......................... (1), and the equation to the conic section of the form y2 = mx + nx........................(2). The equation to AB is y = 0............................. (3). The two lines (1) and (3) may be represented simultaneously by the equation y (Y - ax) = 0, or y2= axy........................... (4). Hence, at A, P, B, the intersections of AP, AB, with the curve, axy = mx + nx2, whence dividing by x, we obtain for the equation to BP, which passes through P, B, ay = m + nx. 326 LINES OF THE SECOND ORDER. 2. If A and B be the extremities of the axis major of a conic section, T the point where a tangent at a point P in the curve, meets AB, Q TR a line perpendicular to AB and meeting AP, BP, in Q, R, respectively; to prove that QT= RT. 3. To find the length of the chord of a conic section, denoted by the equation y2 = 2nx + nx2, the equation to the chord being x y -+ -= 1 If 2c denote the length of the chord, c,_ a (a2+ ").(m2 + 2m(fa + n22 /2) (na2 -_ 2)2 SECTION III. Referred to a Principal Diameter and its Tangent. Focal Properties. 1. From the extremity L of the semi-latus-rectum SL of a conic section, a chord LA is drawn to the vertex A of the diameter through S. A tangent is drawn at L. A straight line MRP is drawn, through any point M in AS, or AS produced, at right angles to AS, meeting the chord AL in R, and the tangent at L in P. To prove that PR is equal to MS. Taking AS, produced indefinitely, as the axis of x, and the tangent at A as that of y, the equation to the curve will be, 2 = (1 + e){2emx - (1 - e) x2}. The equation to the tangent at any point (x, y) is _+_ - tem - (1 - e) x} x' + emX. 1+ e FOCAL PROPERTIES. 327 Now the coordinates of L are em, em (1 + e): hence the equation to PL is y = e(x'+ )........................(1). Again, the equation to AL is =(1 + e)x'........................(2). Hence PR, which is the difference between the ordinates in (1) and (2), is equal to em - x', that is, to MS. De la Hire: Sectiones Conicce, lib. vII. prop. 14. 2. From the extremity L of the semi-latus-rectum SL of a conic section, a tangent LP is drawn. At right angles to the diameter through 5, from any point M in it, a straight line is drawn to meet the tangent LP in P and the curve in Q. The points S, Q, are joined. To prove that SQ = PM. De la Hire: Sectiones Conicce, lib. vIII. prop. 15. 3. Having given the position of the focus S of a conic section, the position and magnitude of the semi-latus-rectum SL, and the position of T, the intersection of the tangent L T with SA produced, A being one end of the diameter through the foci: to determine the position of A, and of the other extremity A' of the axis AA'. Let ST= h, SL = k: then SA k hk A h + k= h - k' De la Hire: Sectiones Conicce, lib. VIII. prop. 21. 4. To prove that a straight line drawn through the focus of a conic section, to the point in which a diameter meets the directrix, will be perpendicular to a tangent at either extremity of that diameter. Leybourn: Mathematical Repository, New Series, vol. ii. p. 95. 328 LINES OF THE SECOND ORDER. SECTION IV. Referred to any two Oblique Diameters. 1. CA, CB, are any two semi-diameters of a conic section. The tangent at A meets CB produced in F, and the tangent at B meets CA produced in A. To prove that, E being the intersection of ArF BA, the triangle AAE is equal to the triangle BrE. Let CA = ac CB = /; then, CA, CB, produced indefinitely, being taken as axes of x, y, respectively, the equation to the curve will be x y2 xy + ~2 12 a2 y2 2 7 7 being some constant. The equation to the tangent at any point (x, y) of the curve is (I \+ Y + ) 1 Hence the equation to the tangent Ar, drawn at the point (a, ), is ay x + r = a. Putting x' = 0, we see that Cr -Y-, and therefore CA. CA = 72. Similarly A. CB = y2. Hence the triangles ACr, BCA, and therefore the triangles AAE, BrE, are equal. 'Eav cC'voOV Tro 4) ' KVflcXOV t7rrepLtepe(a6 EvOelatb Ert4aovoCvatra cV/it7rTGTwaLV, axOwat 86 t'a T rv bc^ov atd/erpot 'v /TrlT'ovct'a raZ cs airTopevaet, ir a c'raa r / yLvo/Jeva Kara KopvO'v TpLyowva. 'AIIOAAQNIOY IIEPrAIOY KWVLKWLV -TO Tpi-rpov' po'-racs ca. 2. To determine the magnitudes and positions of the axes of the conic section ax2 + by2 + 2cxy - 1, o being the angle of coordination. REFERRED TO TWO TANGENTS AS AXES. 329 The magnitudes of the semi-axes are the positive values of r in the equation (ab - c) r4 - (a + b + 2 coso) r2 + sin2 = 0; and (b - x)(x - y cosco) = (a - cy) (y - x cos) is the equation to the two axes. O'Brien: Cambridge Mathematical Journal, vol. IV. p. 101. SECTION V. Referred to two Tangents as Axes. 1. If two conic sections have two points of contact, to prove that they will not meet in any other point. Let the tangents at their points of contact be taken as axes of coordinates. Then, a, b, being the distances of the points of contact from the origin, the equations to the two conic sections will be x2 2xy Y2 x y a —+ + 2- 2- -2 + 1 =0 a mn b a b x2 2xy Y2 x Y -+ - + -2 -2 + = 0. a mn b a b If the two curves meet, we have, subtracting one equation from the other, xy = 0 whence x = 0, and therefore y = b; or y = 0, and therefore x = a. Thus we see that the two points of contact are the only points in which the two curves meet. 'Eav rTcv 7TrpOELtprev)vV ypaptcwv Tve9 KearTa o ao 7tLeta d6rrwrTcvaL dXX\Xwv, oV avap/3\Xovcrv d\XX'Xat9 caO' ~Tepov. 'AIIOAAQNIOT IIEPPAIOT KWVLKwV TO TEap-oyv' 1podraYsL KK'. 2. From a point 0, two tangents OH, OK, are drawn to a conic section: a straight line MQR is drawn, parallel to OK, 330 LINES OF THE SECOND ORDER. to intersect OH in M, and the curve in Q, R. To prove that (MII)'2: MQ.MR:: (OH): (OK)2. Let OH, OK, produced indefinitely, be taken as axes of x, y, respectively. Let OH= a, OK = b. Then, m, n, being arbitrary constants, the equation to the curve will be x2 2xy y2 2x 2y + - + -—. 7 + 1 = 0. a mn b2 a b Let OM = h: then, putting h for x in this equation, we shall have a quadratic in y, the two roots of which are MQQ, MR. Hence, equating the rectangle of these two lines to the last term of the quadratic, we have MQ.MR = b2 (h -) (OK)2. (IHi)2 (OH)2 or (IMH)2: M MQ.MUR:: (OH)02: (OK)2. De la Hire: Sectiones Conicce, lib. III. prop. 26. 3. To find the locus of the centre of a conic section which touches two given straight lines at two given points. The equation to the conic section, the two given straight lines being taken as axes of coordinates, will be of the form a2C'2 + 2cxy + b2y2 + 2ax + 2by + 1 = 0, a and b being given constants. If h, k, be the coordinates of the centre, adh + ck + a = 0, and bk + ch + b = 0. From these two equations we have (c2 - a2b) h = b (ab - c), (c + ab) h = - b; and, similarly, (c + ab) = - a: h k hence b- a~ REFERRED TO TWO TANGENTS AS AXES. 331 which is the equation to the required locus: thus we see that the centre of the conic section lies always in a straight line passing through the intersection of the two given lines, and bisecting the chord which joins the two given points of contact. Gergonne: Annales de Mathenatiques, tom. XI. p. 385. 4. If a conic section be touched by four straight lines, to find the locus of its centre. If we take two of the tangents as axes of coordinates, the equation to the conic section will be of the form adx2 + 2cxy + b2y + 2ax + 2by + 1 = 0. The equation to the tangent, at any point (x, y), is (aCx + cy + a) x' + (by + cx + b) y' + ax + by + 1 = 0. Let this tangent coincide with a line s8' + ay' = a/8........................(1). Then a (cx + cy + a) + ax + by + 1 = 0...........(2), 3 (by + x + b) + ax + by + 1 = 0...........(3). Eliminating x from (1), (2), (3), we shall arrive at an equation a/3 (ab + c) + 2 (aa + b13 + 1) = 0............(4). For another tangent, we shall have, similarly, a'(/' (ab + c) + 2 (aa' + b3' + 1) = 0......... (5). But, if x, y, be the coordinates of the centre of the conic section, rsection, a2x + cy + a = 0.....................(6), b2y + x + b = 0.....................(7). Eliminating a, b, c, from (4), (5), (6), (7), we shall get x y - a3_ - o''I c( 3-l' — 2( 1........ (8) a- a /3 -/,' 2 ( - a') (/3 -/3') for the equation to the locus of the centre, which is therefore a straight line. 332 LINES OF THE SECOND ORDER. COR. It may easily be proved that this line passes through the middle points of the two diagonals of the quadrilateral formed by the four tangents of the conic section. Newton: Principa, lib. I. Lemma, xxv. Cor. 3. Gergonne: Annales de Mathimatiques, tom. XI. p. 382, ann. 1820, 1821. Poncelet: Gergonne, Annales de Matheinatiques, tom. XII. p. 109. Eligius Manderlier: Annales Academice Gandavensis, 1826-28, p. 57. Bobillier: Gergonne, Annales de ]Mathdmatiques, tom. XVIII. p. 362. Hearn: Researches on Curves of the Second Order, p. 37. 5. From a point 0, two straight lines are drawn, touching a conic section in H and K. Two straight lines EPP', EQQ', are drawn, cutting each other in E, the former parallel to OH, the latter to OK; the former cutting the curve in P, P', the latter cutting it in Q, Q'. To prove that EP.EP': EQ.EQ':: (OH)12: (OK)2. If OH OK, produced indefinitely, be chosen as axes of x, y, respectively, the equation to the conic section will be x2 2xy y2 2x 2y - + 72- - 7 = 07 a zn b a b where a, b, denote OH, OK, respectively, m, n, being any constants. Let h, k, be the coordinates of E: then, taking EPP', EQQ', produced indefinitely, as axes of x', y', respectively, we shall have, for the equation to the curve, (X +h)2 2(x'+ h)(y' ~k) (y'+k~)2 _ 2(x+ h) 2c(y'+k) a mn b2 a b Putting x'= 0, the two values of y' in the resulting quadratic will be EQ, EQ', and accordingly EQ.EQ b (2 + 2k k - 2 + 1)..A Q= + -2 n + P.2- - - + \a mn ' REFERRED TO TWO TANGENTS AS AXES. 333 By symmetry we have also EP.EP' = a2 ( 2 2A k2 2h 2k ) ER+ EP +7-y-" +lk ma n b a b EP.EP' (OH)2 Hence EQ.EQ' ( ( ox De la Hire: Sectiones Conicce, lib. III. prop. 29. 6. To find the condition that the straight line x y -+ =1 a I3 may touch a conic section represented by the equation x2 x y a+ 2 x Y - +2-+2 -Y2 1 = 0. a Am n b2 a b The required condition is expressed by the equation /3ab + - ~+ + ) = =0. ab mn \a b 7. From a point 0, two straight lines are drawn, touching a conic section in H and K. A straight line MQRT is drawn, parallel to one of these tangents, cutting the other in M, the curve in Q. T, and the chord HK in R. To prove that MQ.V1 T = (MR). De la Hire: Sectiones Conicce, lib. III. prop. 27. 8. To find the locus of the centres of all conic sections, which touch three given straight lines, and pass through a given point. If two of the given straight lines be taken as the axes of x i, coordinates; then (h, k) being the given point, and - + =1, the equation to the third straight line, the equation to the required locus will be,4hk(2x - h) (2y - k) = {2 (kx - hy - hk) - (2/3x + 2ay - a1)}2. Gergonne: Annales de Mathe/matiques, tom. XI. p. 385. Poncelet: Gergonne, Annales de Mathematiques, tom. XII. p. 111. 334 LINES OF THE SECOND ORDER. 9. To determine the locus of the centres of all the conic sections, which touch two given straight lines and pass through two given points. The two straight lines being taken as coordinate axes, and (h, kc) (.', k')) being the two given points; the equation to the required locus will be {(hy - kx)2 - (h'y - kc'x)2} = 4 {(kx + hy - hk) - (k'x + h'y - h'k')}.(hy - kx).(k'x + h'y - h'') - (h'y - k'x)'.(kx + hy -- zk)}. This equation may be resolved into a double equation of the second degree. If we put (hk' - kh') (hk - 'k') = M the double equation will be (ki + k').[(h - h')2.{(7k+c')x - (h + h')y2- 21ix (2x - (h + h')}].(ck') =~ (h+h').[(k-7')2. {(k + k')x-(h+ A') y}2+ 2i1y2y- (k +-')}].(hh'). Brianchon: Gergonnze Annales de Macthematiques, torn. XI. p. 219. Gergonne: Annales de ilMathematiques, tom. xi. p. 390. Poncelet: Gergonne, Annales de Mathemnatiques, tom. XII. p. 233. Gergonne: Annales de Mathematiques, tom. XII. p. 249. 10. To prove that the straight line which joins the summit of an angle circumscribed about a conic section with the centre of the curve, divides the chord of contact into two equal parts. Frederic Sarrus: Gergonne, Annales de Mathematiques, tom. XII. p. 368. SECTION VI. Referred to a Tangent and Normal. 1. If, in any conic section, be inscribed a series of rightangled triangles, having all of them the summit of the right angle at a given point of the curve; to prove that their hypotenuses will all meet at a single point of the normal drawn through the common summit of all these triangles. REFERRED TO A TANGENT AND NORMAL. 335 Let the tangent and normal at the given point be taken as axes of x and y respectively. Then the equation to the conic section will be ax2 + by2 + 2cxy + Ofy = 0. The equations to the two sides of one of the right angles being y - ax 0 y + - x = 0, the equation to both will be '- -ax) (y+1 x) = O) whence = y2 + ( a ay. Substituting this expression for x' in the equation to the conic section, we get, dividing by y, for the equation to the third side, ( 1 i {a G- a) + 2c[ + (a+ b)y+ 2f= 0. Putting x = 0, we see that 2f Y a+ b' Thus we see that the hypotenuse cuts the normal at an invariable distance from the origin. COR. It hence follows by the theory of poles and polars, that the points of concourse of the tangents at the extremities of these hypotenuses lie all in a single straight line. Fr6gier: Gergonne, Annales de lMfathenatiques, tom. VI. p. 231. 2. If, in any conic section, be inscribed a series of triangles, all having a common summit, the angle of which is bisected by the normal to the curve at the place of the summit: to prove that the sides of these triangles, opposite to this summit, will all meet in the point of intersection of the tangents drawn at the two ends of the normal. The equation to the conic section, referred to the tangent and normal, at the common summit, as axes of coordinates, will be a + by + 2cxy + y = O............... (1). 336 LINES OF THE SECOND ORDER. Also the equation to the two sides, of any one of the triangles, which pass through the origin, will be x2 = Xy.......................... (2). Substituting in the equation (1) the expression for x' given by (2), and dividing by y, we get, for the equation to the third side, (aX + b)y + 2 (ex +f) = 0................ (3). Putting y = 0, we see that - f a value independent of X. Hence the opposite sides of all the triangles intersect the tangent at a distance -- from the point of contact. The tangent at the other end of the normal, as coinciding with the chord (3), when X becomes zero, must also pass through the same point. CoR. Hence it follows, by the theory of poles, that the points of concourse of the tangents at the extremities of these third sides of the triangles, will all lie in a single straight line, viz. the normal itself. Fregier: Gergonne, Annales de Mathimatiques, tom. VI. p. 233. 3. To find the equation to a conic section which has a contact of the third order with the curve ax2 + by2+ 2cxy + 2fy = O...............(1 ) at the origin of coordinates. The equation to a conic section, which has a contact of the first order with (1) at the origin, is ax2 + b,y2 + 2cxy + 2y = 0...............(2). At the intersections of (1) and (2), we have (a,b - ab,) y + 2 (a,c - ac,) x + 2 (a,/- af,) = 0....(3). If the straight line (3), which joins the two points of intersection, pass through the origin, the curves (1) and (2) will have two elements in common at the origin, or the contact will be of the second order: the corresponding condition being a,f= af........................... (4). REFERRED TO A TANGENT AND NORMAL. 337 Again, if the line (3) coalesce with the axis of x, the remaining point of intersection will coalesce with the origin, and thus, the two curves having three elements in common at the origin, the contact will be of the third order. The additional relation is ac = ac........................... (5). The relations (4) and (5) reduce the equation (2) to the form ax + y + 2cxy + 2fy = 0O which is the required equation. Plucker: Gergonne, Annales de Mathdmatiques, tom. xvII. p. 69. 4. If two conic sections touch each other at any point, to prove that they cannot have more than two points of intersection. 'Eav T&rv etpl/&)evOv rypa11/%/Ctv Ttves KaO' Cv ed'ab7rTOWV7a Ia'-epov aXXAjXwv- ov av,4L3\XXovuov avTavsa KcaO e'repa o-/qLeta 7rXeiebova 7 8vo. AIIOAAQNIOY IIEPrAIOY KtWVKWV TO,TrTap-rov IHIpo7TaC KS'. 5. To prove that, for every point of a conic section, there exists a circle which has with the curve, at this point, a contact of the second order. Plucker: Gergonne, Annales de Mathematiques, tom. xvII. p. 71. 6. To prove that a circle cannot have with a line of the second order, a contact of the third order at one of its points, unless this point be at the end of an axis of the curve. Plucker: Gergonne, Annales de Mathematiques, tom. xvii. p. 71. 7. To prove that, if any number of circles touch a conic section at the same point, the chords joining the points of intersection are all parallel to each other. Plucker: Gergonne, Annales de Math4matiques, tom. xvII. p. 71. z 338 LINES OF THE SECOND ORDER. SECTION VII. Referred to any Axes whatever. Centres. If ac, /, be the coordinates of the centre of the curve b (x, y) = ax' + by2 + 2cxy + 2a'x + 2'y + c' = 0, to find the value of b (a, /). Changing the origin of coordinates to the point a, /, we shall have for our transformed equation, a (x' + a)2 + b (y' + 3)2 + 2c (x' + a) (y + 3) + 2a' (x' + a) + 2b' (y' + /) + c' = 0. Since (a, /) is the centre, the coefficients of x' and y' must be zero: hence aa + c3 + a'= 0..................... (1), and b,3 + c b' = o..................... (2). From (1) and (2) we have ad' + b'3 + 2ca/3 + a'a + b'/3 = 0, and therefore b (a, 38) = c' + a'a + 3b'................. (3). From (1) and (2) we may easily get a'b - b'c Va - a'c ca - ab ' c - ab; and therefore, from (3), ab'2 + ba'2- 2a'b'c (a, ) = c' + c2 - ab SECTION VIII. Referred to any Axes whatever. Tangents. 1. To find the condition that the straight line 1.......................... (1) may touch the conic section ax2 + by2 + 2cxy + 2a'x + 2'y + c' = 0. REFERRED TO ANY AXES WHATEVER. TANGENTS. 339 If x', y', be the coordinates of the point of contact, the equation to the tangent is (ax + cy' + a') x + (by' + cx' + 6') y + ax'1 + y' + = 0. Since this equation must coincide with the equation (1), we must have a (ax' + cy' + a') + a'x' + by' + c' = 0, and (by' + cx' + b') + a'x' + b'y' + c' = 0; or (aa + a') x' + (ac + b') y' + aa + c' = 0, and (/3b + V')y' + (/c + a')x' + /3' + c' = 0. But, since the point of contact lies in the line (1), we must have also 8ix' + ay' - a/ = O. Eliminating x' and y' between the last three equations, we shall obtain for the required condition, a"32 (c2 - ab) + 2a2/ (a'c - ab') + 2a/32 (b' - ba') + al (a'2 - ac') + /2 (b' - bc') + 2a/3 (cc' - a'') = 0. Gergonne: Annales de Mathenmatiques, tom. XI. p. 381. 2. To prove that a conic section represented by the equation (lu)" + (mv)5 Jr (nw) = 0................ (1), where u, v, w, are linear functions of x and y, and 1, m, n, are arbitrary constants, is touched by a straight line Xu + /v + vw = 0.....................(2), provided that 1, m, n, X, t, v, are connected together by the equation +I m n +-+- = o0........................ (3). x V Eliminating v between (2) and (3), we get nw = + -. ( + /,) X............... (4). Combining (1) and (4), we have (lu) + (mnv)I + +,). (Xu + jv)~ = 0, z2 340 LINES OF THE SECOND ORDER. and therefore lu + my + 2 (Imuv) = + -) (X + tv),, /~ mX 2 (lmuv)2 = X v + - u x Rev \h/ " [~fT~)= 0? X()u - ( v whence - and therefore, by similarity, X2u V Y2W I n n ' These two equations shew that the lines (1) and (2) meet in one point and one only: hence (2) is a-tangent to (1). It may be observed that, since the equation (2) involves two X x arbitrary parameters, viz. /, -, subject to only one restriction, viz. (3), the tangent line is a general tangent to the conic section. COR. If / = 0 and v = 0, then the equation (2) is reduced to u = 0. Thus it appears that u = 0, and, similarly, v = O, w = 0, are all tangents to (1). The truth of this conclusion is easily seen also by combining u = 0 with (1), whereby we perceive at once that the two lines meet in one and only one point. Hearn: Researches on Curves of the Second Order, p. 36. 3. To find the equation to the tangent at any point of a conic section represented by the equation uv = w............................(1) where u, v, w, are linear functions of x and y. The equations to any point in the curve may evidently be written in the form Xu = w7 v = Xw, X being an arbitrary quantity. REFERRED TO ANY AXES WHATEVER. CHORDS. 341 The equation to any straight line passing through this point may be expressed in the form v - Xw = (Xu - w)...................(2), j/ being an arbitrary quantity. At the intersection of (1) and (2), we have, eliminating v, W2 - XUW = -U (Xuw - uW), or w2 - (X- ti) uw - Xpu2i = 0...............(3). Suppose this equation to be a perfect square: then (X - )2 = - 4X, (xf+ r) = =o, (X + =)2 -; and therefore (2) becomes XV - 2Xw + v = 0.................. (4). Since (3) is a perfect square, (4) meets (1) in only one point, that is, at the point v = Xw, Xu = w: hence (4) is the equation to the required tangent. Salmon: Conic Sections, p. 217. 4. To prove that Xu + tv + vw=0 will be always a tangent to a conic section Ivw + mwu + nuv = 0, where u, v, w, are linear functions of u, v, w, and 1, m, n, are parameters, provided that (iX)* + (m/)i + (nv)? = 0. Hearn: Researches on Curves of the Second Order, p. 34. SECTION IX. Referred to any Axes whatever. Chords. 1. To determine the locus of the middle points of the chords of the curve ax2 + by2 + 2cxy + 2a'x + 2by + c' = 0, 342 LINES OF THE SECOND ORDER. which are parallel to the line x sin -y cos = 0; and thence, the axes being rectangular, to find the positions of the axes of the curve. Let A, k, be the coordinates of the middle point of any one of the parallel chords: let x', y', be the coordinates of either end of this chord referred to (a, k) as origin. Then a(A+x ')2+b (k+y')2+2c(h+x') (k+y')+2a'(h+x')+2b'(k+y')+c'=o. Now for x', y', we may write equal negative values - x' -y' because the chord is bisected in the new origin: hence the sum of the terms involving the first powers of x', y', must be equal to zero, and thus (ah + ck + a') x' + (bk + ch + b') y' = 0. But this chord is parallel to the line xsin - ycos = 0: hence (ah + ck + a') cos 0 + (bk + ch + b') sin = 0, which is the equation to the required locus. COR. Since an axis of the curve intersects its chords at right angles, we shall have, for the determination of the positions of the axes, supposing the coordinate axes to be rectangular, b sin 0 + c cos 0 tan 0 = a cos 0 + c sin whence asin20 + c (1 - cos20) = b sin20 + c (1 q- cos20), (a - b) tan20 = 2c, tan 20 = 2 a - 6 2. In any conic section, if two chords PQ, PR, make equal angles with a fixed chord PIK, and the chord QR be drawn: to prove that QR will pass through a fixed point for all positions of PQ, PR, REFERRED TO ANY AXES WHATEVER. CHORDS. 343 Let PK, produced indefinitely, be taken as the axis of y, and Px, at right angles to it, as the axis of x. The equation to the conic section will be ax2 + by2 + 2cxy + 2ax + 2b'y = 0...........(1). Let the equation to PQ be y = mx; that of PR will be y = - nx, and therefore both PQ and PR will be represented by the equation y = m'22 x.........................(2). From (1) we have ax2 + by2 + 2cxy + 2a'x =- 2'y............ (3), and ax2 + by2 + 2cxy = - 2a'x - 2b'y............(4) and therefore, multiplying the equation (3) by a'x, the equation (4) by b'y and subtracting the latter of the resulting equations from the former, we get a'x (ax2 + by2 + 2cxy + 2a'x) - by (ax2 4- by2 + 2cxy) = 2b'2y... (5). At the intersections of QR with the curve, we have, by (2) and (5), a' (ax + m2bx + 2cy + 2a') - ' (ay + n2by + 2m2cx) = 2m2b'2, which is therefore the equation to the chord QR. This equation is evidently satisfied, for all values of m, by the two equations aa'x + (2a'c - ab') y + 2a'2 = 0, (a'b - 2b'c) x - bb'y - 2b2 = 0, which are the equations to a fixed point. 3. To find the equation to a chord passing through any two proposed points of the conic section represented by the equation UV = w2 u, v, w, being linear functions of x and y. Let the two points be l = }w and =' beingW = proposed consta X, XF, being any proposed constants. 344 LINES OF THE SECOND ORDER. The equation to any straight line through the former point is v - Xw = p (Xu - w), and that to any straight line through the latter point is v - X'w = p' (X'u - w), p and p' being arbitrary parameters. Now these two straight lines will coincide provided that pX = p'X', p - X = p'- X': whence p =- X', p' =- X. The equation to either of the lines then becomes XX'u - ( + ') w + v = 0. Since this line passes through both the proposed points, it is the equation to the chord. COR. If X'= X, the two points coincide and the chord becomes a tangent represented by the equation 2u - 2Xw + v = 0. Salmon: Conic Sections, p. 217. 4. Through any point P in a conic section is drawn a straight line, parallel to a given straight line, intersecting two tangents in T and T', and cutting the chord, which joins the points of contact, in C. To prove that (PC)2 c PT.PT'. Let 0 be the intersection of the two tangents. Let the axes of coor- A dinates be Ox, parallel to the chord / of contact A'A, and Oy, parallel to the line T'TPC. // Then, the equations to OA, OA', AA', being represented by ~/ y= CaI y = ax y = /b that to the conic section will be, X de- / noting an arbitrary constant quantity, (y- ) (y + a) (y- b), REFERRED TO ANY AXES WHATEVER. DIRECTRIX. 345 which, expressed geometrically, becomes PT.PT' = (PC)2, whence (PC)'2 PT.PT'. Carnot: Ge'ometrie de Position, p. 446. 5. If APP' be any chord, drawn through a fixed point A in the axis of a conic section, to meet it in two points P, P'; to prove that, A being the origin of coordinates and the axis of the conic section the axis of x, 1 1 — + - = a constant quantity, X, X1. x,,,, being the abscissae of P, P'. 6. If two conic sections are such that they intercept, on a given straight line, chords the middle points of which are coincident: to prove that the same property will apply to all conic sections which pass through the four intersections of the two conic sections and cut the same given line. Plucker: Gergonne, Annales de Mathematiques, tom. XIX. p. 105. SECTION X. -eferred to any Axes whatever. Directrix. 1. To find the equation to a conic section, of which (h, k) is the focus, e the eccentricity, and of which the directrix, the axes of coordinates being rectangular, is represented by the equation xsina - ycosa + c = 0; and, if A, k, be such that h +ecsina=0, k- elccosa = 0, to find the equation to the curve referred to its axes. The equation to the curve, referred to the existing axes, is (x - h)2 + (y - k)2 = e2 (x sina - y cosa + C)2; and, the two relations between A, k, a, e, c, being adopted, the equation to the curve, referred to its axes, is x' + (1 - e2) y = e22 (1- e2). 346 LINES OF THE SECOND ORDER. 2. To find the equation to the directors of the conic section ax' + by2 + 2cxy = f The required equation to the two directors is {(a + X)l + (b + X) + +X X being given by the equation (a + X) (b + X) = c2. SECTION XI. Referred to any Axes whatever. Conjoint Lines and Circles. 1. Having given the equation to a conic section, and that to a straight line in the same plane, to find the equation to the conjoint line which passes through the origin, and the equation to the corresponding conjoint circle. The expression conjoint lines is used by M. Terquem to signify two straight lines such that, if they be taken as axes of coordinates, the coefficients of the two squares in the equation to the curve become equal. Thus two equal diameters are conjoint lines; the same may be said of the asymptotes of an hyperbola: a principal diameter coalesces with its conjoint line. It is easily seen that two conjoint lines cut a conic section, generally speaking, in four points, situated in a single circle, called a conjoint circle; and, reciprocally, that if four points of a conic section lie in a circle, any two lines joining them two and two, are conjoint lines. Let the equation to any conic section be ax2 + 2cxy + by' + 2a'x + 2'y + c' = 0........(1), and let y be the angle between the coordinate axes. Let the equation to the given straight line be \x + /y + v = 0......................(2). Let the equation to the required conjoint line be X'x + pl'y = 0........................(3). ANY AXES WHATEVER. CONJOINT LINES AND CIRCLES. 347 The equation to the system of the two straight lines will be (Xx +,y + v) ('x +,'y) = 0...............(4). Adding together the equations (1) and (4), we have (a + XX') x2 + (2c 4+ X/'/ + X/fk) xy + (b + fp') y2 + (2a' + X'v) + (2b' + )y + c' = 0......(5). In order that this equation may represent a circle, we must have a + XX'= b + /pA' 2c + X/A' + X'/ = 2 cos y (a + XX'), whence we deduce A' X (b - a) + 2u ([a cosry - c) X = ----— 2 and, (a - b) + 2 (b cosy - c)......... (6). where 2 = X2 -2X/k cosy + 42 Substituting the values (6) in the equation (5), and putting P2a - 2Xcc + OXb = A, we shall obtain kx2 + 2k cosy.xy + ky2 + {2a'82 + Xvi ( - a) + 2/av (a cos y - c)} x + ({2'82 + XAv (a - 6) + 2Xv (b cosy - c)} y + C82 = 0, which is the equation to the conjoint circle. Terquem: Liouville, Journal de MJatkdmatiques, tom. III. p. 17. 2. To find the locus of the centre of the conjoint circle of a given conic section, the two conjoint lines being both supposed always to pass through the origin. If the equation to the conic section be ax2 + 2cxy + by'2 + 2a'x + 2b'y + c' = 0, the required locus will be a straight line of which the equation is (b' - a cos7y)x = (a' - I' cos y)y. Terquem: Liouville, Journal de Mathenmatiques, tom. III. p. 18. 348 LINES OF THE SECOND ORDER. SECTION XII. Passing through given Points. 1. To determine the equation to a conic section which passes through the five points of which the coordinates are 1, - 1; 2, 1; -2, 3; 37 2; - 1, - 3. If a = 0, /3 = 0, 7 = o 8 = O0 be the equations to the sides of the quadrilateral formed by joining in order the first four points, the equation to the conic section will be ary = Xa8 where X = a constant. By giving to x, y, their values at the fifth point, we shall, by this equation, determine X. Now, by the formula (2 - 1) y - (Y2 - Y) X - X2Y1 + XIY2 = 0, we see that a=y-2x+3, /3=-4y-2x+ 8, y=5y+x-13, 8=-2y+3x-5. Hence the equation to the conic section will be (y - 2x + 3) (5y + x - 13) = 2 (2y + x - 4) (2y - 3x + 5). Putting x =- 1, y -3, we have - 58 = -2X x 22, 2X =. Hence the equation to the curve becomes 11 (y - 2x + 3) (5y + x - 13) = 29 (2y + x - 4) (2y - 3x + 5), or 61y - 17xy- 65x2 + 36y + 174x- 151 = 0. 2. To find the locus of the centre of a conic section which always passes through four given points. Let A, A', B, B', be the four points: join AA', BB', 0 being their point of intersection. Take OA, OB, produced indefinitely, as axes of x, y. Let OA = a, OA' = a, OB= /, OB'= = '. Then the equation to the conic section will be x' y +(a-a)+3 = cxy + +- (' - ) + aot /318 c a PASSING THOUGH GIVEN POINTS. 349 Let X, Y, be the coordinates of its centre: then 2X 1 -cY+- (a' - a) = 2Y 1 '- X + (3' -/ ) = O. Eliminating c, we have, for the locus of the centres, 2X2 2 Y12 X Y a - 3' + (a -- A) ( /3 ) - -) = o (..... (1), which is the equation to an ellipse or hyperbola accordingly as aa' and,s/' have different or like signs. COR. 1. The equations to the asymptotes of the hyperbola may be shewn to be _ _ _ _ _ _ = ~ 4 + -p S _ 1, *fa- (1- - (/ss'? (') = 4 | ( (+ )) COR. 2. It is easily seen that the centre of the curve (1) coincides with the middle point of the line joining the bisections of AA', BB'. Brianchon: Gergonne, Annales de Mathenmatiques, tom. XI. p. 219. Gergonne: Annales de Mathematiques, tom. XI. p. 396. Poncelet: Gergonne, Annales de MJJathmnatiques, tom. XII. p. 245. The following is a different solution of the same problem. Let the equations to the sides of the triangle formed by joining, two and two, three of the four given points be Ul = Xcosa, + ysina- - i = 0, u = x cos + y sina2 - 8 = 0, % = xcosa% + y sina - 83 = 0. Then the equation to the conic section will be of the form Xlu2U3 + X2uul + X3uu = 0..............(1), where X, X2 \X,, are arbitrary constants. * Bobillier: Gergonne, Annales de MathJmatiques, tom. xviIi. p. 320. 350 LINES OF THE SECOND ORDER. If we differentiate the equation (1) partially, first with regard to x and then with regard to y, we shall get X(U2 Cos a3+ 3 cos a2) +2(u COSa+ a1+u cos a) +X2(u, COs a,+u2 cOS a)= o 0 X (u2 sin a + u3 sin a2) + X2 (u3 sin al+ u1 sin a3) +X3(u1 sin al+ U2 in al)= 0, two equations which determine the values of x and y at the centre of the conic section. Eliminating X\ between these two equations and effecting obvious simplifications, we shall get 1 _ u2 sin(a, - a) + U sin(a - a2) -- u sin(a, - a,) t - ~2' 3 Sill (a- a) + u1 sin(a2 - a) - 2 sin(a2 - al) or, putting u1sin(a2 - 3) + U2 sin(a - a,) + Uz sin(a - a2) = 2v, X1 _ it1 v-u~sin(a2 - sin - ) 2 2 -t2 sin(a - a 1) By similarity we have also _ 1 1= s- in(a2 - a) \ 3 - v-u3 sin(a1 -a2)' Let mi, I2, rn, be the values of u, u, u,, at the fourth given point. Then, from (1), X + 2 + -= 0, In1 X2 X3 m, m~, ma and consequently, putting for X1 X2, X3, their proportionals, and writing ~,, A, instead of sin (a, - a), sin (a - a), sin (a, - a,), respectively, we have v (+ + 2 + + inI In2 m 3 In2 IT2 m3s as the equation to the required locus, which is therefore a conic section. This very elegant solution of the problem is given by Mr. Hearn, in his Researches on Curves of the Second Order, a work very remarkable for the ingenuity and beauty of its investigations. GIVEN POINTS AND TOUCHING GIVEN STRAIGHT LINES. 351 3. If three conic sections 'pass through the same four points, their diameters, of which the conjugates are parallel to a single fixed line, all pass through one point. Sturm: Gergonne, Annales de Mathematiquestom. XVII. p. 176. Lame: Examen des diffrentes methodes ernloyees pour resoudre les problemes de Geometrie, p. 34. SECTION XIII. Passing through given Points and touching given straight lines. 1. To determine the locus of the centres of all conic sections which touch a given straight line and pass through three given points. Let one of the three points be taken as the origin of coordinates, and let the axes pass through the other two points. Then, h, k, denoting the distances of these two points from the origin, the equation to the conic section will be ax2 + 2cxy ~+ by2 - ahx - bky = 0............ (). Let the equation to the given tangent line be x y -+ =1. Now, generally, in order that a line expressed by this equation may touch a conic section denoted by the equation ax2 + 2cxy + by2 + 2a'x + 2b'y + c' = 0, we must have the condition a'2/ (c' - ab) + 22a3 (a'c - ab') + 2a/2 (b' - ba') + a2 (a'2 - ac') + '2 (b - be') + 2a/3 (cc' - a'') = 0. In the present problem, where a' = - ah b' = - bk c'=0; this condition reduces itself to 3 (aah - bk)2 J ac3 {a3(c'2-ab) + aa (blk-hc) + b/3 (ah-kc)} = 0...(2). 352 LINES OF THE SECOND ORDER. Now if x, y, be taken to represent the coordinates of the centre of (1), we shall have 2ax + 2cy - ah = 0, and 2by + 2cx - bk = 0. From these two equations we have 2cy 2cx h- 2x' k - 2y Substituting these expressions for a and b in (2), we get {3lkx (2x - ) - ahy (2y - k)}2 + a/3 (2x - A) (2y - k).{a/3 (- 2kx - 2hy + hk) + 2ay (2kx + 2hy - hk) + 2/3 (2hy + 2x - hk)) = 0, or {/3kx (2x - A) - aiy (2y - k)12 + a3 (2w - A) (2y - k).(2kx + 2y - hk).(283x + 2ay - a8) = 0, which is the equation to the required locus. Gergonne: Annales de Mathe'matiques, tor. XI. p. 393. The following is a different solution of the same problem. The notation of the second solution of Prob. (2) of the preceding section is retained. Let the equation to the given line be put into the form 11U1 + 12u2 + 13u3 = 0..................... (1) which may always be done by properly determining the constants 1, 12, 1. At the intersection of (1) and the equation to the conic section, viz. X\U2U + X2u3ul + X\u12 = 0, we have l\X2u,2 + 12X12 + (1\1 + lX2 - l3\) u1,2 = 0; it is easily seen that this equation will be a perfect square, provided that \, X2, 3 satisfy the condition (11Xl)t + (2'X2)' + (X3)- = 0, which is therefore the condition of contact. GIVEN POINTS AND TOUCHING GIVEN STRAIGHT LINES. 353 Hence, replacing in this equation X, X2, Xs, by their proportionals, we have for the equation to the required locus, {lut (v - ( elu)} + {13u (v -,u3,)} + {13U3 (v - s3)} = 0. Hearn: Researches on Curves of the Second Order. 2. To find the locus of the centre of a conic section which touches four given straight lines. Let the equations to three of the given lines be ul = x cosal + y sinal - =, = 01 U2 = Xcosa, + y sina, - 8 = 0............(1). Ut = xcosa3 + ysina3 - 3 =0 The equation to the fourth line, c, c2, c3, being constants properly chosen, may be written in the form C Lu + C 2 + C -3 = 0................. (2). The equation to the conic section, which touches the lines (1), will be (X,1) (X )' + ( ) + (X) = o............... (3). This conic section will also touch (2), provided that the parameters Xl, X, X,, are subject to the condition + 2 + -3 = 0.....................(4). C! 2 C3 If we rationalize the equation (3), and then differentiate it, partially, first with regard to x, and secondly with regard to y, we shall obtain two equations for the determination of the values of x, y, at the centre of the conic section, viz. the equation x1Lu1 COS aI + x2U2 cOS a2 - + X 32U cos a = X2 (u2 cOS3 U C 2) + X3 co + 3 (i3 cosal + u1 cos c3) + 1X2X (U cosa2 + %2 cosa,) and an equation which may be derived from this by writing sin throughout instead of cos. From these two equations, by simple trigonometrical operations, we shall arrive at the following formulae: \2 + 2\3 = _ 3 + 3X =: \ + s\ 2 7~I 7~82 3 AA AA 354 LINES OF THE SECOND ORDER. where e, 2, S3, are written for sin(a-a3), sin(a,-al), si (al-a2), respectively. Putting each member of the double equality (5) equal to p, we shall get P 6 U)2 E\ 2 P (V - PE2 ^ ^^ v-= s (V1u1), X2 = (V 322) 3= -E- (V 38), 62e3 631 12 where v = - (u,- + ~22 + s33) Substituting these values of \, 2,1 X\3 in the equation (4), we have, for the equation to the required locus, l+ +! =3 PE2 2 23 ~1 2 C~ 2 C2 C3 Thus it appears that the locus is a straight line. Hearn: Researches on Curves of the Second Order. 3. To find the locus of the centre of a conic section which passes through two given points (u, =0, 3 = 0) and (2=0, 0 3=0), and touches a given straight line 1 + 7U2 = 0 in a given point (u, = 0, u2 = 0). The locus required is a straight line defined by the equation u1 {v - u1 sin(a2 - a,)} = ry2 {v -,2 sin(a - a)}. Hearn: Researches on Conic Sections, p. 34. 4. To find the locus of the centre of a conic section which touches three given straight lines and passes through a given point. The same notation being adopted as in problem (2), and mn1 m 2 m.3 represented the values of u1, 2 u 3, at the given point, the required locus will be a conic section defined by the equation {mI1.((V - ~11)} + {m2s2.(v - %2U2)} + {m3e3.(v - s33)} = 0. Hearn: Researches on Curves of the Second Order. EQUATIONS FROM GIVEN CONDITIONS. 355 5. To find the locus of the centre of a conic section which touches two given straight lines and passes through two given points. Let (a, /), (a', 3'), be the two given points: let the equations to the two given lines be u = alx + by + 1 = 0, v = a2x + bYy + 1 = 0, and that to the line joining the two given points w = a3x + by + 1 = 0. Also let w' represent a3x + b,'y, a,' and b,' being quantities determined by the equations W'a,, = a3 + b3 = (uzv)a, p= wa', p = a3a' + b3'/' = (UV)a,, Ap. Then, putting L = a23 - asb2, M =a,3b - ab,, L' = a2b - a3'b2, 1'= a3b- a1 and Q = a'b - a,b' we shall have, for the equation to the required locus, (Lu - M)2 + 2 Q ((L'u - M'v) w - (Lu - v) w'} = o, an equation of the second order. Hearn: Researches on Curves of the Second Order, p. 42. SECTION XIV. Determination of their Equations from given Conditions. 1. To find the equation to the conic section which passes through the point (h, k) and touches the parabola y = lx at the vertex and at an extremity of the latus-rectum. The equation to the chord joining an extremity of the latusrectum to the vertex of the parabola is y - 2x =0 AA2 356 LINES OF THE SECOND ORDER. Hence the equation to the conic section, passing through both intersections of the lines y2 - lx= 0, y - 2x = 0, must be of the form y - lx + (y - 2x) (Xx + y + v) = 0, X, /, v, being arbitrary constants. But, since the conic section touches the parabola at the vertex, = 0, when x = 0 and y = 0: hence v =0. The equation therefore becomes y - lx + (y - x) (Xx + y) = 0. Also, since the conic section touches the parabola at the end dx of the latus-rectum, we must have -= 1, when x = an, and y =: hence 2p/, + X =0: thus the equation becomes y- lx = _ j (y- 2x)2. But the conic section passes through the point Ah k: hence - Th = - f (k - 2k)2. Hence the equation to the conic section is y - lx _ y- 2" k2 _ h = - 2h ' COR. Clearing the equation of fractions we shall get y2 (4h2 - 4hk + 1h) + 4 (k2 - h) xy -4 - kh) ax... =0. Hence the curve is an ellipse or an hyperbola as (k - l_ )2 < or > (4hk - 4h2 - lh) (k - lh), or (k2- lh) (k- 2h)2 < or > 0. Thus the curve will be an ellipse, if the point (h, k) is within, and an hyperbola, if it is without the parabola. 2. To find the equation to a conic section, "having given that h, k, are the coordinates of its focus, that x cosa + ysina = 8, POLES AND POLARS. 357 is the equation to its directrix, and that e is its eccentricity. Also, if h and k be such that h = e28 cosa, k = e8 sin a to find the axes of the conic section. The equation to the conic section is (x - h)2 + (y - k)2 = e (x cosa + y sin a -). If h = e28 cosac and k = e8 sina, the semi-axes will be eS, (1 - e')e., or eS, (e2 -l)1.e8, accordingly as the curve is elliptic or hyperbolic. SECTION XV. Poles and Polars, 1. If, through a point chosen arbitrarily in the plane of a line of the second order, be drawn a series of secants to the curve; and if, through the two points of intersection of each of them with the curve, be drawn to the curve two tangents, terminated at their point of concourse: the pairs of tangents will form a series of circumscribed angles the summits of which will lie in a single straight line. Let the equation to the conic section be ax2 + by2 + 2cxy + 2a'x + 2by + c' = 0. If (x,, y,), (x,,, y,,) be the coordinates of the two points in which any one of the secants cuts the curve, and (x', y'), those of the point of concourse of the two corresponding tangents; then, by the equations to the tangents at these two points, we shall have (ax + cy, + a')' (by, + C, ')y' a'x, + by, + c' = 0, (ax,, + cy,, + a') x' + (by,, + cx,, + b') y' + a',, + b'y,, + c' = 0. These two equations shew that the two points (x,, y,) and (x,,, y,,) lie in a line (ax + cy + a')x' + (by + ex + V') y + a'x + bVy + c'= O...(1), where x, y, are the current coordinates. 358 LINES OF THE SECOND ORDER. Let A, k, be the coordinates of the point from which all tile secants are drawn: since this point lies in the same line with (X%, y,) (x,,, yj,), we have (ah + ck + ') x' + (bk +- ch + - ') y' + a'h + 'k + c' = 0...(2). This equation shews that the point (x', y') always lies in a single straight line. COR. By virtue of the equation (2), it appears that (h, k) is always a point in the line (1). Hence we may enunciate also the following converse proposition: " If about a line of the second order be circumscribed a series of angles the summits of which all lie in a single straight line situated anywhere in the plane of the curve: the secants drawn to the curve, through the points of contact of the sides of these angles with it, all meet in a single point." In consequence of the relation which subsists between the point (h, k), and the line (ah + ck + a') x' + (bk + c + b') y' + a'h + b'k + c' = 0, the point has been called the pole of the line, and the line has been called the polar of the point. De la Hire: Sectiones Conicce, lib. II. prop. 23, 24, 26, 27. Puissant: Recueil de diverses propositions de Geometrie, p. 167, troisieme edition. Gergonne: Annales de Mathematiques, tom. III. p. 293. Garnier: Geometrie Analytique, p. 161. 2. To prove that the polar is always parallel to a system of chords conjugate to the diameter passing through the pole. The equation to a conic section which has a centre may be put under the form ax + by" + c' = 0, Then, (h, k) being the pole, the equation to the polar will be, by prob. (1), ahx' + bky' + c' = 0................(1). But the equation to any chord, conjugate to a diameter through (h, 7)), is, 2p being any constant, ahx'+ b- y' +2 = 0.................... (2) POLES AND POLARS, 359 the equations (1) and (2) represent two parallel lines: the truth of the proposition is therefore established for central conic sections. Again, since the equation to any conic section without a centre may be taken under the form by2 + 2a'x = 0, we shall have, instead of (1) and (2), the equations a'x' + + +a' = 0................... (3), aIx' + bky' +p = 0.................... (4). From these equations we see the truth of the proposition in regard to conic sections without centres. De la Hire: Sectiones Conicce, lib. II. prop. 25, 28. Rochat: Gergonne, Annales de Mathmmatiques, tom. III. p. 303. 3. If two points P1, P2, be the poles of two lines L1, L,2 to prove that the intersection of L1, L2, will be the pole of the line joining P, P,. Let (h,, kl) be the coordinates of P1, and (h2, k2) those of P,. Then, since P1 is the pole of L1, the equation to L, must be (ah, + ck + a') x + (b/k + ch + b') y + a'hl + b'Vk + c' = 0, or (ax + cy + a') h + (by + cx + b') + c'= 0.....(1). Similarly the equation to L2 must be (ax + cy + a') h2 + (by + cb + b') k, + c' = 0.... (2). Taking now x, y, to represent the coordinates of the intersection of (1) and (2), it is plain that, x', y', representing current coordinates, the two points (h,, kc), (h2, k2), lie in a line (ax + cy + a') x' + (by + cx + ') y' + c' = 0, which is therefore the equation to the line PP2. The form of the equation shews that P1P, is the polar of the point (x, y), that is, of the intersection of Li, L,. Par M. B.***, Abonne: Gerqonne, Annales de Mathematifues, tom. IV. p. 379. 360 LINES OF THE SECOND ORDER. 4. Two points P, Q, are the poles, corresponding to two polars L, MI respectively, in any conic section. To prove that, if P lie in M, Q will lie in L. Any conic section may be represented by the equation xa y2 2x 2y 2 4 J - + 4 ac 2 la p Let (x, Yr), (x2, y2), be respectively the poles P, Q. Then the equations to L, MI will be, respectively, xx, Yy x + x1 Y + Y1 - - =.+ -...............( ), a 2.^2 2. ( a2+ _ + Yl = Y....... x ~*Y. (2). If (x, y,), lie in (2), we have xx yx yx. x1+x2. y~+y2 X. Y1YJ _ f + X2 + 1 + Y+ a2 *2 - a " X a+ b a /l a result which shews that (x2 y2) is a point in (1). 5. To find the equation to the polar corresponding to any proposed pole, relatively to the conic section UV- = w2 U,, w, being linear functions of x and y. Let the point of contact of one of the tangents be (u = w } V = XW then the equation to this tangent will be X2u - 2Xw + v = 0. Let ', v', w' be the values of u, v, w, respectively, at the pole. Then 2u' - 2Xw' + v' 0. But, at the point of contact, XU = WI v - Xw, X2u = v; and therefore u'v - 2ww + v'u = 0. POLES AND POLARS. 361 But this equation will be true also for the contact of the other tangent. Hence, holding good for both points of contact, it will represent, when u, v, w, are regarded as variable, the equation to the chord of contact, or polar belonging to the proposed pole. Salmon: Treatise on Conic Sections, p. 221. 6. If, from any point 0, be drawn two straight lines OAA', OBB', intersecting a conic section in points A, A', B, B'; to prove that the intersection of AB', A'B, and the intersection of AB, A'B' both lie in the polar belonging to 0 as a pole. Let the lines OAA', OBB', be chosen as axes of x, y, respectively. Let OA = a, OA' =a' OB=/3, OB'='. Then, c being some constant quantity, the equation to the conic section will be y /1 1 1\ + cxy + -x + - +a'-y + ) + 13 =0. The equation to the polar of 0 will be (a i )x(+ )- (,+2...............o). But the equations to AB, A'B', AB', A'B, are respectively -+ =, -,+ =, - = 1, - a. a4- a Of c 1' a + J L Since (1) is produced by the addition of either the first and second, or of the third and fourth of these equations, the proposition is established. COR. 1. Since the intersection of the tangents at A, A', and intersection of the tangents at B, B', both lie in the polar of 0, it appears that the intersections of (AB', A'B), (AB, A'B'), and of the two pairs of tangents, lie all four in a single straight line. COR. 2. If we join two and two the reciprocal extremities of pairs of parallel chords of a conic section, the points of intersection of these new chords will all lie on the diameter bisecting the original chords. 362 LINES OF THE SECOND ORDER. A demonstration of this proposition in the particular case of a circle may be seen in De la Hire's Sectiones Conicce, lib. I. prop. 22. Lame: Examen des differentes methodes employdes pour resoudre les problemes de Geometrie, p. 45. Puissant: Recueil de diverses propositions de Geometrie, p. 221, troisieme edition. Frost: Cambridge Mathematical Journal, vol. IV. p. 114. 7. To prove that each of the straight lines u=0, v= 0, w=0, is the polar of the point of intersection of the other two, relatively to the conic section I2u' + m2V2 = n2w2, i, m, n, being any constants. Salmon: Treatise on Conic Sections, p. 205. 8. To find the polar relatively to a conic section uv = W2 where u, v, w, are linear functions of x and y, corresponding to a pole represented by the equations {au = (1v = wy})\ The equation to the required polar is au - 2abw + bv = 0. Salmon: Treatise on Conic Sections, p. 220. 9. The equation to a conic section being 8 = 2u' - +2v2 + n2w2 - 2mnvw - 2nlwu - 21muv = 0, to prove that the equation to the polar of a point where u = u' V = V, W = W' is, dS,dS dS dSalm dv e C dw. 2 Salmon: Treatise on Conic Sections, p. 234. POLES AND POLARS. 363 10. The three sides of a triangle touch a given conic section: two of its vertices move on fixed right lines: to find the locus of the third vertex. Let u = O, v = O, be the equations to the two tangents which pass through the intersection of the two fixed right lines. Then, the equation to the conic section will be of the form uv = W2 those to the two right lines of the form au = v, bhu = v, and the equation to the locus of the third vertex will be 4ab 2 (a + b)2' The required locus is therefore a conic section touching the proposed conic section at the ends of the polar belonging to the intersection of the two fixed lines as pole. Salmon: Treatise on Conic Sections, p. 221. 11. If a point P be the pole and a line Q the corresponding polar in relation to any proposed conic section, to prove that, if P always lie in a second conic section, Q will be always a tangent to a third, and conversely, if Q be always a tangent to a second conic section, P will always lie in a third. Brianchon: Journal de 7 1Ecole Polytechnique, Cahier x. p. 14. O'Brien: Plane Coordinate Geometry, p. 177. 12. From a point P, without a conic section, any two straight lines are drawn each cutting the curve in two points. The points of intersection are joined, two and two, and the points, in which the joining lines (produced if necessary) cross each other, are joined by a line which will in general cut the curve in two points A, B. To prove that PA, PB, are tangents at A, B. R. S.: Cambridge Mathematical Journal, vol. I. p. 32. 13. From any point 0, a straight line OAC is drawn, cutting a conic section in two points A, C, and the polar of 0 364 LINES OF THE SECOND ORDER. in B: to prove that the line 0 C is harmonically divided by the points A, B. De la Hire: Sectiones Conicce, lib. II. prop. 21. 14. To prove that the polars of any given point, relative to all lines of the second order which pass through any four assigned points, all meet in a single point. Bobillier: Gergonne, Annales de Mathematiques, tom. XVIII. p. 362. 15. From a pole P is drawn a straight line PS to the focus of a conic section, cutting the polar in I: the polar intersects the curve in E, F, and the directrix in G. To prove that EI: Fl:: E G: FG. Leybourn: Mathematical Repository, New Series, vol. II. p. 3. 16. From a point is drawn a pair of tangents to a conic section, and a line, bisecting the interior angle between them, is drawn, cutting the corresponding polar in 0: to prove that the intersection of a pair of tangents, belonging to any other polar passing through 0, will lie in the line which bisects the exterior angle between the former pair of tangents. SECTION XVI. Polar Equations. 1. To prove that, in any conic section, if r, r', be focal distances at right angles to each other, and c the semi-latus-rectum, /I. 1\2 fl 1\2 ( - c) + (1- 1) = a constant quantity. \r c) \r Cj The equation to the conic section, 0, r, being the polar coordinates, is c - = 1 + e cos0: r putting I-r + 0, r', for 0, r, respectively, we have - = 1 - esin: r POLAR EQUATIONS. 365 / 2 C \ '2 hence ( — 1) + (- -1) = e2 (r ) +r ) ' /I \2 /1 e2 + - \r c/ \r c) ' 2. A line is drawn from the focus to any point of a conic section, and a circle is described upon it as a diameter: to shew that the locus of the consecutive intersections of all such circles is a circle, except, in a certain case, where it is a right line. Let 0 be the focus, P the place of any point in the circle of which OA is a diameter. Let Ox, p coinciding with a principal diameter of the conic section, be taken as the prime radius vector. Let OP= p, OA = r, L AOx =0 / / L POx = c. Then the equation to the circle is p = rcos ( - 0): but the polar equation to the conic section is C r — 1 + ecos ' where c is some constant: hence the equation to the circle becomes p (1 + ecos O) = ccos ( - 0)............... (1) Differentiating (1) with regard to 0, we have ep sin0 = csin((0 - b), or 0 = (c cos - ep) sin - c sin b cos: but, from (1), p = (ccos0 - ep) cosO + csin4b sinO. Squaring and adding to the last two equations, we get p = (c cos b - ep)2 + c2 sin2 = c2 - 2cep cosqb + e2p2, which is the equation to a circle unless e = 1, in which case 2p cos b = c, which is the equation to a right line. 366 LINES OF THE SECOND ORDER. 3. From any point M in the axis of a conic section, a perpendicular MPP' is drawn, cutting the curve in P: to find the locus of P' under the condition that PP' shall be always equal to the focal distance of P. If the equation to the conic section is r = l+e cos the required locus will be a conic section of the same species represented by the equation 2c sin r 1 + esin20O Lardner: Algebraic Geometry, p. 147. 4. From the focus S of a conic section are drawn n straight lines ri r2. r%...rn terminating in the curve, and dividing it into n parts which subtend equal angles at the focus: RB, R2, R,...Rz are straight lines drawn from the other focus to the same points. To prove that, e denoting the eccentricity of the conic section, and 0 the inclination of ra to the greatest radius vector which can be drawn through S, R1.R12.R...R, _ 1 - 2e cos n + e21 rl.r2.r3...n (1 - e) Herschel: Leybourn's Mathematical Repository, New Series, p. 58. SECTION XVII. Linear Equation. 1. To shew that r =fx + gy + A is the most general equation to a conic section referred to the focus as origin, r being the distance of the point (x, y) from the origin: also to determine the eccentricity and the position of the directrix in terms of f,.A h. LINEAR EQUATION. 367 Let the equation to the directrix be x cos a + ysing = a, and let e be the eccentricity: then, by the definition of a conic section, r = e (x cos a + y sin a - ), an equation of the required form. Comparing this equation with the one enunciated, we have f=ecosa, g=esina, h=-e8: whence e = (f2 + g2), = - (f2 + g92) The equation to the directrix becomes fx + gy + h =0. 2. The focus and directrix of a conic section are given in position. Through the former a line is drawn making with the latter an angle the sine of which is equal to the eccentricity of the conic section. To find the locus of the point where this line meets the curve, the eccentricity being variable. Let OLy be the directrix, S the focus: OSx at right angles to OLy, the axis of x, Oy being y that of y. Let OS = c, OM= x, PM= y, SP= r, e=sin a. Then r = e.PL = ex.......(1), c - x = rsina = er.....(2)..L From (1) and (2),/ r2 = x (c - ), (c - X)2 + y'2= X (c - X), 2x + y2 + c = 3cX the equation to the required locus, which is therefore an ellipse. 3. Having given one of the foci of a conic section and three points in the curve, to find the equation to the directrix. Let the focus be taken as origin of coordinates: then, (x, y,), (x,,, y,), (x,, y,), being the three given points, the equation 368 LINES OF THE SECOND ORDER. to the directrix will be, (y,- y,,) + r,(y,,- y, ) +,,(y,-,,)} + Xy,,r~ +,,y,, + y?.r,, =y {r,(,,-x,,,) +r,,(x,,,-,) r,,,(-x )} +y,,x,,,,+y,,,e,r,, +y,,z,,,, where r, r,,, r,,, are used to represent respectively ( 12+ Y,2), (X,2 + y,,) ) (c,,, + y,, 2). SECTION XVIII. Polar Equation to the Tangent. The polar equation to a conic section, when the focus is the pole, may be written in the form -= 1 + ecosO, r 0 being the inclination of r to the axis containing the focus, and c denoting the latus-rectum. The polar equation to the tangent at any point of the curve, of which a is the angular coordinate, will then be r -= e cos 6 + cos(O - a). This expression for the polar equation to the tangent of a conic section is due to Mr. Davies, by whom it was communicated to the Philosop2ical Mcagazine for 1842, p. 192. 1. If S be the focus of a conic section, and T the intersection of the tangents at the extremities P, p, of any focal chord PSp; to prove that the line ST, joining S and T, is at right angles to PSp. Let a, or + a, be the angular coordinates of P, p, respectively: then the equations to PT, p T, will be respectively -= e cosO + cos(O - a), r - = ecosO - cos(0 - a). r POLAR EQUATION TO THE TANGENT. 369 Hence, at the intersection T of these two lines, cos( - a) = 0, 0-a=+~tq2 77 0 = a ~ r, which establishes the proposition. De la Hire: Sectiores Conicce, lib. viiI. prop. 23. 2. To find the angle subtended at the focus by the tangent drawn to a central conic section from any assigned point. The polar equation to the conic section being C -= l+ ecos0, r that to the tangent at a point, of which the angular coordinate is a, will be c - = e cos 0 + cos (a - 0). r But, x being the abscissa, on the major axis, of any point r, 0, in the tangent, (x - ae) = r cos 0, ~ ~ = + (x - ae) - + cos ( - 0), r T and c = + a (1 - e2) the + or - sign being taken as the curve is the ellipse or the hyperbola: hence a - ex + -- = cos (a - 0)6 which determines the required angle a - 0 in terms of x and r. O'Brien: Plane Coordinate Geometry, p. 156. 3. The angle subtended, at one of the foci of a conic section, by the part of a moveable tangent intercepted between two fixed tangents is always constant. The polar equation to the conic section, when the focus is the pole, being of the form - = 1 + ecos0, rB BB 370 LINES OF THE SECOND ORDER. the polar equations to the moveable and fixed tangents will be - = eo cos(O - c )..................(1), - = e cocos os ( - a')................. (2), - = ecos + cos(0 - a")............... (3), r, a', ca", being the values of 0 at the three points of contact respectively. At the intersection of (1) and (2), cos(O - a) = cos(0 - a'), whence 0 = I(a' + a). Similarly, at the intersection of (1) and (3), 0 = (C-+ a). The angle, subtended by the intercepted portion of the tangent (1) at the focus, will be equal to the difference between these two values of 0, that is, to (a'. a"), a constant quantity. Poncelet: Gergonne, Annales de Mathe'matiques, tom. VIII. p. 4. 4. Two conic sections have a common focus, through which any radius vector is drawn, meeting the curves in P, P', respectively. To find the locus of the point of intersection of the tangents at P, P'. The equations to the conic sections are of the forms C - = 1 + ecos, - = 1 + e'cos(0 + s), and the equations to the tangents at P, P', the angular coordinate of both points being a, will be - ecos0 + cos( - a)....................(1), = ( + ) cos(-a).. (2) -=e'cos(0 + a) + cos(0 - a)........... (2). 7 POLAR EQUATION TO THE TANGENT. 371 At the intersection of (1) and (2), --- = ecosO - e'co s(O + e)...............(3), the equation to the required locus, which is therefore a straight line. COR. 1. The equations to the directors of the two conic sections are - = e cos, r and - = e cos (O + e). r Hence the line (3) passes through the intersection of the two directors. COR. 2. If 1, 1', denote the sines of the angles which (3) makes with the two directors, it may easily be shewn that l_ e' I e' I 5. To circumscribe a triangle about a given conic section, such that each of its summits shall lie in a given line. Let the equations to the three given lines be 81 = r cos ( - e})1,= rcos(-................ ). 83= rcos( (- )J The equations to the three sides of the circumscribing triangle may be written in the form - = ecoso + cos(0 - a)] r - = e cos0 + cos(0 - aX) Conceive the first and second of the lines (2) to intersect in the last of the lines (1). BB2 372 LINES OF THE SECOND ORDER. At the intersection of the first and second of the lines (2), we have =(a +a) C a, + a2 a, - - a2 - = ecos + cosa1 r 2 2 and therefore, by the last of the equations (1), 8 (ecos ~2+ os 2 S- 2 os = cco( - 2 whence, putting tanl-a c = u and tani- a = u2 we have 8 {1 + e + (1 - e) uuJ} = c cosSa - cu1u2 coses + c sins3 (uI + 2), ccoss3 + (1 - e) 8, (1 + e)83 ~3 reUs -- - %U - U2 q- cotas - - c sins + c 3 C si or, as, b3 denoting known quantities, a3U1u2 = u, + u2 + b3. Similarly, in relation to the other two angular points of the triangle, we have au2u3 =u z2 + 3 U+ ^, a2u3u1 = u3 + u1 + b2. Eliminating u2 and %3 between these three equations we shall get a quadratic in u1, and thus u,, u,2 u,, may be determined. The double value of u1 shews that generally there will be two solutions of the problem. A solution of this problem, in the particular case of the circle, may be seen in Mr. Hearn's excellent work entitled Researches on Curves of the Second Order, p. 17. 6. To find the distance of the pole from a tangent to a conic section c r 1 + ecos ' drawn at a point where = a. If p denote the required distance, 1 2 e2c2e - 1 + e2 + 2ecosa' POLAR EQUATION TO THE TANGENT. 373 7. The tangents at the extremities of the major axis of a conic section intersect any other tangent in two points: to prove that the straight lines joining these two points with either focus, are at right angles to each other. 'AIIOAAQNIOT IIEPrAIOT KWvLKwv Td Tptprov' IIpod'ao-is ua'. 8. If, from a focus of a conic section, lines be drawn to the summit of the angle formed by any two tangents and to the two points of contact, the first of these three lines will bisect the angle formed by the two others. De la Hire: Sectiones Conicce, lib. vII. prop. 24. Poncelet: Gergonne, Annales de Mathematiques, tom. VIII. p. 5. De Morgan: Cambridge Mathematical Journal, vol. II. p. 202. 9. When a moveable tangent to a conic section terminates in two fixed tangents, the sum of the angles which the first tangent subtends at the two foci, in the ellipse, and the difference, in the hyperbola, is constant and equal to the supplement of the angle between the two fixed tangents. Poncelet: Gerqonne, Annales de Mathematiques, tom. VIII. p. 8. 10. To find the locus of the intersection of a tangent to a conic section with a straight line drawn through the focus, at right angles to the radius vector of the point of contact. The locus will be the directrix. Chasles: Gergonne, Annales de Mathermatiques, tom. XVIII. p. 273. 11. To prove that two confocal conic sections cannot have more than two common tangents. Chasles: Gerqonne, Annales de Mathematiques, tom. XVIII. p. 273. 12. If a variable angle, circumscribed about a conic section, moves in such a manner as to intercept between its sides, a portion of any fixed tangent subtending a constant angle at the focus, to find the locus of the summit of the variable angle. 374 LINES OF THE SECOND ORDER. The equation to the conic section being - = + ecos0, r9 the required locus will be a confocal conic section, having the same directrix as the original one, and defined by the equation - = coss + ecoso r s being the constant angle. Bobillier: Gergonne, Annales de Mathdmatiques, tom. XVIII. p. 190. 13. The angle between the focal distances of two points of a conic section being given, to find the locus of the intersection of the tangents at these points. Let a denote the given angle; then, the equation to the conic section being - = l + ecos, the required locus will be a conic section defined by the equation C - = cosa + ecosO. r 14. The focal distances of two points of a conic section include a given angle: to find the focal distance of the intersection of the tangents at these points. If 2a denote the given angle between the focal distances r, r', and the equation to the conic section be - = 1 + ecos0, r the square of the required distance is equal to 2rr' c - (1 - e2) rr'.sin' a POLAR EQUATION TO THE CHORD OF A CONIC SECTION. 375 SECTION XIX. Polar Equation to the Chord of a Conic Section. If the equation to the conic section be c = 1 + ecoso0 r the equation to the chord through points, the angular coordinates of which are a + /8 and a - 3, will be - = sec3 cos(0 - a) + e cos0. qr This very useful form of the equation to the chord of a conic section was inserted by Mr. Frost in the Cambridge and Dublin Mathematical Journal, vol. I. p. 68. 1. PSP', QSQ', are two focal chords of a conic section: to prove that a line through S, bisecting the angle PSQ', will intersect the chord QP, produced indefinitely, in the directrix. The equation to the conic section being - = 1 + ecos 0 that to the chord QP will be, a - /, a + 3, being the angular coordinates of P, Q, respectively, c = sec,/.cos(0 - a) + ecos0...............( But the angular coordinate of the line bisecting the angle PSQ' is equal to a +,/ a-/3 7 2 2 2 7T = a- and therefore, at its point of intersection with (1), - = e cos 0, r a result which shews that the intersection lies in the directrix. 376 LINES OF THE SECOND ORDER. COR. If we have given any three points in a conic section and the focus, we can easily construct the directrix, two of its points being readily determinable by the aid of this proposition. Chasles: Geryonne, Annales de Mathematiques, tor. XVIII. p. 275. 2. If a variable angle, circumscribed about a conic section, moves in such a manner as to intercept, between its sides, a portion of any fixed tangent subtending a constant angle at the focus, to find the equation to the chord joining the points of contact of the two sides of the variable angle. Let the equation to the fixed tangent be C - = ecos0 + cos(0 - a)..................(l1), r and those to the sides of the variable angle, = ecos0 + cos ( - X).................(2), r - = ecos0 + cos(0 - u).................. (3). At the intersection of (1) and (2), 20 = a + X, and, at the intersection of (1) and (3), 20 = a + /: hence, s denoting the constant difference in the values of 0 at the two intersections, 2s = X -.......................... (4). The equation to the chord through the points of contact of (2) and (3), the angular coordinates of which are X, /, will be C X - ___ - = see -.cos (0- A) + eos,0 r 2 2 and therefore, by (4), we have for the required equation to the chord, ccosS / + A, c = e coss+ cos + cos ( - ). r 2 COR. This result shews that the chord of contact is always a tangent to a confocal conic section which has the same directrix as the original one. Bobillier: Gerc/onne, Annales de Mathenmatiques, tom. XVIII. p. 190. POLAR EQUATION TO THE CHORD OF A CONIC SECTION. 377 3. Two focal chords of a conic section are drawn and lines joining their extremities are produced to meet in two points P, Q: to prove that, S being the focus, SP is at right angles to SQ. Let ESF, E'SF', be the two focal chords, the angular coordinates of E, E', being a - /, a + 8, and, consequently, those of F,, ' 7 + a -,, r + a +3. The equation to FE' is - = sec/.cos(0 - a) + ecos0, and to FF', 7r + a being substituted for a, - =-sec/ cos (0 - a) + e cos 0. Hence, at P, the intersection of E', FF', cos( - a) = 0........................(1) and therefore 0 = a + (2X + 1) ~ r7 X being any integer, that is, the angular coordinate of P is equal to half the sum of the angular coordinates of E, E', or F, F', together with an odd multiple of I 7r. By similarity of circumstances therefore, 0, denoting the angular coordinate of Q, 0, = - T + a + (2/i + 1) I v = a + (I + 1) 7r, /6 being any integer. Hence, -0 = (/ - X) wr + -~%, which shews that SP is at right angles to SQ. 4. A chord PQ of a conic section passes always through a given point C: supposing PS, QS& CS, to be joined, S being a focus of the curve, to prove that p G QSC tan 2.tan 2 = a constant quantity. 2. 2 The angular coordinates of P, Q, being a - 3, a + /3, the equation to PQ will be of the form -= sec/9.cos(0 - a) + ecos0. 378 LINES OF THE SECOND ORDER. Let r,, 0,, be the coordinates of C: then - = sec/.cos (, - a) + ecos0,.............(1). PSOC t QS- a +.n a +/ - 0, Again, tan 2 tan tan.tan cos(0, - a) - cosS cos(0, - a) + cos3 * But, from (1), cos(0,- a)= cosf.(- - ecosO,). sc~ QS, C - -e cos 8 -1 Hence tan-.tan = 2 c - - ecosO, + 1 which is a constant quantity. Stubbs: Annals of Philosophy, for November, 1843, p. 341. 5. To inscribe a triangle in a given conic section, so that its sides may pass respectively through three given points. Let the equation to the conic section be - = 1 + ecos0: r and let a,1 a2, as, be the angular coordinates of the summits of the inscribed triangle. The equation to the chord joining the first two of these summits is r- ) 2 o( 2 -ecos0) cos 1- a-( 0...) Let rs, 03, be the coordinates of the given point through which this chord passes: then, expanding the cosines and putting tan ~a = u, tanla2 = u, we have (2 w h ( - cos0).(1 + u12) = cos0.(l - u2) + sin 0.(U1 + u) or, a3 7b~, representing known quantities, aU U 1, = U + Ma + b3................. ** (1). POLAR EQUATION TO THE CHORD OF A CONIC SECTION. 379 Similarly, by considering the two other chords we have al2 = u2 + u3 + b6.....................(2), a2u,2 = u + u, + b2..................(3).* From (1) and (3) we see that u + b3 u + b2 m ~3 - __3 ______ a3u -1 3 a2u - and therefore, from (2), a, (u + b2) (U, + 63) = (a,2, - 1) (zu + 6^) + (au - 1) (ul + b2) + bl (a2, - 1) (a3u - 1), and consequently (a,- a2- a- a2a3bJ) u + {2 + a, (b + b3) + lb(a2+ a)- a23- a3b2}l - b, + 62 + b8 + alb62b = 0. From this equation may be determined two values for u, the corresponding values of u2 and u3 being then given by (2) and (3): thus a, a, a3 are ascertained. The double value of u1 shews that there are two triangles which will satisfy the conditions of the problem. The earliest solutions of this problem, in the particular case of the circle, are due to De Castillon and Lagrange, the former being geometrical and the latter trigonometrical. The following is Castillon's history of the problem. "Je vais parler d'un Probleme que je n'ai pas imagind, et que pourtant, dans sa generalite, je n'ai vu propose nulle part. Je dis dans sa gendralte, parce que Pappus dans la Propos. 117, Probl. 40 de sa Collection, en propose un des cas les plus faciles, et en donne une solution qui ne peut pas s'appliquer aux autres cas. Selon mes conjectures, quelque Amateur de la Geometrie des Anciens generalisa le Probleme de Pappus, et, ne l'ayant resolu qu' avec beaucoup de peine, ou, peut-8tre, ne 'ayant point r6solu, il le proposa verbalement a quelqu' un qui couroit la meme carriere. Celui-ci suivit 1'exemple de son Predecesseur; et de main en main le Probleme est parvenu * The forimule (1), (2), (3), coincide in form with those obtained by Lagrange in the particular case of a circle. 380 LINES OF THE SECOND ORDER. jusqu' a moi. II semble que le petit nombre de Gdometres qui le connoissoient, le gardoient pour embarrasser les autres dans les occasions. Je n'en sais de certain que ce qui me regarde: en voici 1'histoire. " Feu Mr. Cramer, celbbre Professeur a Geneve, et digne Membre externe de cet illustre Corps, m'honoroit de son amitie. Ii s'appercut que j'aimois beaucoup la methode Geometrique des Anciens. I1 en prit occasion en 1742 de me proposer le Problgme en question, qu' il falloit resoudre par l'ancienne Analyse. 'Dans ma jeunesse, me dit-il, j'avois le goAt que vous avez: un vieux Geometre, pour essayer mes forces en ce genre, me proposa le Probleme que je vous propose: tentez de le resoudre; et vous verrez combien il est difficile.' " Je 1'entrepris: je trouvai quelques Theorames, qui, a ce que je croyois, m'approchoient du but, mais qui ne purent m'y faire atteindre. Dans la suite de ce Memoire on en trouvera quelques-uns qui viennent a propos. Peu de temps apres m'avoir parle de ce Probleme, Mr. Cramer par ses exhortations m'engagea a donner les Opuscules de Newton; et les soins que cette edition dernandoit, me firent tellement oublier le Probleme dont il s'agit, que je n'y songeai plus. "On m'en fit souvenir en 1755. Un des amis que j'avois a la Haye, etoit Mr. Bouquet, a qui je rendis justice dans le Memoire sur le terme gene'ral des series recurrentes, que je pris la liberte de soumettre au jugement de cette Compagnie, lorsqu' elle me fit l'honneur de me mettre au nombre de ses Membres externes. Depuis j'ai eu le plaisir de voir la bonne opinion que j'avois de mon ami en qualite d'homme de lettres, confirmee par ' illustre Franklin, et le cas que j'en fesois comme militaire, justifie par les eloges de toute 1' Angleterre. " [Mr. Bouquet m' crivit done qu' un Anonyme avoit propos4 a la Haye le Probleme dont il s'agit, comme digne de l' attention des Geomdetres. Mr. Bouquet ajoata qu' il ne doutoit pas de mon courage a l'attaquer, et de mon bonheur a m'en rendre maitre. Les leqons que je devois donner a Utrecht, me laissoient tres-peu de temps. Pendant quelques semaines, je l'employai a cette recherche; m ais sans fruit. L'inutilite de mres POLAR EQUATION TO THE CHORD OF A CONIC SECTION. 381 efforts me piqua sans me d6courager. Comme personne ne declaroit publiquement avoir resolu ce Probleme, je continuai a y travailler, mais seulement quand j'etois de loisir. Je fis bien des tentatives inutiles, je l'avoue. Enfin il me vint dans l'esprit qu' un Theoreme de Pappus s'appliquoit naturellement a la figure que j'avois construite. Je tentai cette application; et j'apercus d'abord qu'elle me donnoit la solution desir6e. C'est ainsi que j'y parvins; mais si tard qu' on ne parloit plus ni de l'Anonyme, ni de son Problame. Je mis donc ma solution parmi mes papiers, oh je viens de la retrouver. J'ai cru que je ferois bien de la publier, pour 6pargner aux Geometres a venir la peine et la perte du temps que ce Problame pourroit leur cofter." De Castillon: IlMmoires de 1'Academie des Sciences et BellesLettres, Berlin, 1776, p. 265, (geometrically for the circle). Lagrange: Ibid., p. 284, (trigonometrically for the circle). Euler: Acta Academice Scientiarum Petropolitance, 1780, pars prior, p. 91, (for the circle). Fuss: Ibid. pars prior, p. 97, (for the circle). Lexell: Ibid., pars posterior, p. 70, (construction of Lagrange's formula for the circle). Oltajano*: Memorie della Societa Italiana, tom. IV. p. 4, 1788, (synthetically for any polygon inscribed in a circle). Malfatti: Ibid., tom. IV. p. 201, (synthetically for any polygon inscribed in a circle). Lhuilier: Elemens d'Analyse Geometrique et d'Analyse Algebrique, p. 277, &c., (geometrically, for a triangle, or any polygon inscribed in a circle). Gergonne: Annales de Mathrmatiques, 1816, 1817, tom. VII. p. 325, (algebraically for the parabola). Carnot: Geometrie de position, p. 384, (algebraically, for the circle, by a method applicable to an inscribed polygon of any number of sides). * Oltajano's solution was presented to the Societa Italiana when he was about sixteen years of age. 382 LINES OF THE SECOND ORDER. LamB: Examen de differentes methodes mploye'es pour resoudre les probl6mes de GJometrie, p. 58, (algebraically, for any conic section). Puissant: Recueil de diverses propositions de Geometrie, p. 125, (algebraically, for the circle, by a method applicable to any conic section). Hearn: Researches on Curves of the Second Order, (algebraically, for the circle). Gaskin: Solutions of Geometrical Problems, p. 166, (algebraically, for each of the conic sections). Salmon: Treatise on Conic Sections, p. 222. 6. If the chord of a conic section, the eccentricity of which is e, subtends at its focus a constant angle 2a, to prove that it always touches a conic section, having the same focus, and of which the eccentricity is ecosa. Booth: Cambridge Mathematical Journal, vol. III. p. 87. Ellis: Ibid., p. 94. Frost: Cambridge and Dublin Mathematical Journal, vol. I. p. 69. 7. Chords are drawn in a conic section, so as to subtend a constant angle at the focus: to find the locus of the intersection of each such chord with the perpendicular upon it drawn from the focus. The equation to the conic section being -= 1 + ecos0O r that to the required locus will be c.(c - 2er cos ) = (1 - e cos2/,).r', which is the equation to a circle, unless ecos3 = 1, when it becomes that of a straight line. 8. If the angle between two focal distances be bisected by a third, which remains fixed in position, to prove that the chords joining the extremities of the two focal distances, as they change their position, always pass through a fixed point in the directrix. Frost: Cambridge and Dublin Mathematical Journal, vol. I. p. 69. INSCRIBED POLYGONS. 383 9. The angular coordinates of the extremities of two chords in a conic section being a + /, a - 3, and a' + /3', a' - ', to find the locus of their intersection, having given that f3' = f, and a' -a =, where e is a constant. The required locus will be a conic section defined by the equation c E - S sec/3.cos - + e cos 0. r 2 Frost: Cambridge and Dublin Mathematical Journal, vol. I. p. 70. 10. A chord being inscribed arbitrarily in a conic section, if, from one of its foci, be drawn radii vectores (1) and (2) to the two extremities of this chord, a radius vector (3) to the point where this chord cuts the directrix belonging to this focus, a radius vector (4) to the summit of the circumscribed angle which touches the curve at the extremities of the inscribed chord, and two other radii vectores (5) and (6) to the points where the sides of this circumscribed angle cut this same directrix; then the radius vector (4) will bisect the angle between the two radii vectores (1) and (2); the radius vector (3) will bisect the angle between the two radii vectores (5) and (6), and the two radii vectores (3) and (4) will be at right angles to each other. Bobillier: Gergonne Annales de Mathcmatiques, tom. xvIII. p. 190. SECTION XX. Inscribed Polygons. 1. If from any point in a conic section perpendiculars be drawn to the sides of a quadrilateral inscribed in it, to prove that the product of the perpendiculars on one pair of opposite sides is to the product of the perpendiculars on the other pair of sides in a constant ratio. 384 LINES OF THE SECOND ORDER. Let the equations to the four sides of the quadrilateral taken in order be =, = u3 0, the general type of the equation to a straight line being taken in the form x cosa + y sina - 3 = 0. Then the equation to the conic section will be of the form Xu\u. + /u2u4 = 0,* X and pL being constants. But u.u3 is equal to the product of the perpendiculars from any point (x, y) in the curve upon the sides ul = 0, u3 = 0: and u2u to that of the perpendiculars on the other two sides. Hence the ratio of the two products is equal to a constant quantity -. Sturm: Gergonne, Annales de Mathe'matiques, tom. xvII. p. 179. 2. If three sides of a quadrilateral, inscribed in a given conic section, pass always through three given points of a given straight line, to prove that the fourth side also will always pass through a given point of the same line. Let the given straight line be taken as the axis of y, and let the axis of x be a diameter conjugate to chords parallel to the given line. Then the equation to the conic section will be 2 = I + mx + nxc.....................(1). Let the equations to the four sides, taken in order, be j = ax + b, y = a'x + ', y = ax + /, y = a'x + i'. Then, p being a constant multiplier, the equation to the conic section must also be (y - ax - b) (y - ax - 3) = p (y - a'x - V') (y - a'x - ')...(2). * This expression for the equation to a conic section circumscribing a quadrilateral was given by Bobillier in the Annales de MathBmatiques of Gergonne. INSCRIBED POLYGONS. 385 Since (1) and (2) must be coincident, we have, by comparing the constant terms and the coefficients of y, pb'/s - b/3 = (1 - p) I, and b + f3 = p (b' + /'); whence, eliminating p, we have, i (b + - - ' - ') = 6b3 (b' + 13') - ' (b + /)....(3). Now, by the conditions of the problem, three of the quantities b, b', /3, ', are known; hence, by (3), the fourth is also determined, and the theorem is established. O'Brien: Plane Coordinate Geometry, p. 178. 3. If two consecutive sides of a hexagon, inscribed in a conic section, are respectively parallel to the sides opposite to them, the other two sides of the hexagon will also be parallel to each other. Let the two sides A, B, of a hexagon ABCDEF, inscribed in a conic section, be parallel respectively to the two sides D, E. Then, if we take the sides A, B, as the axes of y, x, respectively, the equations to A, B, D, E, will be respectively of the forms x=0, y=0, x +a=0, y+b=O. Let the equations to the sides C7 F. and to the diagonal through the points (A, B), (D, E), be respectively u = 0O V = O0 W = 0. Then, since the conic section circumscribes each of the quadrilaterals into which the diagonal divides the hexagon, the equation to the conic section must be indifferently of either of the following forms, uw = Xy (x + a), vw = Px (y + b); where X, /, are constants. Since these two equations must be identical, it follows that the ratio of the coefficients of x2 to that of y2 must be the same in both, and therefore in uw, vw; hence the ratio of the coefficient CC 386 LINES OF THE SECOND ORDER. of x to that of y must be the same in u and v. This conclusion shews that C is parallel to F. Gergonne: Annales de MatheMnatiques torn. IV. p. 79. 4. A triangle is inscribed in a conic section: to prove that the points, in which the sides of the triangle produced meet the tangents at the opposite angles, are in the same straight line. Let the equations to the sides of the inscribed triangle be u = 0......... (1), v = 0.........(2), w = 0.........(3). The equation to the conic section will then be X /z v + + - + = 0......................(4).* U V W Now it is easily seen that the line denoted by v W - 0.......................... (5), /, v is the tangent to (4) at the intersection of (2) and (3); for the equations (4) and (5) combined are equivalent to (2) and (3) taken simultaneously. Again (1) and (5) evidently intersect in a line U V W - + - = 0. X /z V Symmetry shews that this must be a line in which all the three intersections, mentioned in the problem, take place. Carnot: Geometrie de Position, p. 453. Salmon: Treatise on Conic Sections, p. 235. 5. AA'B'B is a quadrilateral inscribed in a conic section, the sides A'A, B'B, produced, intersecting in 0. If a straight line OPIHH'P' be drawn through 0, cutting the conic section in P, P', and the quadrilateral in H, H': to prove that 1 1 1 1 OP OP' O ' OH' * Bobillier: Gergonne, Annales de Math6matiques, tom. xvIII. p. 320. INSCRIBED POLYGONS. 387 Let OAA'7 OBB', be taken as axes of xz y, respectively. Let OA = OA' = ' OB = /3 OB' = 3'. Then, c being any constant, the equation to the conic section is 2 2 f + CSCYJ + B;I (- 0-1 (3 + )8') + 1 = o. aa/c3/3' /(a/3 X y The equation to AB is + - =, a /3 x y and to A'B' + =1 a j3 Let the equation to OPIIHI'P' be y = m. Then, x being the abscissa of ii, 1 1 rn x a /3' Similarly, x' being the abscissa of H', 1 1 m 1 1 1 1 /I11 and therefore - + = + - ) x x a a. /3 At the points P and P' we have 1 f m, )1 1 22 2-{ (a + a ') + ^( +/')} jr t+ cm + =0. Hence, X X', being the abscissae of P, P,' 1 1 1 1 (f 1\ + = -- + - + 1 1 1 1 Consequently + =- +. But it is evident, from the geometry, that X, X', x, x', are proportional to OP, OP', OH7 OH': hence 1 1 1 1 OPf OP' OH+ OH'~ c02 388 LINES OF THE SECOND ORDER. 6. The opposite sides of any hexagon, inscribed in a conic section, intersect in three points lying in a straight line. Let ABCDEF be a hexagon inscribed in a conic section, and let the equations to the alternate sides AB, CD, EFJ be respectively, when multiplied by arbitrary constants, u = 0.........(1); v = 0........(2); w = 0.........(3). The equation to the conic section may be expressed under the form 2 + + w2 - (\ + X-1) VW - (+ -) wu - (v + V-l)UV = 0...(4); for this is an equation of the second order, involving five arbitrary constants, which depend upon the three quantities X + X-1, +, -1, v + v-~l and the two ratios between the three arbitrary constant factors included in u, v, w. To determine the points A and B, put u = 0: then v2 + w2 -(X + X) vw = 0, whence v = Xw or w = Xv; and, in like manner, by putting successively v = 0 and w = O in (4), we shall get the equations which determine the positions of the other angular points of the hexagon. The equations to the points A, B, C D, E, Fi are respectively the following: A.... {}. Iw.....................(5), B v..w. (6), B.......5e...Xv...................(6) (w~*- ~~~~i{ =-\) }* //***@@@4~ (7),.......... Vw 0.....................(7), D..............................(8), E...........................(9), F -V =_ } ( 10). INSCRIBED POLYGONS. 389 Hence (8, 9), (10, 5), (6, 7), the equations to the lines DE, FA, BC, respectively, are, by mere inspection, observed to be = w + vv........................(11) v = vu + Xw......................(12), w = Xv + pu........................(13). Now each of the three pairs of equations (1, 11), (2, 12), (3, 13), evidently satisfies the equation u v w + -+- =.....................(14); X Af v and therefore the opposite sides of the hexagon ABCDEF intersect in three points in a straight line of which (14) is the equation. The demonstration here given of the property of the inscribed hexagon, is due to Mr. Weddle, by whom it was communicated to the Cambridge and Dublin Mathematical Journal, vol. III. p. 285. This beautiful property was discovered by Pascal,* and formed the basis of a complete system of conic sections, written at the age of sixteen, which was never published and of which the manuscript was unfortunately lost. In this treatise he is said by Mersennet to have demonstrated all the propositions of the Conics of Apollonius, emanating in 400 corollaries from this one fundamental theorem.: This work having been sent to Descartes, he could not be prevailed upon to believe it to have been written by one so young, ascribing it either to Pascal's father or to his friend Desargues. The early genius however of Pascal, exhibited in other matters of like nature, renders his title to these discoveries in no degree improbable. Leibnitz,[l in allusion to Pascal, says, " Mr. Perier, son neveu, me donna * Essais pour les Coniques, (Euvres de Blaise Pascal, tom. iv. p. 1. tf Harmonie Universelle. + Pascal himself also ((Euvres, tom. iv. p. 358), in presenting certain works Celeberrimce Matheseos Academia Parisiensi, speaks of this one as " Conicorum opus completum, et conica Apollonii et alia innumera unica fere propositione amplectens; quod quidem nondim sexdecimum r etatis annum assecutus excogitavi, et deinde in ordinem congessi." II Lettres et M. M. tRemond de Montmort, Lettre 2. Opera, tom. v. p. 12, 390 LINES OF THE SECOND ORDER. un jour a lire et a ranger un excellent ouvrage de son oncle sur les coniques, et j'esperois qu' on le publieroit d'abord. On lui auroit conserve par-la Plhonneur d'original, en des choses qui en valoient la peine." In a Lettre a M. Perier, 1676, which may be seen in the (Euvres de Pascal, tom. v. p. 429, Leibnitz has given a slight sketch of the general scheme of this lost treatise: he says that Pascal called his inscribed hexagon the mystic hecxagram. Desargues says that in his time it was called "the Pascal." A great many demonstrations of the property of the mystic hexagram have been given by various mathematicians, both geometrically and analytically. Pascal himself, in his Essais pour les Coniques, has merely enunciated the property in a converse form. The following is a list of some of the demonstrations. Carnot: G6ometrie d Position, p. 452. Gergonne: Annales de Mathematiques, tom. IV. p. 379; 1813, 1814. Dandelin: Gergonne, Annales de Mathematiques, tom. xv. p. 396. Lubbock: Annals of Philosophy, October, 1829. Davies: Annals of Philosophy, July, 1842, p. 37. Rutherford: Ibid., March, 1843, p. 168. Various Mathematicians: Lady's and Gentleman's Diary, 1843. Kirkman: Ibid., 1849. Cayley: One solution (from Chasles), Cambridge Mathematical Journal, Old Series, vol. III. p. 211; another in vol. Iv. p. 18. Rutherford: Davies' Hutton, vol. II. pp. 213-313. Fenwick: Mathematician, vol. I. p. 132. Weddle: Ibid., vol. II. p. 15. Salmon: Treatise on Conic Sections, p. 212, 1st edition. Gaskin: Geometrical Problems, p. 244. 7. AD, BC, are the diagonals of a quadrilateral inscribed in a conic section: from A, C, are drawn lines AF, CF, to any point F in the arc BD, and, from B, D, are drawn lines BE, DE, to any point E in the are AC, meeting AF, CF, respectively in G, H. To prove that the line GH will pass through the intersection of AD, BC. Davies: Philosophical Magazic ne, 1842, vol. XXI. p. 37. INSCRIBED POLYGONS. 391 8. Let P7, P2, P3, Q1, Q2, Q,, be six points lying in a conic section; let the areas of the triangles P1 Q Q3, P1 Qs Q, P1 Q1 Q, be denoted by A,, Bi, 0C and the areas of the triangles, formed by putting P2, P3, successively in the place of P1, be denoted by A27 B2, C,2 A3, B37 C, respectively: then will _ / 1 / 1 \ 1/1 1 \ 1 +7) + ] BC)+ ) -B =0. 9. Let ABCD be a quadrilateral: and let a conic section be described about it and tangents be drawn at A, B7 C, D) forming another quadrilateral. Let the opposite sides of one quadrilateral intersect in P, Q, and of the other in E, F. To prove that P, E, Q, F, are in the same straight line. Carnot: Geomnetrie de Position, p. 453. G. W. H., and A. C.: Cagmbridge and Dublin Mathematical Journal, vol. II. p. 238. 10. If, in the perimeter of any conic section, be taken any six points A, B, C, D, F, and the straight lines AB, AF, be produced to meet the straight lines D (7 DE, produced, in Mti N, respectively: to prove that the three straight lines IMNV BE, FC, all pass through a single point. Carnot: Geomretrie de Position, p. 452. 11. Two conic sections being circumscribed about any given quadrilateral; if, through the extremities of one and the same side of this quadrilateral, be drawn two arbitrary secants; the chords in these curves which terminate in the points in which they are respectively intersected by these secants, will, if produced indefinitely, cut each other in some point on the direction of the opposite side of the quadrilateral. Numerous very interesting consequences have been deduced by Plucker from this theorem and an analogous one respecting any circumscribed quadrilateral. Plucker: Gergonne, Annales de Mathl&matiqies, tom. xvii. p. 39. 392 LINES OF THE SECOND ORDER. 12. If three sides of a variable quadrilateral inscribed in a conic section are always parallel to three given straight lines, to prove that the fourth side will also be parallel to a given line. O'Brien: Plane Coordinate Geometry, p. 178. 13. Having given any convex quadrilateral; 1st, we may always circumscribe about it an infinite number of ellipses, but only two parabolas; 2nd, the locus of the centres of all these ellipses is an hyperbola, the asymptotes of which are respectively parallel to the axes of the two parabolas; 3rd, the conjugates of the diameters of these ellipses, which are parallel to any fixed line, meet all in a single fixed point of the hyperbolic locus of the centres; 4th, the conjugates of the diameters of these ellipses parallel to one of the two asymptotes of the hyperbolic locus of the centres, are parallel to the other asymptote of this hyperbola; 5th, of all these ellipses, the one which approaches nearest to a circle is that of which the conjugate diameters, parallel to the asymptotes of the hyperbola, are at the same time equal conjugate diameters. Gergonne: Annales de Mathematiques, tom. xvIII. p. 100, SECTION XXI. Circumscribed Polygons. 1. If any quadrilateral be circumscribed about a conic section, the point of intersection of the two straight lines which join the points in which the conic section is touched by opposite sides of the quadrilateral, will coincide with the point of intersection of its two diagonals. Let the equations to three sides of the quadrilateral taken in order be u = v = 0 w = 0. Then the equation to the conic section will be (lu)i + (m + (n = 0.* * This equation to a conic section inscribed in a triangle is substantially the same as that given by Bobillier in Gergonne's Annales de Math&ematiques, tom. xvIIi. p. 325. CIRCUMSCRIBED POLYGONS. 393 The equation to any other tangent of the conic section, which we will take for the fourth side of the quadrilateral, will be Xu + /v + vw = 0, X, /t, v) being connected with 1, m, n, by the equation -- + -+- =.......................(1). X /fi V The equations to the points in which the lines u = 0, w = 0, touch the conic section, are, respectively, {* = and {'W i. tmnv = nw) (Mv = A) Hence the equation to the line, which joins the points of contact of two opposite sides of the quadrilateral, is mv = lu + nw.......................(2). Again, the equations to the two diagonals are p/v + vw = 0.........................(3), and Xu + /v = 0.........................(4). Eliminating u, v, w?, between the equations (2), (3), (4), we obtain the equation (1): the two diagonals therefore intersect in the line which joins two opposite points of contact. The quadrilateral being of a perfectly general form, it follows that the intersection of the two diagonals must take place also in the line joining the other two opposite points of contact. The four lines therefore intersect in a single point. Carnot: Geometrie de Position, p. 454. Peschier, Rochat, Ferriot, Fornier: CGergonne, Annales de Mathematiques, tom. III. p. 161. Hearn: Researcles on Curves of the Second Order, p. 52. 2. The sides of a triangle ABC are tangents to a given conic section, CA touching it in P, CB in Q, and AB in B. CA, CB, are fixed in position, AB being variable. To prove that the ratio AP BQ AR BR is a constant quantity. 394 LINES OF THE SECOND ORDER. Let CP, CGQ produced indefinitely, be taken as axes of coordinates, and let the equation to ARB in any position be ' w = ax + by + c = 0. Then, I and m being arbitrary constants, the equation to the conic section will be / iz2 + m/2y2 + w' = 212xwt + 2yw + 21mnxy. - Putting y = 0 in the equation to the conic section, we see that, at the point P, ix=ax+ c - But, at A, putting y = 0 in the equation to AB, we have a Hence AP= c - a (1- a) Again, at R, we have w = 0, lx = vqy, and therefore x - -y =- na + b wma + b Hence, for the length AR, we have, o) denoting the angle xCy7 (AR)2 (ma + ) = (AP).(a2 + 2 - 2ab cos)). Similarly (BR)2 (ma + ) = (BQ)2.( + b2 - 2ab coso). (mb - )2,(AP\2 (BQ^ Hence ( a)2. (AR) - )) ). (J) ) But, the conic section being given, the ratio of the coefficients of x and y in its equation must be invariable for all positions of R: hence I-a -- b = a constant quantity. AP BQ Consequently jAR - R = a constant quantity. CIRCUMSCRIBED POLYGONS, 395 COR. This proposition is an obvious deduction from the theorem that the products of the alternate segments of the sides of a polygon circumscribing a conic section are equal. 3. If any hexagon be circumscribed about a conic section, the three diagonals which join its opposite angles will all pass through a single point. Let the equations to three alternate sides of the hexagon be u =. 0........ (1), = 0.........(3), w.........(5). Then the equation to the conic section will be (lhu) + (mv)~ + (nw) = 0, 1, m, n, being arbitrary constants. The equations to the three other alternate sides will be Xu + /v + vw = 0....................(2), X'u + 2I'v + v'w = 0................... (4), X'u + I'v + v"w = 0...................(6) where the coefficients of u, v, w, are subject to the relations 1 rn n, +- + -.= 0.... (7), X J v 1 m n.(8),=0~~~~~~~~~~~~ ~~~~~(8),, + - +, = 0................... 8 h t fJ V.... I n n. H v'' ~ * * (9). -+ -+ -= o0....................... (9). The equations to the diagonals {(1, 2), (4, 5)}, (2, 3), (5, 6)}, {(3, 4), (6, 1)}, will, as is evident by inspection, be respectively ZU V W +, + - 0 + - == 0 W + v,,,-,v+ vX =0. V, TW+ 0. v +X' +,' FL 396 LINES OF THE SECOND ORDER. If between these three equations we eliminate tt, v, w, by cross multiplication, we shall obtain an equation coincident with ( I - ) (1 4) + - = o: AX L6 l/V v /vh A v v l Xtk Lv v this equation however results from the elimination of 1, m, n, between the equations (7), (8), (9). Hence the three diagonals must all pass through a single point. "M. Brianchon, eleve, m' a remis l'analyse d'un memoire dans lequel il prouve par la geometrie seule, plusieurs proprietes des courbes et des surfaces du second degre, dont quelques-uns lui appartiennent. Ce memoire est destine pour le Journal de 1'Ecole: j'en extrais la proposition suivante: 'si tous les cotes d'un hexagone quelconque touchent une meme courbe du second degre, les trois diagonales, prolongees s'il le faut, se croisent en un meme point."' Hachette: Correspondance sur l'Ecole Polytechnique, tom. I. p. 151. Brianchon: Journal de 1'Ecole Polytechnique, tom. Iv. 44 Correspondance sur l'Ecole Polytechnique, tom. I. p. 307; tom. II. p. 383. Carnot: Essai sur la Theorie des Transversales. Gergonne: Annales de Mathmnatiques, tom. IV. p. 383. Dandelin: Gergqonne, Annales de Mathematiques, tom. xv. p. 396. Davies: Annals of Philosophy, November, 1826. Lubbock: Ibid., October, 1829. Fenwick: Ibid., March, 1843, p. 167. Frost: Cambridqe Mathematical Journal, vol. IV. p. 277. The student is referred also to a memoir by Coste in the Annales de Mathematiques, tom. x. p. 261, where he will find numerous properties of the parabola deduced from Pascal's and Brianchon's general theorems concerning the inscribed and circumscribed hexagons of a conic section. 4. If ABC be a triangle circumscribed about a conic section, a, f/, ey being the points in which the sides BC, CA, AB, touch it, to prove that Aa, B/3, Cy, all pass through a single point. Bobillier: Gergonne Annales de Miatheqnatiques, tom. XVIII. p.323. Salmon: Treatise on Conic Sections, p. 236. CIRCUMSCRIBED POLYGONS. 397 5. Two triangles being the one inscribed and the other circumscribed in relation to a conic section, so that the summits of the inscribed triangle are the points of contact of the circumscribed; the points of concourse of the sides of the inscribed triangle with the respectively opposite sides of the circumscribed, lie all three in a single straight line. Bobillier: GergonneAnnales de Mathenatiques, tom. XVIII. p. 323. Salmon: Treatise on Conic Sections, p. 235. 6. Two conic sections being inscribed in the same quadrilateral; if, in the two sides of one of the summits of this quadrilateral, be taken arbitrarily two points through each of which we draw tangents to the two curves; the point of concourse of the pair of tangents to one of the curves, that of the pair of tangents to the other curve, and the opposite summit of the quadrilateral, will all three lie in a straight line. Plucker: Gergonne, Annales de Mathematiques, tom. XVII. p. 39. 7. If a conic section be circumscribed by a triangle ABC, and lines AA'a, BB'/3 CC'Y, be drawn from the summits A, B, C, to the points of contact a, /, y, of the opposite sides, meeting the conic section in three points A', B', C': to prove that tangents drawn to the curve at the points A', B', C', will intersect each other, two and two, in the lines Aa, Bj3, Cy. Also, to prove that the intersections of the tangents at A', B', C', respectively, with the sides B0, CA, AB, of the circumscribed triangle, will all lie in a single straight line. Leybourn: Mathematical Repository, New Series, vol. II. p. 153. 8. To find the equation to a conic section inscribed in a triangle so as to touch the three sides at their middle points. Taking (x2 - x) y - (y, - ) x - x2y, + xy = O, as the general type of the equation to a straight line, then, the equations to the sides being = 0 v= 0 w= 0 we shall have for the equation to the required conic section u~ + vi + wi = 0. 398 LINES OF THE SECOND ORDER. SECTION XXII. Problems relating to several Curves. 1. If there be, in one plane, any two lines of the second order, such that the principal diameters of the one coincide with the tangent and normal at any given point of the other; the line which joins the points, in which any two conjugate diameters of the one intersect the other, will pass through an invariable point of the given normal. Taking the tangent and normal at the given point as axes of x and y, we shall have for the equation to the one conic section x A2 - + = 1........................ ( ), and, for the equation to the other, ax' + by2 + 2cxy + 2fy = 0...............(2). The equations to any two conjugate diameters of (1) will be la2 y= mx, Y= n X' and the equation to both will be (y-mx) ( + J =0, whence x' = (1 + ) y mJ................(3) Substituting in (2) the expression for xa, given by (3), we have, dividing by y, (a_ m ~ + 2), X+ ( 2 + b) y+ 2.....(4), which will be the equation to the line joining the points in which the conic section (2) is cut by the two conjugate diameters of (1). Putting x = 0 in (4), we see that 2_Q2 y= ~ a2 + b/ 2' which determines a constant point in the normal through which all such lines pass. PROBLEMS RELATING TO SEVERAL CURVES. 399 CoR. By the theory of poles and polars, it hence follows that the pairs of tangents, at the extremities of all lines the equations of which are of the form (4), all intersect in a single straight line. Fregier: Gergonne, Annales de Mathe'matiques, tom. VI. p. 321. 2. If two conic sections (A) and (B) are so placed that the centre of (B) lies in the perimeter of (A); all the chords of (A), which terminate at its intersections with the prolongations of two conjugate diameters of (B), will intersect each other in a single point, situated in the conjugate to that diameter of (B) which is a tangent to (A). Let the tangent to (A), at the centre of (B), be taken as the axis of x, and the diameter of (B), which is conjugate to this tangent, as that of y. Then the equation to (A) will be of the form ax2 + by2 + cxy + fy = 0............(1). Also the equations to any two conjugate diameters of (B) will be y = mx, y= mx, where mm' is equal to some constant quantity k. The equation to both these conjugate diameters will be (y - mx) (y - m'x) = y' - (m + I') x + mm'x2:= - (m + m') xy + kx = 0............(2). At the intersections of (1) and (2), we have, eliminating x', (kb - a)y + {2kc + (m + mi) a} x + kf= O, which is the equation to the chord of (A) joining its intersections with the two conjugate diameters of B. Putting x = 0, we see that a - kb' a result which, being free from m and na', establishes the proposition. Fregier: Gergonne, Annales de Mathenatiques, tom. VII. p. 95. 400 LINES OF THE SECOND ORDER. 3. The equations to two conic sections being Ax' + 2 Cxy + By' + 2A'x = 0, ax2 + 2cxy + by2 + 2a'x = 0; to find the condition that the lines joining the origin with their points of intersection may be perpendicular to each other. Multiplying the former equation by a', and the latter by A', and subtracting, we get (Aa' - A'a) x + 2 (Ca'A - A'c) y + (B' - A'b) y2 = 0, the equation to the two joining lines. The product of the tangents of their inclinations to the axis of x will be equal to Aa' - Aa Ba' - A'b but, if the two lines be at right angles to each other, their product must be equal to - 1. Hence, for the required condition, we have Aa'-A'a =-Bd + A'b, A+B A' aor a+ b a' 4. If in the plane of a conic section 7, we describe any two circles A, A', and two conic sections B, B', of which the former passes through the four points of intersection (real or imaginary) of the former circle and the conic section U, and the latter through the four points of intersection of the latter circle and of the conic section U; to prove that (1), the two conic sections B, B', cut each other in four points which lie in a circle; and, (2), that this circle passes through the points of intersection of the two proposed circles. Let U's equation be u = 0, A's.............. a = A'...............' = 0. Then, X and X' being any constant multipliers, B's equation will be Xu + a = 0, B"s.............. 'u +J r a' = 0. INTERSECTIONS OF CONIC SECTIONS. 401 At the intersections of B and B' we have X - Xa' = 0........................(1), which is of the form of the equation to a circle: the four intersections of B, B', therefore lie in a circle. Again, since the equation (1) is satisfied by all values of x and y which satisfy A's and A"s equations simultaneously, we see that the circle (1) passes through all the intersections of A and A'. 5. To prove that two common chords, real or imaginary, of two confocal conic sections, pass through the intersection of the directors. Any two such conic sections may be represented by the equations c - = 1 + e os( ( + a)...................(1), - =1 + e' cos(0 + a').................... (2). r Now, the coordinates of any point in (1) being (r, 0), those of a coincident point in (2) must be {r, 2n7r +- 0 or {- r, (2n + 1) r + 0}, n being any integer, Hence, where (1) and (2) intersect, we have, either C - C ecos(O + a) - e' cos (2nr + + a'), or + e cos(0 + a) - e' cos(2n7r + r + 0 + a'). r Thus every point of intersection lies in one of the two chords = ecos(0 + a) + e' cos(0 + a'), both of which pass through the intersection of the directors, the equations to which are C C — = ecos(O + a), - = e cos(0 + a'). ~~r r DD 402 LINES OF THE SECOND ORDER. This proposition may be proved also in the following manner. Let the equation to the director of one of the conic sections, the focus being the origin, be lx + my = 8, 1l m, being its direction-cosines. Then, e denoting the eccentricity, the equation to the curve will be x2 + Y2 = e2 (l + my - 8)2. In like manner, the equation to the other conic section will be of the form x2 + y2 = e2 ('x + m'y - ')2. At the intersection of the two curves e (Ix + my - 8) = ~ e' ('x + m'y - 8'), which is therefore the equation to two chords of intersection. The form of the equation shews that both chords pass through the intersection of the directors. 6. A conic section is cut in four points by a circle, and two straight lines, each passing through two of the points of intersection, are taken as axes of coordinates: to prove that the equation to the conic section will be of the form x2 + 2cxy + y + 2a + 2a' 2'y + c'= 0. 7. If any number of circles be described so as to touch a given curve of the second order in one given point: to prove that they will each of them generally cut it in two others; and that, if a chord be drawn through the two points of section for each circle, such chords will be all parallel to one another. 8. If two conic sections meet each other in more than four points, to prove that they will coincide. De la Hire: Sectiones Conicce lib. II. prop. 32. 9. If a parabola, represented by the equation y2= lxI is intersected in four points by any conic section: to prove that, INTERSECTIONS OF CONIC SECTIONS. COMMON CHORDS. 403 Y1, Y, Y3, Y4 being the ordinates of the intersections, Y1 + Y2 + y3 + yX = 0. De la Hire: Sectiones Conicce, lib. v. prop. 31. 10. To prove that two conic sections of Bthe same species, the axes of which are proportional in magnitude and parallel in direction, cannot have more than two points in common. Lame: Examen des diffdrentes methodes employees pour resoudre les problmes de Geometrie, p. 39, SECTION XXIII. Intelsections of Conic Sections. Common Chords. If S= 0 be the equation to a conic section, that to any other conic section cutting it at the ends, real or imaginary, of two chords u = 0, v = 0, will be S= kuv, k being an arbitrary constant. Salmon: Treatise on Conic Sections, p. 199. 1. To find the diameter of a circle described about a semiellipse bounded by its axis minor. Let the equation to the ellipse be x2 2f a+ b =- 1. The equations to the common chord and common tangent, the latter of which is merely a degenerate case of a chord, are x= O, x =a: hence the equation to the circle is included in the equation x~. y" a + - 1 =kxx(x-a). DD2 404 LINES OF THE SECOND ORDER. Since this is the equation to a circle, the coefficients of x' and y2 must be the same: hence 1 1 k= a2 ~b2 and therefore the equation becomes a2 _ 62 x2 _ Y=2 a= +2 4 x, a a2 - b2 the diameter of the circle being therefore equal to ---2. If any number of conic sections have four points in common: in whatever direction parallel diameters be drawn in them, the conjugate diameters will all meet in a single point. Any one of the system of conic sections may be represented by the equation S= kuv, where k is an arbitrary constant, varying as we pass from one of the curves to another; where S = ax2 + by2 + 2cxy + 2a'x + 2b'y + c', = ax + /3y + Y7 and v = a' + 'x y + '; u = 0, and v-= 0, denoting the equations to two common chords of all the curves. Now, if the equation to a conic section be written in the form Ax2 + Byy + 2 Cxy + 2A'x + 2B'y + C' = 0, the equation to a diameter, conjugate to one which is parallel to a chord y = mx, will be Ax + Cy + A' + m (By + Cx + B') = 0: hence, in the present case, if we put for A, B, C, A', B', their values, the conjugate diameter will have an equation of the form U+ kV= O, where U, V, are linear functions of x and y, not involving k. This diameter evidently passes through a point defined by the equations U= O0 V =0. DOUBLE CONTACT. 405 Since these equations are independent of k, this point is invariable in position for every such conjugate diameter. Lame: Memoires de l'Academie dcs Sciences de Paris, Dec. 1816. 44 Gergonne, Annales de MathZematiques tom. VII. p. 233. 3. If a circle cut a conic section, to prove that the chords joining the points of intersection are equally inclined to the axis of the conic section. Leslie: Geometrical Analysis, p. 302. 4. If two conic sections, the axes of which are parallel, intersect one another in four points, to prove that the sides AB, CD, and also the sides BC, DA, of a quadrilateral ABCD, formed by any four common chords AB, BC, CD, DA, are equally inclined to either axis. Elliot: Lady's and Gentleman's Diary, 1851. SECTION XXIV. Double Contact. If two conic sections touch each other at two points, they are said to have double contact with each other. Let u = 0 be the equation to a conic section, and let another conic section touch it at the extremities, real or imaginary, of a chord v = 0: then the equation to the other conic section will be u = k2, k being an arbitrary constant. Salmon: Treatise on Conic Sections, p. 200. 1. To find the equation to an ellipse having double contact with an hyperbola represented by the equation xy = c2, the asymptotes of the hyperbola coinciding with conjugate diameters of the ellipse. 406 LINES OF THE SECOND ORDER. Let the equation to the chord of contact be x Y - + -=1 m m then the equation to the ellipse will be c xy 2 -- 1)...............). Since this ellipse is referred to conjugate diameters, we must have — =0 m k2 - = 0........................... (2). 2.-==- 1 m n These three equations shew that k = 0, - and being finite. m n The equation (1) therefore becomes mn n n or, a, b, being the conjugate semi-diameters, x2 Y2 a-+ =- a and b being, by virtue of the third of the equations (2), connected together by the relation a2b2 = 44. 2. To find the equation to a circle inscribed in a semi-ellipse bounded by its minor axis. Let the equation to the ellipse be x~' yZ a bThe equation to the chord of contact, as is evident from symmetry, will be of the form X =C: CONICAL LOCI. 407 hence the equation to the circle will be included in the equation X2 Y2 a - 1b =-k (x-c)2. Since the circle touches the axis of y, we must have kc = -1; hence c2 (-+ -) = 2cx - \( b 2 a + 1) ax + c2 = 2cx. The curve being a circle, the coefficients of x2 and y' must be equal: hence a2b C a2 - b2 and therefore the equation to the required circle is x2 + y2 (a2 - b2).X. SECTION XXV. Conical Loci. 1. To determine the locus of a point, the algebraical sum of the distances of which from two given straight lines and from a given point, shall be constant. Let the equations to the two given lines be xcose +ysine -8 =0, x co'ss + y sine' - 8' = 0; and (a, b) the coordinates of the given point. Then, by the condition of the problem, the equation to the required locus will be, k denoting a constant, x(coss+cos') + y(sins + sins') - 8 - ' + {(x- a) + (y- b)} = k; or writing c for k + 8 + 8', a for -, and for + 2 aco 2 2xwcosa cosi3 + 2y cosca sinl3 + {(x - a)2 + (y - b)2 c, 408 LINES OF THE SECOND ORDER. and therefore, clearing the equation of the radical sign, (x - a)2 + (y - b)2 = 4 cos'a cos2/,.x2 + 8xy cos22a sin/, cos/3 + 4y' cos2a sin2 -4c (x cosa cosf + y cosa sinfl) + c', x' (4 cos'a cos23 - 1) + 8xy cos' a sin/, cos,3 + y2 (4 cos a sin28 - 1) + 2x (a - 2c cosa coso/) + 2y (b - 2c cosca sin,) = a2 + b2 - c2. The curve will therefore be an ellipse, a parabola, or an hyperbola, respectively, accordingly as 64 cos4 sin'/3 cos28 > 4 (4 cos a cos2I - 1) (4 cosg2 sin'2 - 1), 0 - 4 cos2a, cosc2a ^. Lhuilier: Gergonne, Annales de Mathematiques, tor. II. p. 173. 2. To find the locus of the pole of a given circle, the corresponding polar of which is always a tangent to another given circle. Let the radius of the circle, to which the polar is a tangent, be r: then, the centre of this circle being the origin, the equation to the polar will be xcosO + ysin = r..................... (1). Let the corresponding pole be at the point h, k; then, a, b, being the coordinates of the centre and c the radius of the other circle, the equation to the polar will be x (h - a) + y (k - b) = C2 - a2 - b2 + ha + kb....(2). Since (1) and (2) must be coincident, we have r r o ( - a) = c - a2 - b + ha + kb = si (k - ), cos 0 sin 0 and therefore (e2 _ _ 2 + ha + kb)2 = r2 {(h - a)2 + (k - b)2}, which is the equation to the locus of (h, k). This locus is an ellipse, parabola, or hyperbola, accordingly as r is greater than, equal to, or less than (a2 + b'2), the distance between the centres of the two circles. L' Hospital: Traite Analytique des Sections Coniques, p. 273. Puissant: Recueil de diverses propositions de Geomzetrie, p. 219, troisie'me edition. CONICAL LOCI. 409 3. If a straight line DCP be made to revolve about C, and intersect a curve in as many points P, P2 P,...... as it has 1 dimensions, and if CD* be made always equal to 1 1 1 CR2 CP12+ CR22+.+ to find the locus of the point D, Let the equation to the intersected curve be 1 + ax + by + a'x2 + y2+ cxy +.. =0. Put x =r cos 0, y = r sin 0: then 1 + r(a cos + bsin ) + r2(a' cos20 + b'sin2 0 + c'sin cos 0) +...= O. Hence, by the theory of equations, 1 1 1 C + 2+ CP+......=(a cos + b sin0)2 -2 (a' cos2 0 + b' sin2 0 + c' sin O cos 0). Consequently, putting CD = p, we obtain 1 = (a' - 2a') cos20 + 2 (ab - c') sin cos + (b2 - 2b') sin2, P which is the polar equation to a conic section of which C is the centre. 4. To find the locus of the summit of an angle of given magnitude, circumscribed on a parabola. Let s be the given angle; then, the equations to its two sides being 2m - = cos0 + cos( - a)..................(1), r 2m = cosO + cos( - a')..................(2), we shall have also a'- a = 2sa........................(3). At the intersection of (1) and (2), there is cos(0 - a) = cos( - a'), and therefore 0 = (a + Ct)....................... (), 410 LINES OF THE SECOND ORDER. Again, adding (1) and (2), we get 4m cos a +' - a - = 2 cos6 4 2 cos 6 cos --: r CO 2IC 2 hence, from (3) and (4), we get 2m - = cos 8 + cos, r 2m or r = cos. + cosOc the equation to the locus of the summit of the angle, which is therefore a conic section confocal with the parabola. L' Hospital: Traite Analytique des Sections Coniques, liv. VIII. 5. The distance of a point P from the circumference of a circle is to its distance from a fixed diameter as n to 1: to find the locus of P. The fixed diameter being taken as the axis of y, and a perpendicular diameter as that of x, the locus will be defined by the equation 2 + y2 = (nx + a)2, where a denotes the radius of the circle. 6. From a point 1f in a given straight line AB, a perpendicular MP is erected, such that MP2 ce AiM.MB; to find the equation to the locus of P. Let AB= 2a, MP= y, and, C being the middle point of AB, let CMA= x. The locus will be a conic section defined by the equation y2 = +~ ('2 _ 2), X being a constant quantity. Pappus: Mathematicce Collectiones a Commandino, lib. vII. p. 298. 7. QA 0 is a given angle, 0 a given point: P is a point such that, PO being joined, and PQ drawn parallel to OA, PO: PQ:: n: n. To find the locus of P. CONICAL LOCI. 411 Let PM be drawn parallel to QA) cutting OA in M. Let OA = a OM= x, PM=y, ZLQAO=. /. Then the locus of P will be a conic P section defined by the equation x" (n2 - m2) + n'y2 - 2n2xy cos / / + 2m2ax = m'a'. A M 0 Newton: Arithmetica Universalis, prob. 24. 8. To determine the locus of the poles of a given straight line with respect to a series of circles touching two given straight lines. Let the two given tangents be taken as axes of coordinates, and let the equation to the given polar be + Y=1. a /3 Then, co denoting the angle between the axes of coordinates, the equation to the required locus will be {ax - /3 + (ay - /3s) cosw}2 = (a - 3). [(x + ) {ax -3yJr+(ay-/3x)ccos w+ +a8(x-y) (1 —cos o)]. Poncelet: Gergonne, Annales de Mathernatiques, tom. XII. p. 234 9. Three straight lines revolve about three given points not in the same straight line, and intersect one another in three points: if the loci of two of these intersections be given straight lines, to find the locus of the third. Let O, A, B, be the three points. Let OA, OB, produced indefinitely, be taken as axes of x, y, respectively. Let OA = a, OB = b. Let the lines through A, 0, intersect in the given line y = ax + /, and those through B, 0, in the given line y = a'x + f/': then the locus of the intersection of the lines through A, B, will be a conic section defined by the equation (6 - 3').(aa + /3) xy = [ay + /3 (x - a)}.Ia'bx - 3 (y - b)} 412 LINES OF THE SECOND ORDER. 10. Through two given points C, D, is drawn a straight line DCE, cutting a given straight line BA, produced indefinitely, in c E. Two points F, G, are taken in EA such that AF.BG = c2, --- A c being a constant length. CF, DG, are joined. To find the locus of P, their point of intersection. Taking EAB, ECD, produced indefinitely as axes of x, y, respectively, and putting CE = k, DE = ', EA = a, EB = a', we shall have for the equation to the required locus (2 (Y - k) (y - ') = (k (x - a) + ay]).{k' (x - a') + a'y}. Leybourn: Mathematical Repository, New Series, vol. I. p. 106. 11. The sides BC, CA, AB, of a triangle pass always through three given points (h, k), (h', c'), (h", k"); and the angular points A, B, move along two given right lines Ox, Oy, respectively: to find the locus of C. The lines Ox, Oy, being taken as axes of coordinates, the required locus will be a conic section defined by the equation hi'k' -y) (h - x) k'x - hy hy - kx O'Brien: Plane Coordinate Geometry, p. 176. 12. ABC is a given triangle: to find the locus of a point P, such that, MP, a perpendicular to the side BC, being produced to cut the sides of the triangle BA, CA, produced if necessary, in the points q, r, respectively, (MP)2 oc Pq.Pr. Draw AO at right angles to the line BC, cutting it in O. Take OB, OA, produced indefinitely, as axes of x, y, respectively. Let A 0 =, BO = b, and CO = c. Then the required CONICAL LOCI. 413 locus will be a conic section defined by the equation xZ + (X - 1) Y2 +. h- x + 2y-2 = 0, where X denotes a constant quantity. Carnot: Geommtrie de Position, p. 443. This problem is merely a particular case of one known by the name of Pappus'* Problem, solved by Des Cartes.t The more general problem is stated by Pappus, in the form of a theorem, in the following words, extracted from Commandine's translation: " Si positione datis tribus rectis lineis ab uno, et eodem puncto ad tres lineas in datis angulis rectae linese ducantur, et data sit proportio rectanguli contenti duabus ductis ad quadratum reliquae: punctum contingit positione datum solidum locuin, hoc est, unam ex tribus conicis sectionibus. Et si ad quatuor rectas lineas positione datas in datis angulis linese ducantur; et rectanguli duabus ductis contenti ad contentum duabus reliquis proportio data sit: similiter punctum datam coni sectionem positione continget. Si quidem igitur ad duas tantum, locus planus ostensus est. Quod si ad plures, quam quatuor, punctum continget locos non adhuc cognitos, sed lineas tantum dictas; quales autem sint, vel quam habeant proprietatem, non constat. Earum unam, neque primam, et quae manifestissima videtur, composuerunt ostendentes utilem esse propositiones autem ipsarum hse sunt. Si ab aliquo puncto ad positione datas rectas lineas quinque ducantur rectae lineae in datis angulis, et data sit proportio solidi parallelepipedi rectanguli, quod tribus ductis lineis continetur ad solidum parallelepipedum rectangulum, quod continetur reliquis duabus, et data quapiam linea, punctum positione datam lineam continget. Si autem ad sex, et sit data proportio solidi tribus lineis contenti ad solidum, quod tribus reliquis continetur; rursus punctum continget positione datam lineam. Quod si ad * Pappus: Mathematicea Collectiones a Commandino, lib. vII. p. 165. t Descartes: Geometrie, livre ii. p. 30, &c. 414 LINES OF THE SECOND ORDER. plures quam sex, non adhuc habent dicere, data sit proportio cujuspiam contenti quatuor lineis ad id quod reliquis continetur, quoniam non est aliquid contentum pluribus quam tribus dimensionibus." 13. B, B', are two fixed points in a given straight line Ox; C is a moveable point in the straight line Oy. To find the locus of c a point P, such that, the straight lines BC, BP, B'C B'P, being drawn, the angles CBP, CB'P, are always of given o t A' *' magnitudes. Take Ox, Oy, as axes of x and y, and let OB = b, OB' = b' x Oy = c, L CBP = a, L CB'P = a. Then the required locus will be a conic section defined by the equation (b - + ( x a) cosa - cos(co - a') (b' - x) sina' + y sin (co - a), ct t (b - x) cosa - ycos(t + a) } - ( - x) sina + y sinm( + a) L' Hospital: Traite Analytique des Sections Coniques, p. 281. O'Brien: Plane Coordinate Geometry, p. 179. 14. Having given one side of a triangle and the product of the tangents of the adjacent angles, to find the locus of the vertex. Let + m denote the given product and 2a the given side: then, this side being taken as the axis of x, and its middle point as the origin of rectangular coordinates, the required locus will be a conic section represented by the equation 2= + m(a - X). Lardner: Algebraic Geometry, p. 116. 15. From a point P, in one given straight line, is drawn PM at right angles to another given straight line AMx: from A, which is a given point, is drawn A Q to a point Q, in PM, such that AQ = MP: to find the locus of Q. ENVELOPES. 415 If the equation to the former given line, referred to Ax, and Ay, at right angles to Ax, as axes, be y = ax + /3, then the polar equation to Q, A being the pole and Ax the prime radius vector, will be 1 - acos0' Lardner: Alqebraic Geometry, p. 150. 16. A straight line of given length moves parallel to itself in a given direction, and two other straight lines revolve round given points and pass always through the extremities of the first line: to find the locus of the intersection of the revolving lines. Let the axis of x be taken parallel to the moving line, the axis of y always bisecting this line. Let (a, b), (a', b'), be the two given points; and suppose the length of the moving line to be 2c. Then the equation to the required locus will be ' -6 (m -a) y- b a Y - b' x- a - a. SECTION XXVI. Envelopes. 1. A, B, are the centres of two equal circles, and AP, BQ, are two radii which are always perpendicular to each other: to find the curve which is always touched by the right line PQ. Let C, the middle point between A and B, be taken as the A C B origin of coordinates, CB, produced indefinitely, being the axis of x. From C draw CR at right angles to PQ. 416 LINES OF THE SECOND ORDER. Let AC=c=BC, L PAB =, L CB= +, CR=8, AP= a=BQ. Then the equation to PQ is xcos' + y sin- = 8 = Iatcos(- a cos( ) + sin ( - 0), also 2c cos = a{cos(r - 0) - sin ( - 0)}. Hence, squaring and adding these equations, we have 4(x cosf + y sin )2 +- c' cos2' = 2a2...........(1), and therefore (x cos' +- y sin)2 = a - 2cos2 + sin2. 2 2 -This equation represents a straight line always touching a conic section x2 y_ a2 _ 2c + " 2 COR. If 2c2 = a2, the equation (1) becomes (x cos + y sin *)2 = C2 sin2', which shews that PQ always passes through a point x = 0 y = c, 2. To find the envelop of a right line such that the product of perpendiculars drawn to it from two given points may be constant. LEMMA. If ) 0,) denote the reciprocals of the intercepts of the axes of any conic section from the centre made by any tangent, then a2 b202 = 1........................ (1) For, let x', y', be the coordinates of the point of contact: then, the equation to the conic section being x2 y2 a2 ~ 6 = 1, we shall have for the equation to the tangent, axx b2 at + 6 -1, and therefore -= a + = 60. a b ENVELOPS. 417 Hence, by the equation to the curve, a22 + b20 = 1. The equation (1) has been called by Mr. Booth, in an ingenious treatise entitled " On the Application of a New Analytic Method to the Theory of Curves and Curved Surfaces," the tangential equation to the conic section. The general tangential equation of a conic section, given in his work, is of the form a,12 + b02 + 2cq0 + 2a'q + 2'(9 = 1, c being zero in the case of a parabola. In the present problem, let the line joining the two given points be chosen as the axis of x, its middle point being taken as the origin of coordinates. The equation to the moveable right line is of the form lx + Oy = 1. Let a represent the distance of either of the given points from the origin, and 8, S', their distances from the moveable line. Then. 1 - a1 1 + fna (v2 + 0)' X= (9v + 02)~' and therefore, c2 being taken to represent the constant value of 8S', we have, as the tangential equation to the envelop, 1- n^2a2 2 + 02 (a2 + c) 12 ~+ c2 = 1. Comparing this equation with (1) we see that the required envelop is a conic section the semi-axes of which are (a2 + c2)~ and c. For other exemplifications of the Method of Tangential Coordinates the student is referred to Mr. Booth's treatise. 3. To find the curve, which is touched by all the chords of a central conic section, which subtend a right angle at the centre. If the conic section be represented by the equation ax2 + by2 = c, EE 418 LINES OF THE SECOND ORDER. the required locus will be a circle concentric with the ellipse and having a radius equal to (a + b Querret: Gergonne, Annales de Mathematiques, tom. xv. p. 197. Un Abonne: Ibid., tor. xv. p. 199. 4. Through the centre of a conic sectidn are drawn two straight lines at right angles to each other, one meeting the curve in P, the other the minor director in B: to prove that the line PR envelops a circle the diameter of which is, in the case of the ellipse, the minor axis, and, in that of the hyperbola, the transverse axis. Booth: Annals of Philosophy, 1845, vol. II. p. 541. 5. A given angle moves along a given right line, one side always passing through a given point; to find the equation to the curve which is always touched by the other side. The given right line being taken for the axis of x, and a perpendicular to the given right line, through the given point, as that of y, then, the ordinate of the given point being 38, and the given angle being denoted by tan-m, the equation to the required curve will be (y + mx - )2 = 4m3 (my - ), which belongs to a parabola. Booth: Application of a New Analytic Method to the Theory of Curves and Curved Surfaces, p. 9. 6. The vertex of a right angle moves along a given circle, and one side passes through a fixed point: to find the envelop of the other side. Let the centre of the circle be chosen as the origin of coordinates, the axis of x passing through the fixed point: let c represent the distance of the fixed point from the centre of the circle, a the radius of the circle. ENVELOPS. 419 Then the required envelop will be a conic section of which the tangential equation is a292 + (a2 - c2) 02=. Booth: Application of a New Analytic Method to the Theory of Curves and Curved Surfaces, p. 5. 7. To find the envelop of the conic section lvw + mwu + nuv = 0, u) v, w, being given linear functions of x and y, and 1, m, n, being arbitrary parameters subject to the following relation, (la)c + (m)3) + (ny) = o. The required envelop is the straight line au + /3v + yw = 0. Hearn: Researches on Curves of the Second Order, p. 34. 8. To find the envelop of the conic section (lu+) + (mv)= + (nw)o = 0, u7 v, w, being given linear functions of x and y, and 1, n, n, denoting arbitrary parameters subject to the following relation, viz. m n -+ - + - = 0. a /3y The required envelop is the straight line defined by the equation acu + v + ryw = 0. Hearn: Researches on Curves of the Second Order, p. 36. 9. To find the envelop of the base of a triangle inscribed in a conic section, two sides of which pass through fixed points. Let w = 0 be the equation to the chord passing through the two fixed points; u = 0, v = 0, those to the tangents at its extremity. Then, the equations to the lines joining the fixed points to the intersection of u = 0, v = 0, being au == v bu = v EE2 420 LINES OF THE SECOND ORDER. the required envelop is a conic section represented by the equation (a + b)2 uV = -. W. 4ab Salmon: Treatise on Conic Sections, p. 222. For additional examples on the subject of conical envelops the student is referred to Mr. Salmon's Treatise on Conic Sections. 10. Conic sections are inscribed in the same quadrilateral: to prove that the polar of any point in their plane envelops a conic section. Booth: On the Application of a New Analytic Method to the Theory of Curves and Curved Surfaces, p. 6. SECTION XXVII. Similar Curves. 1. To prove that two general equations of the second order ax2 + 2cxy + by2 +......= 0, ax + 2c,xy + b,y2 +......= 0 will represent similar curves, if c - ab c2- a,b, (a + b)2 (a, + b,) Changing the axes of the second curve through an angle 0, its equation will become a, (x' cos - y' sin 0) + 2c, (x' cos - y' sin ) (x' sin + y' cos0) + b, (x' sin 0+ y cos 0)2 +....... = 0; and therefore, in order that the two curves may be similar in form, 0 being supposed to be of such a magnitude that they are similarly situated in regard to their coordinate axes, we must SIMILAR CURVES. 421 have, X being some arbitrary quantity, 2Xa = a, + b + (a, - b,) cos20 + 2c, sin20.......(1), 2Xb = a, +, - (a - b,) cos20 - 2c, sin20.......(2), 2Xc = - (a, - b,) sin20 + 2c, cos2................(3). Hence we have (1) + (2),...... X (a + b) = + b,...........................(4), (1) - (2)...... X (a - b) = (a, - b,) cos2 + 2c, sin2...(5), (3 )2 + (5)2..... {4 + (a, - ),,)'2 + 4c,2....(6), (6) - (4)2...... X (c2 - ab) = 2 - a,.....................(7), (72 - ab c,2 - ab, (7) + ~4)...... 2 -(a + b) = (a, + b,)2 This condition of similarity is ascribed by Mr. Salmon to Mr. Jellett. Salmon: Treatise on Conic Sections, p. 194. 2. GH is the chord of an ellipse, which touches a concentric, similar, and similarly placed, ellipse, ' at the extremity B of one of its axes. From B are drawn any two chords /G BP, BQ, in the smaller ellipse, equally inclined to the axis through B, and / from H are drawn, in the larger '. U ellipse, two chords HE, HF, parallel respectively to BP, BQ. To prove that, denoting BP or BQ by r, HE by r', and HF by r", r' + r" = 2r. Let 0 denote the inclination of the chord BP to the axis through B; then, 2b denoting this axis, and 2a the other, we have by the polar equation, a(1, sin0) r= 2b cos..............(1) Again, by the polar equation to the larger ellipse, H being the pole, 2a,, 26,, denoting its axes, and h, k, the distances BB', BH, (1 - a "- sin2) r' = 2 (b - I) cos0 + 2k I sin, I I~~~~~~a 422 LINES OF THE SECOND ORDER. or, the two ellipses being similar, and b, - h being equal to b, ( a- a2 sin2 r' = 2bcosO + 2k 2 sin.....(2). Again, putting - 0 for 0 and r" for r', we have ( a sin20 r" = 2bcos0 - 2k sin.....(3). From (2) and (3) we have (1 ^-l sin" (+ r") = 4 cos0, and therefore, by (1), r' + r" = 2r. This proposition is due to Clairaut and serves as the basis of his investigations respecting the figure of the planets on the hypothesis of homogeneity of substance. Clairaut: Theorie de la Figure de la Terre, p. 158. Puissant: Recueil de diverses propositions de GeomJtrie, p. 156, troisieme &dition. 3& To prove that two curves of the second order, not of the same kind, cannot be similar. De la Hire: Sectiones Conicce lib. VI. prop. 1. To prove that all parabolas are similar curves. De la Hire: Sectiones Conicce, lib. VI. prop. 3. 4. If the inclination of a system of parallel chords in one central conic section to their diameter be the same as that of similarly related chords and diameter in another central conic section, and if these diameters are to each other as their parameters, to prove that the two conic sections are similar. De la Hire: Sectiones Conicce, lib. VI. prop. 2. 5. Two similar unequal ellipses AEB, aeb, have similar diameters AB, ah, and centres C, c, in the same straight line A T, the two conjugate diameters being parallel. From the centres C, c, are drawn the semi-diameters CE, ce, parallel SIMILAR CURVES. 423 to each other; E, e, are joined, and Ee is produced to meet Cc, A C B ca c b produced, in T. To shew that CT: cT:: EC: ec. De la Hire: Sectiones Conicce, lib. vI. prop. 6. 6. To shew that all hyperbolas, which have the same asymptotes, are similar curves. De la Hire: Sectiones Conicce, lib. VI. prob. 8. 7. If, through the common summit A of two similar curves of the second order, the greater axes of which coincide in direction, any two chords are drawn cutting one of the curves in B, B', and the other in D, D', to shew that AB: AB':: AD: AD'. Puissant: Recueil de diverses propositions de Geometrie, p. 154, troisieme edition. 8. Any number of concentric, similar, and similarly situated ellipses or hyperbolas, are intersected by a line parallel to a directrix in P, P', P",...: to prove that the extremities of diameters respectively conjugate to the diameters through P, P', P",... are in a line perpendicular to the directrix. 9. Any number of ellipses, concentric, similar, and similarly placed, are intersected by an hyperbola xy = c) in points P, P' P",...: to prove that the extremities of the diameters respectively conjugate to the diameters through P, P. P',.. are situated in an hyperbola XY =- - C 424 LINES OF THE SECOND ORDER. 10. Two similar ellipses have a common vertex, the directions of their major axes being opposite; a common tangent meets them in P, Q, and a perpendicular to their major axes through the common vertex meets PQ in 0: to prove that OP= OQ. SECTION XXVIII. Miscellaneous Problems. 1. To find the locus of the summit of a moveable angle, of invariable magnitude, the two sides of which always touch a conic section. By properly choosing the axes, the equation to the conic section may be written in the form y2 a + bx = 0........................(1). Let h, k, be the coordinates of the intersection of the two sides of the angle in any position: then the equation to either of its sides will be y -k = m(x - )....................... (2). Eliminating y between (1) and (2), we shall obtain a quadratic in x, the two roots of which must be equal: this condition will, after performing the requisite reductions, give us the equation 4 (ah2 + bh) m2 - 4 (2ahk + bk) m + (4akC2 - b2) = 0. Representing the two roots of this equation by m, m', we shall have 2ahk + bk, 4ak - b2 m + m = - 4mm' = 2 ah + bh ' ah2 - bh 2 2 (k2 + ah' + bh) and therefore (m - m') = ( ah - + ) (ahP + bh), 4a (h2 + k2) + 4bh- b2 ~and ~ + msm - 4 (ah2 + bi) MISCELLANEOUS PROBLEMS. 425 But, the magnitude of the moveable angle being constant, - m' = X (1 + mm'), where X represents a constant quantity. Hence, eliminating m and m' from the last three equations, we have, for the equation to the required locus, 16b2 (k'1 + ah' + bh) = X' {4a (Ph + k2) + 4bh - 2}2. Poncelet: Gergonne, Annales de Mthatheatues, tom. VIII.. 201. 2. To find the locus of the middle points of a system of chords in a conic section, which all pass through a given point. Let a diameter through the given point, as origin, be taken as the axis of x, and a straight line, parallel to its conjugate, as the axis of y. The equation to the curve will then be of the form y' = ax2 + 2bx + c.......................(1). The equation to one of the system of chords will be y = mx............................. (2). At the intersections of (1) and (2), there is (a - n2) ox + 2bx + c = 0. The abscissa x' of the middle point of (2) will be equal to the semi-sum of the roots of this equation; hence = - a........................(3). Also, y' being the ordinate of the middle point, y'= mx'..............................(4). Eliminating m between (3) and (4), we obtain for the equation to the required locus, y12 = ax'1 + bx' which is therefore a conic section similar to the proposed one and similarly situated, passing through its centre and through the given point. Durrande: Gergonne, Annales de Mathenatiques, tom. IX. p. 122. 426 LINES OF THE SECOND ORDER. 3. To determine the positions of the foci of a conic section of which the equation is given. The distance of the focus (a, 3) of a conic section, from any point of the curve, is always a linear function of the coordinates x, y, of the point. Let the equation to any conic section, referred to a system of rectangular coordinates, be axi + by + 2cy2a'x + 2b'y + c' = 0........(1). Then, by the nature of the focus, {(x - a)2 + (y - 8)2} = gx +h y + k........ (2), where g, h, k, are constants. Since the equations (1) and (2) must be identical, we have, making use of an arbitrary multiplier X, (- a)2+ (y-/3)2- (g + y + k)2=X (ax2+ by2+ 2cxy + 2a'x+ 2b'y+c'), and, equating the coefficients of like variables, we get six equations, 1 - g2 = Xa - a - kg = Xa', 1 - h2 = Xb, - - kh = Xb', -gh = X+c a' + 32 - 'P = Xc', involving the six unknown quantities a, /3, g, h, c, X. If we eliminate g, h, k, X, we shall obtain two equations of the second degree in a, 3. From these two equations we may obtain four pairs of values of (a, 3): two only of the pairs, however, are possible. Ex. Take the case of the ellipse referred to its axes, when its equation will be X2 y2 A- - B= 1 =1 Then a=~, b=C. - c=O, a'=O, b'=O, c'=-1: Then ~A=2 6- 2 c-O MISCELLANEOUS PROBLEMS. 427 and the six equations become 1 - = A2 a + kg = 0, - h2 = B-2 3 + kh = 0, gh = 0, as + 32 - i' = - X. Hence we obtain two systems of values, g=0, h= 0, 1i-. ff 1 a (B"2- A2)J, g = 2 (A42- B2)j, k=B, k=A, a = O 8 = 0, /8 =- (B- A2) =- a -(AS- B ). Since A and B have different values, it follows that one, and one only, of these systems of equations will give possible values to a and l3. Bret: Gergonne, Annales de Mathimatiques, tom. VIII. p. 317. 4. To prove that the difference between the squares of the reciprocals of any two coincident semi-diameters of two conic sections, which have the same minor directors, is constant. Booth: Annals of Philosophy, 1845, vol. II. p. 545. APPENDIX. 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