■MMxacoa mm HflMHBKH&si ',y«;;.v |m inaeoS hSbBBBOB •';■'":■.■ -v .■,:■•:'■'■■ . ■'■■-:■:'."..'■*•'•' ; -. :. ••••."•.■■■■ "'■■'•■■''■■ .'.'//■:-■■■■ BB Mua MBBBBBijIB ,v','.'. .* ' '■ ■'.'.■:.. •••■«■•■•■• ■'"■■'v.*"'--"' fflSKHl liiliiW i ■ ;V-JfI i , .Si I Cfavenfcon (Drees levies ANALYTICAL GEOMETRY JOHNSTON JSonbon HENRY FROWDE Oxford University Press Warehouse Amen Corner, E.C. Qtew $orft MACMILLAN & CO., 112 FOURTH AVENUE AN ELEMENTARY TREATISE ON ANALYTICAL GEOMETRY WITH NUMEROUS EXAMPLES BY W. J. JOHNSTON, M.A. LECTURER IN MATHEMATICS IN THE UNIVERSITY COLLEGE OF WALES BOSTON OOLLEO« JjBRARV 4 CHESTNUT BJ14,, m<* SS; MATH, DEPT^ AT THE CLARENDON PRESS 1893 PRINTED AT THE CLARENDON PRESS BY HORACE HART, PRINTER TO THE UNIVERSITY 1505bU PREFACE This book is to some extent an amplification of lecture notes; my special aim has been an easy and gradual develop- ment of the principles of the subject. An account of the elementary properties of the Ellipse, Parabola, and Hyper- bola precedes the discussion of the general equation of the second degree. I have had in view the requirements of two classes of readers. First, a limited course of the subject is read by most students in the University Colleges of this country. Such a course, confined to selected paragraphs, with the exercises appended thereto, is indicated at the end of the table of contents. Secondly, as chapters on Trilinears, Reciprocal Polars, and Projection are included, it is hoped that the book will be useful to candidates for mathematical honours ; and especially as an introduction to the writings of Dr Salmon. It seems superfluous to state that I am under great obliga- tions to the works of Dr Salmon; I would also make grateful mention of Dr C. Taylor's Conies, and the tract on Conies and Curves by the Rev. W. H. Laverty. The Exercises at the end of each chapter have been mainly selected from the examination papers of different vi Preface Universities. I have freely availed myself of Mr Miller's kind permission to extract problems from the Educational Times. Many problems have been manufactured to illustrate special points ; and for several of the most interesting I am indebted to Professors Curtis, S.J., Genese, and Purser. In endeavouring to simplify the treatment of the subject, I found that some novelties were unavoidable. Thus the determinant form of the equation of the join of two points is used throughout, the elements of the determinant notation being explained in § 20 ; my experience with beginners justi- fies this innovation. The generality of the usual trapezium proof of the formula for the area of a triangle in terms of the co-ordinates of its vertices is invalid without a more detailed examination than is usual of the consequences of the con- vention of signs. A more rigorous proof has therefore been given. I have included the usual methods of tracing a conic whose Cartesian equation is given ; but further methods, which seem to possess some advantages, are given in §§ 294, 296. Professor Purser's investigation of the equation of a dia- meter (§§ 228, 255, 303) is the simplest that I know of. The use of the invariants 0, 6' (§§ 443, 444) has not been emphasised in any previous textbook; this method is due to Professor W. S. Burnside. Further applications are given in his paper in vol. 8 of the Quarterly Journal. I have ventured to indicate (§§ 479, 480) an elementary method of projecting metrical properties, which was sug- gested to me by a paper of Clifford's on Analytical Metrics. The formula of § 481 had previously been deduced from other considerations by Professor Curtis. I have merely suggested my own method; further development is easy. As an instance of its use I may mention that the equation Preface Vll of the conic-tangent to three conies, all the conies passing through two given points, is thereby at once deducible from Dr Casey's equation of the circle-tangent to three circles. I have further to acknowledge my great indebtedness to my former mathematical master, Mr R. C. J. Nixon, from whom I have received many valuable suggestions, and to whose advice as to the arrangement of the book much of its value is due. My best thanks are due to my colleague Professor Genese for many important hints ; and to another colleague, Mr A. W, Warrington, whose generous assistance has been of the utmost value to me in preparing the work for the Press. W. J, JOHNSTON. University College of Wales, Aberystwyth. April, 1893. CONTENTS CHAPTER I. PAGE Co-ordinates i CHAPTER II. The Straight Line . . . .28 CHAPTER III. The Straight Line (continued) , . . . . . . 51 CHAPTER IV. Applications to Geometry . . . 75 Loci ............ 79 Loci found by elimination 85 CHAPTER V. Equations with Linear Factors. Abridged Notation ; Ranges and Pencils ........... 93 Abridged notation .......... 103 The line at infinity ......... 107 Harmonic ranges .......... 108 Cross ratios ........... 109 Involution . . . . . . . . . . .114 x Contents CHAPTER VI. PAGE The Circle ......'....,. 121 Power of a point .......... 135 CHAPTER VII. Transformation of Co-ordinates ....... 166 Invariants ........... 17 1 CHAPTER VIII. The Parabola ........... 177 * CHAPTER IX. The Ellipse 205 CHAPTER X. The Hyperbola ........... 244 CHAPTER XI. General Equation of the Second Degree 265 Reduction to Canonical Form ...... 265-276 Condition for rectangular hyperbola ...... 272 Equation of asymptotes ......... 272 Equation of tangent at (x'y') ........ 279 Diameters ........... 280 Poles and Polars .......... 282 Segments of chord .......... 287 Chords of intersection with a circle are equally inclined to the axes 288 Carnot's Theorem .......... 289 Power of a point with respect to a conic ..... 289 Equation referred to tangent and normal ; Fregier s Theorem . 290 Contents XI PAGE Two conies have a common self-conjugate triangle . . .291 The system S — A S' = o ........ 291 All rectangular hyperbolas circumscribing a triangle pass through the ortho-centre ......... 292 Equation of conic passing through two given points on each axis of co-ordinates ......... 293 Equation of conic touching axes ....... 294 Five conditions determine a conic ...... 297, 298 Nine-point conic . . . . . . . . - • . . 298 Tangential equation ......... 299 Equation of pair of tangents . . . . . . . . 300 Equation of director circle ........ 300 Equation to determine eccentricity ...... 304 Equation of axes .......... 305 Centre locus of conies inscribed in a quadrilateral .... 307 CHAPTER XII. Polar Equation of a Conic referred to Focus ; Confocal Conics . 308 Confocals 315 CHAPTER XIIL Abridged Notation ; Miscellaneous Propositions . 3 2 4 Abridged notation of the straight line .... 324-328 The system S - A LM = . . . 328 The system S - A L 2 = , 329 Interpretation of LM = R 2 . 33o Interpretation of (Xy = A/38 . • 33i The focoids . 334 Determination of foci . 336 Normals .... , 34o Curvature .... 342 Similar conics . 344 Invariants . 347 One-one correspondence 348 Plane projection . 35© xii Contents CHAPTER XIV. PAGE Trilinear Co-ordinates 362 Equation of straight line in terms of perpendiculars p, q, r from vertices of triangle of reference ...... 3^4 Areal co-ordinates ........-• 3^7 Relation connecting the perpendiculars p, q, r . . - . 3^8 General equation of second degree ...... 37 1 Condition for parabola .....-.•• 373 Condition for rectangular hyperbola ...... 373 Equation of asymptotes ........ 373 Conditions for a circle 373 Circumscribing conic ......... 374 Inscribed conic . . . . . . . . . . 377 Equation of nine-point circle ........ 379 Feuerbach's Theorem ......... 380 Pascal's Theorem 381 Brianchon's Theorem 382 Equation of conic referred to a self-conjugate triangle . . . 383 Self-conjugate circle 384 Equation of conic touching four given lines 384 Equation of conic passing through four given points . . . 385 Conic referred to two tangents and their chord of contact . . 386 Cross ratio of range in which four fixed tangents meet any fifth tangent is constant ... ..... 387 Invariants ........... 388 Faure's Theorem 389 CHAPTER XV. Envelopes 400 Tangential co-ordinates ......... 403 CHAPTER XVI. Methods of Transformation 411 Reciprocal polars . . . . . . . . . .411 Projection ........... 418 Projection of metrical properties . . . . ... . 423 Contents XUl The following limited course is recommended to beginners. The Straight Line and Circle in Rectangular and Polar Co- ordinates : — Chap. i. §§ 1-8, 10-13, 15-24, 26-40, 44. n. §§ 45-59, 61-67. in. §§ 68-83. iv. §§ 94-99, 102, 103, 107. v. §§ 112-114, 123-125. vi. §§ 149-165, 168-186, 197-199. II. The Following Additional Portions of Chapters I-VII should BE READ BEFORE READING THE CHAPTERS ON CoNICS \ — lap. 1. §§ 14, 2 5- „ 11. § 60. „ in. §§ 84, 86, 89. „ IV. §§ 100, 101, 104-106, 108-111. „ v. §§ 116-121, 126, 129-137. „ VI. §§ 166, 167, 187-191, 200, 202 „ VII. §§ 204-208. III. The Central and Focal Properties of Conics; Poles and Polars, etc. Chap. vm. §§ 213-234. „ ix. §§ 237-262. „ x. §§ 266-285. „ xi. §§ 287-303, 306-311, 313, 314, 316-318, 320, 324. „ xii. §§ 326, 328-331, 336, 338-340. ANALYTICAL GEOMETRY CHAPTER I. CO-ORDINATES POSITION DETERMINED BY CO-ORDINATES § l« Algebra may be applied to deduce relations between geo- metrical magnitudes. For example, combining the statements (i) The area of a rectangle is measured by the product of the lengths of its sides, and (2) (a + b) (a - b) = a 2 - b 2 , we deduce the theorem : The rectangle contained by the sum and difference of two lines is equal to the difference of the squares on those lines. As another example let us solve this problem : Divide a given line AB into two parts such that the square on one may be twice the square on the other. Suppose that X is the point required. X B Put a for AB, and x for AX. Then x 2 = 2 (a - x) 2 .-. x = V^2 (a — x) a V2 / .-. x = — - — 7- = a (2 — v 2) 1 + v 2 v 7 B Analytical Geometry [»• § 2. In other geometrical inquiries we have to consider position. Thus the locus of a point which moves subject to an assigned condition may be required. It is not at once obvious how Algebra may be applied to such questions. The method of co-ordinates is an adequate one for the purpose. This method was first in- troduced by Descartes in his Geometrie, 1637 ; we shall now give an account of it. § 3. In a plane take two fixed lines (or axes), X'OX, Y'OY. Then through any point P in the plane drawing PN parallel to YOY' to meet X'OX in N ; to every position of P correspond definite lengths of ON, PN. Conversely, if the lengths of C ON, PN are given, the point P is determined. We have merely to measure along OX the given distance ON, and then the other distance NP on a line through N parallel toOY. Or thus, if we are given ON = a and PN = b; measure a distance ON = a along OX, and a distance OM = b along OY: parallels through N, M to the axes intersect in the point re- quired. § 4. The two fixed lines OX and OY are called the axes of co-ordinates or simply the axes ; their point of intersection O is called the origin. OX is called the axis of x and OY the axis ofy. If the axes are at right angles they are said to be rectangular, otherwise they are oblique. O N and N P are called the co-ordinates of P ; O N is called its abscissa or its ' x/ and N P its ordinate or its ' y/ The point whose co-ordinates are a and b may be referred to as 5-] Co-ordinates 3 'the point (x = a, y = b)'; or briefly, 'the point (a, b)': the, abscissa being always named first. § 5. The straight lines XOX' and YOY' divide the plane into four compartments. The following convention serves to distinguish points in the several compartments. Distances measured along OX are considered positive and along OX' negative ; distances measured along OY are considered posi- tive and along OY' negative. Accordingly, if along OX we measure ON = ON'=a and OM = OM'= b, and through N, N', M, M' draw II s to the axes meeting in P, Q, R, S : then the co-ord's of P are x = + y = + X = - y x = — a) = - b) Zli\ y X = y ziii Thus, as in Trigonometry, the abscissa of a point is positive or negative according as the point is on the right or the left of Y'OY, and its ordinate is positive or negative according as the point is above or below X'OX. We shall presently make some further remarks on the use of the signs + and — to indicate opposite directions of measurement along any line. B 2 Analytical Geometry [6. § 6. To illustrate this, Suppose the axes rectangular; and let us 'plot' the points (3, 2), (- 3, 4), C- 4> - 2), (2, - 3). Qr -M N -S Take any convenient unit of length. Then the point (3, 2) is P (-3, 4)»Q » » (~4> - 2 )»R ( 2 > -3)»S What are the co-Orel's of N in the figure ? ^7z.y. (3, o) ; for to reach N from O we measure 3 units to the right so that the abscissa of N is 3 : while the distance to be measured upwards or downwards is zero, so that its ordinate is o. What are the co-ord's of L? Ans. ( — 3, o). What are the co-ord's of M ? Ans. (o, 3). What are the co-ord's of the origin O ? Ans. (o, o). THE CONVENTION OF SIGNS § 7. If distances measured in either direction along a straight line be considered positive, distances measured in the opposite direction are considered negative. If A and B are two points on a straight line, then in Modern Geometry AB not only stands for the distance between A and B, but also implies that this distance is measured in the direction from A to B. If then AB = + 5, we have BA = - 5. Generally AB = - BA. It follows that if O, A, B are three points anyhow placed on a B straight line, then AB = OB - OA (1) 9-] Co-ordinates For a point may travel from O to B by two steps : first from O to A, then from A to B. Thus OA +. AB = OB. Subtract OA from both sides, .-. AB = OB — OA . . (2) Cor' — Let O be the origin of co-Orel's and A, B two points on the axis of x. Let OA = x 2 and OB = x 2 so that the co-ord's of the points A, B are (x 15 o), (x 2 , o). Then AB = x 2 — x x (3) § 8. To illustrate this I 1 ! 1 I ' ■ ' O A B 1 1 iii'i A OB 1 1 ' ' ' B O In the first figure AB = 6, OA = 2, OB = 8 ; 6 = 8-2 „ second „ AB = 6, OA = — 4, OB = 2; 6=2 — (— 4) „ third „ AB = 6, OA = - 8, OB = - 2; 6 = (- 2) - (- 8) Thus in each case the equation AB = OB — OA is verified. § 9. Ex. 1. If A, B, C are three points in a straight line AB + BC + CA = o. For if a point travels successively from A to B, from B to C, and from C to A, the total distance traversed is zero. Or thus : Let a, b, c be the distances of A, B, C from a fixed point O on the line ; then AB = b — a, &c. .*. AB + BC + CA = (b - a) + (c - b) + (a - c) = o. Ex. 2. Let the distances of four points A, B, C, D anyhow situated on a straight line from a fixed point O on the line be a, b, c, d ; then the identity (d - a) (b - c) + (d - b) (c - a) + (d - c) (a - b) = o gives AD . CB + BD . AC + CD . BA = o. Analytical Geometry [,«, Suppose now that the four points are in the order A, B, C, D. Then since -£ ^ CB = - BC, BA = we obtain A B -AB, BD . AC = AD . BC + CD . AB, where all the lines and rectangles are positive *. DISTANCE BETWEEN TWO POINTS § 10. To find the distance between two points whose co-ordinates are given. Suppose that the axes are rectangular. Let the two points be P, (x^), and Q, (x^). Draw PM, QN parallel to OY, to meet OX in M, N and PR parallel to OX to meet QN in R. Then PQ 2 = PR 2 + QR 2 . But PR = MN = ON - OM = Xg x x and QR = QN - PM = y 2 - y t .-.PQ 2 = (x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 §11. We may express this result in words : The square of the distance between two points is equal to the square of the difference of their abscissae together with the square of the difference of their ordinates. The order in which we take the differences is indifferent, as they are afterwards squared. Y K*2 < -y 2 ) F R O IN n r J > * A graphical proof of this theorem is given in Nixon's Euclid Revised: p. 104. 3-] Co-ordinates § 12. The proof in the last § is perfectly general, in virtue of equation (2) of § 7. As an Exercise the learner may go through the proof for the case where P is in the first compartment and Q in the third. § 13. Ex. 1. Find the distance 6 between the points (2, —3) and ( — 3, 4). B 2 = [2 - (- 3)] 2 + [- 3 - 4? = 5 2 + (- 7) 2 =25+49 = 74_ .-. 8 = >/74 The learner should illustrate this by a diagram. Ex. 2. Find the distance b between (x^) and the origin. g 2 = ( Xi _ )2 + (y x _ o) 2 = Xl 2 + yi 2 . This is also obvious from the figure, § 10; OP 2 = OM 2 + PM 2 = x x 2 + y, 2 . Ex. 3. Find the co-ord's of a point equidistant from the three points (2, - 2), (8, - 2), (8, 6). Let its co-ord's be x, y. .-. (x - 2) 2 + (y + 2) 2 = (x - 8) 2 + (y + 2)2 = (x - 8) 2 + (y - 6) 2 . Cancel x 2 + y 2 .-. — 4X + 4y + 8 =— 16 x + 4y + 68 =— i6x — 12 y + 100 Solving these we get x = 5, y = 2. Ex. 4. Find the condition that the point (x, y) may be at a distance 3 from the point (4, 5). 3 2 = 9 = (x - 4) 2 + (Y - 5) 2 /. x 2 + y 2 — 8x — ioy + 32 = o. This equation then is satisfied by the co-ordinates of every point on a circle whose centre is (4, 5) and radius 3. Ex. 5. Express that a point P (x, y) may be equidistant from A (1, 2) and B (3, 4). PA 2 = ( X - i) 2 + (y - 2)2 PB 2 = (x - 3 ) 2 + (y - 4) 2 - Equating these values and simplifying we get x + y — 5 = o. We know from Elementary Geometry that P must lie on the line which bisects AB at right angles : hence the co-ordinates of every point on this line satisfy the equation x + y — 5 = o. 8 Analytical Geometry [14. § 14. Suppose that in § 10 the axes are oblique, the angle between them being to. Then PRQ = ONQ = 180 - co .'. PQ 2 = PR 2 + QR 2 - 2 PR . QR cos PRQ = PR 2 + QR 2 + 2 PR . QR cos co; ... PQ 2 = (x 2 - xj 2 + (y 2 - yj 2 + 2 (x 2 - x x ) (y 2 - y x ) cos co. Cor' — The distance of (x, y) from the origin is Vx 2 + y 2 + 2 xy cos co. Note — Oblique axes are rarely used as the formulae are more complicated ; and in future we shall usually assume that the axes are rectangular. DIVISION OF JOIN OF TWO POINTS IN A GIVEN RATIO § 15. To find the co-ordinates of the point which divides the join of (x 2 y,) and (x 2 y 2 ) in a given ratio m : n. Let (x, y) be the co-ord r s of the required point R, so that PR : RQ = m : n. Draw II s to the axes through R, P, Q as in the figure. Then PT : TV = PR : RQ = m : n 5 or x — x x : x 2 x = m : n .-. n (x — xj = m (x 2 — x) 7- J Co-ordinates . mx 2 + nx 1 m + n Similarly y = m * + . n * 17 m + n These formulae are applicable whether the axes are rectangular or oblique. CV — The co-ord's of the mid point of PQ are x = x, + x 2 _ Yl + y 2 2 ' • 2 § 16. If PQ is to be divided externally in R so that PR : QR = m : n ; then since QR = — RQ we have PR : RQ = m : - n. Thus by changing n into — n in the formulae of § 15 we obtain the required co-ord's _ mx 2 - nx t _ my 2 - ny t x — , y — • m — n m — n § 17. Ex. 1. The co-ord's of A are (3, — 1) and of B are (6, 4) ; find the co-ord's of the point of trisection of AB nearest A. Here m = 1, n = 2, . x -_ * 2 + 2X x = 6 + 2 (3) 2 3 3 4 y 2 + 2y x = 4 + 2(- 1) = 2 3 3 3 Ex. 2. A, B being the same points as in Ex. 1, AB is produced to C so that BC = AB ; find the co-ord's of C. Here AC = 2 BC = - 2 CB, AC : CB = 2 : - 1, .*. m = 2, n = — 1 ; and 2 Xo — X-i ,~ x = a _ i = 2 x 2 - x x = 2 (6) - 3 = 9 2 y, - y, y = \ _ / = 2y 2 - Yl = 2 (4) + 1 = 9 IO Analytical Geometry ['7. Ex. 3. Prove that the joins of the mid points of the opposite sides of a quadrilateral and the join of the mid points of its diagonals meet in one point and bisect each other. ( x sys) Let ABCD be the quad', C x i Yi) tne co-ord's of A, &c. Then the co-ord's of P the mid point of AB are Kxi+Xa); | (yj + y 2 ) ; of Q the mid point of CD are |(x 3 + x 4 ), |(y 3 + y 4 ); of T the mid point of PQ are \{^ + Xc + Xo + X, j- = \ (Xj + x 2 + x 3 + x 4 ), 2 2 and \ ( yi + y 2 + y 3 + y 4 ). We shall get the same values for the co-ord's of the mid points of the joins of AD, BC and of AC, BD. ACxxYi) C(x 3 y 3 ) Ex. 4. A, B, C are three points whose co-ord's are given, BC is bisected in D and on AD is taken DG - \ DA. Find the co-ord's of G. The abscissa of D is 2 ( X 2 + X 3) 'i .'. the abscissa of G is 2 X |(x 2 + x 3 ) + x ± Xi T Xo T Xo 2 -(- 1 Similarly its ordinate is yi + y 2 + y 3 We see then that the three medians of a triangle meet in one point, this point being the point of trisection of each furthest from the vertex. This point G corresponds to the centre of gravity of a physical triangular lamina. 7-] Co-ordinates it Ex. 5. The angle O of a triangle OB A is a right angle, M is the mid point of AB : prove OM = |AB. Take OA for axis of x and OB for axis of y. Then the co-ord's of A are (a, o) and of B (o, b), .•. of M are - and — ; 2 2 OM /a 2 W _ I Va 2 + b 2 = I AB. Exercises 1. Find the length of the join of (1, — 3) and (o, —2). Ans. a/2. 2. The vertices of a triangle are (— 1, — 1), (2, 3) and (o, 2). Plot these points and find the lengths of the sides of the triangle. Ans. 5, y^, V IO « 3. Show that the points (1, 2), (1, 6), (V12 + 1, 4) are the vertices of an equilateral triangle. Ans. Side = 4. 4. Find the co-ord's of the mid points of the sides of the triangle in Ex. 2. Ans. (i, 1), (1, a*), (-&*)• 5. Find the co-ord's of the centre of its circum circle. Ans. (2|, — |). 6. Find the point of trisection nearest (1, 2) of the join of (1, 2) and (7» - 13)- Ans. (3, — 3). 7. The co-ord's of P are (4, 1) and of Q (9, 8) : find the co-ord's of a point R in PQ produced such that (1) QR = PQ, (2) QR = 2 PQ. Ans. (14, 15), (19, 22). 8. Show that (1, 2) lies on the join of (—4, 5) and (11, — 4). In what ratio does it divide this join? Ans. 1 : 2. 9. Show that the following four points are the vertices of a parallelogram : C- i. ~ 2), (2, - 1), (5, 2), (2, 1). 10. Show that the following four points are the vertices of a square: (—2, 5), (2, 2), (5, 6), (1, 9). 12 Analytical Geometry [.8. 1 1 . Prove by co-ordinates that the join of the mid points of two sides of a triangle is half the third side. 12. M is the mid point of the side BC of a triangle ABC : prove by co-ordinates that AB2 + AC 2 = 2 AM 2 + 2BM 2 . determinants; area of a triangle § I8> In the more advanced parts of the book we shall assume that the reader is familiar with the properties of determinants, which are proved in most modern books on Algebra. For the convenience of the junior reader we shall explain here the elements of the determinant notation; this notation is often useful in abbreviating formulae, and a knowledge of it often facili- tates numerical calculations. § 19. Def— The symbol a b c d means ad — ■ be, and is called a determinant of the second order. = 3 (- 1) - (- 4) (- 2) = -3 - 8 = - 11. Ex. 3 - 4 — 2 — 1 Again 1 — 2 3 o = o - (- 6) = + 6. N.B. Observe that the diagonal term ad always comes first. § 20. Def— The symbol a d g b e h c f k means a e h -b d g f k + c d g e h = a (ek - hf) - b (dk - gf) + c (dh - ge) and is called a determinant of the third order. Observe that the determinants of the second order which are multiplied by a, — b, c are got from the original determinant by scoring out the row and column containing a, b, c respectively. 22 •] Co-ordinates A 3 § 21. Ex. = 2 — II — 2 - 5 i 2 -5 -i - 15 — II — 2 -(-5) + (- 4 ) 1 - i - 15 — II — 2 - i - 5 — ii i = 2 [- I 5 - (+ IO)] - (- 5) [- I - (+ 55)] - 4 [2 - (+ 165)] = — 50 — 280 + 652 = 322. After a little practice it will be found unnecessary to set down more than the following steps : The det/ = 2 (- 25) + 5 (- 56) - 4 (- 163) = —50 — 280 + 652 = 322. § 22. To find the area of a triangle in terms of the co-ordinates of its vertices. Let the vertices be A (x^), B (x 2 y 2 ), C (x 3 y 3 ) Through A, B, C draw II s to the axes. Then A ABC = A AHB + A BHC + A CHA = |DHDEB + iaBHCF + inCHKG = i[pGAEF _ oADHK] = J(AE. AG - AD. AK) = i [(x 2 - Xi) (y 3 - y x ) - (x 3 - x x ) (y 2 - y t )], multiplying out we get 2 area = x x (y 2 - y 8 ) - y x (x 2 - x.) + x 2 y 3 - x 3 y 2 . G cCxgYs) F K H \ s(x2y 2 ) / Ktoti D I 1 14 Analytical Geometry [23- area = But this is the expansion of x x y x x 2 y 2 x 3 y 3 x x y x x 3 y 3 This formula should be remembered. Cor' — Let the vertex A be at the origin : then x 1 = o, y 1 = i(x 2 y 3 -x 3 y 2 ). = o, area = 4 o x„ o y 2 y 3 The learner should prove this case independently. § 23. Ex. Find the area of the triangle whose vertices are (1, 2), (—2, 3), (5, 6). 2 area = 121 - 2 3 1 5 6 1 = I(~ 3) - 2(- 7) + l(~ 27) = -3 + 14-27 = - l6. Thus the area contains 8 square units. In § 41 we shall account for the negative sign. § 24*. Ex. Find the area of the triangle (2, 3), (— 4, 1), (— 1, 2). = 2 (- 1) -3 (-3) + 1 (-7) =-2 + 9-7=* o. 2 area = 2 3 1 -411 — 121 We infer that the three points are collinear. Similarly three points (x^), (x 2 y 2 ), (x 3 y 3 ) are collinear if *1 yi 1 x 2 y 2 1 *3 y 3 1 = o. 26.] Co-ordinates I .5 Exercises 1. Evaluate the determinants : I 2 4 2 3 2 — 1 1 I 2 3 » — 2 4 j 1 2 — 1 I 3 4 - 3 - 4 — 1 1 2 Ans. I, o, 14. 2. Evaluate the determinants : 6 7 - 3 4 3 1 3 — 1 2 5 2 2 > 3 6 > - 5 2 4 2 4 1 2 5 7 - 3 6 Ans. - 9 X > I2 > 16. 3. Find the areas of the triangles whose vertices are i°. (2, 1), (3, - 2), (-4,-1) 2 . The origin and (2, 3), (4, - 5) 3°. (2, 3), (4, - 5), (- 3. - 6) Ans. 10, n, 29. 4. Show that (4, 5), (—2, 3), (16, 9) are collinear. § 25. If the axes are oblique and inclined at an angle co we obtain as before 2 area ABC = □ GAEF - □ ADHK, but area of o GAEF = AE . AG sin GAE = (x 2 — x x ) (y 3 — y^ sin co, &c. area of A ABC = \ sin 00 y 3 LOCUS OF AN EQUATION. § 26. An equation connecting the co-ordinates x and y repre- sents a geometrical locus. The learner will see this clearly after he has considered the following instances. i6 Analytical Geometry [»r. X — 2 — i O I 2 3 4 y + 5i 2^ z 2 I 2 I ~ 2 I ~~2 i 2 2± § 27. Take for example the equation 2y = x 2 — 3X+1. Here if we assign any value to x we can determine a corresponding value of y. Thus if x = 3 we get y = |. We may form a table of corresponding values of x and y : &c. The co-ordinates then of the points (— 2, 5|), &c. satisfy the equation. Let us ■ plot ' these points. It will be seen that the points are arranged according to a continuous law, and we may draw a curve through them on which in- termediate points will lie, and by plotting intermediate points, as for instance x = if, y = - f, we may draw the curve with greater accuracy. We see then that there exists a curve such that the co-ordinates of every point on it satisfy the equation 2y = x 2 — 3X + 1 ; accordingly this equation represents that curve locus. § 28 ■ Suppose we wish to find the points P and Q where the curve cuts OX ; the distances OP, OQ might be found by measurement to a close degree of approximation. Or we may calculate OP and OQ thus, y = o for each of these points. But their co-ord's satisfy the equation 2y = x 2 — 3X + 1. Put then y = o in this equation. .-. x 2 — 3X + 1 = o. Thus OP = .382, OQ = 2.618, PQ = V5 = 2-236. 3I-] Co-ordinates 17 8 29 , As another example consider the equation y = 2 x + 3. Form a table and plot the points as before : x — 2 —1 0123 y -1 13 5 7 9 It is seen that the points are ranged on a straight line. Sec. Note — The equation y = 2 x + 3 is of the first degree in x and y : it will be proved in Chap. Ill that such an equation always represents a straight line. § 30. Generally an equa- tion represents the locus of a point which moves so that its co-ordinates satisfy the equa- tion. If we assign any value a to x we may determine one or more corresponding values of y, say y=b ) y=b' ) y= b", ... Thus the points x = a\ x = a ) x = a) y=b)' y = b')' y = b"j' &c - lie on the locus. We may thus obtain as many points on the locus as we please. § 31. Exs. 4 and 5, § 13, afford further illustrations. In each of these examples a point moves subject to an assigned condition ; we have replaced the condition by a relation between the co-ordinates x, y of C i8 Analytical Geometry [3= a point on the locus. Thus the resulting equation in Ex. 4 represents a certain circle, and that in Ex. 5 a certain straight line. 8 32 • Conversely we may give a geometrical interpretation to an equation containing x or y or both. Ex. 1 . Thus the equation x = y expresses that if P is any point on the locus then PN = ON. By Euclid Book I, we may see that the A locus of P is the bisector of YOX. Ex. 2. y = 3 expresses that the ordinate of any point on the locus is 3. Hence y = 3 represents a straight line parallel to OX at a distance 3 from it. Ex. 3. Again, x 2 + y 2 = r 2 ex- presses that the distance of (x, y) from the origin is r : hence this is the equation to a circle whose centre is the origin and whose radius is r. § 33. Assuming, as stated in § 29, that a simple equation represents a straight line ; we see that to draw a straight line whose equation is given it is sufficient to plot two S' points on the locus and join them. B It is best to get the points where the line cuts the axes. Thus to draw the line 2X— 3y + 6 = o- Putting y = o we get x = — 3. Thus ( — 3, o) is a point on the line. This is the point A in the figure. Putting x = o we get y = 2. Thus (o, 2) is a point on the line. 34-] Co-ordinates 19 This is B in the figure. Hence the required line is AB. Again, draw the line 3 x + 2 y = o. We see that this is satisfied if x = o, y = o ; thus the line passes through the origin. If we give x any other value we can find y. Thus put x = 2, then y = — 3. The required line is OC. INTERSECTIONS § 34-. The co-ordinates of a point of intersection of two loci satisfy both their equations ; accordingly we may obtain the points of intersection of two loci by solving for x, y from their equations, considered as simultaneous. Ex. 1. Find the point of intersection of the straight lines 3X + 4y = 11, y - x = 1. Solving these, x = 1, y = 2. Ex. 2. Find the points of intersection of 2y = x 2 — 3 X + 1 and 2 y + x = 4. Solving these we find (—1, 2^) and (3, |). The learner should draw diagrams to illustrate these Examples. Exercises 1. Plot the loci of the equations y 2 = 4x, x + y-4 = o, 2X-3y + 6 = o, 3x-5 = o, xy = 24, x 2 + y 2 = 25 2. Find the p't or p'ts of inters'n of the loci i°, x + y- 4 = o, 3 y-2X + 3 = o 2 , x + y - 4 = o, 3 x-4 = o 3°, x 2 + y 2 = 25, 3 x + 4 y = 25 4°, y 2 - 4 x, y - 6 x L y x ,_ y ab ab v a b ' b a (ab ab \ a - *!)' aTb/' c 2 so Analytical Geometry [35- 3. Find the distance between the two p'ts of inters'n of x 2 + y 2 = 25 and 4 x — 3 y = 7. Am. iff. 4. Find the intercept which the locus 4(x — 2) 2 = y+i cuts off on the axis of x. Ans. 1. 5. Find the intercepts which the locus ofx 2 + y 2 — 4X — 8y — 5 = cuts off on OX and OY. Ans. 6, 2 V21. § 35. The degree of an equation is the greatest sum of the indices of x and y in any term. Thus 2X+3y+4 = o is of the first degree. xy+5x+6 = o is of the second degree, for the sum of the indices in the term xy is 1 + 1 = 2. x 2 y + y 2 = x is of the third degree, for the sum of the indices in the term x 2 y is 2 + 1 = 3. Vx -f- Vy = 1 when rationalized gives x 2 — 2 xy + y 2 — 2X — 2 y + 1 = 0; this is an equation of the second degree. Equations are classified according to their degrees. The general equation of the first degree is ax+by+c = o, where a, b, C stand for constant numbers. It will be shown in Chap. Ill that this represents a straight line. The general equation of the second degree is ax 2 +2hxy+by 2 +2gx+2fy+c = o. It will be shown hereafter that this represents a conic section. POLAR CO-ORDINATES § 36. The conception in the method of rectangular co-ordi- nates is that we may reach a point P by first travelling a definite 38.] Co-ordinates 21 distance ON along a fixed line OX, and then another definite distance NP parallel to another given direction OY. We may reach the point P by another method. Rotate OX through an angle O into the position OK and on OK measure off a definite length O P. If OP = r and POX = 6, then r, 6 are called the polar co-ordinates of P. These co-ordinates evidently determine the position of P. r is called its radius vector and 6 its vectorial angle. § 37. 6 is positive or negative according as we suppose that OX revolves in a counter-clockwise or in a clockwise direction. In the figure it is positive. r is positive if it is measured along the line which is the new position of OX after the rotation; i.e. if it is measured along OK. Negative values of r are got by measurements along OK produced backwards, i.e. along OK'. OX is called the initial line. § 38. Ex. i. Plot the points (3, o), (3, -J > ( 3> —y These are A, B, C in the figure. (Page 22) Ex. 2. Plot the point ( — 3, —J . This is C in the figure. (Page 22) 22 Analytical Geometry [39- Here instead of measuring the distance 3 along OK it is measured along OK'. (2 TT \ > 3j* This is C in the figure. Ex. 4. Plot the point (—3, if). Ans. A in the figure. Ex. 5. Plot (3, 2 mr+ —J , where n is any integer. Ans. B in the figure. § 39- To transform from rectangular to polar co-ordinates or vice versa. 4 1 ■ r 1 / X N From the right-angled triangle O PN we have x = r cos # ) y =s rsin 3 These express x, y in terms of r 5 0. Again r 2 = x 2 + y 2 } and ^ = tan<9 f x ) :. r = Vx 2 + y 2 ^zrrtan" 1 ^ X These express r, # in terms of x, y. (*) 4 o.] Co-ordinates ^3 §40. To express the distance between two points in terms of their polar co-ordinates. P (»\0i) Q (noi) Let the points be Then PQ 2 = OP 2 + OQ 2 - 2 OP. OQ cos POQ, a PQ 2 = r, 2 + r 2 2 - 2 r x r 2 cos (0 1 - 2 ). Otherwise : let the rectangular co-ordinates of P be (x t y x ) and of Q, (x 2 y 2 ). Then x x = r x cos 6 X } x 2 = r 2 cos 2 ) y L = r 1 sin6 1 ) y 2 = r 2 sin 2 ) .-. PQ2 = ( Xl - x 2 ) 2 + ( Yl - y 2 )2 = (r t cos B t — r 2 cos 2 ) 2 + (r x sin X — r 2 sin 2 ) 2 = r^ (cos 2 X + sin 2 X ) + r 2 2 (cos 2 2 + sin 2 2 ) -2r 1 r 2 (cos $! cos 2 + sin X sin 2 ) = r*! 2 + r 2 2 - 2 r x r 2 cos (0 X - 2 ). Exercises 1 . Plot the points whose co-ord's are ( 3 '!)- ( 3 --f)' (-*t)' (- 3, -3' Obtain their rectangular co-ord's. -<• (*£• ^f 2 > (f ; =¥*)' d- ^> (r¥* : ^) 2. Find the polar co-ord's of the p'ts whose reef co-ord's are (1,1), (-1,2), (-3,3), (-4* -4). Jm.(V2,?y, Wh -tan-^2); (3^,^); (4 V*, 5 ^) • 24 Analytical Geometry [ 4 .. 3. Find the distance between the points ( 4, — J and ( 2 V2, - ) • Ans. 2 {V$— *)• 4. Show that the origin and the points (4, *-—\ (4, — \ are the vertices of an equilateral triangle. 5. Express in polar co-ord's the eq'n, x 2 — y 2 = a 2 . Ans. r 2 cos 2 = a 2 . 6. Express in rect' co-ord's the eq'n, r 2 sin 2 = 2 a 2 . Ans. xy = a 2 . 7. What loci are represented by the equations r = 5, = - — ? § 4|. In fig', §40, area OPQ = \ OP. OQ sin POQ = \r x r 2 sin (0 X - 2 ). This is positive or negative according as 1 — 2 is positive or negative. The reader will see on reflection that if we go round the triangle OPQ in the order in which the points O, P, Q are mentioned ; then if this order is clockwise X — 2 is positive ; otherwise it is negative. We can now account for the sign of the determinant in § 22. Take A for origin and AE for initial line in fig' of that § . Let the polar co-ord's of B be (r x X ), and of C (r 2 2 ). Then x 2 — x x = r x cos X \ x 3 — x x = r 2 cos 2 y 2 - yi = r i sin 0i J ' y 3 - yi = r 2 sin 2 = (x 2 - x x ) (y 3 - y x ) - (x 3 - xO (y 2 - y x ) = r x cos X . r 2 sin 2 — r 2 cos 2 . r x sin X = - rjrasinC^ - 2 ). Thus x. Yi 1 y 2 1 y 3 1 Yi 1 y 2 1 y 3 1 is positive or negative according as in tra- versing the perimeter of the triangle ABC from A to B, from B to C and from C to A the order is counter-clockwise or clockwise. Thus area ABC = ± \ yi y 2 y 3 , the upper or lower sign being used according as the order is counter-clockwise or clockwise. <4-] Co-ordinates 25 § 4-2. The expression at the beginning of the last §, area OPQ = %r x r 2 sin (Q x - 2 ) gives area OPQ = |(rj sin Q x . r 2 cos 2 — r 2 cos Q x . r 2 sin 2 ) = K x 2 yi-x 1 y 2 ). This gives the area of the triangle whose vertices are (o, o), (x x y t ), (x 2 y 2 ), and is positive if the order of (o, o), (x 1 y 1 ) } (x 2 y 2 ) is clockwise. S 4-3. To find the area of a polygon whose vertices are (x x yj, (x 2 y 2 ), Fig. (1). Fig. (2). In fig. i, area ABODE = OAB + OBC + &c. = \ [ x 2Yi - Xxy 2 + x 3 y 2 - x 2 y 3 + &c] In fig. 2, area ABODE = OAB + OBC - OCD - ODE + OEA But area ODE = - \ (x 5 y 4 - x 4 y 5 ) and area OCD = - |(x 4 y 3 - x 3 y 4 ). Thus in all cases, area of polygon =' | [x 2 y x - x x y 2 + x 3 y 2 - x%y 3 + x 4 y 3 - x 3 y 4 + &c], the order of the suffixes being cyclic. § 44-. To find the area of the triangle whose vertices are (r.ft). {rAl (rA)- Let the vertices be P, Q, R. Then area PQR = OPQ + OQR - OPR. 26 Analytical Geometry [44. But A OPQ = iOP . OQsin POQ = \ r\ r 2 sin (0 x -0 2 ), similarly A OQR - J r 2 r 3 sin (0 a -0 t ), AOPR = ir 3 r 1 sin((9 1 -^), = — Jr. i\ sin (0.-00- .-. area PQR = | [r x r 2 sin (# x — 2 ) + r 2 r 3 sin {0 2 —0 3 ) + r s r x sin (0 t -0j], the order of the suffixes being cyclic. Cor' — We shall obtain the condition that the three points are collinear by equating the preceding expression to zero. Exercises on Chapter I 1. Show that (a, b), (b, a), (3 a — 2 b, 3 b — 2 a) are collinear. 2. O is the origin, A is (r x X ) and B is (r 9 2 ); show that the polar A co-ord's of the point where the bisector of AOB meets AB are r = ^TT cos 2 & " 6 ^> 6 = * Cft + to r l "*" r 2 3. A, B, C are collinear points : if P is any fourth point on this line, prove PA 2 . BC + PB 2 . CA + PC 2 . AB + AB . BC . CA = o. 4. If P is any point not on the line ABC ; the preceding equation is still true. 5. G is the intersection of the medians (or centroid) of a triangle ABC : if O is any other point, prove OA 2 + OB 2 + OC 2 = GA 2 + GB 2 + GC 2 + 3 GO 2 . 6. Prove that A GBC = \ A ABC. 7. AB, AC are adjacent sides of a O, and AD the conterminous diag 7 : if the sign of an area LMN be negative or positive according as the order of L, M, N is clockwise or not ; prove that A PAD - A PAC -f A PAB, where P is any point in the plane of the □. 44-] Co-ordinates 27 {Note— Take AB, AC for axes ; let AB = h, AC = k ; co-ord's of P (xy): then area PAD = — | sin a> = — I sin a) (hy — kx), &c. x y 1 001 h k 1 This gives a proof of the fundamental theorem of moments in mechanics.] 8. The condition that the point (xy) may be within the triangle (x 1 y 1 ), (x 2 y 2 ), (x 3 y 3 ) is that the following three det's have the same sign : x y *i Yi x 2 Ya X x 2 *3 y y 2 y 3 y y 3 yi 9» Find the area of the triangle whose vertices are (a/*! 2 ), 2 3/Xi), (a^ 2 2 , 2a/^), (a^ 3 2 , 2a^ 3 ). Ans. a 2 (/*! - M 2 ) (ju 2 - Ms) (Ms - Mi)- CHAPTER II. THE STRAIGHT LINE EQUATION TO A LINE § 4-5. To find the equation to a straight line parallel to the axis of x. B Let it cut OY in B, put OB = a. Then the ordinate of every point in the line is a : hence its equation is y = a. Similarly the equation to a straight line parallel to OY is x= b The equation to the axis of x is y = o. For the ordinate of every point on OX or its "y" = o. Similarly the equatiori to the axis ofy is X = o. The Straight Line 29 § 4-6. To find the equation to a straight line through the origin. Let OD be the line, 6 its inclination to OX, tan 6 = m. Then if P be any point (x, y) on the line PN ON = tan 6 = m, i.e. — = m x .-. the required equation is x y = mx. Example. Find the equation of the join of (4, 3) to the origin. Here m = tan 6 = f .-. the equation is y = f x, or 4 y = 3 x. § 4-7. To find the equation to any straight line. Y Let AB be the line, cutting the axes in A and B. Let the inclination of the line to OX be 6 ; put OB = c, tan 6 = m. Let P (x, y) be any point on the line. Draw BD II OX. 30 Analytical Geometry [48. PD Then g= = tan 6 .-. PD = BDtan<9 .-. PM - OB = OM tan 6. That is y — c = x tan 0, y = xtan 6 + c. Thus the required equation is y = mx + c. Thus the equation to any line is y = mx + C, where m is the tangent of its inclination to the axis of x, and c its intercept on the axis qfy. § 48. If two points on a line are known the line is determined. Thus two conditions determine a straight line. If two conditions are assigned which a straight line is to fulfil these will give two equations which will enable us to find the constants m and c. Ex. 1. Find the equation to a line through (o, 3) and inclined at 6o° to OX. Here C = 3, m = tan 6o° = ^3 ; thus the required equation is y = x V3 + 3. Ex. 2. Find the equation to a line through (— 1, 2) inclined at 150 to OX. One condition is m = tan 1 50 = — . V3 Another is got by expressing that y = mx + c is to be satisfied by x = — 1, y = 2. This gives 2 = m (— 1) + c 2 = (--7i) ( - 1) + c = ^ + I .'. C = 2 T .'. eq'n required is y = . rX + 2 7- Vz Vl 5c] The Straight Line 3 l We have drawn fig' to illustrate ; here OB = 2 - V3 Ex. 3. Find the equation to the join of (1, 2), (3, 4). Let it be y = mx + c. This is to be satisfied if x = 1, y = 2, .*, 2 = m + c ) Also if x = 3, y = 4, .-. 4 = 3 m + c ) Solving these simultaneous equations m = 1 , C = 1 . Thus the line required is y = x + 1. General formulae for the solution of such examples will be given presently. § 49. If two lines are parallel they have the same inclination to OX; hence parallel lines have the same m. Ex. Find the equation of a line through (1, — 2) parallel to y = 3X — 4. The m of the line is 3. .'. its equation is y = 3 x + C. To find c we express that this is satisfied by x = ijy=— 2 .-. - 2 = 3 + c, c = - 5. Thus the eq r n req r d is y = 3 x - 5. § 50. Any simple equation Ax+ By + C = o, represents a straight line. For the equation gives By = - Ax- C. 32 Analytical Geometry [51. I. Suppose that B is not = o. Then dividing by B we get A C y = -B- x -B ; this may be written y=mx + C, if we put A C m = -B' C = "B Thus the equation A x + By + C = o represents a straight line passing through the point fo, — ■ ^J and inclined to OX at an angle tan -1 f — -=-) . II. Let B = o : then Ax + C = o, C •'• X== -A Q This represents a straight line parallel to OY at a distance — -r* N.B. A, B, C here stand for any numerical quantities, signs included. Thus take the equation -3y + 4X-5 = o- Here A = 4, B = - 3, C = - 5, m B -3 3 Or we may go through the process, thus - 3y + 4X - 5 = o ••• 3y = 4X-5 •'• y = fx-f Thus — 3Y + 4X — 5 = represents a line through (o, — f), and inclined at tan- 1 4 to OX. § 51. We may give another proof that Ax+ By+C = o represents a straight line. Let (Xj y,) 3 (x 2 y 2 ), (x 3 y 3 ) be any three points on the locus. 5i.J The Straight Line 33 Then Ax x + By x + C = 0, Ax 2 + By 2 + C = o, A x 3 + B y 3 + C = o. We may eliminate A, B, C from these equations. Thus, subtract the second equation from the first and the third from the second. This gives A(x 1 -x 2 )+ B( yi -y 2 ) = o A (x 2 - x 3 ) + B (y 2 - y 3 ) = o. Multiply the first of these by y 2 — y 3 and the second by y x - y 2 and subtract ; then divide by A. •'• (*! - x 2 ) (y 2 - y 3 ) - (x 2 - x 3 ) (y x - y 2 ) = o i.e. = o Xi yx i x 2 y 2 1 x 3 y 3 1 Thus the area of the triangle whose vertices are (x^), (x 2 y 2 ), (x 3 y 3 ) is zero ; and .-. the three points are collinear. Exercises 1 . Draw the lines x = 2, y=3, 3 x + 5 Y = °, y-x = o, y+x = o. 2. Draw the lines y + x = i, y= 3 x + 4, y = 3X-4, y = ~3x + 4- 3. Write the equations of lines through the origin inclined respectively at 45 , 6o° and 120 to OX. Ans. y = x, y = x V3, y = — x Vz- 4. Find the equation to a line through (1, 2) at 6o° to OX ; also to a line through (o, - 3) at 45 to OX. Ans. x V3 - y = Vz ~ 2 , y = x - 3. 5. Reduce the equation 2 x — 3 y + 4 = o to the form y = mx + c. Ans. m = £ , c = £ . 6. Find the equation to a line through (1, 2) parallel to y = 2 x + 3. Ans. y = 2 x. D 34 Analytical Geometry [52. EQUATION IN TERMS OF INTERCEPTS § 52. To express the equation to a straight line in terms of its intercepts on the axes. Let AB be the line cutting the axes in A and B. Let OA = a, OB = b. Let P (x, y) be any point on the line. Y -^B N "^\P O I A A^. By similar triangles PM_OB MA~OA a — x a .-. ay = a b — b x .\ ay + bx = ab x y Divide by a b : thus hr=i- ' a b This is the required equation. § 53. Alternative Proofs : I. Join OP. 53-] The Straight Line $5 Then 2 A OBA = A OBP + A OPA .-. OA . OB = OB . PN + OA . PM .-. ab = bx + ay x y II. Let the intercepts which Ax + By + C = o cuts off on the axes be a, b. To get a put y = o : this gives Ax + C = o, orx = — -r- . A That is a = — r- A C Similarly by putting x = o we get b = — =• Now the equation Ax+ By+ C = o may be written -Ax- By = C A B or ~c x -c y = I x y x y or ^- -\ ^— = 1, i.e. — Kf=i. (-D (-D Ex. In fig' page 36 the eq'n x v To AB is - + - = 1, or 3X + 4y = 12 4 3 X V To A'B, + - = 1, or 3X — 4y + i2 = o ~4 3 x v To A' B', — + -=-r =1, or 3x+4y+i2=o — 4 —3 X V To AB', - + -*— = 1, or — 3X + 4y + 12 = o 4 —3 D 2 3* Analytical Geometry 54- Exercises 1 . Find the equation to a line across the third quadrant cutting off intercepts 4, 5 from the axes. Ans. 5x+4y + 20 = o. 2. Find the intercepts cut off on the axes by 3X-4y = i2, y = x V3 + 4, Ix + my = i. ^.4,-3; -7^4; ],~- 3. Find the line through (3, 5) which cuts off equal intercepts on the axes. Ans. x + y = 8. 4. Find the line through (3, 3) which forms with the axes a triangle whose area is 18. Ans. x + y = 6. STANDARD FORM § 54-. To express the equation to a straight line in terms of p and 06 where p is the length of the perpendicular from the origin 011 the line and OL is the angle which this perpendicular makes with the axis of x. Let AB be the line, ON = p, NOX = a. Let P be any point on the line ; x = OM, y = PM its co-ord's. 57-] The Straight Line 37 Draw MR X ON, and PS ± MR. Then PMS = 90 - RMO = 06; OR = OM cos 06 = x cos a, NR = PS = PMsin PMS = ysin06; p = ON = OR + RN = x cos a + y sin a. Thus the equation to the line is x cos 06 + y sin 06 — p = o. § 55 . Alternative proof. x v The equation to the line is -=r-r- + ~= = i (§52). UA OB But p = ON = OA cos OC ,. OA= J- cosCX Also p = ON = OB cos NOB = OB sin a sin a Substitute these values of OA and OB : then the equation becomes x _ A y f_p_) (JL.) \cos a/ \sm a/ = I, or x cos a + y sin (X = p. § 56. The form x cos 06 + y sin at = p of the equation to a straight line is called the ' standard ' form. It will be assumed as a convention that p is always positive, and also that 06 is always positive, i. e. measured round in the positive direction from OX. § 57. With this understanding, it will be found on examination that whatever be the position of the line the proofs in Arts. 54 and 55 are perfectly general, and that in all cases the equation is x cos OC + y sin OC = p. The generality of the proof in Art. 55 depends on this, that whatever be the 38 Analytical Geometry L58- magnitude of OC, OA and cos OC have always the same sign, and also OB and sin OC have always the same sign. Example. If in fig' the Eu- clidean angle XON - 120 , then OC = 240 , also p = 3; and the equation to A B is x cos 240 + y sin 240 = 3 or or x + y V3 + 6 = 0. § 58. We have obtained the equation to a straight line in the following forms, jO } y = m X + c 2 , Ax + By+ C = o x . y 3 ' a + b 4°, x cos 06 + y sin 06 = p (the standard form). If an equation is given in one of these forms we may reduce it to any of the others. x v Ex. Express - + £ = 1 in the form y = mx + c. a b y x b u Here r- = l , y = x+b. b a J a Thus m = , c = b. a 59-] The Straight Line 39 § 59. To reduce the equation Ax + By + C = oto the form x cos 06 + y sin 06 — p = o. Divide by >/A 2 + B T ; then ABC • x + -7^== • y + V A 2 + B 2 7 A 2 + B 2 V A 2 + B 2 Now since the sum of the squares of — and /A a + B 2 /A 2 + B 2> = C P " " VA 2 + B 2 If C is positive, then changing the signs of all the terms the equation may be written -Ax-By-C = o; the preceding method is then applicable. In words : To reduce the equation of a straight line to the standard form : if necessary change the sign of every term so that the last term is negative, then divide by the square root of the sum of the squares of the coefficients of x and y. Ex. 1. Reduce y = A x + A to tne f° rm x cos a + y sin a — p = o. Here 5X-i2y + 7=o Changing the signs, — 5 x + 12 y — 7 = Divide by V(- 5) 2 + 12 2 or 13 40 Analytical Geometry [60. Thus p = A, cos OC = - t^, sin OC = f|- As the cosine of a is negative and its sine positive, a must be an angle in the second quadrant, i. e. between 90 and 1 80 . Ex. 2. Reduce x-y/3 + 6 = o to the form x cos OC + y sin OC — p = o. Here — x + y V$ — 6=0. Divide by V(- i) 2 + (V3) 2 = aA + 3 = V4 = 2 ; thus - \ x + y - 3 = o. a/2 Thus p = 3 ; also cos a = — f , sin a = - — , or CX = 1 20 . The learner should draw diagrams to illustrate these examples. Exercises 1. Express the following equations in the standard form : x + y = 3, x^3 + y + 6-o, -xV3 + y+6 = o, xv / 3-y+6 = o, x >/3 + y — 6 - o, 5 x — 12 y + 20 = o. Ans. OC = 45°, p = 4^ 5 OC = 210 , p = 3 ; OC = 330 , p = 3 ; a = 150°, p = 3 5 OC = 30 , p = 3 ; cos a = - T % sin OC = if, p - ff . 2. Express in the standard form x y y = mx + c, - + <-=!. a b m 1 r» -<4«.y. =- x + — ==- y — Vi + m 2 Vi + m 2 V 1 + nrr bx ay ab + — == = o. Va 2 + b 2 Va 2 + b 2 Va 2 + b 2 PROJECTIONS 8 60i If the properties of projections are assumed, the statement of the proof in § 54 may be simplified. Def f — If A' be the foot of a 1 from a point A on a line OP, then N is called the projection of A on OP. 6o.] The Straight Line 41 Def'—liN, B' be the projections of two points A, B on OP then A' B' is called the projection of AB on OP. Now let OC be the angle between OP, AB : then drawing A' G || AB we see that A' B' = A' G cos OC ■-= AB cos OC. If A, E are the terminals of a broken line AB, BC, CD, DE ; then A' E' =-. A' B' + B' C + C'D'+ D' E', the signs being taken into account. Or, projection of AE = sum of projections of AB, BC, CD, DE. If OC, /3, y, 8 are the angles which AB, BC, CD, DE make with OP; then the projection of AE = AB cos OC + BC cos (3 + CD cos y + DE cos 8. In the figure it will be seen that cos (X, cos /3, cos y are positive while cos 8 is negative. We may now prove the equation of § 54. The projection of OP on ON = sum of projections of OM, MP. (fig', § 54) But the projection of OP is p, that of OM is x cos OL, and of MP is y sin OL .-. p = X cos OC + y sin a. 42 Analytical Geometry [6, INTERSECTION OF TWO LINES § 61. The co-ordinates of the point of intersection of two straight lines are got by solving the equations simultaneously for x and y. We here remind the reader of a theorem of Algebra, which will give a general formula. If (ax+ by + cz = o ta'x + b' y + cz = o (0 Then X y z b c b' c' c a c' a' a b a 7 b' To remember this formula we may use this mnemonic : Imagine the coefficients written twice in succession, thus a b c a be a' b' c' a' b' c' Then the second and third columns give the denominator of x ; sliding forward one column we get the denominator of y ; and sliding forward another we get the denominator of z. If now we put z = i, we see that the intersection of the lines ax+by + c = o, a' x + b' y + c' = o is given by x ^_ y T be c a a b b' c' I c' a' a' b' Ex. The intersection of 2X— 3y + 6 = o, x — y + i = o is given by x _ y i -36 622-3 —11 11 1 —1 x y or - = - = 1, 3 4 Zn 62.] The Straight Line 43 § 62. To find the condition that three lines may be concurrent. The values of x, y got by solving two of the equations must satisfy the third. Ex. Show that the lines x + y+i^o, 2 x - y - 7 = o, 4X + y — 5=0 are concurrent. Solving the first two we get x = 2, y - - 3. This point lies on the third line if 4(2) -3- 5=°. o r 8 = 3 + 5- Q.E.D. Or we may obtain a general formula. Let the three lines be ax+by + c = o, a'x + b'y + & — o, a"x + b"y + c" = o. From the last two equations we obtain x y T 1 = k say. b' c' c' a' a' b' b" c" c" a" a" b" .-. x = k (b' c" - b" c'), y = k (c' a" - c" a'), 1 = k (a' b" - a" b') Substituting these values in ax + by + C = o, and dividing by k, we obtain the req'd cond'n a (b' c" - b" c') + b (c' a" - c" a') + c (a' b" - a" b r ) = o or a b c a' b' c' a" b" c" = o 44 Analytical Geometry [63. § 63i To find the condition that the lines ax+by + c = o, a'x+b'y + c' = ?nay be parallel. They must have the same m : this gives a _a^ b " b'' .-. -r = Y-.i and a b' - a' b = o. b b Thus the coejj's ofx and y must be in the same ratio in both eq'ns. Ex. 1. 6 x — 4 y + 1 =0, 9X— 6y— 7 = 0, are parallel, for 6 9 , _ . -x = — -r , each being: = — - • -4 -6' & 2 Ex. 2. ax + by + c = o, ax + by + c' = o are parallel. Ex. 3. Find the equation of a line through (2, — 1) parallel to 3X-5y+i = o. The required equation is of the form 3X-5Y + k = o. To find k substitute x = 2, y = — 1, .-. 6 + 5 + k = o, k = — 11. Ans. 3X— 5y— 11=0. Exercises 1 . Show that the lines x y x y a b b a _ + _ = !, t- + ^= 1, x-y are concurrent. Ans. The point of intersection is ( r , | \a + b a + b/ 2. Find the equation to a line through (1, — 2) parallel to 3X + 4y + 6=o. Ans. 3* + 4y + 5 = o. 65-] The Straight Line 45 3. Find the area of the triangle whose sides are x — y = o, x+3y = o, x + y + 4 = o. Ans. 8. 4. Find the area of the triangle whose sides are 3x + y-7 = o, x+7y+n-o, x - 3 y + 1 --- o. Ans. 10. EQUATION TO JOIN OF TWO POINTS § 64. Suppose we take two points (x x y ± ), (x 2 y 2 ) on a straight line The m of the line = tan# CL ~DL = yi - y 2 Thus the m of a line = the difference between the y's of any two points on the line divided by the difference between their x's. § 65- To find the equation to the join of two points (x^), (x 2 y 2 ). Let (x, y) be any other point on the line. The m of the line = ^"-^ (fig', § 64) It also = x x - x 2 y-yi y-y, X Xj Xj X 2 This is the required equation. (0 4 6 Analytical Geometry [66. If we multiply up and transpose we get x(y 1 -y 2 )-y(x 1 -x 2 ) + x 1 y 2 x 2 yx = o = o (») x y i x x y x i x 2 y 2 i This is another form of the required equation. Both forms should be remembered. The form (i) is intuitive, if we think of the figure. We might also write down (2) at once from the consideration that the area of the triangle whose vertices are (xy), (x^), (x 2 y 2 ) is zero. § 66. Ex. 1. Find the equation to the join of (1, — 2) and (—3, 4). y+2_ —2 — 4 1 + 3 Applying (1) it is or x — 1 y + 2 x — 1 or 2 (y + 2) + 3 (x — 1) = o or 3X+2y+i=o. {Note — We may verify this : We see that 3x+2y + i=oisa simple eq'n satisfied both when x = t y = — 2 and when x = — 3, y = 4] Or thus, applying (2) the req'd eq'n is x 1 -3 y —2 4 or x (- 6) - y (+ 4) + 1 (— 2) = o or — 6x — 4y— 2— o or 3 x + 2 y + 1 = o ; as before. The form (2) of the equation of the join will usually be found the most convenient in practice. Ex. 2. The equation to the join of (a, o) and (o, b) is = o, 66.] The Straight Line 47 bx — ay + ab = o, x y or - + f- = i, a b We have thus another proof of the equation of § 52. Exercises 1 . Find the equations to the joins of i°. The origin and (3, 4) 2 . The origin and (— 3, 4) 3°. (4, 2) and (- 3, 9) 4°- (-3, -i)and(4, -8) 50. (2, 3) and (4, 3) 6°. (a, b) and (-a, - b) 7°. (a, b) and (b, a) Ans. i°, 4X-3y = o. 2 , 4X + 3y = o. 3 , x + y-6 = o. 4 , x + y + 4-0. 5°> y = 3- 6°, bx — ay = o. 7 , x + y = a + b. 2. Find the equations to the sides of a triangle whose vertices are (2, 1), (3, - 2), (- 4, - 1). Ans. 3X+y— 7 = 0, x + 7 y + 11 = o, x— 3 y + 1 = o. 3. Find the equations to the medians of the same triangle. Ans. x — y — 1 = 0, x + 2 y + 1 = o, x — 13 y — 9 = 0. 4. Show that the following three points are collincar : (- 3, -9)> C 1 * - O* ( 2 > 0- Find the ratio of the segments into which these points divide the line. Ans. 4:1. 5. Find the equations to the diagonals of the rectangle formed by the lines x = a, x = a', y = b, y = b'. Ans. (b - b') x - (a - a') y + ab r - a' b =0, (b - b') x + (a - a r ) y - ab + a' b' == o. 48 Analytical Geometry [67. 6. Find the equations to i°, the join of the origin to the intersection of y=2X-3, 3y+x + 4 = o; 2 , the join of ( — 4, 7) to the intersection of y = 3X - 1, x = y- 1. Ans. i°, 5 y + 11 x = o. 2 , x + y — 3 = o. 7. Find the equations of the lines through the intersection of y = 3X-1, x = y-i parallel to OX and to 4X + 5y + 6 = o. Ans. y = 2, 4 x + 5 y = 14. 8. Find the equation of a line joining the intersection of 3y + 2x+3 = o, 2y+ x+2-^0 to that of y + 2X + 3 = o, 2y + 3x + 4 = o. Ans. y + x + 1 = o. LINE THROUGH A FIXED POINT § 67. Suppose that any line is drawn through a point (h, k); y — k then if (x, y) be any point on the line its m is r- • Thus _ . =m, or y — k = m (x — h) denotes a line through (h, k), and inclined to OX at an angle whose tangent is m. By giving m a suitable value the equation y— k=m(x— h) may be made to represent any one of a pencil of lines through (h, k). Ex. The line through (— 1, 2) at 150 to OX is y - « = - -7= ( x + *)• (See Ex. 2, § 48.) 67.] The Straight Line 49 Exercises on Chapter II 1. Find the equation of a line through (3, 4) at 75 to OX. Ans. y - 4 = (2 + V3) (x - 3). 2. Through the vertices of the triangle whose sides are x + 2y-5=o, 2x + y-7^o, y - x - 1 = o parallels are drawn to the opposite sides : find their equations. Ans. x — y — 2=0, x + 2y — 8=0, 2x + y — 4 = 0. 3. Find the equation of the join of (a, b) to the intersection of x y x y abba Ans. b 2 x — a 2 y + (a — b) ab = o. 4. Two lines can be drawn through (6, — 4) each of which forms with the axes a triangle whose area is ^ ; find their equations. Ans. 3X + 4y = 2, i6x + 27y + i2=o. 5. Show that the area of the triangle contained by the axis of y and the straight lines y = m l x + c l5 y = m 2 x + c 2 is 2 (m 2 - mi) [Note — If lines meet in P and cut OY in B, C; 2 area = BC . abscissa of P.] 6. Deduce that the area of the triangle formed by the lines y = m 1 x + c 1 , y=m 2 x + c 2 , y = m 3 x + c 3 is 1 (C a ~ c s) 2 + 1 (C3 ~ Ci) 2 + T (Qi ~ C2) 2 2 m 2 — m 3 2 m 3 — m 1 2 vr\ x — m 2 7. Obtain also the following expression for the area 1 m 1 1 m 2 1 rm 2 (nri! - m 2 ) (m 2 - m 3 ) (m 3 - m x ) \Note — We may reason thus : Calling the determinant in the numerator A ; then area = o if A = o. Also area = 00 if 0^ — m 2 or m 3 - m 3 or m 3 = tr\ 1 . E 5° Analytical Geometry Further rrij, m 2 , m 3 are ratios while c lt C 2 , C 3 are lines. Hence as the area must be of the second degree in c r , c 2 , C 3 we may assume it = k A 2 -r (mj. - m 2 ) (m 2 - m 3 ) (m 3 - 11^) where k is a numerical constant. We may determine k by taking the three lines y = o, y = x, y=- X +2, which form a triangle whose area is I. We thus get k = |.] 8. Show that the area of the triangle formed by the three lines Ax + By + C = o, A'x + B'y + C A" x + B" y + C A A 7 A" B B' B" C C C" + (AB'~ BA') (A 7 B"- B'A") (A" B - B"A) I. Show that the area of the triangle whose sides are xv xv x n y x y a b - sec 6 — -f- tan d = i a b is ab. 10. Find the area of the triangle whose sides are x - MiY + a Mi 2 = °> x - /x 2 y + aju 2 2 = o, x - yot 3 y + aju 3 2 = o. Ans. — (Mi - M2) (M2 ~ Ms) (Ms ~ Mi)- CHAPTER III THE STRAIGHT LINE— (continued) ANGLE BETWEEN TWO LINES 68. To find the angle between the lines y = mx + c, y = m'x + c'. Here m = tan m' = tan 0' By Euclid I. 32, e = $ +• 9 f /.

(1) 1 + mm v ; Cor" (1) — If the lines are ||, (j) = o .*. m — m' = o or m = m' This agrees with § 49. Cor' (2) — If the lines are ± (j) = 90 , tan (f) = 00 .-. 1 + mm' = o ...... (2) E 2 52 Analytical Geometry [69. Cor' (3) — If the lines are ax + by + c = o, a'x + b'y + c' = o. a a' Then m = — t-> m' = — b b' the lines are _L if 1 ♦ (-1K-6)- Multiply up by bb' .*. the lines are -L if aa' + bb' = o (3) The formulae (1), (2), (3) should be carefully remembered. From (3) we see that Two lines are at right angles if PRODUCT COEFf's OF X + PRODUCT COEFf's OF y = O Cor" (4) — Generally, if = aa' + bb' § 69. Ex. 1. Find the angle between x-yV3+ I= o, x + y>/3-2=o. Here m = — T , m' = 7- V3 V3 I T m - m' Vz Vs 2 2 /- .-, tan$ = = = -J- -=- - = V3, ^ 1 + mm' 1 V3 3 3 ?t.] The Straight Line 53 Ex. 2. 6x + 4y + 5==o and 2 x — 3y + 7=0 are at right angles. For 6(2) + 4(— 3) = 12 — 12 =0. Ex. 3. Ax + By + C = o and Bx — Ay + C = o are at right angles. For A (B) + B (- A) = AB - AB = o, .-. &c [Cor' (3)] Similarly v ax + by + c = o and r + c' = o J a b are at right angles. § 70. To find the equation of a line through (h, k) perpendicular to y = mx + c. By § 67 the required equation is of the form y — k = m'(x — h). Also 1 + mm' = o [§68, Co/ (2)] 1 This gives m' = — m and the eq'n req^d is y — k = (x — h) Ex. Find the equation of a line through (6, 7) perpendicular 3X+ 2y+4 = o Here 2 y = _ 3 x — 4 /. y = - fx - 2 .-. m = - f m' = - £ = ■+ I to m and the eq'n req'd is Y - 7 = t ( x - 6 )> or 2X-3y+9 = o. § 71. To find the equation of a line through (h, k) perpendicular /o Ax+By + C = o. This case may be reduced to the last. 54 Analytical Geometry [72. A Here m = — ^ 1 B .-. m' = — x m A Thus the equation required is y - k = ^ (x - h) x — h y — k or -flT = -B- The student will find it convenient to remember this equation. Thus taking the example in § 70 we at once write down the equation of the J_ x 6 v — 7 — , or 2X — 3V + Q = o as before. 32 § 72. Or we may proceed thus. Bx — Ay = constant represents a J_ (§ 69, Ex. 3). To find the constant express that the eq'n is satisfied by the co-ord's of the given p't (h, k) ; .'. eq'n req'd is Bx — Ay = Bh — Ak. Thus taking the Ex. in § 70 ; the eq'n req'd is 2X-3y = 2(6)-3 (7), or 2x-3y+9-o. Exercises 1 . Find the angles between the pairs of lines i°>5y-3x + i=o, y-4X + 2=o 2 , 2 x — 3 y = o, 6x + 4y+7 = o 3°, y + x = o, (2 + V3) y - x = o 4°, y — kx -rr o, (1 — k) y — (1 + k) x = o Am. 45 , 90 , 6o°, 45 . 2. Find the equations to i°, a line through (1, 2) perpendicular to 3X+4y + 5=o 2 , „ „ (o; -1) „ ,, x + y = 1 Am. 4X — 3y + 2 = o, x — y— 1=0. 73-] The Straight Line 55 3. Find the equation to a line through the origin -L Ax + By + C = o Ans. Bx = Ay. 4. Find the equation of a line through (2, 3) J_ the join of (1, 2), (- 3, _ 14) Ans. x + 4y = 14. 5. Find the lines through the origin inclined at 45 to y + x V$ = o. Ans. y + x (2 - >/ 3 ) = o, y - x (2 + >/3)- 6. Show that the acute angle between the lines 2 2 2x + 3y = 4, 4X-5Y=6 is sin- 1 —7= v 533 POSITIVE AND NEGATIVE SIDES OF A LINE § 73 . To find the ratio in which the join of (x 1 y x ), (x 2 y 2 ) is divided by Ax + B y + C = o. Let the required ratio be m : n. Then the values mx 2 + nx t i my 2 + ny x m + n " m + n (§ 1 5) must satisfy Ax+ By+ C = o. Substitute these values and multiply up by m + n. :*. A (mx 2 + nxj + B (my 2 + ny x ) + C (m + n) = o .-. m(Ax 2 + By 2 + C) = - n(Ax x + By a + C) m _ Ax t + B y x + C n~ Ax 2 + By 2 + C 5^ Analytical Geometry [74- Cor' — If (x 1 y t ) and (x 2 y 2 ) are on the same side of Ax + By -t- C = o then their join is cut externally, and m : n is negative (§ 16) ; .-. AXj + By x + C and Ax 2 + By 2 + C have the same sign. If (x x y x ) and (x 2 y 2 ) are on opposite sides their join is cut internally ; m : n is positive, ■'• A x x + B y 2 + C and Ax, 4- By 2 + C have opposite signs. § 74. Thus, two points (x^), (x 2 y 2 ) are on the same or opposite sides of a line Ax+ By + C = o according as the results of substituting the co-ordinates of the points in the sinister side of the equation have like or unlike signs. Ex. How are the origin and the points (i, i), (i, 3) situated with reference to the line 3X — 4y + 5 = o? Substitute (o, o), (1, 1), (1, 3) successively for (x, y) in the expression 3X — 4y + 5 : the results are +5, +4, —4. Hence the origin and (1, 1) are on the same side of the line, and (1, 3) on the opposite side. § 75. We see now that a line Ax+ By + C = o divides the plane of the axes into two compartments such that the ex- pression Ax + By + C = o is positive for all points in one compartment (this may be called the positive compartment), and negative for all points in the other (the negative compartment). If we write the equation — Ax— By — C = o; then the positive and negative compartments are interchanged. 7 6.] The Straight Line 57 LENGTH OF PERPENDICULAR § 76. To find the length of the perpendicular from (h, k) on Ax + By + C = o Let RS be the line Ax + By + C = o . (i) N the foot of the _L The equation to PN is (§ 71) x — h __ y — k Y /P(hk) R l*K O S X B ■(») The co-ord's x, y of N are got by solving (1), (2). Put each member of (2) = A .-. x = AA + h, y=AB+k . Substitute these values in (1); ... a (A A + h) + B (A B + k) + C = o .-. (A 2 + B 2 ) A + A h + B k + C = o Ah + Bk + C (3) A = - (4) A 2 + B 2 Now PN 2 = (x - h) 2 + (y - k) 2 = (AA) 2 + (AB) 2 by( 3 ) = A 2 (A 2 + B 2 ) Hence substituting for A its value from (4) and taking the Ah + Bk + C square root, PN = + 7A 2 + B 2 58 Analytical Geometry [77. § 77. We have seen that the numerator is positive or negative according as (h, k) is one side or other of the line. (§ 74.) It is important to remember that the absolute length of the J_ is Ah+Bk+C VJK A + B 2 Thus after, if necessary, bringing all the terms of the equation to one side, ive substitute the co-ordinates of the point, and divide by the square root of the sum of the squares of the coefficients of x and y. Ex. 1. Find the length of the perpendicular from (2, 3) on 3x + 4y— 20=0 Here p = 3. 2 +4-3 -so 2 V3 2 + 4 J 2 Ans. The required length is - Ex. 2. Find the length of the perpendicular from the origin or 3X + 4y+ 20 ^--0 3. o + 4. o + 20 20 Here p = , = — =4 V32 + 42 5 Exercises 1. Find the ratio in which the join of (1, 2), (3, —2) is cut by 4* + 5y- 6 - °- Ans. The join is cut in a point of trisection. 2. Show that (1, —2), (o, 1), and the origin are on one side of 4 x + 5y - 6 = °; and (1, 1) on the opposite side. 3. Show that the origin and (|, |) are within the triangle whose sides are 5 y_ 4 x=i, x-3y = 9, x + 9y = 9. 77] The Straight Line 59 4- Find the lengths of the -L s from (i, 3), (2, —3) and the origin on 4X-5Y + 6 = o. z 2Q 6 5. The length of the JL from the origin on h (x + h) + k (y + k) = o is Vh 2 + k 2 ; , x y . hk and on — + f = 1 is h k \/h 2 + k 2 6. The vertices of a triangle are (3, 8), (12, 2), (—4, — 6) ; find the equations to the J_ s from the vertices on the opposite sides. Show that they cointersect in (4, 6). Ans. 2 x + y = 14, x + 2 y = 16, 3 x — 2 y = o. 7. Find the lengths of these J_ s . Ans 2I 24 5 6 V5 V5 V13 8. Find the equations to the X s to the sides of the same triangle through their mid points ; show that they cointersect in (3^, — 1). Ans. 2x + y = 6, 2X + 4y = 3, 6x — 4y = 25. 9. Find the parallels to 12X - 5y + 1 = o ata distance 2 from it. Ans. 12 x — 5 y + 27 = o, 12 x — 5 y — 25=0. 10. Find the points on 3 x = y + lata distance 2 from 12 x — 5y + 24 = o. Ans. (1, 2), t*± t 54). 11. Find the equation to the join of the feet of the X s from the origin on 3x-4y+25=o, i2X+5y=i69; find also the length of this join. Ans. x — 15 y + 63 = o, V'226. 12. Show that the origin is inside the triangle whose vertices are (3, 4), (2, - 3), (- 2, - 2±). 6o Analytical Geometry [78- § 78. The formula of § 76 gives the length of the ± from x y on viz. it is x cos oc + y sin 06 — p = o, x cos a + y sin a — p . , , . . J = ± (x' cos a + y' sin 06 - p). vcos 2 a + sin 2 a We add an independent geometrical proof of this important formula. § 79. To find the length of the perpendicular from (x'y') on x cos ol + y sin a — p = o Y > \JN' B \^N \p(x'y') / a\ MT\ ^\ O A\ A'^ ^x B s. \j(l b" ^N^'^^/^ /d\ (x>0\ A'\ Fig (2) A X Let AB be the given line, ON = p, NOX = a. Through P, (x'y') draw A' B' || AB and PM ± AB. If ON or ON produced meet A'B' in N', then ON' is ± A'B' (Euclid I. 29). If p' = ON' the equation to A' B' is x cos 06 + y sin 06 — p' = o. Expressing that this equation is satisfied by the co-ord's of P, (viz. x'y') x'cosa + y'sin 06 — p' = o .-. p' = ON' = x' cos a + y' sin 06. So.] The Straight Line 61 In fig. i, PM = NN' = p'— p = x' cos 06 + y' sin 06 — p In fig. 2, PM = N'N = p — p' = p — x' cos a — y' sin 06 Thus the required length is ± (x' cos 06 + y' sin 06 — p). The upper or lower sign is to be used according as (x'y') is on the opposite side of the line to the origin, or on the same side. This agrees with § 74. BISECTORS OF ANGLE §80. To find the equations to the bisectors of the angles between Ax + By + C = o, A'x + B'y + C = o. N, \ /C^ _,'''' P -'"/ N M M .,--""" /N\ \ \ P The dotted lines in the diagram represent the bisectors. 62 Analytical Geometry [80. Now the J_ s PM, PN from any point P (x, y) on either bisector on the lines are equal. .*. the equation to one bisector is Ax+By+C = . + A'x + B'y + C ,, VA 2 + B 2 a/ A' 2 + B / 2 and to the other is Ax+By + C _ A'x + B'y + C VAT+W VA' 2 + B' 2 (2) To distinguish between these. Let both square roots have the positive sign ; and let the equa- tions he so written that C, C are negative. Then the origin is on the negative side of both lines. If now (i) is satisfied Ax+By + C and A'x + B'y + C must have the same sign, i. e. any point which satisfies (i) lies on the positive side of both lines, or on the negative side of both. Accordingly (i) represents the bisector of that angle in which the origin lies. Ex. Find the bisectors of the angles between 3x + 4y-9 = o, i2x-5y + 6 = o. Let the second equation be written — i2x + 5y — 6 = 0. Then the absolute terms — 9, — 6 have the same sign, so that the origin is on the negative side of both lines. The bisector of that angle in which the origin lies is 3X + 4y- 9 = + - 12X + 5y - 6 5 ~ 13 or 99 x + 27 y — 87 = o. The other bisector is 3X + 4y- 9 = _ - 12 x + sy - 6 5 J 3 or 3X— uy+2i=o. 8..] The Straight Line 63 Co/ — If the equations be given in the form x cos a + y sin a — p = o, x cos a' + y sin a' — p' = o ; then x cos 06 + y sin 06 — p = x cos 06' + y sin a' — p' is the bisector of that angle in which the origin lies, and x cos 06 + y sin 06 — p = — (x cos 06' + y sin 06' — p') is the bisector of the other angle. Exercises 1 . Find the bisectors of the angles between 3x-4y + 7 = o, i2X + 5y-5=o. Ans. 21 x -t- 77 y = 116, 33 x — 9 y + 22=0. 2. Find the lengths of the J_ s from the origin on the bisectors of the angles between x cos OC + y sin OC = p, x cos (3 + y sin {3 = p'. ArJ P ~ p/ P+ P ' ' 2 sin \ (a - (3) 1 cos \ (OC— /3) 3. Show that the origin lies on the bisector of the acute angle between the lines 3X - 4y = 5, 5X+i2y=i3. LINE DRAWN IN GIVEN DIRECTION § 81. The following form of the equation to the straight line is sometimes useful. Let the line be inclined a to OX at 6. Let F(h, k)be a fixed point on the line ; P (x, y) any other point on the line. Let FP = r. 64 Analytical Geometry [82. From the figure we see that x — h = FG = r cos 6 y- k = PG = rsintf. Thus the equation to the line may be written in the form x — h _ y — k cos 6 sin Ex. Through the point D (2, 3) a line is drawn inclined at 6o° to OX and meeting 4X+5y+6 = oinE; find the length DE. x 2 V 3 The equation to DE is ^- n = . _ - n = r cos 00 sin 00 /. x = 2 + r cos6o° = 2 + - , 2 y = 3 + r sin 6o° = 3 + r (1) If x, y are the co-ordinates of E, the intersection of the two lines, then the equations (1) and 4X + 5y + 6 =0 (2) are simultaneously true. Substituting then the values of x, y from (1) in (2) 4 ( 2+ D +5 ( 3+ ^ r ) +6=o - 29 58 (5 V3 - 4) This gives DE = r = 2 + §_V3 59 § 82. To find the equations to the lines through (h, k) which are inclined at a given angle (p to the line y = mx + c. Let AB be the line y = mx + c. 83-] The Straight Line 65 The req'd eq'ns are y — k = (x ~ h) tan \js . . . . y — k = (x — h) tan \\/ .... (vide fig') ; tan y]/, tan \|/ are to be determined. By Euclid \js = + (j) . tan 6 + tan (b m + tan (b .-. tan \U = -x '-r — r - i — tan u tan

.*. tan \// = tan (# - (f>) = i + m tan 0i)> (2^, 1 + |j Ans. r x = r cos (0 — ± ). 4. Find the length of the X from the origin on the join of (rift), (r 2 2 ) r x r 2 sin (0 2 - X ) ^^i - v rj 2 + r 2 2 — 2 r^ cos (0 2 — #1) 85-] The Straight Line 5. Find the equation to the join of (3 cos OC, 2 OC), (3 cos 2 OC, 3 Of) Ans. r — 3 cos (9 — OC). 6. Find the equation of the line through (XM -L a= rcos(0-a). Ans. r sin {OC - 0) = r x sin (a - ^). 67 OBLIQUE AXES § 84. As in Art. 45, y = a, x=b represent straight lines parallel to the axes. § 85. To find the equation to any straight line. Y/ Let it cut the axes in A, B. Let 6 be its inclina- tion to OX, co the angle between the axes. A 0~ Let P be any point in the line ; x = OM, y = PM its co-ord's. Let OB = c PD OB sin/9 Then That is DB OA" sin(©-0) y — c sin x sin {co — 6) s\nO y- c = ^. sin (co — 6) F 2 68 Analytical Geometry [86. Putting — = m, this becomes sin (co — 6) y = mx + c This is the required equation. The signification of c is. the same as for the case of rectangular axes, viz. it is the intercept on OY. /, , sin 6 m no longer means tan u, but Co/— tan sin (co — 0) m sin co i + m cos co § 86. As in § 50 or 51 we may show that any simple equation represents a straight line. As in § 52 it is shown that the equation in terms of the inter- cepts on the axes is x y a b As in § 53 the intercepts on the axes by Ax + By + C = o C C are "A'-B § 87. To find the equation to a straight line AB in terms of p, OL, p where p is the perpendicular on the line from the origin and Oi, p are the angles which this perpendicular makes with OX, OY. 89.] The Straight Line 69 Let P be any point on the line ; OM = x, PM = y its co-ord's. Then,. as in § 60, the projection of OP on ON = sum of pro- jections of OM, MP on ON. ."•. p = x cos 06 + y cos /3 Or we may supply the proof in full, thus : draw M R parallel to AB and PS parallel to ON. Then OR = OM cos 06 = x cos 06; RN = PS = PM cos SPM = y cos /3 .-. x cos a + y cos /3 = p § 88. Or thus : The equation to A B is x y OA ' OB P HR- P But OA = — £-, OB = cos a cos p .-. x cos 06 + y cos /3 = p, as before. § 89. If the axes are oblique no modification is required in the investigations of §§ 61-63. As in § 64 we get — — = m, where m has now the same meaning as in § 85. Thus the formulae (1), (2) of § 65 for the equation of the join °f ( x i Yi)' ( x 2 Y2) are a PP^ ca ble when the axes are oblique. ?o Analytical Geometry [90. § 90. As in § 79 it is proved that the length of the _L from x cos a + y cos p — p = o is + (x' cos (X + y' cos /3 — p) §91. To find the length of the perpendicular from (x'y') 077 ax + by + c = o Let p(xy; ax -f by + c = o meet the axes in D, E. Then OD = ° E ="E DE 2 = OD 2 + OE 2 - 2OD . OE cosco c 2 . C 2 2C 2 = ~, + o - b 2 ab COSG) .-. DE = -7- Va 2 + b 2 — 2 ab cos co ab If the required length PN = P, then 2 area of A PED = P . DE c / v a 2 + b 2 — 2 ab cos = P. ab co (1) Again, the co-ord's of D are ( , o I , and of E are / o\ v a J 92.] The Straight Line 7* .-. 2 area PED = -f sin CO X' c a y 1 o 1 o - B I « 2 S) = ± sm»( B x +i / +aB ) = + sin co . — r- (ax" + by' + c) Comparing (1) and (2), (ax' + by'+ c) . sin ft) Va 2 + b 2 — 2 ab cos co P = + (2) (3) § 92. To reduce ax + by + c = o to the form x cos oc + y cos /3 — p = o Infig / §9i, p . DE = 2 area ODE = OD . OE sin co, also cos 06 — Qpr , cos /3 = —^ Using then the values of OD, OE, DE in § 91 we shall get c sin co cos a = cos /3 = Va 2 + b 2 - 2 ab — a sin co COS ft) Va 2 + b 2 — 2 ab — b sin co COS ft) Va 2 + b 2 — 2 ab cos co The equation is then transformed by multiplying by — sin ft) -/a 2 + b 2 — 2 ab cos ft) This is also obvious from (3), § 91. Analytical Geometry [93- § 93 . If (b be the angle between two lines y = mx + c, y = m'x + c' then with the notation of § 85 . z, /,, j tan — tan #' (h= 0-0', tan <2> = z ^ v r 1 + tan tan 0' /, m sin co /,, m' sin co Also tan = , tan a = 1 + m cos co' 1 + m' cos co Substituting and reducing we obtain (m — m') sin co tan© = 1 + (m + m') cos co + mm" Cor' (1) — The lines are f| if m — m' = o, and J_ if 1 + (m + m') cos co + mm' = o Cor' (2) — If the two lines are ax + by + c = o, a'x + b'y + c' = o, a , a b> m= -b- then m = — ^, m' = — u and we may substitute these values in the preceding formulae, This gives • _ (a'b — ab') si nco ^ "" aa' + bb' — (ab' + a'b) cos co Thus the lines are || if a'b — ab' = o, and X if aa' + bb' = (ab' + a'b) cos co Exercises 1. If a) = 30°, prove that y + x = i is inclined at 105 to OX. 2. The lines y + x = o, y — x = o are at right angles, whatever be the angle between the axes. 93-] The Straight Line 73 3. If go = 6o°, find the equation to the J_ from (i, i) on 2 x + 3y+4 = ; find also the length of this JL . 2 Vl Ans. 4X -y- 3 = 0, — jf. 4. Find the equation to a line through (h, k) perpendicular to the axis of x. Ans. x + y cos go = h + k cos go 5. Prove that the lines y = mx + C, y = m'x + c' are equally inclined to the axis of x in opposite directions if I I 1 ; = — 2 COS O) m rrv Exercises on Chapter III 1. Find the length of the perpendicular from (b, a) on x y a b a 2 + b 2 - ab Ans. — Va 2 + b 2 2. Interpret the equation sin 3 6 = 1. Ans. Three lines through origin inclined at 30 , 150 , 270 to OX. 3. Find the centre of the inscribed circle of the triangle whose sides are 3x-4y = o, 8x+i5y = o, x = 20. Ans. (13, 1). 4. The vertices of a triangle are (1, 2), (2, 5), (3,4); show that the co-ordinates of the centre of the inscribed circle are (V5> 4)« 5. Show that the area of the parallelogram whose sides are Ax + By + C = o, Ax + By + C + y = o, A'x + B'y + C = o, A'x + B'y + C + / = o yy' !«; '—L AB' - A'B 74 Analytical Geometry 6. Find the lines through (i, — 2) which are inclined at 45 to 3X-4y + 5 = o. Ans. 7x — y = 9, x + 7y+ 13 = 7. Show that the angle between 7y = 17X + 1, y =x + 2 is cos -1 ^f. 8. Show that the equation to the join of the points whose polar co-ordinates are (2 c cos OC, OC), (2 c cos /3, /3) is 2 c cos (3 cos a = r cos ((3 + OC — 6) 9. Prove that the equation to the straight line through (r x 0-^ perpendicular to = a cos + b sin 6 r n b cos 6 — a sin 9 is l - i" b cos 0j — a sin X 10. Find the length of the perpendicular from (3, — 4) on 4 x + 2 y = 7, the axes being inclined at 6o°. Ans. f. 11. From the point P (h, k) perpendiculars PM, PN are drawn to the A axes. If the axes are inclined at a) show that MN = sin 00 Vh 2 + k 2 + 2 hk cos co [Note— OM = h + kcosco, ON = k + h cos co]. 12. With the notation of the last question, the equation to the perpen- dicular from P on MN is hx — ky = h 2 — k 2 If this perpendicular meet the bisectors of the angles between the axes in R, S, prove that OR - 2(h + k)cos- } OS = 2(h - k)sin-, RS = 2 MN cosec co 2 2 CHAPTER IV APPLICATIONS TO GEOMETRY PROOF OF GEOMETRICAL THEOREMS § 94. We shall now give some examples to illustrate the manner of applying the method of co-ordinates to the investigation of Geometrical Theorems. 8 95. Ex. i. Prove that the perpendiculars from the vertices of a triangle ABC on the opposite sides are concurrent. Take the mid point O of BC for origin and OC for axis of x. Let the co-ord's of A be OD = h, AD = k; let BO = OC = a Thus C is (a, o) „ B „ (-a, o) ,, A „ (h, k) The eq'n to AC is x y i h k i a o i = o, or kx - (h - a) y - ak = o 7 6 Analytical Geometry [ 9 6. The eq'n to the _|_ from B (— a, o) on this is x + a y (0 (2) (3) k - (h - a) or (h — a) x + ky + (h — a) a = o Similarly we obtain the eq'n of the JL from C on AB (h + a) x + ky — (h + a) a = o The equation of the _L from A on BC, i. e. of AD is x= h To ascertain whether the three lines (1), (2), (3) meet in one point we must solve two of the eq'ns for x, y and ascertain if the resulting values of x, y satisfy the third. From (2), (3) we obtain x = h a 2 - h 2 These values satisfy (1). Q.E.D. Cor' — Let M be the point of intersection of the perpendiculars (the orthocentre). We have its y, viz. DM = ( a + h) (a - h) = BD.DC k AD .-. DM . DA = BD . DC 9 96. Ex. 2. Show that the medians of ABC are concurrent. ''a + h k\ The mid point of AC is f ' The eq'n to the join of this to B ( — a, o) is x y 1 a + h k = o 2 2 — a 01 or kx — (3 a + h) y + ak = o Similarly the join of the mid point of AB to C is kx + (3 a — h) y — ak = (1) (2) 99-] Applications to Geometry 77 The third median is AO ; its equation is y k x = f, ' or h y - kx = ° • ' (3) Solving (2), (3) we get h k x = -, y = -; 3 3 these values satisfy (i). Q.E.D. § 97. Ex. 3. The perpendiculars to the sides of ABC at their mid points are concurrent. The J_ to AC through its mid point is a + h k x- y-- 2 J 2 a-h , , , . a 2 - h 2 - k 2 " or (a — h) x — ky = o Similarly the _L to AB through its mid point is , , . . a 2 - h 2 - k 2 (a + h) x + ky + = o The J. to BC through its mid point is x = o These three lines concur in the point X r-. O \ h 2 + k 2 - a 2 > y = 2k CHOICE OF AXES § 98. We may select any two lines in our figure for axes of co-ordinates. Much of the elegance and brevity of a solution depends upon a judicious choice. To illustrate this remark we add two other proofs of the theorem of § 96. § 99. First Proof. Take OC for axis of x and OA for axis of y (fig' § 95) ; these axes are usually oblique. Let OC = a, OA = b. 78 Analytical Geometry [ioo. Then C is (a, o), B is ( - a, o) and A is (o, b) The mid point of AC is ( - , - J , and the join of this to B is = o, or bx — 3 ay + ab = o . x y i a b - — i 2 2 — a o i Similarly the join of the mid point of A B to C is bx + 3 ay — ab = o The third median is AO ; its equation is X = o . . The three lines (i), (2), (3) concur in the point b x = o, y=- This proof is somewhat shorter than that in § 96. (0 (2) (3) § 100. Second Proof. Take any lines as axes of co-ordinates; let the co-ordinates of A be (x 1 y 1 ), of B (x 2 y 2 ), and of C (x 3 y 3 ) The mid point of BC is 'X 2 + : /X2 + X3 y 2 + y s \ The join of A to this point is x = o y 1 xi yi 1 x 2 + x 3 y 2 + y 3 J 2 2 or x (2 y x - y 2 - y 3 ) - y (2 Xj - x 2 - x 3 ) + x l (y 2 + y 3 ) - yi (x 2 + x 3 ) = o By symmetry we may write the eq r ns to the other two medians x (2 y 2 - y 3 - yO - y (2 x 2 - x 3 - x x ) + x 2 (y 3 + y x ) - y 2 (x 3 + x x ) = o x (2 y 3 - y 1 - y 2 ) - y (2 x 8 - x x - x 2 ) + x 3 Cyi + y 2 ) - y 3 (xi + x 2 ) = o CO (2) (3) [02.] Applications to Geometry 79 It will be found that (i), (2), (3) concur in the point ^(x 1 + x 2 + x 3 ), |(y 1 + y 2 + y 3 ) Although longer, this solution has the advantage of symmetry. § 101. Ex. 4. The mid points of the three diagonals of a quadrilateral ABB' A' are collinear. B f Take OA, OB as axes. Let OA = a, OB = b, OA' = a', OB' = b' Then A is (a, o) and B' is (o, b') A T .*. mid p't of AB' is (- , -\ Sim'y mid p't of A'B is (- , -) x y Again, eq'n to A B is - + f- = 1 & ^ a b " A B " a 7 + P " ' Solving these we get the co-ord's of E and thence those of the mid p't of OE, viz. aa' (b' - b) bb' (a - a') 2 (ab' - a'b) ' 2 (ab' - a'b) It may now be verified as in § 24 that the three points are collinear. Or thus : Observe that the values just obtained are the co-ord's of the p't /a b'\ /a' b\ which divides the join of 1 - , — I and I — > — J in the ratio \2 2/ \2 2/ ab':-a'b (§16) LOCI § 102. In § 13, Ex. 5, we have found the equation of the locus of a point equidistant from two given points (1, 2) and (3, 4). In this case we know from Elementary Geometry that the locus is the line which bisects at right angles the join of the given points. 8o Analytical Geometry [103. This might be verified in the present instance by forming the equation of the line through the mid point of the join of (1, 2) and (3, 4) perpendicular to this join. In most cases it is easy to translate a geometrical condition into an equation in x and y. Wc now give further examples. 8 103. Ex. 1. A and B are two given points; find the locus of P if PA 2 - PB 2 = a constant - k 2 Take the mid point O of AB for origin, and OB for axis of x. Let AB = 2 a. Let P, (x, y) be a point on the locus. Then A is (— a, o), and B is (a, o) ... PA 2 = (x + a) 2 + y 2 , PB 2 = (x - a) 2 + y 2 (§ 10) /. PA 2 - PB 2 = 4 ax k 2 .-. the equation to the locus is 4 ax = k 2 , or x = — 4 a Thus the locus is a straight line 1 AB. Ex. 2. Find locus of P if PA 2 + PB 2 = k 2 . Using fig' and notation of Ex. 1, the equation to the locus is 2 x 2 + 2 a' 2 + 2 y 2 = k 2 k 2 or x 2 + y 2 = — — a 2 J 2 This expresses that the distance of x, y from the origin is or the locus is a circle centre O and radius V?- ">4.J Applications to Geometry 81 Ex. 3. Find the locus of P if PA = n PB. Here PA 2 = n 2 PB 2 ; /. eq'n to locus is (x + a) 2 + y 2 = n 2 [(x - a) 2 + y 2 ] It will be seen hereafter that this represents a circle. Exercises 1. A point moves so that its distance from (1, 2) = its distance from (3, — 4) ; find its locus. Ans. The straight line x — 3 y = 5. 2. The co-ord's of A are (3, o) and of B ( — 3,0) ; a point P moves so that PA 2 + PB 2 = 50 ; find the locus of P. Ans. The circle x 2 + y 2 = 16. 3. The co-ord's of A are (— a, o) and of B are (a, o) ; if PA = 2 PB find the eq'n to the locus of P. Ans. 3 x 2 + 3 y 2 — 10 ax + 3 a 2 = o 4r. Find the eq'n to the locus of a point which moves so that its distance from the origin = twice its distance from the axis of x. Ans. The two lines x + y V3 = °« 5. Find the eq'n to the locus of a point which moves so that its distance from the axis of y = its distance from the point (1, o). Ans. y 2 — 2 x + 1 = o. § 1 04*. Ex. 4. Find the equation to the locus of P if PA + PB = constant = 2 a. As before take the mid point O of AB for origin and OB for axis of x. Let AB = 2 c ; then A is (— c, o) and B is (c, o) The equation to the locus is V(x + c) 2 + y 2 + V(x - c) 2 + y 2 = 2 a or V(x + c) 2 + y 2 = 2 a - V(x - c) 2 + y 2 82 Analytical Geometry IO: Square both sides, cancel and transpose ; we get 4 a V(x — c) 2 + y 2 = 4 a 2 — 4 ex Divide by 4 and square again : a 2 (x - c) 2 + a 2 y 2 = (a 2 - ex) 2 This gives (a 2 - c 2 ) x 2 + a 2 y 2 = a 2 (a 2 - c 2 ) Divide by (a 2 - c 2 ) a 2 , and put a 2 - c 2 = b 2 x 2 y 2 — + — — 1 a 2 b 2 This equation represents an ellipse ; a curve which will be discussed in Chap. IX. 8 105. Ex. 5. A, B, C are three given points : find the locus of P if PB 2 + PC 2 = 2 PA 2 . Take any two lines as axes of co-ordinates. Let the co-ord's of A be (x x yj), of B (x 2 y 2 ) and of C (x 3 y 3 ). Let P be a point on the locus ; x, y its co-ord's. Then PA 2 = (x - x x ) 2 + (y - yO 2 , PB 2 = &c, PC 2 = &c. and the equation to the locus is (x - x 2 ) 2 + (y - y 2 ) 2 + (x - x 3 ) 2 + (y - y 3 ) 2 = 2 (x - Xl ) 2 + 2 (y - yi ) 2 or (4x 1 -2x 2 -2X 3 )x + C4yi-2y 2 -2y 3 )y = 4*i 2 + 4yi 2 -x 2 2 -y 2 2 -x 3 2 -y 3 2 This equation being of the first degree in (x, y) represents a straight line. Otherwise thus. Take the mid point of BC for origin and OC for axis of x (see fig' § 95) ; let BC = 2 a, OD = h, AD = k. Then C is (a, o), B is ( — a, o) and A is (h, k). Thus PB 2 = (x + a) 2 + y 2 , PC 2 = (x - a) 2 + y 2 , PA 2 = (x - h) 2 + (y - k) 2 and the equation to the locus is 2 x 2 + 2 a 2 + 2 y 2 = 2 (x — h) 2 + 2 (y — k) 2 or 2 xh + 2 yk = h 2 + k 2 — a 2 This represents a straight line J_ OA. We have here another instance of the advantage of a judicious choice of axes. IO 7-] Applications to Geometry 83 § 106. Ex. 6. Find the locus of a point P such that if PM, PN be the perpendiculars on the axes the sum of PM, PN is constant and = a. Here PM = y, PN = x ; and the word sum in the question being under- stood to mean algebraic sum the equa- tion to the locus is x + y = a This is the equation to a straight line whose intercepts on the axes are OA = a, OB = a. M r O As an illustration of the force of the words in italics we give the following discussion by Elementary Geometry. On the axes cut off OA = a, OB = a, and let P be a point on AB between A and B. Then since OA = OB AAA OAB = OBA= MPA .-. PM - MA .'•. PM + PN = MA + OM = OA Again, taking a point P' on AB produced P' N' = BN' and R M' + (- R N') = R M' - R N' = R M' - N' B = OB Thus it is the arithmetical difference of the perpendiculars from R that is equal to a. § 107. Ex. 7. Find the locus of a point whose distances from the lines ax + by + c = o, a'x + b'y + c' = o are in a given ratio m : n. By § 76 the locus consists of the two straight lines ax + by + c a'x + b'y + c' n . — J = + m >/a 2 + tf Va /a + b' 2 G 2 84 Analytical Geometry [108. Ex. 8. A, B, C, D are given points ; find the locus of a point P which moves so that A PAB + A PCD = constant = k 2 . Draw PM _L AB and PN J_ CD (the reader can easily supply the figure) Let AB = h, CD = h'. Let the eq'n to AB be x cos a + y sin OC — p = o „ „ CD „ x cos Off + y sin a' - p' = o Then AB.PM + CD.PN = 2 k 2 But PM = x cos a + y sin a - p, PN = xcosa' + y sin a' - p' (§ 78) Thus the locus is the straight line h (x cos OC + y sin OC — p) + h' (x cos 0L f + y sin OL? — p') = 2 k 2 Note — As in § 106, by sum is meant algebraic sum. CASES IN WHICH FORMULAE ARE NOT IMMEDIATELY APPLICABLE § 108. Ex. 9. A straight line moves parallel to the base BC of a given triangle ABC, and cuts the sides in P, Q; BQ, PC are joined: find the locus of their point of intersection T. Take OB, OC for axes. Let OB = b, OC = c. We cannot as in the preceding Examples replace at once the geometrical statement by a relation between x, y the co-ordinates of T. We may proceed thus : Let the co-ord's of T be (OC, /3) Then the eq'n to CT is x y 1 oc p I OCT, The eq'n to OB is y = o = 0, or (/3~c)x-ay + ac^o lop.] Applications to Geometry 85 Combining these eq'ns we get the coord's of P, ( , o ) \C - (3 / Similarly we get the co-Orel's of Q, ( o, — | 3 \ b - Oi) Thus the eq'n to PQ is x(c-/3) y(b-q ) _ ac + /3b 1 x y Now express the condition that this is || BC, whose eq'n is 7- + — = I. This gives b (c - /3) _ c (b - a) ac ~ J3b This is a relation between (X, f3 the co-ordinates of T : we may now replace Of, /3 by x, y and the equation to the locus of T is b (c - y) = c (b - x) ex by This reduces to (ex — by) (ex + by — be) = o The factor ex + by — be = o is the equation to BC ; rejecting this irrele- vant factor the locus is the straight line through O and the mid point of BC, ex — by = o N.B. We may account for the irrelevant factor thus. If we take any point in BC and join this to C and B, and if the joins meet OB, OC in P, Q ; then PQ coincides with BC and .\ fulfils the condition of being parallel to it. In this sense then BC is part of the locus. LOCUS FOUND BY ELIMINATION § 109. We may solve the last example in another way. Take as before OB, OC for axes. Let OB = b, OC = c. Then putting OP = /xb we have OQ = jjlC. The eq r n to CP is 4r + ~ = I (1) H lib c v ; BQ is £ + ¥- = r (2) b JJLC v ' 86 Analytical Geometry [no. Now T is on CP and also on BQ; thus each of the eq'ns (i), (2) states a fact about the co-ord's of T depending on the individual value of /jl. If we eliminate fxfrom (1), (2) we obtain a relation between x and y which is independent of /jl: this relation is the equation to the locus. fjL is eliminated by subtracting the eq'ns : this gives (!-*)(-=)- Thus the locus is b c We shall frequently in the sequel obtain the equation to a locus by elimina- tion : the reader should consider the above illustration carefully. § HO. Ex. 10. A and B are fixed points one on each of the axes : if A' and B' be taken on the axes so that OA' + OB 7 = OA + OB ; find the locus of the intersection of AB' A'B. Let OA = a, OB - b ; put AA' = k The eq'n to AB' is y X a + b - k = 1 O A A' X or bx + ay + (a - x) k = ab (1) The eq'n to A'B is — ^-r + £ = 1 ^ a + k b or bx + ay + (y - b) k = ab (2) The equation to the locus is got by eliminating k from (1), (2). (1) - (2) gives (a-x-y+ b)k = o /. the locus is the straight line x + y = a + b III. Applications to Geometry 87 Ex. ir. The extremities A, B of the hypotenuse of a right-angled triangle ABC move on two rectangular axes OX, OY ; find the locus of C. A A Let AB = h, OBA = co, ABC = a. Then BC = h cos OC Let x, y be the co-ordinates of C. Project OC and OBC on OX and OY (§ 60) ; .*. x = h cos co — h cos a cos (a) + OC) (1) y = h cos OC sin (co + OC) ......... (2) The eq'n to the locus is obtained by elim'g co from (1), (2). Now o) = (o) + a) — a, h cos co = h cos (co + OC) cos OC + h sin (co + OC) sin OC Substitute this expression for h cos co in ( 1 ) ; thus X = h sin OC sin (co + OC) (3) From (2), (3) y = x cot OC ; or the locus is a straight line through O. POLAR CO-ORDINATES § Ml. If a straight line revolves round a fixed point and we require the locus of a point on the revolving line whose position on that line is defined by any law ; it is advantageous to use polar co-ordinates. Ex. 1. A straight line which revolves round a fixed point O meets a given line PM in P ; on OP a point Q is taken such that OP . OQ - k 2 , a constant ; find the locus of Q. Draw OM X the given line. 88 Analytical Geometry in. A Let OM = a, OQ = r, QOM = 0. Then OP. r = k 2 , r Also k 2 a - OP cos 6 = — cos 9 r O M Thus the equation to the locus is ar = k 2 cos 6 We may write this ar 2 = k 2 (r cos 0) or a(x 2 + y 2 ) = k 2 x By Chap. VI. this equation represents a circle on ON as diameter, where N k 2 is a point on OM such that ON = — . Ex. 2. One vertex O of a triangle OBC whose angles are given is fixed; another vertex B moves on a given line BM ; find the locus of the third vertex C. But and or or Take O as origin and OM _L the given line as initial line. A Let OM = a, OC = r, COM = 0. A OB = OC A Then a = OB cos BOM. sin y sin/3 BOM = 6 -OC sin y the eq'n to the locus is a = r - — ~ cos (0 — Oi) sin p a sin/3 siny = r cos cos OL + r sin sin (X ; x cos a + y sin a = a sin /3 sin v the locus is a straight line. Ill •] Applications to Geometry 8 9 Ex. 3. A straight line revolves round a fixed point O and cuts two given lines DE, FG in R, S; find the locus of a point P on the revolving line such that 2 1 1 OP ~ OR + OS Let (r, 0) be the polar co-ordinates of P. Let the equation to DE be a i x + D i y + c x = o and to FG a 2 x + b 2 y + c 2 = o .'. the polar equation to DE is a x r cos + b x r sin + c x = o That is Similarly a, cos 6 + bi sin 6 + — = o r - -— cos 6 sin 1 OR — cos sin 1 p-= = cos d sin d OS Co Cq c i b '2 C 2 Thus the polar equation to the locus of P is given by 2 2 1 1 r " OP = OR + OS = _(^ + ^ C os0-(^U^) S in0 \0 X C 2 / \C! C 2 / Multiply up by r and replace r cos 0, r sin by x, y : we thus get \C! C 2 / \C X Ca/- 7 This may be written ajX + b t y 4- Cj a 2 x + b 2 y + c 2 + — o Ci c 2 By Chap. V it is seen that the straight line represented by this equation passes through the intersection of the given lines. 9° Analytical Geometry Exercises on Chapter IV A 1. ACB is a triangle in which C is aright angle. Squares ACED, BCGH are described as in the figure to Euclid I. 47. Show that BD, AH meet on the perpendicular from C on AB. 2. PQ, RS are parallels to adjacent sides of a parallelogram. Show that PR, QS meet on a diagonal. 3. The base AB of a triangle is given. Taking ( — a, o) and ( + a, o) as the co-ord's of A, B, find the equation to the locus of the vertex C i°, If cot A + m cot B = k 2 , If A - B = D 3 , If B = 2 A 4°, If m AC 2 + n BC 2 - k 2 Ans. i°, The straight line (1 — m) x — ky + (1 + m) a = o 2 , x 2 — 2 xy cot D — y 2 = a 2 3°, 3 x 2 — y 2 + 2 ax = a 2 4°, (m + n) (x 2 + y 2 ) + 2 (m - n) ax + (m + n) a 2 -= k 2 4. Points P, Q are taken on the sides AB, AC of a triangle ABC such that AP.AQ- BP.CQ; find locus of mid point of PQ. [Take AB, AC as axes ; let AB = a, AC = b. If mid point of PQ is (xy) then AP = 2 x, AQ = 2 y ; .\ 4 xy = (a — 2 x) (b — 2 y) .*. locus is the straight line 2 bx + 2 ay = ab] 5. Given base and sum pf areas of a number of triangles with a common vertex P : show that the locus of P is a straight line. 6. From a point P perpendiculars PM, PN are drawn to the axes (which A include to); if locus of P is a straight line show that locus of mid point of MN is a straight line. [Note— If P is (xy) and mid point of MN is (hk) : then OM — x + y cos a) = 2 h, ON = y + x cos a) = 2 k] Applications to Geometry 91 7. With the notation of the last question, if OM + ON is given (= k), find the locus of P. Ans. The straight line (x + y) (i + cos co) = k. 8. A parallel to the base of a triangle meets the sides in B', C ; R, S are fixed points on the base : show that the locus of the intersection of B' R, C / S is a straight line. Note— Let B be (-a, o), C ( + a, o), A (h, k), R (r, o) and S (s, o). Let eq'n to B' C be y - A. We find that eq'n to C S is A [y (h + a) + k (s - x)] = ky (s + a) „ B'R „ A[y(h-a) + k(r-x)]--ky(r-a) Elim' A : locus is (r - a) [y (h + a) + k (s - x)] = (s + a) [y (h - a) + k (r - x)] 9. A and B are fixed points : if PA, PB intercept a constant length c on a given line, find the equation to the locus of P. Note — Let AB meet given line in O ; take given line and OAB as axes. Let OA = a, OB = b. Ans. c (a — y) (b — y) = (a — b) xy 10. A line AB of constant length (= c) slides between the axes, which A include a) : show that the equation to the locus of the orthocentre of the triangle OAB is x 2 + 2 xy cos a) + y 2 = c 2 cot 2 co 11. On a line which revolves round the origin and cuts the lines ax + by + c = o, a! x + b'y + c r = o in P, Q a point R is taken such that OR = OP + OQ: show that the equation to the locus of R is (ax + by + c) (a'x + b'y + c') = cc' 12. Show that the orthocentre, centroid and circum centre of a triangle are collinear. 9 2 Analytical Geometry 13. OIJ, OLM are given straight lines ; I, J are fixed points; IL, JM meet at P. Prove that P describes a straight line if a b OL + OM ~ *' where a and b are constants. 14. The opposite sides of a quadrilateral meet at P, Q; if the internal bisectors of the angles P, Q are at right angles, prove that their intersection lies on the join of the mid points of the diagonals. [Note — Take the bisectors as axes ; then sides of quad' are y = mx + c, y = — mx — c, x = /xy+k, x=— /uty— k We find that mid points of diag's are / [Ac mk \ / — fxc — mk \ \i — m jh ' i — m ft/ ' \i + m^' r + m y.) ' these lie on the line y/x — m kAi c] CHAPTER V EQUATIONS WITH LINEAR FACTORS. ABRIDGED NOTATION; RANGES AND PENCILS CASE OF FACTORS § (12. An equation which splits into factors of the first degree represents straight lines. Ex. i. What is represented by x 2 - 5 xy + 6 y 2 = o ? Here (x - 2 y) (x - 3 y) = .*. either X — 2 y = o or x — 3 y = o Thus the coord's of every point on either of the lines x - 2 y = o, x - 3 y = o satisfy the given equation ; which .*. represents this pair of straight lines. Ex. 2. What is represented by x 3 — 6 ax 2 + 1 1 a 2 x — 6 a 3 = o ? Here (x — a) (x — 2 a) (x - 3 a) = j .\ the equation represents the three lines x = a, x *= 2 a, x = 3a Ex, 3. What is represented by 2xy-8x+3y-i2=o? 2xy-8x+ 3y-i2 = y(2x + 3)-4(2x + 3) = (y - 4) (2 x + 3) Ans. The two lines y*=4> 2 x + 3 = o. 94 Analytical Geometry [T13. Ex. 4. Show that the equation 6x 2 — xy— y 2 — x + 3y — 2 =0 represents two straight lines. Arrange according to powers of x, and solve for x. .-. 6 x 2 — (y + 1) x — y 2 + 3 y — 2 = o y + 1 + V(y + i) 2 + 24(y 2 - 3y + 2) x = 12 _ y + 1 ± V25V 2 - 70 y + 49 12 = y + T ± ( 5 y - 7) 12 y- 1 2 -y or 2 3 Thus the given equation implies that either y - 1 2 - y x = or x = ; 2 3 i. e. it represents the two lines 2 x - y + 1 = o, 3x + y — 2 = 0. Note — The quantity under the radical, viz. 25 y 2 - 7°y + 49 turned out to be a perfect square : had it been otherwise the equation would not represent straight lines. § 113. The homogeneous equation of the second degree ax 2 + 2 hxy + by 2 = o represents two straight lines through the origin. Dividing by x 2 , b (0 +2h (x) +a = ° y If m 13 m 2 are the roots of this quadratic in - -h+vV-ab , -h-Vh 2 -ab m, — ; and m = ; II 6.] Linear Factors 95 y y Thus the equation implies that either — = rrij or - = m 2 ; i. e. it represents the two lines y = n\ x x, y = m 2 x. § H4-. If is the angle between the lines ax 2 + 2 hxy + by 2 = o tan 6 = mi ~ m2 (§ 68) 2 VV — ab b 2 >/h 2 — ab -(■♦» a + b Cor' (i) — The lines are X if a + b = o Co/ (2) — Any two lines at right angles through the origin may be represented by x 2 + A xy — y 2 = o § H5. If the axes are oblique tan Vh' 2 — ab a + b — 2 h cos a) Cor' — The lines are J_ if a + b — 2 h cos to = o § 116. To find the equation to the bisectors of the angles between the lines ax 2 + 2 hxy + by 2 = o The bisectors of the angles between y— m 1 x = o, y — m 2 x=o = o y-m,x y — m 2 x y— nrijX , y- m 2 x / " ■ ' Vi + m x 2 V 1 + m 2 2 V 1 + m x 2 vi + m 2 2 g6 Analytical Geometry [117. Multiplying, the equation to the pair of bisectors is (y- m t x) 2 _ (y- m 2 x) 2 _ q 1 + m/ 1 + m 2 2 or (m 2 2 — m/) (y 2 — x 2 ) + 2 (m, m 2 - 1) (m, — m 2 ) xy = o or (m 2 + rrij) (y 2 - x 2 ) - 2 (nr^ m 2 - 1) xy = o 2 or — ^(y 2 -x 2 )- 2 (g- i)xy = o x 2 — y 2 xy *'• a- b "" TT This result should be remembered. 8 \\7. To find the condition that the general equation of the second degree ax 2 + 2 hxy + by 2 + 2gx + 2fy + c = o should represent a pair of lines. Rearrange, ax 2 + 2 (hy + g) x + by 2 + 2 fy + c = o Solving for x, ax + hy + g = ± Any + g) 2 - a (by 2 + 2 fy + c) = ± V(h 2 - ab)y 2 + 2 (gh - af)y + g 2 — ac The expression under the radical is a perfect square if (h 2 - ab) (g 2 - ac) = (gh - af) 2 or abc + 2 fgh — af 2 — bg 2 — ch 2 = o . . . (1) This is the required condition ; it should be remembered. II 9-] Linear Factors 97 § 118. The quantity abc + 2 fgh - af 2 — bg 2 - ch 2 is called the discriminant of the expression * ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c The discriminant is usually denoted by A. The condition (1) then is A = o We have also A = a h g h b f g f c Ex. Determine A so that x 2 + Axy — 8 y 2 + 12 y — 4 = o may represent a line-pair. Here a = i f = 6 b = -8 g = o c = — 4 h = \\ .-. A = 32 - 36 + A 2 = A 2 - 4 Put A = o ; A = + 2 or — 2 § 119. If Ix + my = 1 and ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = o . then evidently ax 2 + 2 hxy -f by 2 + 2 (gx + fy) (Ix + my) + c(Ix + my) 2 = o 4 (0 (3) * The notation ax 2 + 2 hxy + by 2 +2gx + 2fy + c = o will be invariably used to represent the general equation of the second degree. If we introduce the linear unit z = 1 so as to render the expression homo- geneous it assumes the symmetrical form ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2 hxy = o ; this remark will assist the learner to remember the notation. H gS Analytical Geometry [119. ,2 Since the values of x and y which satisfy (1) and (2) also satisfy (3), the locus represented by (3) passes through the points of intersection of the loci represented by (1), (2). Also since (3) is homogeneous and of the second degree, it represents a pair of lines through the origin: .-. (3) represents the pair of lines joining the origin to the inters'ns of the loci represented by (1), (2). Ex. The line-pair which joins the origin to the inters'ns of 2 x + y = 6, x 2 + y 2 = x + 5 y + 6 is or reducing, xy = o ; i. e. the axes. Exercises 1 . Show that 3 x 2 + 2 xy — 3 y 2 = o represents two lines at right angles ; and x 2 + 2 xy + y 2 — x — y — 6 — o two parallel lines. 2. Interpret the equations (x - 3) (y - 4) = °> x 2 - 3 ax + 2 a 2 = o, xy = 0, x 2 - y 2 = o, xy — 2 x + 3 y — 6 = o, 6 xy + 2 bx + 3 ay + ab = o. 3. Show that the angle between the lines 6x 2 — xy — y 2 = o is — • 4 4. Show that 3x 2 + xy— 2y 2 + x + 6y — 4 = represents two lines in- clined at tan -1 5. 5. Show that the angle between the lines x 2 — 2 xy sec 6 + y 2 = o is 0. 6. Determine A. so that 6xy — 2x + Ay+5 = o may represent straight lines. Ans. K — — 15. 7. Determine A so that x 2 + Axy + y 2 -5x-7y + 6^o may represent straight lines. Ans. \ = f or ^. 120. Linear Factors 99 8. Find the line-pair joining the origin to the intersections of y + x = 2, x 2 + y 2 - 2 x — 4 y - 31 = o. Ans. 31 x 2 + 74xy + 35 y 2 = o. 9. Show that the lines joining the origin to the intersections of y = A(x- 4 ), y 2 = 4* are at right angles. Ans. The line-pair is A. (x 2 — y 2 ) — xy = o. 10. Find the bisectors of the angle between the lines 3 x 2 + 4 xy - 5 y 2 = o. Ans. x 2 — 4 xy — y 2 = o. 1 1 . Find the bisectors of the angle between the lines 3 x 2 + 8 xy + 3 y 2 = o. Ans. x 2 — y 2 = o. '"'"" 12. Show that the lines ax 2 + 2 hxy + by 2 = A (x 2 + y 2 ) are equally inclined to the lines ax 2 + 2 hxy + by 2 = o \_Note — They have the same bisectors of angles.] IMAGINARY POINTS AND LINES § 120. Further examples. Ex. 1 . What is represented by x 2 + y 2 = o ? Since the square of any positive or negative number is positive, x 2 + y a is never negative ; it can only be zero if both x = o, y = o. It was .'. formerly the custom to say that x 2 + y 2 == o represented the point x = o, y = o. Another account may be given : x 2 + y 2 is the product of x + yV-i, x-yV-i Thus we may say that the equation represents the two imaginary straight lines x + y V— 1=0, x — y V — 1=0 These imaginary lines intersect in the real point x = o, y = o. H 2 ioo Analytical Geometry 122. Ex. 2. Interpret 4 (3 x - y - i) 2 + 3 (x + y - 3) 2 = o The only real values of x, y which satisfy this are given by 3X-y-i = o, x + y-3 = o, i.e. x = i, y = 2 If .*. we do not admit imaginary lines the equation represents the point (I, 2). If imaginaries are admissible, then factorizing, we see that the equation represents the two imaginary lines 2 (3X-y- i) + V-3(x + y-3) = o 2 (3X - y - i) - V-3(x + y- 3) =o These intersect in the real point, (x = i, y = 2). § 121 . Let us now recur to the investigation of § 113. If h 2 — ab > o the values obtained for m v m 2 are real and unequal. „ h 2 — ab = o „ „ real and equal. „ h 2 — ab < o „ „ imaginary. Thus the two lines through the origin represented by ax 2 + 2 hxy + by 2 = o are real and different, coincident, or imaginary according as h 2 — ab > = < o § 122 • A homogeneous equation of the n th degree ay 11 + bxy 11-1 + cx 2 y n " 2 + ...+ kx^y + lx n = o . . . (1) represents n straight lines through the origin. Dividing by x n , + 1 = y Let the roots of this equation in - be rrti, m 2 , m 3 , .., The equation then represents the n straight lines y= rrtiX, y = m 2 x, ... 124- ] Linear Equations 101 Note — The sinister side of (i) = a (y - rriiX) (y - m 2 x) (y - m 3 x) ... Similarly it is proved that the equation a(y-y 1 ) n +b(y-y 1 ) n - 1 (x-30 + c (y - yi) n ~ 2 (x - Xl ) 2 + ... + l (x - Xl ) n - o represents n straight lines through (x^). LINE THROUGH INTERSECTION OF TWO GIVEN LINES § 123. Consider the equation Ax + By + C + A (A'x + B'y + C) = o . . (i) I. This is of the first degree in X, y ; .-. it represents a straight line. II. It is satisfied if Ax + By + C = o and A'x + B'y + C = o : thus (i) represents a line through the inters'n of Ax + By + C = o, A'x + B'y + C = o By giving A a suitable value (i) may be made to represent any such line. § 124-. Thus, to find equation to join o/^y^) to intersection of Ax + By + C = o, A'x + B'y + C = o, A is determined by Ax x + B yi + C + A (A'x x + B'yi + C) = o Substitute the value of A so determined in (i) ; .'. the eq'n req'd is Ax + By + C A'x + B'y + C Axj+ By x + C ~ A'x ± + B'yj. + C 102 Analytical Geometry [125. Note — The form I (Ax + By + C) + m (A'x + B'y + C) = o by dividing by I and putting -y = A is reducible to (1). § 125. Ex. Find equation of line _L 3X+4y + 5=o through inters'n of 3 x-y-i = o, x + y-3 = o. A line through this inters'n is 3X-y-i + A(x + y-3) = o or (3 + A) x + (A — 1) y — 3 A — 1 = o This isJ_3X + 4y+5 = o, if 3 (3 + A) + 4 (* - 1) = o [§ 68, Cor' (3)] .-. A = — |, and the eq'n req'd is 3 x-y-i-^(x + y-3) = o or 4X — 3y + 2=o Exercises 1. Find equation of join of (1. —2) to inters'n of 3*-4y + 5=o> 2x-3y + 4 = o Ans. x = 1. 2. Find join of same intersection to origin. Ans. 2 x — y = o. 3. Find line through same inters'n JL iox+ 7 y = 8 Ans. 7x— ioy + i3 = o 4. Find the equation of join of origin to inters'n of x y x y a b b a Ans. x = y. 5. Find the equation of join of origin to intersection of Ax + By + C = o, A'x + B'y + C = o Ans. (AC - A'C) x + (BC - B'C) y = o 12 8.] Linear Equations 103 6. Find the equations to diag's of □ whose sides are 3 x- 3 y=i, 4*-5y = 6, 3x-2y = 2, 4 x-sy = 3 Ans. 13 x = 11 y + 9, 5* = y- 7. Find the area of this O Ans. -f-. ABRIDGED NOTATION § 126. It is often convenient to represent such an expression as Ax + By + C by a single symbol. Thus we may put u = Ax + By + C, v = A'x + B'y + C The proposition of § 123 may then be stated thus. If u = o, v = o represent straight lines then u + Av = o represents a line through their intersection ; and by giving a suitable value to A it may be made to represent any such line. Thus u + Av = o represents a ray of the pencil through the intersection of u = o, v = o. § 127. Greek letters are used as abbreviations for expressions of the form x cos 06 + y sin 06 — p. We put 06 = x cos 06 + y sin 06 — p, /3 = xcos/3 + ysin/3— p', &c. Thus 0C + /3 — o, Oi — /3 = o are the bisectors of the angles between 06 = o, p = o (§80, Cor') No ambiguity results in practice from the double signification of the symbols 06, /3, &c. • § 128. If we write at full length the equation 06= k/3 it becomes x cos 06 + y sin 06 — p = k (x cos 13 + y sin/3 — p') . . . (1) io4 Analytical Geometry [129. Let P, (x, y) be any point on (1), and draw PM, PN respec- tively ± Oi = o, /3 = o. Then (1) expresses that PM = kPN PM PN •'* CP CP C M « = ° r Sln0 = kS ' m( P Thus 06 — k /3 = o divides the angle between a = o, /3 = o into parts whose sines are in the ratio k : 1 . Again, u — kv = o where u = Ax + By + C, v = A'x + B'y + C may be written Ax + By + C VA'z + B^ A'x + B'y + C VA2 + B2 VA 2 + B 2 VA /2 + B'2 , VA' 2 + B /2 . or a = k . — — ==r- . )3 \/A 2 + B 2 Thus u — kv = o divides the angle between u = o, v = o into parts whose sines are in the ratio k Va /2 + B'2 VA 2 + B 2 : 1 CONDITION FOR CONCURRENCE OF THREE LINES § 129. If the equations to three lines u=o, v = o, w = o are connected by an identical relation lu +mv+ nw E o; the three lines are concurrent. For w — • u .v; n n thus the equation to the third line is I m . u n n . v = o or la + mv = o i3°«] Linear Equations 105 This passes through inters'n of u = o, v = o (§ 123). Q.E.D. Ex. 1. The lines x — y + 1 = o, x + y-3=o, 2 x — 3y + 4 = o are concurrent. For multiplying the first equation by 5, the second by — I, and the third by — 2, and adding, the result is identically zero, or 511— v— 2 w = o Note — The beginner will see from this example that the proposition is sufficiently obvious. Thus multiplying the first equation by 5, and the second by — 1 and adding, we get 4X— 6 y + 8 = o, or 2X — 3 y + 4 = o, which is the third equation. Thus the values of x, y which satisfy both of the first two equations must satisfy the third. Ex. 2. The medians of the triangle (x x y x ), (x 2 y 2 ), (x 3 y 3 ) are concurrent. Let u, v, w stand for the sinister sides of equations (1), (2), (3), § 100. The medians are u = o, v = o, w = o. Adding the three equations we find that the sum vanishes identically ; or u + v + w = o. Q.E.D. LINE THROUGH A FIXED POINT § 130. We have seen that the line u + A v = o, where A is a variable parameter, passes through a fixed point, viz. the intersec- tion of u = o, v = o. Thus if the equation to a line contain a variable parameter A in the first degree, the line passes through a fixed point. Ex. 1. A straight line moves so that the sum of the reciprocals of its intercepts on the axes is constant ; show that it passes through a fixed point. The equation to the line is x y - + £ =-- i (1) a b v J io6 Analytical Geometry [i3°- Also - + .- — constant = k a b We may eliminate r from (i) by means of (2) ; thus k- and (1) becomes x - + y a J ( k -a) = (x — y) + ky — 1 = o (2) Whatever be the value of - this equation is satisfied if x - y = o, ky — 1 = o ; i. e. the line passes through the fixed point x = k' y = Ex. 2. A straight line moves so as always to have two of the three points ( x i y0> ( x 2 y2)j ( x 3 ys) on one s ^ e °^ ** an( ^ ^e third point on the opposite side; and the sum of the JLs on the movable line from the first two points = the _L from the third point : show that the line passes through a fixed point. Let its equation in any position be xcosCX + y sin a — p = o (1) Then the algebraic sum of the J_s = o .-. (x x cos a + yj sin a - p) + (x 2 cos a + y 2 sin a - p) + (x 3 cos OC + y 3 sin OC — p) = o or (Xj + x 2 + x 3 ) cos a + (y x + y 2 + y 3 ) sin a - 3 p = o . • . (2) Eliminate p from (1) by means of (2) ; thus the equation to the movable line becomes sin OC = o (x - x ' + * + Xs ) cos a + (y - * * * + y » ) si, or ( X - X - + * + * ) + (y- * + h + y^ tana = o Whatever tan OC may be, this passes through the fixed point i (x 2 + x 2 + x 3 ), i (y x + y 2 + y 3 ) 32.] Linear Equations 107 THE LINE AT INFINITY 8 131* The intercepts which Ax + By + C = o cuts off on the axes C G are ~ A ' ~ B (§ ^ C . I. If A is very small, — ^ is numerically very great ; or the point where the line cuts OX is very distant from O. The line Ax + By + C = o is then nearly parallel to OX. C Let now A = o, then — = oo ; and the line A Ax + By + C = o is parallel to OX. This agrees with § 45. C C II. If both A and B are very small, both the intercepts — — , — = are numerically very large. Let now A = o, B = o ; then the intercepts are infinite ; or the line is altogether at infinity. We see then that if a line cuts off infinite intercepts on both axes its equation is o.x+ o.y+ C = o. This statement is usually abbreviated thus : The equation to the line at infinity is constant = o. § 132. Parallel lines intersect at infinity Let the equations to two lines be Ax + By + C = o (1) Ax' + By' + C = o 1 (2) The equation to the line at infinity is o.x + o.y + C" = o (3) The condition that the lines (1), (2), (3) may be concurrent is ABC = o (§ 62) A' B' C o o G" or (AB' - A'B) C" - o or AB' - A'B - o But this is cond'n that the lines (1), (2) may be || (§ 63) io8 Analytical Geometry [133. Otherwise, solving (i), (2), _ BC - B'C _ CA' - C'A X ~ AB' - A'B ' y ~ AB' - A'B When the lines are |j , AB' - A'B = o ; .-. x = 00 , y = a HARMONIC RANGES § 133. Def' — Four collinear points A, B, C, D form a har- monic range if AB AD BC~CD (l) That is AC is divided internally in B and externally in D in the same ratio. B and D are said to be harmonically conjugate to A and C. A BCD For example, the internal and external bisectors of any angle of a triangle divide the opposite side harmonically (Euclid VI. 3). From (1) we deduce: I. Since CD -- DC, ,. g-S-gg— 1. II. Multiplying up, AD . BC = AB . CD. Or, rectangle under whole line and middle segment — rectangle under extreme segments. III. AB, AC, AD are in harmonic progression ; or — , ---^, -— - are in ,,i i Ab /\o /\U arithmetical progression. For let AB = x, AC = y, AD = z. x z Then (1) gives y - x z - y •'• x (z - y) = z (y -. x) Divide by xyz, 1 1 _ 1 1 y~z x~y 211 or - = - + - y x z That is, - , - , - are in A. P. xyz 1 35«] Cross Ratios 109 IV. Similarly DC, DB, DA are in H. P. Also V. If M be the mid point of AC, MB.MD = MC 2 . This is proved in Nixon's Euclid Revised : page 360 VI. From V. we see that if the points A, C are given then B, D are on the same side of M ; and if B approaches M indefinitely D moves off to infinity. Let 00 denote the point at infinity on AC; then the points AMC 00 form an harmonic range. § 134. The points on OX determined by ax 2 + 2 hx + b = o (1) a'x 2 + 2 h'x + b' = o (2) are harmonic if ab' + a'b — 2 hh' = o. Let the roots of (1) be OA = CX, O ± £— ^ =• f OA' = a'; and of (2) OB - 3, u * * A B OB' = /3'. The condition is AB' . BA' = AB . A' B' or (/3' - a) (a' - (3) = 03 - a) (/3' - a') or 2 /3/3' + 2 (xoi! = (a + (3) (a' + ft) b' b / 2hW 2h / \ v + 2 a = (-irA-^; .-. ab' + a'b - 2hh' = o * CROSS RATIOS § 135. Def' — The Cross Ratio (or Ankarmomc Ratio) in which two points A, C on a line are divided by two other points B, D on the same line is the ratio of the ratios in which AC is divided by B, D and is denoted by {ABCD}. rAr.^^, AB AD AB.DC Thus {ABCD} ^^-^^^-^ If {ABCD} = — i, the range is harmonic. * The beginner may omit the rest of this Chapter. no Analytical Geometry [137. Ex. If {ABCD} = - 1, prove {BDAC} = 2. (See figure, § 133) By Art. 9, Ex. 2, AD . BC + BD . CA + CD . AB = o Divide by AD . BC ; BD CA AB DC _ ■"" I- DA'BC _ BC*AD"° or 1 - {BDAC} - {ABCD} = o .-. {BDAC} = 1 - {ABCD} =2. § 136. Def' — The Cross Ratio (or Anharmonic Ratio) in which two lines OA, OC of a pencil are divided by two others OB, OD is sin AOB sin AOD ... , , r _ .„_. — — =r^r~ -r- — — ^-a and is denoted by {O . ABCD}. sin BOC sin DOC J l ) § 137. If a pencil of four lines whose vertex is O be cut by any transversal in A, B, C, D ; then the cross ratio of the range {ABCD} is constant and = that of the pencil. Let the ± from O on AD = p. (fig' page in) _, AB A AOB Then BC = ABOC ( EuchdVLl ) = \ OA . OB sin AOB -=- \ OB . OC sin BOC OA sin AOB OC " sin BOC (1) . ., , AD OA sin AOD Similarly — = — .^ m ^ (2) Divide by (1) by (2) ; thus AB AD _ sin AOB sin AOD BC "*" DC ~~ sin BOC ^ sin DOC .-. {ABCD} = {O. ABCD}. Q.E.D. (See Euclid Revised', page 324.) 38.] Cross Ratios in Note (i) — Since sine of an angle = sine of its supplement, it is indifferent on which side of the vertex O any ray of the pencil is cut by a transversal ; e. g. in fig' {ABCD} = jA'B'C'D'}. Note (2) — It will be remarked that in our diagram AD : DC and sin AOD : sin DOC are both negative. Note (3) — Draw a transversal A"B"C" parallel to OD. This will meet OD in a point at an infinite distance which we shall denote by 00 . Thus {O.ABCD} =§ ^-^ Now A" 00 A" 00 C" A"*) oTC 77 = ; if we put A"C" = p, C" 00 = q, „3JlP = _f I + P , \ = _ I , fo r P=o q \ q/ q A" B" .-. {O.ABCD} = -r^77 B"C Thus if the pencil is harmonic A^C" is bisected in B". § 138. Ex. The joins of a point O to the vertices of a triangle ABC meet the opposite sides in D, E, F ; FE meets BC in D' : prove that {BDCD'} is a harmonic range. Let AO meet EF in D". Then {BDCD'} = {O.BDCD'}. ii2 Analytical Geometry [139- Notice the points in which the rays of this pencil are cut by the transversal EF; thus {BDCD'} = {ED"FD'j = {A.ED"FD'} = {CDBD'}. Now ICDBD'} - 95 i.5^ - ?P^ 5P - 1- (BDCD'} .\ {BDCD'} 2 = 1 .-. {BDCD'} = + 1 But {BDCD'} cannot = + i, for this would imply that D, D' coincide; .-. {BDCD'} = - 1. Q.E.D. § 139. The cross ratio of the pencil u — o, v = o, u = kv, u = k'v is r- f Let u E ax + by + c, v = a'x + b'y + c' Let 6, & f be As which u = kv makes with u = o, v = o; and (f), (j) A As which u = k'v makes with u = o, v = o. sin , A /a /2 + b /2 , e ON sin 6 , ■ A /a /2 + b' s and — F; = k' V— = r-s- v a 2 . 52 sinc^/ sin sin <£ k Q.E.D. " sin0' * sin<£' k 7 Cor f (1) — The four lines u = o, v = o, u + kv = o, u — kv = o form a harmonic pencil. Cor' (2)— The lines x = o, y = o, x + ky = o, x — ky = o form a harmonic pencil. As x + ky = o and x — ky = o are equally inclined to the axis of x we infer — Cor' (3) — If a pencil is harmonic and two alternate rays are at right angles, they bisect the angles between the other two. Hi.] Cross Ratios 113 § 14*0. The cross ratio of the four lines I — n r — m Iv, u = mv, u = nv, u = rv is n — m I — r Put u E u — Iv, V=u- mv, W = u - nv Then (n - m) U + (I - n) V + (m - I) W = o ,\ W = o, or u = nv is equivalent to (n-m)U + (Ir-n)V = o or U + !=-HL v = o (1) n — m ' Similarly the equation U = rv may be written ■ 1 I — r w U + V = o (2) r — m v ' We have then the four lines U = o, V = o, and (i), (2) whose cross ratio is (§ 139) |_ n I- r _^ -h Q.E.D. n — m r — m § 141. The lines a'u 2 + 2 IVuv + b'v 2 = o are harmonically conjugate to the lines au 2 + 2 huv + bv 2 = o if ab' + a r b — 2 hr/ = o Let au 2 + 2 huv + bv 2 = a (u — Iv) (u — mv) a'u 2 + 2 h'uv + b'v 2 = a r (u — nv) (u — rv) Then the pencil is harmonic if (I - n) (r - m) (n - m) (I - r) ^ 9 4 ; or if 2 nr + a ml = Ir + In + mr + mn = (I + m) (n + r) b' b / 2 h\ / 2 h'\ 2 a^ + 2 a = (-^A"^") or if ab' + a'b - 2 hh' = o. Q.E.D. Cor' — In general, if x is the cross ratio of the pencil (I _ n ) (r - m) x = (n - m) (I - r) 1 ii4 Analytical Geometry [142. x — 1 (I — m) (r — n) x + 1 (I + m) (r + n) — 2 {\m + nr) Reducing, we get /x - iV _ (h 2 - ab) (h /2 - a'b') \x + 1) " (ab' + a'b - 2 hh') 2 § 142* Ex. Find the equation to the bisectors of the angles between the lines ax 2 + 2 hxy + by 2 = o A the axes being inclined at go Let the required equation be a'x 2 + 2 h'xy + b'y 2 = (1) Since bisectors are at right angles a' — 2 h' cos go + b' = o (2) Since lines and bisectors form a harmonic pencil a'b — 2 hh' + ab' = (3) We can eliminate a', h', b f linearly from (i), (2), (3) : .'. req'd eq'n is x 2 xy y 2 1 - cosa) 1 b - h a or x 2 (h — a cos go) — xy (a — b) + y 2 (b cos go — h) — o INVOLUTION § 143. Def — Let O be a fixed point on a straight line, and let A, A' ; B, B' ; C, C' ; be pairs of points on the line such that OA . ON =-- OB . OB' = OC . OC' - &c. = constant = k 2 : then the points A, A r , B, B r , &c. form a system in involution. The point O is the centre Points such as A, A' are called conjugate points. The point conjugate to the centre is at infinity. A point which coincides with its conjugate is called a double point or a focus. § |44. If F is a focus, OF 2 - k 2 , .-. OF = + k. Thus there are two foci F, F' at equal distances on each side of the centre. Since OF 2 = OA . OA r , the range F'AFA' is harmonic (§ 133, V). Thus any two conjugate points and the two foci form a harmonic range. M7-] Cross Ratios n r 5 If two conjugate points A, A' are on opposite sides of the centre k 2 - OA . OA' is negative ; the foci are .'. imaginary. § 145. The cross ratio of any four points of the system is equal to that of their conjugates . Let A, B, C, D be any four points of the system, A', B', C, D' their conjugates. k 2 Let OA = OC, OB = £, OC = y, OD - b. Then OA' = — , &e. 1 S y - (3 b - a /k 2 k 2 \ /k 2 k 2 \ Also f a'bwi - ^"^V"^ _ (fl-«)(y-*) Also { A B C D } - ^ ^ ^ ^ - (y _^ )(8 _ a) \y p/\b ol) .-. {ABCD} = {A'B'C'D'} Cor'— {ABCA'} = {A'B'C'A} This relation enables us to ascertain whether six points A, B, C, A', B', C are in involution. § 146. If two pairs of conjugate points are given, the system is com- pletely determined. For let ft, 01' ; /3, &' be their distances from any fixed point on the line,, x the required distance of the centre from this point Then (OC - x) (a' - x) = (/3 - x) (£' - x) .-. (a + a' - /3 - /30 x = aa' - /3/3' This equation determines the centre. § 147. If a system of points in involution be joined to any point P we obtain a pencil of lines in involution. The cross ratio of any four lines of the pencil = that of their conjugates and any transversal is cut by the pencil in a system of points in involution (§ 137). There are two lines of the pencil which coincide with their conjugates ; these are called focal lines. The two focal lines and any two conjugate rays form a hrrmonic pencil (§ 144). I 2 n6 Analytical Geometry [i 4 8. § 148. To find the condition that the three line-pairs au 2 + 2 huv + bv 2 = o, a'u 2 + 2 h'uv + b'v 2 = o, a"u 2 + 2 h"uv + b"v 2 = o should form a pencil in involution Let the equation to the focal lines be Au 2 + 2 Huv + Bv 2 = o These are harmonically conjugate to each of the above : Ba - 2 Hh + Ab =0 Ba' - 2 Hh' + Ab' = o Ba" - 2 Hh" + Ab" = o Eliminate B, H } A : the required condition is .\ h b h' b' h" b" Exercises on Chapter V X . The sides of a parallelogram are 4 x + 5y = 6, 4*+5y=9> 5* + 4y = 6, 5* + 4y = 9- Show that its area = 1, and find equations to its diagonals. Ans. x = y, 3x4- 3y = 5 2. Show that the equation of the join of the iriters'n of a x x + b x y + c 1 = o, a^ x + b 2 y + c 2 = o to that of a 3 x + b 3 y + c 3 = , a 4 x + b 4 y + c 4 = is a x x + b x y + c x + A (a 2 x + b 2 y + c 2 ) = where A is determined by A a 2 b 2 c 2 a £ b 3 c 3 a 4 d 4 c 4 + »i *>i c x ag b 3 c 3 a 4 b 4 c 4 = 3. Show that the lines bx 2 — 2 hxy + ay 2 = o are at right angles to the lines ax 2 + 2 hxy + by 2 = o 148.] Exercises on Chapter V 117 4. Show that one of the lines ax 2 + 2 hxy + by 2 = o will coincide with one of the lines a'x 2 + 2 h'xy + b'y 2 if 4 (ah' - a'h) (hb' - h'b) = (ab' - a'b) 2 Find also the condition that a line of the first pair should be perpendicular to a line of the second pair Ans. Write b, — h, a for a, h, b in preceding cond'n (by Ex. 3). 5. Find the lines represented by x 3 — 3X 2 y — 3xy 2 + y 3 = o [Subst' r cos 0, r sin 6 for x, y ; deduce tan 3 = 1 ; = 15 , 75 or 135 ] 6. Find the lines represented by m (x 3 - 3 xy 2 ) + y 3 - 3 x 2 y = o [Note —deduce tan 3$ = ml 7« Show that the lines a 2 x 2 + 2 h (a + b) xy + b 2 y 2 = o are equally inclined to the lines ax 2 + 2 hxy + by 2 — o 8. Show that (h* + k 2 - 1) (x 2 + y 2 - 1) = (hx + ky - i) 2 represents a line-pair. 9. Find the lines through (OC, /3) perpendicular to the lines ax 3 + bx 2 y + cxy 2 + dy3 = o Am. d (x - a) 3 - c (x - a) 2 (y - (3) + b (x - a) (y - /3) 2 - a (y - /3) 3 = c 10. Prove that the product of the perpendiculars from (x'y') on the lines ax 2 + 2 hxy + by 2 = o ax' 2 + 2 hxy + by /2 V(a- b) 2 + 4 b 2 n8 Analytical Geometry 11. Prove that the distance between the points where the line x cos OL + y sin OL = p meets the lines ax 2 + 2 hxy + by 2 = o is 2 p Vh 2 — ab a sin 2 (X — 2 h sin OC cos OC + b cos 2 OC 12. The line cuts the lines lx + my + n ax 2 + 2 hxy + by 2 = o, making angles OC, /3 with them. Prove that (a _ b) lm - h (I 2 - m2) tan OL + tan /3 = 2 al 2 + 2 him + bm- 13. Find the conditions that the polar equation A tan + B sec 6 = 1 should represent the bisectors of the angles between the straight lines whose polar equation is A sec 6 + B tan Q — 1 Ans. 1 + A 2 = 2 B 2 , (1 - B 2 ) 2 + 2 AB = o 14. If the equation ax 2 + 2 hxy + by 2 + 2gx + 2fy + c = o represent straight lines, the co-ord's of their inters'n are determined by any two of the equations ax + hy + g - o, hx + by + f = o, gx + fy + c = o {Note — By § 117 the lines are ax + hy + g = ± V{Z7) ; adding these ax + hy + g = o] 1 5. Find the condition that the line Ax + By + C = o should pass through the intersection of the lines ax 2 + 2 hxy + by 2 + 2gx+ 2fy + c = o Ans. a h g =0 h b f ABC Exercises on Chapter V 119 16. A straight line moves so that the algebraic sum of the perpendiculars on it from n fixed points = o: show that the line passes through the fixed point f| (Xi + X 2 +...+ x n ), i ( Yl + y 2 +...+ y n ) [A r ote — These values may be briefly expressed -Sx, i2y. n n J The fixed point is the mean centre of the given points.] 17. More generally, if m 1 times the first _!_ + m 2 times the second _L + &c. = o ; the line passes through the fixed point /2 mx 2 my\ V2rr7' Trn7 {Note — This point is the mean centre for the syste?n of multiples xr\ 1 , m 2 , m 3 , ... See Euclid Revised, p. no.] 18. Given three fixed lines meeting in a point, if the three vertices of a triangle move one on each of these lines, and two sides of the triangle pass through fixed points, prove that the third side passes through a fixed point. A 19. If the axes are inclined at u>, show that x 2 + 2 xy cos (o + y 2 cos 2 o> = o represents lines inclined at 45 , 135 to the axis of x. 20. Find the equation to the locus of a point which moves so that the product of the perpendiculars from it on the four lines ax 4 + bx 3 y + cx 2 y 2 + dxy 3 + ey 4 is constant (= 8 4 ). Ans. ax 4 + bx 3 y + ex 2 y 2 + dxy 3 + ey 4 = 8 4 V\a — c + e) 2 + i^b — d) 2 [Note— Let ax 4 + bx 3 y + cx 2 y 2 + dxy 3 + ey 4 = e (y - xx\ 1 x) (y - m 2 x) (y - m 3 x) (y - m 4 x) . . . (1) The locus is XI y ~ mX = S 4 V 1 + rrv or ax 4 + bx 3 y +...= e8 4 VH(i + m 2 ) Again in the identity (1) substitute successively (1, V ' — 1) and (1, — V — 1) for x, y and multiply the results : /. (a - c + e) 2 + (b - d) 2 - e 2 II (1 + m 2 )] 120 Analytical Geometry 21. Determine k so that two of the lines 2 xy (ax + by) — k (x 3 + y 3 ) = o may be at right angles. Am. k = o, a + b. \_Note — We must have 2 xy (ax + by) — k (x 3 + y 3 ) = (x 2 — Axy — y 2 ) (^x + z^y) Multiply out dexter and equate coefFs : we get eq'ns to determine fx, V, A, k] 22. Find the conditions that two of the lines ax 4 + bx 3 y + cx 2 y 2 + dxy 3 + ey 4 = o should bisect the angles between the other two. Ans. 3 a + 3 e + c = o, 2 (a — e) 2 (a + e) = (b + d) (be + da) [Note— Put ax 4 + bx 3 y + ... = (Ax 2 + 2 Hxy + By 2 ) f x 2 — — xy — y 2 ) Multiply out and equate coeff's ; then elim' A, H, B.] 23. The condition that one of the lines ax 3 + 3 bx 2 y + 3 cxy 2 + dy 3 — o may bisect the angle between the other two is a (b + d/ - d (a + c) 3 - 3 (a + c) (b + d) (b 2 + bd - ac - c 2 ) CHAPTER VI. THE CIRCLE EQUATION TO A CIRCLE § 14-9. To find the equation to a circle whose centre is (a, b) and radius r. Let P (x, y) be any point on the circle, C its centre. P (xy) Then CP = r .-. CP 2 = r 2 .-. (x-a) 2 +(y-b) 2 =r 2 (§ 10) o x This is the required equation. S (50. Note these particular cases. i°. When the centre is the origin, a = o, b = o. The equation is X 2 + y 2 = r 2 This is also obvious from the figure. 122 Analytical Geometry [ x 5 ; 2°. When the circle passes through the origin and its centre is on OX. Let the centre be (a, o) Then the radius is a. .'. the equation is (x - a) 2 + (y - o) 2 = a 2 or (x — a) 2 + y 2 = a 2 or x 2 + y 2 = 2 a x N D X Othenvise, by Euclid III. 35, PN 2 = ON.ND - ON. (2 a - ON) i.e. y 2 = x (2 a — x) ; x 2 + y 2 = 2 a x 4 . When the circle passes through the origin and its centre is on OY. Y Let the centre be (o, b). Then the equation is x 2 + (y - b) 2 = b 2 or x 2 + y 2 = 2 b y § 151. Ex. 1. Let the centre be (3, — 4) and the radius 5. The equation to the circle is (x- 3 ) 2 + (y + 4) 2 = 5 2 ; or x 2 + y 2 — 6x + 8y = o Ex. 2. What locus is represented by 9X 2 + 9y 2 - 42 x + 36 y = 59? 53-] The Circle 123 Divide by the coeff ' of x 2 , viz. 9, ... x 2 -f y 2 - v 4 x + 4Y = t Rearrange, x 2 — ~ x + y 2 + 4 y = — Complete the squares (as in solving quadratics) : x 2 - ~* + a? + y 2 + 4y + 2 2 = *£■ + ^ = 16 •'• (x - £) 2 + (y + 2) 2 = 4 2 Thus the locus is a circle, centre (•£, — 2) and radius = 4. CONDITIONS THAT AN EQUATION REPRESENT A CIRCLE § 152. To discuss the locus x 2 + y 2 + 2 gx + 2 fy + c = o Rearrange, x 2 + 2 gx + y 2 + 2 f y = — c Complete the squares, x 2 + 2 gx + g 2 + y 2 + 2 fy + f 2 = g 2 + f 2 - c; .-. (x + g) 2 + (y + f) 2 = g 2 + f 2 - c Compare this with (x - a) 2 + (y - b) 2 = r 2 .-. a=-g, b=-f, r 2 = g 2 +f 2 -c Thus the locus is a circle, centre (— g, — f ) and radius = -/g 2 + f 2 - C § 153. We notice then that an equation of the second degree in rectangular axes represents a circle if i°, there is no term in xy and 2 , the coeff' 's of "x 2 and y 2 are equal. If the coeff' of x 2 is not unity we may divide by it ; thus the general form of the equation to a circle is x 2 + y 2 + 2 gx + fy + c — o The centre and radius of this circle are determined in § 152. 124 Analytical Geometry ['54- CONDITIONS DETERMINING A CIRCLE § 1 54-. Three conditions determine a circle. For let the circle be x 2 + y 2 + 2 gx + 2 fy + c = o. Each condition gives a relation between g, f, c. We have .\ three eq'ns to determine g, f, c. Ex. Find the circle through (o, o), (3, o), (3, 4). Let its eq'n be x 2 + y 2 + 2 gx + 2 fy + c = o Express that coord's of each of the given points satisfy this eq'n / c = o .*. < 9 + 6g + c = o v25 + 6g + 8f+c=o Solving these, C=o, 2 g = — 3, f = _ 2 .-. req'd eq'n is x 2 + y 2 — 3X — 4y = o INTERSECTIONS § 155. The points of intersection of a line and a circle are obtained by solving the simultaneous equations (§ 34). Ex. 1. Find the inters'ns of the line y = x + 1 and the circle x 2 + y 2 — 2x — 3y + 3 = o Eliminate y ; we find 2X 2 — 3x4-1=0 .*. x = 1 or I and y = x + 1 = 2 or -*. Am. (i, 2) and (|, f) We have here two distinct p r ts of inters'n. Ex. 2. Find the inters'ns of x + y = 7 and x 2 + y 2 - 2 x Eliminate y. We find x 2 — 6x + 9 = o, or (x — 3) 2 = o ••• x = 3, y = 7 - x = 4 Am. (3, 4). Here the two p'ts of inters'n coincide : in fact the line touches the circle. 4y - 3 I55-] The Circle 125 Ex. 3. Find the p'ts of inters'n of x + y = 12 and x 2 + y 2 = 25 Solving these, x = I (12 ± ^^94), y = 1 (12 + V^^) Here the line does not meet the circle. It is however proper to say that it meets the circle in the two imaginary- points whose co-ord's have just been assigned. The beginner should illustrate these examples by figures. Exercises X. Find the equations to the following circles : i°, centre is (i, — 2) and radius = 3 2 , centre (a, b) and radius Va 2 + b 2 3 , on join of (1, — 3), (3, 5) as diameter ,. 4 , touching OY at O and on left of OY ; radius = 4 5 , in first quadrant touching OX, OY ; radius = c Ans. i°, x 2 + y 2 — 2 x + 4y = 4 2 , x 2 + y 2 = 2 ax + 2 tyy > 3°, x 2 + y 2 — 4 x — 2 y — 1 2 = o 4 , x 2 + y 2 + 8x = o 5 , x 2 + y 2 — 2 ex — 2 cy + c 2 = o 2. Find the centre and radius of each of the circles x 2 + y 2 — 4X + 4y-i=o x 2 + y 2 + 2x — 6y = ' o 4 x 2 + 4 y 2 + 1 2 ax — 6 ay — a 2 = o sec OL (x 2 + y 2 ) — 2 ex — 2 cy tan (X = o Ans. (2, -2), 3; (-1, 3), Vio; (- — , ?—)>—'> (ccosa, csina), c 3. Find the circle through the origin which cuts off intercepts (3, 4) on the axes. Ans. x 2 + y 2 = 3X + 4y 126 Analytical Geometry [155- 4. Find the circle through the three points (o, o), (h, o), (o, k). Ans. x 2 + y 2 = hx + ky 5. What is represented by x 2 + y 2 - 6x - 8y + 25 = o? [Here (x - 3) 2 + (y - 4) 2 = o; a circle centre (3, 4), radius o: i.e. a point-circle *] 6. Find the circle through (o, o\ (2, 3), (3, 4). Ans. x 2 + y 2 = 23 x — ii y 7. Find the circle through (2, 3), (4, 5), (6, 1). Ans. 3 x 2 + 3 y 2 = 26 x + 16 y — 61 8. Find the circle on the join of (x x y x ), (x 2 y 2 ) as diameter. [Here centre is \ (*i + x 2 ), |(y + y 2 ) and (diamV = (x, - x 2 ) 2 + (y x - y 2 ) 2 .'. eq'n is [x - i(xi + x 2 )] 2 + [y - f (yi + y 2 ^] 2 = i [(Xi - x 2 ) 2 + (y x - y 2 ) 2 ] or reducing, (x - x x ) (x - x 2 ) + (y - y,) f y - y 2 ) = o.] 9. Find the points where the line 3x + y = 25 cuts the circle x 2 + y 2 = 65 Ans. (7, 4) and (8, 1). 10. Find the inters'ns of the line y = x + 1 and the circle x 2 + y 2 = 23 x - 11 y Ans. (2, 3), (3, 4). 11. Find the inters'ns of 4 x + 3 y = 35, x 2 + y 2 - 2 x - 4 y = 20 Ans. The line touches the circle at (5, 5). * The factors of (x - 3) 2 + (y ~ 4) 2 are x - 3 ± (y - 4) V~i : .•. the given equation may also be regarded as representing two imaginary lines x - 3 + (y - 4) ^^ = °> x - 3 - (y - 4) V- 1 « o. These intersect in the real point (3, 4). •155.] The Circle 127 12. Find the length of the chord which x 2 + y 2 -5x-6y + 6 = o intercepts on OX. [Put y = o, .♦. x 2 - 5 x + 6 -- o ; x = 2 or 3. .*. intercept = 3 — 2 = 1. See § 28] 13. Find the chords which the circles in Ex. 2 intercept on OX. Ans. 2 V5, 2, a^io, 2Ccos(X. 14. Find the chord which the circle x 2 + y 2 = 4 x + 2 y + 20 intercepts on the line 3X + 4y + 5 = o. [Centre of circle is (2, 1) and its radius 5. Draw a figure, let C be centre and let line cut circle in P, Q. Draw CM 1 PQ. Then CM = 1 from (2, 1) on (3 x + 4 y + 5 = o) = 3 ; .-. PQ = 2 PM = 2 VCP 2 - CM 2 = 2 Vzf- 9 = 8] 15. Find the chord which x 2 + y 2 =2x + 3y — 3 intercepts on y = x + 1 Ans. —j- V2 16- Find the chord which x 2 + y 2 = 2 x + 2 y + 23 intercepts on y = x + 3. Ans. V82 17. The co-ord r s of A are (— a, o) and of B (+ a, o) ; find the locus of a point P which moves so that PA = 2 PB. [Let P be (x, y) ; then PA 2 = 4 PB 2 .-. (x + a) 2 + y 2 - 4 [(x - a) 2 + y 2 ] or 3X 2 + 3y 2 — 10 ax + 3 a 2 = o. Ans. Locus is a circle, centre (f a, o), radius $ a.] 128 Analytical Geometry [156. 18. If PA = nPB ; find the locus of P. Ans. A circle, centre | — a, o), radius — ^ a. \n 2 — 1 / n 2 — 1 A 19. If APB = a constant /3 ; find the locus of P. Ans. The circle x 2 + y 2 — 2 ay cot j3 = a 2 . 20. Find the locus of a point which moves so that the sum of the squares of its distances from the sides of a square (side = a) is constant = 2 k 2 . Ans. Taking two adjacent sides of square as axes, the circle x 2 + y 2 — ax — ay + a 2 — k 2 = o 21. Find locus of a point which moves so that sum of squares of its dis- tances from sides of an equilateral triangle = constant = k 2 . [Let two vertices of A be ( — a, o), ( + a, o) so that third vertex is o, a \/3.] Ans. Locus is the circle 6 x 2 + 6 y 2 - 4 ay ^3 = 4 k 2 - 6 a 2 DEFINITION OF TANGENT § 156. Def (1) — If P is a point on a curve, then if P be joined to another point Q on the curve, the limiting position of the chord PQ when Q moves up to P so as ultimately to coincide with it is called the tangent at P. Def (2) — The normal at P is a line PG through P per- pendicular to the tangent. The following geometrical illustration will assist the beginner to understand the preceding Def ' (1) I57-] The Circle 129 In the case of a circle, centre C, AAA (X + ft + y = 180 (Euclid I. 32) But A A OC = y (Euclid I. 5) A A \ 2 a + /3 = 180 A A ■. a + ±£ = 90 A A .. a = 9 o° -±p Now let O move up to coincidence with P. Then ultimately A A fi = o; .'. a = 90 . That is, the angle between CP and the tangent at P is a right angle. This agrees with Euclid III. 16 EQUATION OF TANGENT § 157. To find the equation of the tangent at (x'y') to the circle x 2 + y 2 «2 Let P be the point (x' y') ; Q (x"y") an adjacent point on the circle. The eq'n to PQ is // y - y = y - y x — x' '" x' — x" (I) If Q approaches P indefinitely, then ultimately y r — y" =0, x' — x" = o and the dexter side of (1) assumes the indeterminate form §. But K 13° Analytical Geometry [158. if we make use of the fact that Q always remains on the circle we can find a determinate limiting value. Since P, Q are on the circle x /2 + y/2 _ r 2 x "2 + y »2 _ p 2 Subtract, .-. x /2 - x" 2 + y' 2 - y" 2 = o .-. (x'- x") (x' + x") = - (/ - y") (/ + y") y^-y^ _ x' + x " •• x'-x"~ / + y" Thus (1) becomes y — y' x' + x" ,T + 7^ = o x — x' y + y Now put x" = x' and y" = y', and multiply up .-. y' y — y' 2 + x' x — x' 2 = o .-. xx' + yy' = x /2 + y' 2 = r 2 , since (x'y') is on the circle Thus the equation of the tangent at (x'y') is xx' + yy' = r 2 Ex. The tangent to x 2 + y 2 = 25 at (3, 4) is 3X + 4y = 25 § 158. The normal is the J- to the tangent through x'y';. its equation is . y — y' x — x' or x'y — y'x = o Thus the normal passes through the origin. § 159. The result of § 157 may be obtained by assuming Euclid III. 16. The normal is the join of P (x'y') to the centre (o, o). Its equation is .*. y y' ,f > or x'y — y'x = o i6o.] The Circle 131 The tangent is the J_ to this through P (x'y') ; its equation is .\ y — y' x - x' or y f (y — y') + x' (x — x') = o or x'x + y'y = x /2 + y' 2 - r 2 , as before. § 160. To find the equation of the tangent at (x'y') to the circle x 2 + y 2 + 2 gx + 2 fy + c = o Let Q (x"y") be an adjacent point on the circle. The equa- tion of PQ is ,- , ,, y -y _ y -y / T x x-x' ~~x'-x" w Since P, Q are on the circle x /2 + y/ 2 + 2 g X / + gfy/ + C _ Q x //2 + y //2 + 2gx // + 2fy // + c _ Q Subtract, .-. (x' - x") (x' + x" .+ 2 g) + (/ - y") (/ + y" + 2 f) = o y' — y" x' + x" + 2g Thus (1) becomes x'-x" /■+/' + af es y-y' = — x' + x" + 2g x — x' y , + y // +2f Now put x" = x', y" = y' ; the equation of the tangent at ;xy)is.-. y-y" _ x' + g x - x 7 ~ y' + f This equation may be reduced. Multiply up, .-. (x - xO (x 7 + g) + (y -/)(/ + f) = o . . . (2) ■•• xx' + yy' + gx + fy = x /2 + y' 2 + gx' + fy' Add to both sides gx' + fy' + c K 2 132 Analytical Geometry [161. .-. xx' + y/ + g(x + x') + f (y + /) + c = x /2 + / 2 + 2gx' + afy' + c = o, since (x'y) is on the circle. Thus the equation to the tangent at (x' y') is xx' + y/ + g(x + x') + f (y + y') + c = o § 161 ■ Otherwise, assuming Euclid III. 16. The normal CP is the join of the centre P ( X Y) C ( - g, - f) [§ 152] to (xy), .*. its eq'n is y 1 / 1 = o -g -f I or x (/ + f ) - y (x' + g) + gy' - fx' = o The tangent PT is the ± to this through (x'y') ; its equation is .'. (x - x') (x' + g) + (y - yO (y' + f) = o This is equation (2) of the last §, and is reduced as before. § 162. The equation of the tangent at (x'y') xx' + yy' + g (x + x / ) + f (y + /) + c = o may be remembered by this rule : — Write the equation to the circle thus. xx + yy + g (x + x) + f (y + y) + c = o ; then dash one letter in each term. Ex. Find the tangent at (1, 2) to x 2 + y2_2x-3y + 3 = o Write this xx + yy - (x + x) - f (y + y) + 3 = o The req'd eq'n is x(i) + y (2) - (x + 1) - f (y + 2) + ,3 = o; or reducing, y = 2 164.] The Circle 133 CONDITION OF TANGENCY § 163. To find the condition that the line y = mx + n may touch the circle x 2 + y 2 = r 2 Eliminate y; .-. x 2 + (mx + n) 2 = r 2 .*. (1 + m 2 ) x 2 + 2 mnx + n 2 — r 2 = o This quadratic determines the abscissae of the points of inter- section. The quadratic has equal roots, i. e. the line touches the circle, if (1 + m 2 ) (n 2 — r 2 ) = m 2 n 2 or n 2 = r 2 (i + m 2 ) or n = + r Vi + m 2 Cor* — Whatever m is, the line y = mx + r Vi + m 2 is a tangent to the circle x 2 + y 2 = r 2 § 164. The method in the last § is applicable to any curve. In the case of the circle however the following method is easier. It follows from Euclid that a line touches a circle if _L on line from centre = radius. /. condition is that length of X from (o, o) on [y = mx + n] is r ; n i.e. ■-.:.... . = r + v 1 + m 2 or n = ± r Vi + m 2 , as before. Ex. 1. Find the tangents to x 2 + y 2 «= 25 which are parallel to 3 y - • 4 x = o Any parallel is 3y~4X + k=o 134 Analytical Geometry [165. First method. Eliminating y, x + ( 3 ) = 25 or reducing 25 x 2 - 8 kx + k 2 - 225 — o This quadratic in x has equal roots if 64k 2 = 4. 25. (k 2 - 225) Solving this, k = + 25 and the required tangents are 3y-4X + 25 = o, 3Y-4X- 2 5=o Second method. Length of _L from centre, viz. (o, o) on [3 y - 4 x + k = o] = 5 ; .-. == 5 and k = + 25, as before. dh 5 Ex. 2. Find the equations of the tangents from the origin to x 2 + y 2 — 4x — 4y + 7 = Here centre is (2, 2) and radius = 1. Any line through the origin is y = mx ; this will be a tangent if J_ from (2, 2) on [y — mx — - o] = 1 2 — 2 m — = 1 + Vi + m 2 .*. (2 — 2 m) 2 = 1 + m 2 Solving this, m = i (4 + V7) and the required tangents are y = £ (4 + V7) x, y = i (4 - V7) x LENGTH OF TANGENT § 165. To find the length of either tangent from (x'y') to (x - a) 2 + (y - b) 2 - r 2 = o Let C be the centre, T the point (x' y'), P the point of contact. Then CPT = rt / Z; ... TP 2 = TC 2 -CP 2 = TC 2 - r 2 6 5 .] The Circle J 35 But TC 2 Tp2 (x' - a) 2 + (/ - b) 2 , (§ 10). (x' - a) 2 + (/ - b) 2 - r 2 Thus the square of the tan- gent = result of substituting * the given co-ordinates in the sinister side of the equation, pre- pared if necessary by dividing by the coeff of x 2 . Ex. Find the length of the tangent from (3, 4) to 3X 2 + 3y 2 + 2x+y + 1 =0 The ' prepared ' equation is Length req'd = V3 2 + 4 2 + (f) 3 + (i) 4 + i = ^IT Exercises 1. Find the equation of the tangent at (5, 4) to x 2 + y 2 — 4X — 6 y + 3 = o Ans. 3X + y = 19 2. Find the tangent and normal to x 2 + y 2 = 25 at (3, - 4). Ans. 3X - 4y = 25, 4X + 3y = o * Def — The power of a point (x'y r ) with regard to a curve S = o is the result of substituting x', y' instead of x, y in S. The power of T (x'y 7 ) with regard to the circle (x - a) 2 + (y - b) 2 - r 2 = o is (x' - a) 2 + (y r - b) 2 - r 2 This = CT 2 — r 2 ; it is .*. positive, zero, or negative according as T is outside, on, or inside the circle. If the point T is outside, its power = square of tangent from T (§ 165). i 3 6 Analytical Geometry [165. 3. Find the tangent at the origin to x 2 +y 2 + 4X + 5y = o Ans. 4 x + 5 y = o 4. Find the tangent at the origin to x 2 + y 2 + hx + ky = o Ans. hx + ky = o 5. Find the tangent and normal at (— 2, 3) to x 2 + y 2 - 2x - 4y = 5 Ans. 3x-y + 9 = o, x+3y = 7 6. Find the tangent at (3, 2) to (x - i) 2 + (y + 2) 2 = 20 Ans. x + 2y = 7 7. Find the tangents to x 2 + y 2 = 36 which are inclined at 6o° to OX. Ans. y = x Vz + 12, y = x>/3 - I2 8. Find the tangents to x 2 + y 2 — 2 x — 2 y + 1 =0 which are parallel to x = y. Ans. x — y + V 2 = ° x + y = 2 + V2 x 2 + y 2 — 2x— 2y + i =0 9. Show that the line touches the circle and determine the point of contact. Ans. (1 + \ +/2, 1 + \ V 2 ) 10. Find the tangents to x 2 + y 2 — 4x — 2y+ 1 =0 which are parallel to OX ; also the tangents from the origin. Ans. y=3, y + i=o; x = o, 3X + 4y = o. 11. Find the length of the tangent from (1, - 2) to 3X 2 + 3y 2 -x-4 = o; also of the tangent from the origin to 2X 2 +2y 2 + 5X + 6y + 8 = o Ans. -v/M, 2. 167.] The Circle 137 12. Find the length of the tangent from (13, 8) to x 2 + y 2 + 22 x — 2 y = 278 Ans. 15. 13. If the tangent from (h k) to x 2 -r y 2 = 3 be twice the tangent from the same point to . x 2 + y 2 '= 3X + 6 prove that h 2 + k 2 = 4 h + 7 14. Find the condition that the line x y — +-—=£ I h k may be a tangent to the circle x 2 + y 2 = r 2 Ans. h 2 k 2 = r 2 (h 2 + k 2 ) 15. Any tangent to a circle whose radius is r meets the tangents at the ends of a diameter AB in Q, R : prove that AQ . BR = r 2 TANGENTS FROM A GIVEN POINT § 166. We may obtain the equation to the pair of tangents from (x'y') to the circle x 2 + y 2 = r 2 thus — Let (hk) be a point on either tangent from (x'y'). Form the eq'n to the join of (x'yO, (hk) ; it is x (y' — k) — y (x r — h) + x'k — y'h = o The X on this from the centre (o, o) is r, i. e. the join touches circle if x'k - y'h » r + V(y f - k) 2 + (xf - h) 2 Writing x, y instead of h, k we obtain the required equation, r 2 {(x - xO 2 + (y - yO 2 } = (xy r - x r y) 2 Another method is deduced in the next §. § 167. We may find the ratio m : n in which the join of two points ( X V)> (x'VO is divided by the circle x 2 + y 2 = r 2 , thus — ^, , mx" + nx' my r/ + ny' ,. The values x = , y = — — (§ 15) must satisfy x 2 + y 2 = r 2 i 3 8 Analytical Geometry [168. .-. (mx" + nx') 2 + (my" + ny') 2 = r 2 (m + n) 2 ... m 2 ( x //2 + y //2_ P 2) + 2 mn ( X / X " + y/y//_ r 2) + n 2 ( x /2 + y /2_ p 2) = G . This quadratic determines the values of m : n corresponding to the two points where the join cuts the circle. Cor' — The quadratic has equal roots, i. e. the join touches the circle if ( X '2 + y'2 _ r 2) ( x //2 + y // 2 _ p 2) = ( x / x // + y y/ _ P 2)2 This is true if (x // y // ) is any point on either tangent from (x'y'). Writing then x, y instead of x", y", we obtain the equation to the pair of tangents from (x'y'), viz. (x' 2 + y' 2 - r 2 ) (x 2 + y 2 - r 2 ) = (xx' + yy' - r 2 ) 2 JVb/^ — The preceding method is due to Joachimsthal. POLES AND POLARS § 168. Def — If L, M are the points of contact of tangents from P, the line LM is called the polar of P, and P is called the pole of LM. L(hk) lv x y,) To find the equation to the polar of (x'y') with respect to the circle x 2 + y 2 = r 2 . Let (hk), (h'k') be the points of contact of tangents from (x'y')- The equation of the tangent at (hk) is xh + yk = r 2 . But (x / y / ) is a point on this tangent, x'h + y'k = r W 171.] The Circle 139 Similarly x'h'+y'k'=r 2 (2) Now consider the straight line whose equation is xx'+yy'=r 2 (3) By (1), (hk) is a point on this line By (2), (h'k') „ .*. (3) is the equation to the join of (hk), (h'k')- Therefore the required equation is xx' + yy' = r 2 8 169, If (x'y') are real this equation represents a real straight line ; if (x'y 7 ) is inside the circle the points (hk), (h'k') are imaginary, but their join is the real straight line xx' + yy' = r 2 § 170. The equation of the tangent at (x'y') is xx' + yy' = r 2 , with the implication x /2 + y' 2 = r 2 The equation of the polar is of the same form as the equation of the tangent, but the values of (x'y') are unrestricted. Thus if a point is on the circle its polar coincides with the tangent. This may also be seen geometrically. If in fig' § 168 P approaches L along the line PL (the point L remaining fixed), then M will approach L along the circle ; and as P approaches to coincidence with L the polar LM approaches coincidence with the tan- gent LP. S 171. To find the pole of the line Ax + By + C = o (1) Let it be (x'y 7 ) ; the polar of this point is xx' + yy' — r 2 = o (2) d) and (2) must represent the same line; x' / _ r 2 * '* A ~ B ~ C 14° Analytical Geometry [172. Ar 2 . Br 2 ••• x '=-ir' y , = --c- These valties of x', y' are real if A, B, C are real. Thus every line has a real pole. § 172. If a point P lie on the polar of Q, then Q will lie on the polar of P. Let P be (x'y'), and Q (x"y")- The polar of P is xx' + yy' = r 2 (1) The polar of Q is xx" + yy" = r 2 (2) Express that (x / y / ) lies on (2) ; . x / x // + y'y" = r 2 But this is also the condition that (x"y") lies on (1). Q.E.D. 8 173. Straight lutes are drawn through a fixed point P cutting a circle in R and S ; the tangents at R and S meet in Q : to find the locus of Q (see figf, § 168). We see here that P lies on the polar of O ; .*. Q lies on the polar of P (§ 172) i. e. the locus of Q is LM the polar of P. The proposition may be otherwise stated thus : If a line revolve round a fixed point P ; then its pole moves on a fixed line, viz. the polar of P. Conversely, if a point P move on a fixed line, then its polar will pass through a fixed point, viz. the pole of the fixed line. For let Q be the pole of the fixed line. By hypothesis P lies on the polar of Q ; .*. Q lies on the polar of P (§ 172) : i. e. the polar of P passes through the fixed point Q. § 174. If the polars of K and B intersect in C, then C is the pole of /\3. The polar of A passes through C, .*. the polar of C passes through A (§172). Similarly the polar of C passes through B : .-. AB is the polar of C. Q.E.D. : 7 6.] The Circle 141 8 175. To deduce a construction for the polar of a point P with respect to a circle. Let O be the centre of the circle. Fig. (1) Fig. (2) Take O for origin and OP for axis of x. Let the co-ord's of P be (x', o), and the equation to the circle 2 _ r 2 or x* + y* = r Then the polar of P is xx' + y . o = r 2 xx' = r 2 r 2 This is a line parallel to the axis of y at a distance — f We have then this construction for the polar of P : Join OP, let OP meet the circle in A. Take a point N on OP such that OP. ON = OA 2 . Then a -L through N to OP is the polar. 8 176. If the equation to the circle be x 2 + y 2 + 2 gx + 2 fy + c =s o. it is proved as in § 158 that the polar of (x'y 7 ) is xx' + yy' + g (x + x') + f (y + /) + c = o. This is of the same form as the equation of the tangent. Exercises 1. Find the polars of (1, 2), (o, 1), (1, — 1) with respect to x 2 + y 2 = 3 Ans. x + 2y = 3, y=3, x~y=s. 142 Analytical Geometry [177. 2. Find the poles of 2 x — 3 y = 5, x + y = i, x cos a + y sin (X = p with respect to x 2 + y 2 = 50 . . /50 cos a 50 sin a\ ^*w. (20, - 30), (50, 50), f ^— , : — — 1 3. Find the points of contact of tangents from (1, 2) to Ans. (o, 2), (f, f) x 2 + y 2 = 4 4. Find the condition that the polar of ( h k) with respect to x 2 + y 2 = 9 may touch x 2 + y 2 = 6 y Ans. h 2 + 6 k = 9 5. Find the condition that the polar of (hk) with respect to x 2 + y 2 = a 2 may tonch (x - a) 2 + (y - /3) 2 = r 2 Ans. (ah + p k - a 2 ) 2 = r 2 (h 2 + k 2 ) RADICAL AXIS § 177. Let the equations of two circles be x 2 + y 2 + 2 gx + 2 fy + c = o . . . . (1) x 2 + y 2 + 2 g'x + 2 fy + c' = o . ._ . . (2) and let us interpret the equation x 2 + y 2 + 2gx + 2fy+ c = x 2 + y 2 + 2g / x+ 2fy + c' . (3) Any values of x, y which satisfy (1) and (2) also evidently satisfy (3) : •'• (3) represents a locus passing through the inters'ns of (1), (2). Again, (3) reduces to 2(g-g0x + 2(f-f)y + C-c r = o ... (4) This equation represents a straight line. Accordingly, (3) represents the join of the points of inters'n (real or imaginary) of the circles (1), (2). I8i J The Circle 143 § 178. Def — The join of the points of intersection (real or imaginary) of two circles is called their radical axis. Equation (3), or (4), represents the radical axis of the circles (0. (*)■ Although the circles may not intersect in real points, we see from (4) that the radical axis is always a real line. Put S = X 2 + y 2 + 2 gx + 2 fy + c S / =x 2 +y 2 +2g , x+2 fy + c' so that S = o, S' = o are the prepared eq'ns to two circles. Then the equation to the radical axis is S - S' = o § 179. The equation S — S' = o expresses (§ 165) that the square of the tangent from (x, y) to S = o is equal to the square of the tangent from (x, y) to S' = o : thus tangents drawn to two circles from any point on their radical axis are equal. 8 180. The radical axis is perpendimlar to the join of the centres. For the join of the centres is x y 1 = o -g - f 1 -g' -f 1 or x (f ' - f ) - y (g' - g) + gf ' - g'f = o This is JL the radical axis, 2 (g - gO x + 2 (f - V) y + c - c' = o [§ 68, Cor' (3)] § 181. One of the circles may be a point-circle. Thus the radical axis of the circles (x - a) 2 + (y - b) 2 - r 2 = o . . . . , . (1) (x - a') 2 + (y - b') 2 = o . (2) is the line 2 (a' - a) x + 2 (b' - b) y + a 2 + b 2 - a /2 - b /2 - r 2 = o t . (3) Accordingly the line (3) [the radical axis of the circle (1) and the point (a'b')] is the locus of a point which moves so that the tangent drawn from it to the circle (1) is equal to its distance from the point (a'b'). 144 Analytical Geometry [182. § 182. Suppose two circles to intersect in P, Q so that PQ is the radical axis. Let Q approach coincidence with P : then the circles touch and PQ becomes the common tangent. Thus if two circles touch, their radical axis is the common tangent at the point of contact. Ex. Determine c so that the circles x 2 + y 2 — 3 x — 4y — c = o, x 2 + y 2 — 4 x — 3y— c = o may touch. Subtracting, the radical axis is x — y = o At the points where this meets the first circle 2x 2 — 7x — c = o: this eq'n has equal roots if 49 + 8 c — o, or c = — ~ . § 183. Let the prepared equations to three circles be S = o, S' = o, S" = o Taking these in pairs, the radical axes are S - S' = o, S' - S" = 0, S" - S = o It is evident that any values of (x, y) which satisfy two of these equations will satisfy the third. Thus the three radical axes of three circles taken in pairs are concurrent. The point of concurrence is called the radical centre. CO-AXAL CIRCLES § 184. Let the equations to two circles be S = o, S' = o where S = x 2 + y 2 + 2 gx + 2 fy + c S r = x 2 + y 2 + 2g'x 4- 2 fy + c' and let us interpret the equation S-AS'=o i88.] The Circle 145 Writing it at full length we see that it represents, a circle (§ 153). Again, any values of (x, y) which render S = o and S' = o will also render S - A S' = o. Thus S - X S' = o represents a circle passing through the points of intersection of S = o, S' = o. If we choose A suitably S - A S' = o will represent any such circle. § 185. Def — A system of circles passing through two fixed points is called a co-axal system. The points may be real or imaginary. Their join is the common radical axis of any two circles of the system. If S = o, S' = o are two circles of the system, any other is S - AS' = o (§ 184). § 186. S - A S' = o signifies (§ 165) that the square of the tangent from (x y) to S = o is A times the square of the tangent from (x y) to S' = o. Therefore if a point move so that the tangents from it to two given circles are in a constant ratio, its locus is a co-axal circle. § 187. If S — o be one circle of the system and u = o the radical axis, then any other circle of the system is S + Au = o (1) where A is a numerical constant. For the radical axis of (1) and S = o is Au = o, or u = o. Q.E.D. § 188. A co-axal system may be simply represented thus. Take the join of the fixed points for axis of y, and the mid point of this join for origin. Let the fixed points be (o, + c). The common radical axis is x = o. x 2 + y 2 — c 2 — o is evidently one of the circles. Therefore by § 187 the equation x 2 + y 2 + 2 Ax — c 2 = o (1) represents any circle of the system. c is the same for all the circles ; A. varies from circle to circle. L 146 Analytical Geometry [189. 8 189. Equation (i) of the last § may be written (x + X) 2 + y 2 = A 2 + c 2 If A 2 + c 2 = o, this reduces to a point- circle. If .*. A = + c V — 1, the circle reduces to one of the points (+cV-i,o), These points are called the limiting points of the system. If the circles intersect in real points, c is real and the limiting points are imaginary. If the circles intersect in imaginary points, i. e. if then the limiting points are real, viz. they are ( + k, o). Ex. The polar of a limiting point is the same for every circle of the system. For the polar of (k, 6) with respect to x 2 + y 2 + 2 A x + k 2 = o is kx + A (x + k) + k 2 — o or (x + k) (A + k) = o, or x + k = o : i. e. the line through the other limiting point || radical axis. Exercises 1 . Find the radical axis of the circles x 2 + y 2 + 4X + 5y = 6 and x 2 + y 2 + 5x + 4y = 9 Ans. x — y = 3. 2. Show that the circles x 2 + y 2 + 4 x + y = 3, x 2 + y 2 -x-y = i, x 2 + y 2 + 14 x + 5 y ^ 7 are co-axal. 3. Find the radical centre of the circles x 2 + y 2 + 3 x = o, x 2 + y 2 + 5 y = 1, x 2 + y 2 + 2 x + 2 y = 1 Ans. (3, 2) 4. Find the radical axis of the circles . (x - a) 2 + (y - b) 2 = c 2 , (x - b) 2 + (y - a) 2 = c 2 ; and determine c so that the circles may touch. Ans. x — y = o, — T (a — b). 190.] The Circle i47 5. Find the length of the common chord of the circles x 2 + y 2 — 4X — 2 y — 20 = o, x 2 + y 2 + 8 x + 14 y = o. [The radical axis is i2x + i6y+2o = o, or 3X + 4y + 5=o. The chord which the first circle intercepts on this line is obtained, Exercise 14, page 127.] Ans. 8. 6. Find the length of the common chord of the circles in Ex. 4. Ans. V4 c 2 — 2 (a — b) 2 7. Find the equation to a circle through the points of intersection of the circles in Ex. 1 and through the point (1, 2). [The req'd eq'n is of the form x 2 + y 2 + 4 x + 5 y - 6 + A (x 2 + y 2 + 5 x + 4y — 9) = o. Expressing that this is satisfied by x= 1, y = 2, 13 + 9 A = o.] Ans. 4 x 2 + 4y 2 + 29 x + 7 y — 63 = o 8. Find the circle through the origin and the inters'ns of the circles x 2 + y 2 + 2 gx + 2 fy + c = o, x 2 + y 2 + 2 g'x + 2 f 'y + o' = o Ans. (c' — c) (x 2 + y 2 ) + 2 (gc r — g'c) x + 2 (fc' — f'c) y = o 9. Find the locus of a point which moves so that tangents from it to x 2 + y 2 = 3, x 2 + y 3 = 3 x + 6 are in the ratio 2:3. Ans. The circle 5 (x 2 + y 2 ) + 1 2 x = 3 CENTRES OF SIMILITUDE § 190. Def — If the join AB of the centres of two circles is divided internally in O' and externally in O in the ratio of the radii : then O' is their internal centre of similitude, and O their external centre of similitude. The following properties are easily proved by Pure Geometry (see Euclid Revised, p. 343). Analytical proofs are indicated in the Exercises. ~ l 2 148 Analytical Geometry [191. (1) The two direct common tangents pass through O and the two transverse common tangents through O'. (2) Any line through either centre of similitude is cut similarly by the circles. Thus if a line through O cut the circles in P, Q, P', Q' then OP:OP'= OQ : OQ' = r: r', where r, r" are the radii. (3) The six centres of sim'de of three circles taken in pairs lie in threes on four straight lines (called axes of similitude). * § 191. Ex. 1. Find the external common tangents to the circles y£ + y 2 — 26 x — 12 y + 201 = o, x 2 + y 2 — 12 x — 4 y + 39 = o The centres are (13, 6), (6, 2) and the radii are 2, 1 . .-. the external centre of similitude is (§ 16) 2.6 — 1. 13 2.2—1.6 2—1 ' 2—1 or (- I, - 2) Any line through this is y+2 = m(x + i) If this touch the first circle, the J_ on it from (13, 6) = 2 8 — 14 m = 2 + Vi + m* Solving this, m = | or ^ .•. the common tangents are y+2=-|(x + i), y+2=^(x+i) or 3x- 4y = 5, 5X- 12 y = 19 r 9 2.] The Circle 149 Exercises 1 . Find the direct common tangents to the circles x 2 + y 2 — 4 x — 2 y + 4 = o, Ans. y = 2, 4X — 3y = 10. x 2 + y 2 + 4X + 2y — 4 =0 2. Find the transverse common tangents to the same circles. Ans. x - 1, 3X + 4y = 5. 3. Find the direct common tangents to the circles x 2 + y 2 — 6 x — 8 y = o, x 2 + y 2 — 4X — 6y = 3 Ans. x + 2 = o, y + 1 = o. 4. Show that the tangents from the origin to the circle (x - a) 2 + (y - /3) 2 = r 2 touch the circle (x - Aa) 2 + (y - A/3) 2 = (A r) 2 5. Also, if any line through the origin cut these circles in P, Q, P f , Q', prove OP : OP' = OQ : OQ' = 1 : A [Writing p cos 6, p sin 9 for x, y in the equation to the first circle it becomes p 2 - 2 p (a cos 6 + /3 sin 6) + (X 2 + /3 2 - r 2 = o Solve this quadratic in p ; its roots are OP, OQ ; &c] Note — Exercises 4, 5 afford proofs of properties (1), (2), § 190. 6. The centres of three circles are (Of/3), (a'/3')> (&" ft") and their radii are r, x J , r ff . If these circles are taken in pairs, show that the three external centres of similitude lie on the line r p/ p// x — (3 f3' /3" 111 r r' r" y = a a' a" 1 1 1 r K r" /3 /3 r /3- a a' a /r ANGLE OF INTERSECTION § (92. If two circles whose centres are A, B intersect in P, then since the tangents at P are _L AP, BP the angle of inter- section is APB. I v5° Analytical Geometry |>93- A Let APB = Oi, AB = 8, AP = r, BP = r' then 8 2 = r 2 + r' 2 — 2 rKcos 06 If the circles be S = x 2 + y 2 + 2 gx + 2 f y + c =0 S' = x 2 + y 2 + 2gfx + 2fy + c' = o then r 2 = g 2 + f 2 - c, r' 2 = g /2 + f 2 - c', 5 2 = (g - g') 2 + (f - f? .*. 2 rr' cos a = 2 gg' + 2 ff — c — c' C0/ — The circles cut orthogonally if 2 gg' + 2 ff — c — c' =0 § 193. ^ « variable circle cut the two circles S = o, S' = o at constant angles OC, 8 then it cuts any co-axal circle S + k S' = o at a constant angle y. Let the variable circle be x 2 + y2 + 2 Gx + 2 Fy + C = o, R its radius. S + k S' = o when ' prepared ' is s + ker' f + kf c + kc' x 2 + v 2 + 2 s -o- x + 2 j— y + r = o a i + k i + k ' i+k Let r" be its radius. The formula of § 192 gives _ er + kg' _f + kf c + kc' _ .-. 2R r"cosy (1 + k = 2Gg + 2 Ff — c - C + k [2 Gg' + 2 Ff - c' - C] = 2 Rr cosOC + 2 kRr'cos/S .-. (1 + k) r" cos y = r cos (X + W cos /3 This formula shows that y is constant. § 1 94-. ^4 system of circles which cuts orthogonally two given circles has a common radical axis. Let the given circles be x 2 + y 2 + 2 gx + 2 fy + c - o, x 2 + y 2 + 2 g / x + 2 f y + c' = o t 9 6.] The Circle 151 and x 2 + y 2 + 2Gx + 2 Fy + C = o (1) one of the orthogonal circles. ,.\ 2 gG + 2 f F - c — C = o (2) 2 g>G + 2f F - c' - C = o (3) Eliminate G, F linearly from (1), (2), (3). Thus (1) is replaced by x y x 2 + y 2 + C g f -c -C gf f _c'-C + C or x y x 2 + y 2 g f - c g r -c' x y 1 g f - 1 g r f -1 = o If C varies this represents a circle of a co-axal system whose common radical axis is x y g f = o g> f -i i. e. the join of the centres of the given circles. § 195. Otherwise thus : Let L, M be the limiting points of the system determined by the given circles S, S r . Let P be any point on the radical axis. Then PL = PM = tangent from P to any of the circles. .•. a circle, centre P, radius PL, passes through M and cuts each of the circles orthogonally. The system passing through the two fixed points L, M is .'. orthogonal toS, S'. Cor' — The centre of the circle which cuts three given circles orthogonally is their radical centre. § 196. Ex. Find the circle which cuts orthogonally the three circles x 2 + y 2 + 2 gx + 2 f y + c =0 x 2 + y 2 + 2 g r x + 2 f y + c r = o x 2 + y 2 + 2 g"x + 2 f 'y + c" = o Let the orthotomic circle be x 2 + y 2 + 2 Gx + 2 Fy + C = o (1) '52 Analytical Geometry |>97- Then 2gG +2fF — c -C=o 2 g'G + 2 f F - c' - C = o 2 g"G + 2 f / F-c"-C = o . From (i), (2), (3), (4) we may eliminate linearly G, F, C. The required equation is .*. x 2 + y 2 X y 1 — c g f — 1 -c' g' f — 1 -c" g" p — 1 = o (2) (3) (4) POLAR CO-ORDINATES § 197. To find the polar equation to a circle. In x 2 + y 2 + 2 gx + 2 fy + c = o, substitute rcos#, rsin# for x, y. This gives the required equation, viz. r 2 + 2 r (g cos + fsin#) + c = o If the circle pass through the origin c = o; dividing by r the equation becomes r + 2(gcos# + f sin 0) =0 § 198. Another method. To find the polar equation to a circle, centre (r x 0J, radius = 8. PO0) Let C be the centre, P any point (r 6) on the circle. Then CP 2 = OP 2 + OC 2 -20P.OCcosCOP; I99-] The Circle 153 .-. 8»=i*+ 1V-2 rr-! cos (0-0J . . . (i) This is the equation required. § 199. The equation (i), viz. r 2 - 2 rri cos (0 - X ) + r x 2 - 5 2 = o is a quadratic in r. Thus for each value of it determines two values OP, OP' of r. By the theory of quadratics the product of the roots is r-j 2 — b 2 ; this is independent of 0. Thus OP . OP = ry 2 - g 2 = OC 2 - CP 2 . This agrees with Euclid III. 35, 36. Ex. 1. What locus is represented by r = a cos ? Multiply by r ; . •. r 2 = a (r cos 0) That is, x 2 + y 2 = ax ; thus the locus is a circle. Otherwise thus. Along the initial line measure OA = a. Let P be a point on the locus ; join PA. Then OP = OAcos0 A ,-. OPA = rt. I .'. locus is a circle on diameter OA. Ex. 2. Show that r = 2 a cos + 2 b sin represents a circle ; find its radius and the polar co-ordinates of its centre. Multiply by r, .'. r 2 = 2 a (r cos 0) + 2 b (r sin 0) .-. x 2 + y 2 = 2 ax + 2 by .-. (x - a) 2 + (y - b) 2 = a 2 + b 2 This represents a circle, centre (a, b), and radius = Va 2 + b 2 By § 39, (2), the polar co-ordinates of the centre are (r = Va 2 + b 2 , = tan- 1 -) Ex. 3. Two circles intersect in O ; through O any line is drawn cutting the circles in P, Q ; find locus of mid point of PQ. Take O for origin ; let the equations of the circles be x 2 + y 2 + 2 gx + 2 fy = o, x 2 + y 2 -1- 2 g'x + 2 f y = o 154 Analytical Geometry 200. Or, in polar co-ordinates (§ 197) r + 2 g cos 6 + 2 f sin 6 — o, r + 2 g 7 cos , + 2 f sin = o A Now let OPQ be inclined at 6 to OX .-. OP = — 2 g cos 6 — 2 f sin 0, OQ = — 2 g 7 cos — 2 f sin Let R be mid point of PQ ; let OR = r. Then 2 r = OP + OQ. Thus the locus is 2 r = — 2 (g + g') cos — 2 (f + V ) sin 6 or r 2 + (g + g / ) (r cos 0) + (f + f ) (r sin 0) = o or x 2 + y 2 + (g + g') x + (f + f ) y = o ; another circle through O. OBLIQUE CO-ORDINATES § 200. To find the equation to a circle whose centre is (a, b) a?id radius r ; the axes being inclined at co. If (x, y) be any point on the circle, the distance between the points (x, y) and (a, b) is r. Thus by § 14 the required equation is (x — a) 2 + (y — b) 2 -f 2 (x — a) (y — b) cos co = r 2 . . (i) If this be expanded, we see that the terms of the second degree are x 2 + 2 xy cos co + y 2 . Thus the most general equation of the second degree represent- ing a circle is h (x 2 + 2 xy cos co + y 2 ) + 2 gx + 2 fy + c = o . . (2) We may find the co-ordinates of the centre and the radius of (2). Thus divide (2) by h, and comparing its coefficients with those of (1) we get g f r^ = — a — b cos co, 7- = — b — a cos co, h h -r = a 2 + b 2 + 2 ab cos a) — r 2 h Solving these we obtain the values of a, b, r. 202 .] The Circle 155 CO-ORDINATES EXPRESSED BY A SINGLE PARAMETER § 201. (r cos 6, r sin 6) is evidently a point on the circle x 2 + y 2 = r 2 . We may call this ' the point 6.' Ex. 1. The equation to the chord joining the points OC, /3 is x y 1=0 r cos OC r sin OC i r cos (3 r sin /3 1 or reducing, x cos % (OC + ft) + y sin £ (a + /3) = r cos |(0£ - /3) Ex. 2. Put /3 = (X in the preceding: the tangent at OC is .'. x cos OC + y sin OC = r Ex. 3. Find locus of mid points of chords of x 2 + y 2 = r 2 drawn through a fixed point (hk). Let (r cos OC, r sin OC), (r cos /3, r sin /3) be the extremities of one of these chords. We have to eliminate OC, /3 from h cos |(a + /3) + k sin |(a + /3) = r cos|(a - /3) 2 x = r (cos OC + cos /3) 2 y = r (sin OC + sin /3) The result is x 2 + y 2 = hx + ky MISCELLANEOUS PROPOSITIONS § 202 • If O is the centre of a circle, and A, B two points, then OA : OB = X from A on polar of B : X from B on polar of A {Salmon) Let the circle be x 2 + y 2 = r 2 , let A be (x^) and B (x 2 y 2 ). Let the Xs be AM, BN. The polar of B is xx 2 + yy 2 — r 2 .-. AM = (x x x 2 + yiy 2 - r 2 )/^ 2 + y 2 2 .-. AM .OB = XiX 2 + y x y 2 - r 2 = BN .OA by symmetry; .-. &c. i56 Analytical Geometry [203. § 203. If the polars of the vertices of a A ABC form a A A' B' C ; then AA', BB', CC are concurrent. Let the circle be x 2 + y 2 = r 2 . Let A be (x^), B (x 2 y 2 ), C (x 3 y 3 ). The equation to A' B' is xx 3 + yy 3 = r 2 (1) A' C „ xx 2 + yy 2 = r 2 (2) As in § 124, the equation of the join of (x^) to inters'n of (1), (2), is (xx 3 + yy 3 - r 2 )/( Xl x 3 + y^ - r 2 ) = (xx 2 + yy 2 - r 2 )/( Xl x 2 + y x y 2 - r 2 ) (XjX 2 + y x y 2 - r 2 ) (xx 3 + yy 3 - r 2 ) - (Xi x 3 + y! y 3 - r 2 ) (xx 2 + yy 2 - r 2 ) = o say u = o. If by symmetry we write down the equations of BB', v = o and of CC, W = o, we observe that u + v + w = o identically :. &c. (§ 129) Exercises on Chapter VI 1 . Find the circle through the origin and the inters'ns of the line 2X+3y + 4 = o and the circle x 2 + y 2 + 3X + 4y+ 2=0; also the circle through (1, 2) and the same intersections. Ans. 2x 2 + 2y 2 +4X + 5y = o ; 2 x 2 + 2 y 2 = y + 8 2. Find the circle whose diameter is the common chord of the circles x 2 + y 2 — 23 x + 11 y = o, x 2 + y 2 — 12 x + 11 = o Ans. x 2 + y 2 — 5X — 7y+ 18 = 3. Given base of a triangle, and ab sin (C — Oi), where OC is a given angle, find locus of vertex. [Let ab sin (C — Oi) = b 2, ; take side AB as axis of x, A being origin] Ans. The circle x 2 + y 2 — ex — cy cot OL + b 2 cosec OC = o 203-] The Circle 157 4. A and A' are two points moving along a given line connected by a rod of length a, B and 3' two other points connected by a rod of length b, moving along a line at right angles to the first. Show that if A and B are connected by a third rod of length c, the middle point of A' B' will describe a circle. Ans. Taking the given lines as axes, the equation to the locus is (2 x — a) 2 + (2 y — b) 2 = c 2 5. Two segments AB, CD of a given line subtend equal angles at P ; find the locus of P. Ans. Taking the given line as axis of x; if OL, /3, y, b are the distances of A, B, C, D from origin, locus is the circle (a - (3 - y + b) (x 2 + y 2 ) + 2 (fly - ocb) x + yb (oc - (3) - 0C(3 (y - b) = o 6. A and B are fixed points ; R, S are variable points on AB such that AR 2 + RS 2 + SB 2 = c 2 = constant. If PRS is an equilateral triangle find the locus of P. Ans. Taking A as origin, AB as axis of x, AB = a ; 2 x 2 + 2 y 2 — 2 ax 7^- = c 2 — a 2 7. Prove that the diameter of the circle through the origin and (/>i^i), (p 2 2 ) * s Vp? + p 2 2 - 2 pj p 2 cos (d x - 6 2 ) /sin (0 X - 2 ) 8. A point moves so that the square of its distance from the base of an isosceles triangle = rectangle under its distances from the sides : show that its locus is a circle. [Take mid point of base for origin and base for axis of x. Let vertices be (a, o), (— a, o), (o, h).] Ans. h (x 2 + y 2 ) + 2 a 2 y = ha 2 9. A, B, C, &c. are given points ; a point P moves so that PA 2 + PB 2 + PC 2 + &c. = constant : show that its locus is a circle whose centre is the mean centre of the given points. 10. If rri! PA 2 + m 2 PB 2 + m 3 PC 2 + &c. = constant, the locus of P is a circle whose centre is the mean centre of the points A, B, C, &c. for the system of multiples m lt m 2 , &c. i 5 8 Analytical Geometry 1 1 . A point moves so that the sum of the squares of its distances from the sides of a regular polygon is constant : show that its locus is a circle. [Equation to locus is (x cos a + y sin OL — p) 2 ( a + V>ysin( a+ V>p] 2 Then x cos = constant coefP of xy = sin 2 a + sin ( 2 OC + —J + sin ( 2 OC + — 1 + ... = o * Also coeff' of x 2 — coefF of y 2 = COS 2 £X + COS I 2 OC + ?— 1 + COS ( 2 & + 1 + ...= OJ 12. On a line which revolves round a fixed point O and meets a given circle in P a point Q is taken such that OP.OQ = constant = k 2 ; show that the locus of Q is a circle except in the case when O is on the given circle, when the locus is a straight line. [Proceed as in § in, Ex. i. The locus of Q is the inverse (see Euclid Revised, p. 351) of the given circle. Accordingly the inverse of a circle is a circle, unless the centre of inversion is on the given circle, when the inverse is a straight line.] 13. Show that the inverse of a straight line is a circle (see § in). 14. A point moves so that the square of its distance from a fixed point varies as its distance from a fixed line : show that its locus is a circle. 15. The co-ord's of the ends of the base of a triangle are (a, o), ( — a, o) : if * the vertical angle C is given, show that the locus of the orthocentre is the circle x 2 + y 2 + 2 ay cot C = a 2 16. A and B are fixed points on rectangular axes such that OA = OB = a; A A a point P moves so that OPA = OPB : show that the equation to the locus of P is ( x — y) (x 2 + y 2 — ax — ay) = o Interpret this. * See Nixon's Trigonometry, page 328. The Circle i59 17. Show that the equation to the tangents from the origin to the circle (x - a) 2 + (y - /3) 2 = r 2 is (/3x - ay) 2 = r 2 (x 2 + y 2 ) 18. Show that the equation to a segment of a circle through (x 1 y 1 ), (x 2 y 2 ) containing an angle OC is (x - xj) (x - x 2 ) + (y - y x ) (y -y 2 ) = ± cota x y i xi yi i x 2 y 2 ! A [Express that joins of (hk) to (x^) and to (x 2 y 2 ) include OC: then write x, y for h, k.] 19. The area of the triangle formed by the two tangents from (hk) to X 2 + y2 = r 2 and their chord of contact is r (h2 + k 2 - rV h 2 + k 2 20. The circle whose diameter is the join of (r x 6-^), (r 2 2 ) is r 2 = r x r cos {6 — flj) + r 2 r cos (6 — 6 2 ) — r x r 2 cos {O x — 0%) [Express that PA 2 + PB 2 = AB 2 , where P is (r 0), A (r\ X ), B (r 2 2 ).] 21. Find the equation of the chord joining the points (2 a cos OC, OC), (2 a cos /3, /3) on the circle r = 2 a cos 6 Ans. 2 a cos /3 cos a = r cos (/3 + (X — 0) 22. Find the tangent to the circle r = 2 a cos at the point (2 a cos OC, OC). Ans. 2 a cos 2 a = r cos (2 a — 6) [Put /3 = OC in preceding result] 23. Find the condition that the line - = a cos 6 + b sin r may touch the circle r = 2 c cos 6 Ans. b 2 c 2 + 2 ac = 1 160 Analytical Geometry 24. If x 2 + xy + y 2 + 2 x + 3 y = o represent a circle, show that co = 6o°. Determine the centre and radius. Ans. (-1, -f); | a/2I 25. Find the equation to a circle whose centre is (i, — 2) and radius 5, the axes being inclined at 120 . Ans. x 2 — xy + y 2 = 4X — 5 y + 18 26. The axes being inclined at an angle ft), find the equation to the circle through the origin which intercepts lengths OL, (3 on the axes. Ans. x 2 + 2 xy cos oi + y 2 = (X x + (3 y 27. If the intercepts in the last question are connected by the relation la + m/3 - 1 where I, m are constants, show that the circle passes through the fixed point / ! ™ ^ \l 2 + 2 Im cosco + m 2 ' I 2 + 2 Im cosft) + m 2 / A 28. A circle touches the axes (which include a)) at points A, B such that OA = OB = b : show that its equation is (x + y — 6) 2 = 4xy sin 2 — 29. Show that K + k = X touches the circle in the last question if [Combining eq'ns of line and circle x + y— 5= ±2 sin— Vxy we get x + y ± 2 sin - Vxy = b (r- + .} or x (1 - A ± 2 sin " Vxy + y (1 - ^ j -- o : Express that the roots of this quadratic for \/x : Vy are equal.] The Circle 161 30. Find the equation to the locus of a point whose polar with regard to x 2 + y 2 = a 2 touches (x - a) 2 + (y - /3) 2 = r 2 Ans. (ax + /3y - a 2 ) 2 = r 2 (x 2 + y 2 ) 31. Show that the polar of a point with regard to any circle of a co-axal system passes through a fixed point. [The polar of x'y' with respect to x 2 + y 2 + 2 Ax - c 2 (§ 188) is xx' + yy' + A (x + x') = c 2 The fixed point is determined by x + x' = o, xx' + yy' = c 2 ] 32. If A B is a diameter of a circle, prove that the polar of A with respect to any circle which cuts the first orthogonally will pass through B. 33. Show that the orthotomic circle of x 2 + y 2 = a 2 , (x - b) 2 + y 2 = a 2 , x 2 + (y - c) 2 = a 2 is x 2 + y 2 — bx — cy + a 2 = o {Note — Centre of req'd circle is radical centre ; its radius = tangent from radical centre to one of the circles.] 34. Find the circles through (i, 2), (1, 18) touching the axis of x. Ans. x 2 + y 2 + iox — 20 y + 25 = o, x 2 + y 2 — 14X — 20 y + 49 = o 35. Find the circle through (2, o) which cuts x 2 + y 2 = 4 and x 2 + y 2 --= 2 y + 8 at right angles. Ans. x 2 + y 2 — 4X + 4y + 4 = o 36. Show that a circle can be inscribed in the quadrilateral whose sides are x = o, y = o, x cos a + y sin a = p, x cos/3 + y sin/3 = p', if p (1 + sin /3 + cos /3) = p' (1 + sin OC + cos OC) 37. Find the system of circles which cut orthogonally each of the circles x 2 + y 2 + 2 A x — c 2 = o where A is a variable parameter. Ans. x 2 + y 2 + 2 \x y + c 2 = o, where jx is a variable parameter. M i6z Analytical Geometry 38. Show that any common tangent to two circles is bisected by their radical axis, also that the common tangent subtends a right angle at either limiting point. [Note — If A, B are p'ts of contact, L a limiting p't, C point where AB meets radical axis: then CA = CB = CL (§ 179)] 39. The equation to the circle whose diameter is the join of the centres of similitude of two circles S = o, S' = o is S _ S^ r 2 p/2 — ° where r, K are the radii of the circles. [Note — This is the circle of similitude of the two circles. See Euclid Re- vised, page 336. We see then that tangents to S, S' from any point on their circle of similitude are in the constant ratio r : K.] 40. Find the locus of a point P at which two circles subtend equal angles. Ans. The circle of similitude. 41. Show that the circles S S ~ ± -D = ° r r cut at right angles, [Note — The centres of these circles are the centres of similitude.] 42. Show that the circles in the last question bisect the angles between S = o, S' = o. 43. Show that the equation to the circle through the three points (xiYi), (^2)* (XsYs) is = o x z + y x y 1 X1 2 + Yi 2 x x y x 1 x 2 2 + y 2 2 x 2 y 2 1 x 3 2 + y 3 2 x 3 y 3 1 [Note — Let circle be x 2 + y 2 + 2 gx + 2 fy + c = o Then x x 2 + yy 2 + 2 gx x + 2fy x + c = o, ...= o, ...= o. Elim'g f, c] 44. Find the condition that (x^), (X2y 2 ), (x 3 y 3 ), (x 4 y 4 ) may be con- cyclic ; interpret geometrically. Ans. (1) Write x 4 , y 4 instead of x, y in preceding result. (2) If A, B, C, D are four concyclic points and O any other point, OA 2 . area BCD + OC 2 .areaABD = OB 2 .areaACD + OD 2 . area ABC The Circle 163 45. Show that the Nine-points' circle of the triangle whose vertices are ( 2a 's)' ( 2b '5)' ( 2C '0 passes through the origin. 46. Show that the radical axis of a circle and a point bisects the distance between the point and its polar with regard to the circle. 47. A and B are points on the circle x 2 + y 2 = a 2 : if AB subtend a right angle at a fixed point (x'y'), show that the locus of the mid point of AB is the circle (x - x') 2 + (y - yO 2 + x 2 + y 2 = a 2 {Note— Let A be (a cos a, a sin (X) and B (acos/3, asin/3). Eliminate (X, /3 from (x' — a cos a) (x' — a cos/3) + (y r — a sin a) (y' — a sin /3) = o (Page 126, Ex. 8) 2 x = a cos OC + a cos (3, 2 y = a sin (X + a sin /3] 48. Show also that locus of the foot of perpendicular from centre of x 2 + y 2 = a 2 on AB is the same circle ; and that locus of intersection of tangents at A, B is the circle (x /2 + y /2 - a 2 ) (x 2 + y 2 ) - 2 a 2 x'x - 2 a 2 y'y + 2 a 4 = o 49. Two chords of a circle cut at right angles; show that tangents at their extremities form a quadrilateral whose vertices are concyclic : and deduce that the problem 'to inscribe a quadrilateral in a circle whose sides shall touch another given circle ' is either indeterminate or impossible. {Note— This follows from Ex. 48] 50. Show that the points where the line Ax + By + C = o cuts the circle x 2 + y 2 + 2 gx + 2 fy + c are concyclic with the points where the axis of x cuts the circle x 2 + y 2 + 2 g'x + 2 f f y + c' = o if 2 C(g-g') = A(c-c') {Note— The lines y = o, Ax + By + C = o must meet on radical axis (§ 183)] M 2 164 Analytical Geometry 51 . The equation to the real common tangents to the circles x 2 + y 2 = 2 ax, x 2 + y 2 = 2 by is 2 ab (x 2 + y 2 — 2 ax) = (by — ax + ab) 2 52. Show that the locus of the centre of a circle which cuts two fixed circles orthogonally is their radical axis. 53. The centres of three circles are A, B, C and their radii r u r 2 , r 3 : show that the circles are co-axal if BC . rv 2 + CA . r 2 2 + AB . r 3 2 + BC . CA . AB = o 54. Prove that if t x , t 2 , t 3 are the tangents from any point to three co-axal circles whose centres are A, B, C, then BC . I? + CA .t 2 2 + AB . t 3 2 = o 55. A variable circle cuts two given circles at constant angles OC, (3 ; show that it cuts their radical axis at a constant angle y given by 6 cos y = r cos OC — r 7 cos /3 where r, K are the radii of the given circles and b the distance between their centres. 56. Deduce that the perpendicular p from the centre of the variable circle on the radical axis varies as its radius R. {Note — p = R cos y] 57. Show that three circles are cut at equal imaginary angles by one of their axes of similitude. [Note — Observe that J_s from centres on an axis of sim'de are proportional to radii.] 58. A circle which cuts two given circles at constant angles touches two fixed circles. 59. Find locus of centre of a circle which cuts three given circles at equal angles. Ans. A perpendicular from radical centre on one of the axes of similitude. {Note — Obtain eq r n of locus by elim'n from three eq'ns like that in § 192 ; observe that radical centre, being centre of orthotomic circle, is a point on locus, and that axis of sinrMe is a circle of infinite radius cutting given circles at equal angles.] The Circle 165 60. One vertex of a rectangle is fixed (x'y') ; two other vertices move on the circle x 2 + y 2 = a 2 : find the locus of the fourth vertex. Ans. The circle x 2 + y 2 = 2 a 2 — x' 2 — y r 2 61. Find locus of intersection of two straight lines at right angles, each of which touches one of the two circles (x - a) 2 + y 2 = b 2 , (x + a) 2 + y 2 = c 2 , and prove that the bisectors of the angles between the straight lines always touch one or other of two fixed circles. [Note — Two _L tangents are (x — a) cos + y sin = b, y cos — (x + a) sin = c ; elim' 0, locus is (x 2 + y 2 - a 2 ) 2 = (bx + cy + ab) 2 + (by - ex + ac) 2 The bisectors are (x — a + y) cos + (y + x + a) sin = b ± c, which touch (x — a + y) 2 + (y + x + a) 2 = (b + c) 2 , or x 2 + (y - a) 2 = \ (b + c) 2 , x 2 + (y + a) 2 = | (b - c) 2 ] 62. The vertices of a triangle are (x^), (x 2 y 2 ), (x 3 y 3 ). If the lengths of its sides are a, b, c, show that the co-ordinates of its incentre are axx + bx 2 + cx 3 ay x + by 2 + cy 3 a+b+c ' a+b+c A A [Note — Let bisector of A meet BC in D, and bisector of C meet AD in O. Then BD : DC - c : b, and ab AO : OD - b : CD = b : .- = b + c : a; apply § 15,] 63. Show that the co-ordinates of the limiting points of the circles x 2 + y 2 + 2 gx + 2 fy + c = o, x 2 + y 2 + 2 g'x + 2 f 'y + c r = o are given by the equations x = - g + A ^ v . _ i+Xff 1 + A ' Y i+A where X is either root of the equation (g + AgO 2 + (f + AT) 2 = (1 + A)(c + Ac') CHAPTER VII TRANSFORMATION OP CO-ORDINATES TRANSFERENCE OF ORIGIN § 204. It is sometimes useful to refer points to a new pair of axes. To transform to parallel axes through a given point (hk). Let O' be the new origin ; OR = h, RO'=k its co-Orel's. Let OM =x, PM = y be the co-ord's of any point P referred to the old axes ; 0'M'=x', PM' = y' its co-ord's referred to the new axes. We see from the figure that x = OM = O'M' + OR = x' + h y = PM = PM' + RO' = y' + k Thus if, in the equation to a curve which expresses a relation between x and y, we substitute x' + h for x and y' + k for y we obtain a relation between x' and y'; i. e. this is the equation to the curve referred to the new axes. Ex. The equation to a curve is 5 x 2 + 6 xy + 5 y 2 - 38 x - 42 y + 93 = o What is its equation referred to parallel axes through (2, 3) ? Transformation of Co-ordinates 167 Substitute x' + 2 for x and y' + 3 for y and reduce ; then 5X /2 + 6xy + 5y /2 = 8 We may now suppress the accents : and the req'd eq'n is 5X 2 + 6xy + 5y' 2 = 8 RECTANGULAR AXES TURNED THROUGH AN ANGLE § 205. To change from a set of rectangular axes to another set through the same origin inclined to the first set at an angle 0. Let the old co-ord's of a point P be OM =x, PM = y; and the new co-ord's OM' = x', PM' = y'. Project the broken line OM'P on OX; this projection = the projection of O P, .*. (§ 60) x = x'cos# — y'sin^ (1) Again, project OM'P on OY, .-. y = x'sin# + y'costf (2) (1) and (2) are the required equations expressing the old co- ordinates in terms of the new. 206. The equations (1), (2) may also be obtained thus. Join OP; let OP = r, POX' = Then x = OM = r cos POM = r cos (0 + ) 170 Analytical Geometry [208. Solving these equations, for x, y we have the required formulae expressing the old co-ord's in terms of the new : y' cot co X = X y = y cosec co § 208. Suppose that we take the point (h, k) as origin of polar co-ordinates, the initial line being parallel to OX. By drawing a figure we see at once that x = h + r cos 6 y = k + rsii These formulae are sometimes useful. os 61 \x\0 \ The above are the most useful cases of transformation. The formulae in- vestigated in the next § are rarely used. §209. To transform from one set of oblique axes to another, retaining the same origin. Let OX, OY be the old axes, OX', OY' the new axes. A XOY = co. A Let Also let XOX' = (X, YOY'-/3. Let the old co-ord's of a point P be x = OM, y = PM ; and its new co-ord's x r = OM', y' = PM'. Draw PN ± OX. The broken lines OMP, OM'P have equal projections on any line. Project these broken lines on PN ; .-. y sin a) = x' sin OC + y' sin (co - /3) 2ii.] Transformation of Co-ordinates 17 1 Similarly by projecting on the _L from P to OY we obtain x sin co = x' sin (to — (X) + y' sin /3 Cor' — If we change to axes through the point (h, k) parallel to OX', OY'; then from § 204 it follows that the equations expressing the old co-ordinates in terms of the new are of the form x = h + lx' + my', y = k + IV + m'y' DEGREE OF TRANSFORMED EQUATION § 210. We have just seen that however the axes are changed the old co- ordinates are expressed in terms of the new by equations of the form x = h + lx' + my', y = k + IV + m'y' The degree of an equation is unaltered by transformation of co-ordinates . For when the above expressions for x, y are substituted any term Ax r y s (whose degree is r + s) becomes A (h + lx' + my') r (k + IV + my) s ; and if this be expanded no term is of a degree higher than r + S. Thus the degree cannot be raised by transformation. Neither can it be depressed: for if it could the degree of the new equation could be raised by returning to the original axes. INVARIANTS § 211. Suppose that we are transforming from axes inclined at an angle co to other axes through the same origin inclined at an angle a/, and that on making the substitutions of § 209 (which are of the form x = lx' + my', y = I'x' + my), the expression ax 2 + 2 hxy -f by 2 becomes a'x' 2 + 2 h'x'y' 4- b'y' 2 ; we proceed to investigate relations between the coefficients a, h, b, a', h', b r . There are two expressions for the square of the distance of any point from the origin ; if P is the point whose old co-ordinates are x, y and new co-ordinates x', y', OP 2 = x 2 + 2 xy cos 00 + y 2 , and C P 2 = x /2 + 2 x'y' cos oS + y' ! . ijz Analytical Geometry 211. ,1 Therefore when the substitutions are made x 2 + 2 xy cos co + y 2 must become x /2 + 2x'y' cos co' + y' Also ax 2 + 2 hxy + by 2 becomes a'x' 2 + 2 h'x'y' + d'/ 2 Therefore ax 2 + 2 hxy + by 2 + X (x 2 + 2 xy cos co + y 2 ) . . (1) becomes aV 2 + 2 hVy' + by 2 + X (x' 2 + 2X y cos co' + y' 2 ) . (2) Whatever be the value of X the expression (1) is identically equal to (2) ; if we suppose X a constant and its value so chosen that (1) is a perfect square, then (2) is a perfect square for the same value of X. (1) is a perfect square if (a + X) (b + X) = (h + X cos co) 2 a + b — 2hcoso)x , ab — h 2 or X 2 H :— = A -1 ^~2 — = ° sin' co sin 2 co Similarly (2) is a perfect square if a '+ b'- 2h'cosco / . a'b'-h' 2 A 2 H : — - — -. A + . „ — — = o smW sin 2 co' Since these quadratics have the same roots, we see that a + b - 2 h cos co __ a' + b' — 2 h' cos co' sin 2 co sin 2 co' ab-h 2 _ aV - h' 2 and sin 2 co ~ sin 2 co' If both sets of axes are rectangular, co = co' = 90° .-. a + b = a' + b' and ab - h 2 = a'b' - h /2 212 ,] Miscellaneous Exercises 173 §212. Functions of the coefficients which are equal to the same functions of the new coefficients obtained by transformation are called invariants. We have then obtained two invariants of the expression ax 2 + 2 hxy + by 2 , a + b — 2 h cos to , ab — h 2 viz. r-s and — — — . sin- 2 a) sin 2 co When the axes are rectangular the invariants are a + b and ab — h 2 Miscellaneous Exercises 1 . The equation to a line referred to rectangular axes OX, OY is y = 3X + 2; find its equation referred to OX, OY' where 0Y r makes an angle of 6o° with OX. Ans. 6 x + (3 — V3) y + 4 = o 2. If x, y are the co-ord's of a point referred to rectangular axes; find its co-ord's referred to the two lines whose equation is x 2 _ y 2 a 2 ~ b 2 Ans. I Va 2 + b 2 fc - 0, \ Va 2 + b 2 (- + ?\ 3. Show that (x - a cos a) 2 + (y - a sin a) 2 = k 2 (x cos OC + y sin a - a) 2 represents two straight lines ; and that the bisectors of the angle they include are y = X tan OC, x cos OC + y sin OL = a 4. If the formulae for transformation to a new pair of axes with the same origin be x = mx' + ny 7 , y = mV + n'y 7 ; prove that nn'(m 2 + m /2 — 1) = mrr/Cn 2 + n /2 — 1) [Note— x / 2 + 2 x /y/ cos tf + y/ 2 _ x 2 + 2 xy cos a) + y 2 = (mx' + ny r ) 2 + &c. Equate coeff's of x /2 , y /2 and elim 7 cosco.] 174 Analytical Geometry 5. If lx + my + n = o, Lx + My + N = o represent the same line, referred to axes with a common origin but inclined at angles co, 12, prove that (|2 + m 2 _ 2 | m coso)) sin 2 12 - (L 2 + M 2 - 2 LM cos 12) sin 2 a) \JSfote — n = N ; compare expressions for length of _L on line from origin.] 6. If a straight line meet the sides of a triangle ABC in X, Y, Z, then the product of the ratios (BX : XC) (CY : YA) (AZ : ZB) = - i [Note— Let A be (x^), B (X2y 2 ) and C (x 3 y 3 ); and let the equation to XYZ be ax + by + c -- o Then BX XC ax 2 + by 2 + c (§ 73), &c. ax 3 + by 3 + c This is Menelaus' Theorem. See Euclid Revised, p. 321] 7. The three sides of a triangle pass each through one of three collinear points, and two of its vertices move on fixed lines : show that the third vertex describes a straight line. 8. Show that the lines (ah/ - a'h) x 2 + (ab' - a'b) xy + (hb' - h'b) y 2 = o are harmonic conjugates of the lines ax 2 + 2 hxy + by 2 = o and also of the lines a'x 2 + 2 h'xy + b r y 2 = o 9. If the circles (y — b) 2 + (x - a) (x — a') = o (y - B) 2 + (x - A) (x - A') = o touch each other, prove that either (B - b) 2 + (A - a ) (A' -30 = or (B - b) 2 + (A - a') (A' - a ) = o 10. Determine the equation of the circle intersecting the circle x 2 + y 2 +3x + 5y + 2 = o in the chord x + 2y — 3 -— o Miscellaneous Exercises 175 and the circle x 2 + y 2 — x + y — 2 = in the chord 2X + y + 6=o; and examine under what circumstances such a problem is possible. Ans. 3X 2 +3y 2 + 5x + 7y+i8 = o {Note — The two lines must meet on the radical axis of the circles.] 1 1 . Determine A so that x „ „ r 0C + 8 OL + 8 0C- 81 A (x 2 + y 2 — a 2 ) + x cos + y sin — — a cos - [OL + y . OL + y 0L-yl x cos J- + y sin '- - a cos '- 2 2 J may represent a pair of straight lines. , CX-/3 . y-OL Ans. A = sm sin 2 2 12. The equations of two circles taking a centre of similitude as origin may be written x 2 + y 2 — 2 ax + a 2 cos 2 y = o, x 2 + y 2 — 2 a r x + a /2 cos 2 y = o : find that of the circle passing through the four points of contact of the common tangents from the origin. Ans. x 2 + y 2 — (a + a') x + aa' cos 2 y — o 13. If 2 hxy + 2gx + 2fy + c = o represent two straight lines, show that 2fg = ch, and that these lines and the axes form a parallelogram whose diagonals are x y x y 1 f g f g h 14. Find the locus of a point such that the two pairs of tangents drawn from it to the circles x 2 + y 2 + 2 gx = o, x 2 + y 2 + 2 fy = o may form a harmonic pencil. Ans. The two parallel lines gx — fy = + fg. 176 Analytical Geometry 15. If S = o, S' = o be two circles whose radii are r, r r , prove that their internal centre of similitude is the centre of S S' r + P = °' and their external centre of similitude the centre of S S' JT = ° r r Thence infer that the six centres of similitude of three circles lie three by- three on four right lines. 16. A, B, C are three points on a circle; show that the feet of the perpen- diculars from any point O on the circle on the sides of the triangle ABC are collinear. [Take O as origin of polar co-ord's ; let eq'n of circle be r = 2 a cos 6 Let OC, ft, y be the vectorial As of A, B, C. The eq'n of BC is 2 a cos ft cos y = r cos (ft + y — 0) The eq'n of the _L on this line from O is o = r sin Q3 + y — 0) The co-ord's of foot of -L from O on BC are .\ (2 a cos /3 cos y, ft + y) Similarly the feet of the other _Ls are (2 a cosy cos OC, y + OC) and (2 a cos OC cos ft, OC + ft) These points evidently lie on the line 2 a cos OC cos ft cos y = r cos (OC + ft + y — 0) This line is Stmson's Line ; see Euclid Revised, p. 168]. 17. One diagonal of a complete quadrilateral is altogether fixed, the second diagonal is a segment of a fixed right line ; show that if one extremity of the third diagonal describe a right line, the other extremity describes another right line. CHAPTER VIII THE PARABOLA § 213. A Conic Section or a Conic is the locus of a point which moves so that its distance from a fixed point is to its distance from a fixed line in a constant ratio e : i. Thus if (see fig' § 214) S is the fixed point, KK' the fixed line, and PM the -L on this line from a point P on the curve ; then SP = ePM The fixed point S is called the focus and the fixed line KK' the directrix ; e is called the eccentricity. The curve is called a parabola if e = 1, an ellipse if e < 1, and a hyperbola if e > 1. The properties of the curves obtained by cutting a cone on a circular base by a plane were first investigated by the Greek Geometers, Apollonius, &c. ; any such section, as will be seen hereafter, is one of the curves just defined. Ex. 1. Find eq'n to a parabola whose focus is (1, 2) and directrix 3X + 4y + 5 = o Express that distance of a point (x, y) on the curve from focus = its distance from directrix. .-. V(x-i)* + (y- 2 )* = 3X+ + 4y + 5 Squaring and reducing, the req'd eq'n is 16 x 2 — 24 xy + 9y 2 — 8ox — i4oy + 100 = o N 178 Analytical Geometry [214. Ex. 2. The eq'n to an ellipse whose focus is (1, o), directrix x - 4 = o, and eccentricity |, is V (x — i) 2 + y 2 = |(x — 4) ; or reducing, 3 x 2 + 4 y 2 = 12 EQUATION TO PARABOLA § 214. Let S be the focus, KK' the directrix, P a point on the curve. Draw PM, SX ± KK'. Bisect SX in A ; then by def A is a point on the curve. Take AS and the JL to AS through A as axes of co-qrd's. Put AS = a, and let AN = x, PN = y be the co-ord's of P. SP= PM .-. SP = NX /. SP 2 , or SN 2 + NP 2 = NX 2 i. e. (x - a) 2 + y 2 = (x + a) 2 .-. y 2 = 4 ax This is the simplest form of the equation to a parabola. § 215. Defs' — The point A is called the vertex, and the line SX 1 the directrix is called the axis of the curve. Cor's — The co-ord's of the focus are (a, o). The equation to the directrix is x = — a, or x + a = o 2i8.] The Parabola 179 Again, SP = PM = NX = AN + XA .-. SP = x + a FIGURE OF THE CURVE § 216. From y 2 = 4 ax we deduce y = + 2 Vax Thus to each value of x correspond values of y which are equal in magnitude and of opposite sign. If then p is the image of P with respect to the axis, p is also a point on the curve. .". the curve is symmetrical with respect to the axis. If x is negative, y is imaginary. Thus the curve lies wholly on the same side of the axis of y as the point S. If x is very great so is y. The curve .*. widens out indefinitely. The axis of y is the tangent at the vertex. This is proved in § 221. § 217. The double ordinate LSL/ through the focus is called the latus rectum. The latus rectum = 4 a. For by def SL = X from L on directrix = SX = 2a; .-. LL/=2SL = 4 a INTERNAL AND EXTERNAL POINTS § 218. If (x, y) are the co-ord's of an internal point Q (see (%' §214). y 2 - 4 ax = QN 2 - 4 a . AN = QN 2 — PN 2 , which is negative. Thus the y 2 — 4 ax of an internal point is negative. Similarly the y 2 — 4 ax of an external point is positive. Of course the y 2 — 4 ax of a point on the curve is zero. N 2 i8o Analytical Geometry [219. § 219. Ex. 1. If a chord PP' passes through a fixed point O on the axis and PN, P' N' are drawn -L the axis. Prove that AN . AN' is constant. Let AO = h, .*. co-ord's of O are (h, o) The eq'n to any line PP' through O is y = m (x — h) If we combine this equation with y 2 = 4 ax we obtain the co-ord's of P, P'. Eliminating y, m 2 (x - h) 2 = 4 ax .-. x 2 -( 2 h +^)x+ h 2 =o If the roots of this are x x , x 2 then XiX 2 = h 2 i.e. AN. AN' = AO 2 Ex. 2. Trace the curve x 2 = — 4 ay. The curve passes through the origin. Giving any value to y there are two values of x, viz. ± V — 4 ay That these may be real, y must be negative. The curve is evidently a parabola of which the origin is the vertex, and axis of x tangent at the vertex ; the curve lies below the axis of x. The co-ord's of the focus are (o, — a) and the equation to the directrix is y - a The learner should draw the figure. Ex. 3. Trace the curve (x — i) 2 = 2 (y + 3). Changing to parallel axes through (1, — 3) this becomes x 2 = 2 y As in Ex. 2 we see that this eq'n represents a parabola having for tangent at vertex the new axis of x, and its axis along the positive direction of the new axis of y. Its latus rectum or 4 a = 2, .*. a = |. 21 9-] The Parabola 181 The co-ord's of the focus with reference to the new axes are (o, |) and the equation to the directrix is y = -i Returning to the original axes, the focus is (i, — 2|) and the directrix is y = - 3| These details will be clear from a figure, which the learner should draw. Exercises 1. Is the point (i, 2) inside or outside the parabola y 2 = 8x? 2. Find the equation to a parabola whose focus is (1, o) and directrix 3* = 4y Ans. (4X + 3y) 2 = 50 x - 25. 3. The three vertices of an equilateral triangle are on the parabola y 2 = 4 ax, one of them being the vertex : find the length of its side. Ans. 8 a V$ 4. Find the points where the line y = 2 x — 4a cuts the parabola y 2 = 4 ax Ans. (a, — 2 a), (4a, 4a) 5. Show that the line y = x + 3 touches the parabola y 2 — 8x — 8 =0 Ans. The point of contact is (1, 4). 6. Find the length of the chord which the parabola y 2 = 8x intercepts on the line 2 x + y = 8 Ans. 6 «/$. 7. Find the co-ordinates of the focus, the equation of the directrix, and the length of the latus rectum in each of the parabolas y 2 + 4 ax = o, x 2 = - 4 ay, (y - i) 2 = 4 (x - 2), x 2 + 4 y + 8 = o. Ans. (- a, o), x = a, 4a; (o, - a), y = a, 4a; (3, 1), x = 1, 4; (o, - 3). y + 1 = o, 4- 182 Analytical Geometry [220. 8. Show that x 2 + 2 ax + 4 by + a 2 + 4 be = o represents a parabola whose focus is (-a, - c - b), and directrix y + c = b. 9. PSQ is a focal chord ; PA meets the directrix in M. Prove that QM is parallel to the axis of the parabola. 10. A chord PQ of the parabola y 2 = 4 ax subtends a right angle at the vertex. Prove that PQ passes through the fixed point (4 a, o). 11. The vertices of a triangle are three points on the parabola y 2 — 4 ax = o whose ordinates are y x , y 2 , y 3 . Show that its area is ^ (yi - y 2 ) (y 2 - y 3 ) (y 3 - yO 12. Show that the polar equation of a parabola, the vertex being pole, is r sin 2 d — 4 a cos 6 13. Two chords through the vertex, whose lengths are r, K, are at right angles. Prove that ra r 3 = io a 2 (ra + r 3) EQUATION OF TANGENT § 220. To find the equation of the tangent at (x'y'). If (x'y'), (x"y") are two points on the curve, the equation of their join is , ,, y - y _ y - y , x x-x' "x'-x" {1) But since (x' y r ), (x r/ y") satisfy the equation to the curve y' 2 = 4 ax' and y" 2 = 4 ax" ... y'2_ y //2 = 4 a( x / -x ,/ ) !22. The Parabola 183 y' - /' 4 a x'-x" y' + y" .-. by substitution (1) becomes y - y' 4 a x - x' " y' + y" or y (y' + y") - 4 ax = y' 2 + y'y" ~ 4 ax' = /y" (2) since y' 2 = 4 ax' In (2) put y" = y', x" = x' : it becomes 2 yy' - 4 ax = y' 2 = 4 ax' Thus the required equation is yy'= 2 a(x + x') / 3 ) This equation should be remembered, § 221. Cor' (1) — The tangent at the vertex (o, o) is 0=2 ax, or x = o ; i.e. the axis of y. Cor' (2) — The equation of the chord obtained above, viz. y (y' + y") = 4 ax + y'y" is sometimes useful. Cor' (3) — The normal is the line through (x'y') J_ yy' == 2 a (X + X r ) y y' x x' Its equation is .*. - — ~- = ^ y' -2a or 2 a (y - y') + y' (x - x r ) = o § 222. If we eliminate y between y = mx + c and y 2 = 4 ax we get (mx + c) 2 = 4 ax or m 2 x 2 + (2 cm — 4a)x + c 2 = o . . . (1) 184 Analytical Geometry [223. This equation determines the abscissae of the points where the line y = mx + c cuts the curve ; and since the equation is a quadratic we see that the line meets the curve in two points ; these may be real, coincident, or imaginary. If the points are coincident (1) has equal roots : the condition for this is , o o (cm — 2 a) 2 = c 2 m 2 or 4 a 2 — 4 acm = o or c = a/m Again, substituting this value of c in (1) it reduces to (mx — a/m) 2 = o .-. the abscissa of the point of contact is given by mx — a/m = or x = a/m 2 . Substitute this in y 2 = 4 ax .-. y = 2 a/m Thus whatever he the value of "m, the line a y = mx H m — 2' — j * This result should be remembered. 8 223. Otherwise thus. Compare yy r = 2 a (x + x') or 2 a 2 ax' y = —x + — -j- y/ y> and y = mx + c This gives 2 a 2 ax' m — r> — y' / These enable us to express c, x r , y f in terms of m. Thus 2 a .2a y r m "5-] The Parabola 185 x' = y C = the same results as before. 72 ^ 4a m 2 ax' 2 a z 72a y' rrv / m a m § 224. Or thus. Since (5?)' '- *• (s?) (a 2 a\ — 2 j — I is a point on the parabola y 2 = 4 ax ; o 2 3. and substituting — 2 , — for x', y f in the equation of the tangent yy' = 2 a (x + x') it reduces to y = mx + — J m / a 2 a\ We may call the point ( — ^ » — ) on ^ e parabola 'the point m.' GEOMETRICAL PROPERTIES § 225. We shall now deduce some properties of the parabola. Def — If N is the foot of the ordinate at any point P of a curve, and the tangent and normal at P meet the axis of x in T, G respectively, then NT is called the subtangent and NG the sub- normal. I. The subtangent is bisected at the vertex. Let (x'y') be the co-ord's of P. .*. tangent at P is yy' = 2 a (x + x') To get the intercept of this line on the axis of x put y = o i86 Analytical Geometry [225. .-. x + x' = o or x' = — X i.e. TA = AN Co/ — We deduce this construction for the tangent at P : — Draw PN X the axis, measure AT = NA and join PT. II. The subnormal is constant. Put y = o in eq'n of normal 2a(y-/) + y'(x - x') = o ; thus its intercept on the axis of x is given by — 2 ay' + y' (x — x') = o or x - x' = 2 a i.e. AG -AN, or NG = 2 a III. SP = ST For ST = AT + a = AN + a by I.; .-. ST = x' + a = SP (§215) IV. If PM is drawn _L the directrix then the tangent bisects SPM. We have just proved SP = ST .-, SPT = STP (Euclid I. 6) = TPM (Euclid I. 29) V. The foot of the _L from the focus on the tangent lies on the tangent at the vertex. Let SM, PT meet in Y. Then in the As SPY, MPY we have SP = PM 1 PY = PY A A SPY = MPY .-. SY = MY and SYP = MYP = a right angle; i. e. Y is foot of ± from focus on tangent. Also since SY = MY and SA = AX, AY is || XM. 225.] The Parabola 187 VI. If the tangents at Q, Qf intersect in T, a parallel to the axis through T bisects QQ'. Let the || to the axis meet QQ' in V (fig' § 231). Let the coord's of Q be (x'y') and of Q' (x"y")- The equation to QT is yy'= 2a(x + x') (1) The equation to Q'T is yy" = 2 a (x + x") (2) The co-ord's of T are got by solving these eq'ns for x, y. By subtraction y (y'-y") = 2 a(x'-x") = i (y' 2 - y" 2 ) .*. ordinate of V = ordinate of T = J (y' + y") .'. V is mid point of QQ'. Exercises 1. Find the equations of the tangent and normal at L (fig', § 214). Ans. y = x+a, y + x=3a 2. The tangent at P meets the directrix in Z ; prove that PZ subtends a right angle at S. 3. Y is the foot of perpendicular from focus on tangent at P; prove that SY 2 = SA . SP 4. Find co-ord's of the inters'n of tangents at (x'y'), (x ,f y f/ ). Ans. yy/4a, (y r + y")/ 2 5. If J, J' are points on the axis equidistant from the focus, the difference of squares of J_s from J, J' on any tangent is constant, and = 2 a . J J'. 6. P (x'yO and Q (x f/ y") are two points on the parabola y 2 = 4 ax : prove that their join passes through the focus if y'y" + 4 a 2 = o i88 Analytical Geometry [226. 7. Show that tangents at the extremities of any focal chord meet at right angles on the directrix. 8. Show that the tangent at any point meets the directrix and latus rectum at points equidistant from the focus. 9. Find the locus of the foot of the perpendicular from the focus on the normal. Ans. The parabola y 2 = a (x — a). 10. Prove that the parabolas /2 _ cut at an angle y 5 = ax, x 2 = by tan- 1 j. 1 3 a 3 b 3 (a 3 + b 3 ) 11. Tangents are drawn to the parabola y 2 = 4 ax at points whose abscissae are in the ratio /x : 1 ; show that the locus of their intersection is the parabola y 2 = (^ + yT*Y ax § 226* Ex. 1. To find locus of foot of JL from focus on tangent. (See § 225, V.) Any tangent is The _L to this from focus (a, o) is y = mx + — J m y = (x - a) J m v To eliminate m, subtract the eq'ns; this gives (m -f — ) x = o m/ x = o i. e. the locus is the tangent at the vertex. Ex. 2. To find locus of inters 7 n of tangents at right angles. Let the tangents be then y = mx + — , y = m'x + — , J m J m' ,' = mm' = — 1 We have to eliminate m, m r from these three eq'ns. 22 7.] The Parabola 189 Subtract second eq'n from first and divide by m — m' a ••• o = x 01171' Using then the third eq'n, x = — a i. e. the locus is the directrix. TANGENTS FROM A GIVEN POINT § 227. Any tangent is a y = mx H m This will pass through a given point (x'y') if y = mx H m or mV-my'+a = o (1) This quadratic gives two values of m, and .*. in general two tangents can be drawn from a given point (x'y'). If the roots of (1) are m, m / then the two tangents are y = mx + — , y = m'x + A m m The roots of (1) are real, coincident, or imaginary, according as y /2 — 4 ax' > = < o The two tangents are .-. (§ 218) real, coincident, or imaginary according as (x'y') is outside, on, or inside the parabola. A Ex. 1. Find (f> between tangents from (x'y') y/ From (1) m + m / = — f i a mm' = -7 x' .-. (m - m') 2 =* (m + m') 2 - 4 mm' y 7 2 — 4 ax' - ^2 , m — m' .*. tan

. This may be written y 2 + (x - a) 2 = (x + a) 2 sec 2 This evidently (§ 213) represents a hyperbola having the same focus and directrix as the parabola and whose eccentricity = sec (j). Exercises 1. Find co-ord's of the point of inters'n of tangents at the points m, m'. a Z 1 I \ Ans. -. , a I — + -—. I mm' \m my 2. Find the co-ord's of the point where the directrix is intersected by the _L from the inters'n of y = nfijX + m. a a y = rrioX + — , on y = itux + — J 2 m 2 m rric Ans. x = - a, y- a(i + i + ^- + j^^-J 3. Deduce that the orthocentre of the triangle formed by any three tangents lies on the directrix. [The symmetry of the above result shows that each _L intersects the direc- trix in the same point.] 4. Two tangents to the parabola y 2 = 4 ax make angles 6, & with its axis. Find locus of their intersection, i°, If tan tan Q' — constant = A 2 , If cot + cot 6' = constant = A. 3 , If sin sin f = constant = X Ans. i°, the straight line a = A.X 2 , the straight line y = A a 3°, the circle x 2 + y 2 - 2 ax = a 2 (1 - A 2 ) / A 2 jo.] The Parabola 191 DIAMETERS § 228< To find the locus of the mid points of a system of parallel chords. A Let the chords be inclined to the axis at 0. Let (x' y') 3 (x" y") be the extremities of one of the chords ; (xy) its mid point. Then 2 x = x' + x", 2 y = y' + y" \ . . . (1) v' — v" Also ' __ y = tan 6 (2) Since (x'y'), (x^y^) are on the parabola y /2 = 4 ax', y //2 = 4 ax r/ By subtraction y /2 — y //2 = 4 a (x r — x") ••• (/-y // )(y , + y // ) = 4a(x / -x / o Divide by x' — x /r ; then from (i), (2) we see that 2 y tan 6 = 4 a .-. y = 2 a cot # This is the eq'n to the locus. It is .\ a straight line || the axis. § 229> Def's — The locus of mid points of a system of parallel chords of a conic is called a diameter. The chords which a diameter bisects are called its ordinates. We have proved then that the diameters of a parabola are straight lines parallel to the axis. §230. The tangent at the end of a diameter is parallel to its ordinates. In fig' § 231 Q(y is one of the chords bisected by the diameter PV Let QQ' move parallel to itself until V comes to P. Then ultimately the bisected chord becomes the tangent at P. Q.E.D Or thus. By § 228 y = 2 a cot is diameter bisecting chords || y = x tan 0. 192 Analytical Geometry [231. The co-ord's of its extremity satisfy the equations y = 2 a cot 0, y 2 = 4 ax .*. they are (a cot 2 0, 2 a cot 0) The tangent at this point is y. 2 a cot = 2 a (x + a cot 2 0) which is || y = x tan 0. Q.E.D. § 231 . To find the equation of a parabola referred to a diameter and the tangent at its extremity as axes. Let the new axes be PX', PJ. Let QVQ' be one of the chords bisected by the diameter, so that the co-ord's of Q are PV = x, QV = y. Draw QN J_ AS. Let P be the point m, (§ 224). Then m = tan where = JPX / ; and the co-ord's of P referred to the old axes are a/m 2 = a cot 2 and 2 a/m = 2 a cot Project the broken line QVPA on AN and NQ /. . AN = y cos + x + a cot 2 NQ = y sin + 2 a cot !32.] The Parabola 193 But QN 2 = 4 a . AN .*. (y sin 6 + 2 a cot 0) 2 = 4 a (y cos + x + a cot 2 0) which reduces to y 2 sin 2 6 = 4 ax (1) Again, SP = a + a cot 2 (§ 215) = a/sin 2 0. Putting SP = a' ', (1) becomes y 2 = 4a'x This is the required equation. Cor' — As in § 220 the tangent at (x'y') is yy' = 2a'(x + x') Putting y = o we get intercept on axis of x : X + x' = o .'. X = — x' i. e. if Q be (x r y r ) we deduce (see fig / ) TP = PV § 232* Ex. Any diameter of the parabola meets the tangent at P, a chord PQ and the curve in I, V, J respectively : prove that IV : IJ - QP : VP Take for axes the diameter through P and the tangent at P; let the eq'n to the parabola be y 2 = 4 a'x. Let eq'n to PQ be y = mx, and to IJ, y = c. Then co-ord's of Q are determined by y 2 = 4 a'x, y = mx V ,. „ y = c, y = mx »> >> >> » 5» i) >5 JJ y =c, 4 a'x .-. co-ord's of Q are (4a'/m 2 , 4a'/m) ,, V „ (c/m, c) „ „ J „ (c 2 / 4 a', c) QP VP y ofQ ~ y of V = 4* mc IV xofV 4 a' IJ x of J mc 194 Analytical Geometry [233. Evidently Also ^7 = ^H = — » •'• &c - IJ x ot J mc POLES AND POLARS § 233. Def — If P, Q are the points of contact of tangents from T, then PQ is called the polar of T, and T is called the pole of PQ. To find the equation of the polar of{x!y'). Let (hk) be the point of contact of either tangent from (x'y') to y 2 = 4 ax Express that the tangent at (hk) viz. ky = 2 a(x + h) passes through (x' y') .-. ky' = 2 a (x' + h) This equation expresses that (hk) lies on the straight line yy' = 2 a (x + x') As both points of contact lie on this line, it is .*. their join. The equation of the polar of (x'y') is .\ yy' = 2 a (x + x') 8 234-. I/P lies on the polar o/Q., then Q lies on the polar of P. Let P be (x^) and Q (x 2 y 2 ). The polar of Q is yy 2 = 2 a (x + x 2 ) P (x 2 y x ) lies on this if yi y 2 = 2 a (x 2 + x 2 ) The symmetry of this equation shows that it is also the condition that Q lies on the polar of P. Cor' — The polar of the focus (a, o) is x + a = o, i.e. the directrix; .*. tangents at the ends of a focal chord intersect on the directrix. 236.] The Parabola 195 NORMALS 8 235. If we substitute a/m 2 , 2 a/m for x' ', y f in the equation of § 221, Cor* (3), we obtain, after reduction m 3 y + (x — 2 a) m 2 — a = o (1) This is the equation to the normal at the point m. This may be expressed differently. If the normal is y = fj. x + c then m = since normal is _L tangent. Substituting this value of m in (1) the equation of the normal becomes y = jux — 2ajut — a//, 3 (2) Ex. Find locus of point of inters'n of normals at right angles. A normal through (h k) is y *= l*x — 2 a fA — a jj? where jjl is one of the roots of the cubic k = juh-2aju-a^ 3 (1) Let the roots of this cubic be /Xi , ju 2 , jtx 3 . Then ^ M2 M-3 = — k /a Bu * Mi M2 = ~ I t if two of the normals are at right angles. .-. /x 3 = k/a Substitute this value of ju 3 instead of jjl in (1) : then writing x, y for h, k we obtain the locus required, viz. y 2 = ax — 3 a 2 § 236. Suppose we wish to find the normals which can be drawn through a given point (,hk). Let (Xj: y x ) be the foot of one of the normals. Express that normal at (x 2 y x ) passes through (hk) : .-. 2 a (k - y x ) + y 1 (h - aq) = o .-. za(k -y x ) + y^h -y x 2 /4a) = o or yj 3 + 4 a (2 a — h) y x — 8 a 2 k = o This cubic in y x has three roots y f , y rt , y'" ; .*. three normals can be drawn. Cor r — As the term in y x 2 is absent, y' + y" + y' n = o 2 196 Analytical Geometry Ex. If chords are drawn parallel to y = mx, find locus of intersection of normals at their extremities. Let (x'y'), {x"y") be the extremities of one of the chords. Let the normals at these points meet in (hk) : let (x"'y f ") be the foot of the third normal from (hk). Then [§ 221, Cor' (2)] y' + y" = 4 a /m .*. (by preceding Cor') y'" = — 4 a/m Thus (x'"y"') is a fixed point ; and the locus is the normal at (x'"y'"). Exercises on the Parabola [Unless otherwise implied, the equation of the parabola in these questions is y 2 = 4 ax.] 1. If p, p' are perpendiculars from the extremities of a focal chord on tangent at vertex, prove pp' = constant = a 2 2. Find equation of locus of intersection of tangents inclined at 6o°. Ans. 3 x 2 — y 2 + 10 ax + 3 a 2 = o 3. Find equation of chord joining the points m, m'. Ans. 2(mm'x + a) = y(m + m') 4. Find locus of intersection of tangents inclined at complementary angles to the axis. Ans. The latus rectum, 5. The join of a point P on the parabola to the vertex cuts the perpen- dicular from the focus on the tangent at P in R ; find equation of locus of R. Ans. y 2 + 2 x 2 = 2 ax 6. The equation of the parabola referred to its axis and latus rectum as axes of co-ordinates is y 2 = 4a (x + a); and any tangent is x cos a + y sin a + a/cos OL — o Exercises on the Parabola 397 7. Show that the line y = m (x + a) + a/m touches the parabola y 2 = 4 a (x + a) 8. Find locus of intersection of rectangular tangents to the confocal parabolas y 2 - 4a(x + a), y 2 - 4 a' (x + a') Ans. The straight line x + a + a' — o 9. If R is the mid point of a chord PQ, show that the polar of R is parallel to PQ. [Note — Taking diam' through R and tangent at its vertex as axes ; if co-ord's of R are (h, o), then eq'n of PQ is x = h : and the polar of R is x + h = o*] 10. The equation of the tangents from (hk) may be written in either of the forms h (y - k) 2 - k (y - k) (x - h) + a (x - h) 2 = o or (k 2 - 4 ah) (y 2 - 4 ax) = [ky - 2 a (x + h)] 2 [Note — Proceed as in §§ 166, 167.] 11. If the intercept of the tangents from P on tangent at vertex is constant : prove that locus of P is an equal parabola. [Note— If Pis (hk), intercept = Vk 2 - 4 ah.] 12. Find equation of chord whose mid point is (hk). Ans. (y — k) k = 2 a (x — h) [Note — By Ex. 9, chord is || yk = 2 a (x + h).] 13. Find locus of mid points of chords through (x'y'). Ans. The parabola y (y — y') = 2 a (x — x') [Note — Express that chord whose eq'n is obtained in Ex. 1 2 passes through (x'y') ; then write x, y for h, k.] * This may also be seen geometrically ; the polar of R is the join of the poles of PQ and the diameter through R : the latter pole is the point at infinity on PQ. 198 Analytical Geometry 14. Given base and area of a triangle ; find locus of orthocentre. Ans. If (+ a, o), (— a, o) are extremities of base and area = ap/2 ; locus is the parabola x 2 + yp = a 2 15. If P, Q, R are three points on the parabola whose abscissae are in geometrical progression ; prove that the tangents at P, R intersect on the ordinate of Q. 16. The area of the triangle formed by three tangents is half that of the triangle formed by joining their points of contact. 1 7 . SY is the perpendicular from the focus on the tangent at P : show that locus of centre of circum circle of the triangle SYP is the parabola y 2 = a (2 x — a) 18. Show that the length of the chord of contact of tangents from (hk) is V(k? + 4 a 2 ) (k 2 - 4ah)/a 19. Find equation of lines joining vertex to points of contact of tangents from (hk). Ans. hy 2 — 2 x (ky — 2 ax) 20. Find locus of mid points of focal chords. Ans. The parabola y 2 = 2 a (x — a) 2 1 . PQ is a double ordinate of a parabola ; the join of P to the foot of the directrix cuts the curve in P'. Show that P'Q passes through the focus. 22. Find locus of mid point of PG. Ans. The parabola y 2 = a (x — a) 23. Find locus of intersection of normals inclined at complementary angles to the axis. Ans. The parabola y 2 = a (x — a) 24. Show that the locus of the mid point of the intercept on a variable tangent between two fixed tangents is a straight line. 25. C x iyi)j (x 2 y 2 ), ... are the vertices of a re-entrant quadrilateral whose sides touch the parabola. Prove that x x x 3 = x 2 x 4 and y x + y 3 = y 2 + y 4 Exercises on the Parabola 199 26. Normals are drawn at two points on opposite sides of the axis whose abscissae are in the ratio 1:4. Show the locus of their intersection is the curve 27 ay 2 = 4 (x — 2 a) 3 27. A perpendicular A p from the vertex to the tangent at P meets the curve in q. Prove that A p . A q = 4 a 2 28. The tangents at P, P r meet in T. Prove that TP 2 : TP' 2 = SP : SP' and ST 2 = SP. SP 29.. If the normals at P, Q meet on the curve; prove that PQ passes through a fixed point on the axis of the parabola. [Note — Let normals at P, Q meet in R : let R, P, Q be the points rrij, m 2 , m 3 . The m's of the normals through (xy) are the roots of m 2 (my — 2 a) + m 2 x - a = o In this eq'n write a/nr^ 2 , 2 Q.jrr\ 1 for x, y ; .-. m 1( m 2 , m 3 are the roots of „/m \ m 2 2 m 2 ( 1 + — 5 - 1 = o Hence m 2 , m 3 are the roots of 2 m 2 + — + 1=0; mi .-. m 2 m 3 = |, and join of m 2 , m 3 [Ex. 3, p. 196] passes through ( — 2a, o).] 30. The circle whose diameter is the join of the points mj , m 2 meets the parabola again in the points m 3 , m 4 : show that 1 1 = + 4 m 3 m 4 rri! m 2 [Note — The circle is (Ex. 8, p. 126) (x - a/rri! 2 ) (x - a/m 2 2 ) + (y - 2 a/m^ (y - 2 a/m 2 ) «= o .*. mj, m 2 , m 3 , m 4 are the roots of (a/m 2 - a/rri! 2 ) (a/m 2 - a/m 2 2 ) + (2 a/m — 2 a/mj) (2 a/m — 2 a/m 2 ) = o : .'. m 3 , m 4 are the roots of ( — + — ) (— + — ) + 4=0] \m rx\J \m m 2 / 2co Analytical Geometry 31. The circle whose diameter is a chord PQ meets the parabola again in R, S : prove that if PQ passes through a fixed point on the axis so does RS. {Note — Deduce from Exercises 3, 30 that distance between fixed points is 4 a.] 32. The sides of a quadrilateral inscribed in a parabola are respectively inclined at angles Oi, /3, y, b to the axis : prove that cot OC + cot y = cot /3 + cot b Hence infer that if three of the sides are parallel to given lines the fourth side is parallel to a given line. \Note — Let parameters of vertices be m 1} m 2 , m 3 , m 4 . It follows from Ex. 3, page 196 that 1 1 „ -, 2 cot OC = — + — ; &c] m 1 m 2 33. Find the co-ordinates of the second point in which the normal at the point m meets the parabola. a (2 m 2 + i) 2 — 2 a (2 m 2 + 1) Ans. —3 , m 2 m [Note — Elim r x from y 2 = 4 ax, m 3 y + m 2 (x — 2 a) — a = o (§ 235) This gives m 2 y 2 + 4 am 3 y — 4 a 2 (2 m 2 + 1) = o ; a quadratic in y the product of whose roots is — 4 a 2 (2 m 2 + 1) /m 2 . One root is 2 a/m ; &c] 34. Find equation of locus of mid points of normal chords. Ans. y 4 + 8 a 4 = 2 ay 2 (x — 2 a) 35. A circle cuts the parabola y 2 = 4 ax in four points whose ordinates are y 1} y 2 , y 3 , y 4 ; prove that yi + y 2 + y 3 + y 4 = o {Note — Elim' x from y 2 = 4 ax, (x - a) 2 + (y - /3) 2 = r 2 ; this gives a biquadratic in y which wants the term in y 3 .] Exercises on the Parabola 201 36. The circle through the feet of the three normals from any point passes through the vertex. [Note — Let y lt y 2 , y 3 be ordinates of feet of normals; y 4 the ordinate of fourth point in which circle meets parabola. Then (§ 236, Cor') yi + y 2 + y 3 = ° : from preceding Ex. it follows then that y 4 = o ; &c] 37. If P, Q, R are the feet of the normals from T : prove that SP.SQ.SR = a ST 2 . [Note — Let T be (h, k) : the normal at (x'y 7 ) passes through (hk) if 2 a (k — yO + y' (h — x') = o Write x, y for x r , y' ; we infer that feet of normals lie on the curve 2 a (k — y) + y (h — x) = o But they also lie on y 2 = 4 ax. Eliminate y; we get (x + 2 a — h) 2 x = ak 2 Now if SP — p and x is abscissa of P, p = x + a: .-. (p + a - h) 2 (p - a) = ak 2 The roots of this cubic in p are SP, SQ, SR ; .-. SP.SQ.SR = a[k 2 + (a - h) 2 ] = aST 2 ] 38. Find the equations of the common tangents to the parabola y 2 = 4ax and the circle 2 (x 2 + y 2 ) = 9 ax. Ans. i2y + i6x + 9a = o 39. Show that the polar of any point on the circle x 2 + y 2 = ax with respect to the circle x 2 + y 2 = 2 ax touches the parabola y 2 = 4 ax 40. Find locus of poles of tangents to the parabola y 2 = 4a'x with respect to the parabola y 2 — 4 ax Ans. The parabola a'y 2 = 4 a 2 x 203 Analytical Geometry 41. Two parabolas have a common axis, but different vertices; show that the portion of a tangent to either intercepted by the other is bisected at the point of contact. 42. Find locus of poles of chords of the parabola whose mid points lie on a fixed line Ax + By + C = o Ans. The parabola A (y 2 — 2 ax) + 2a(By + C) =o 43. Find locus of intersection of tangents which form with tangent at vertex a triangle of constant area A. Ans. The curve x 2 (y 2 — 4 ax) = 4 A 2 44. The locus of the intersection of equal chords of a parabola drawn in fixed directions is a straight line. 45. A chord PQ of a parabola, which is normal at P, meets the axis in G ; y is the ordinate of P : prove that area SPQ = PG 4 /(4 ay) 46. PQ is a normal chord of the parabola : prove that the locus of the centroid of SPQ is the curve 36 ay 2 (3X — 5 a) — 81 y 4 = 128 a 4 47. Prove that the locus of a point such that normals to the parabola y 2 = 4 ax at its intersections with the polar of the point meet on the parabola is the straight line x = 2a 48. If the normals at three points P, Q, R on the parabola are concurrent ; and if P f ' , Q! ' , R' are three other points on the parabola such that PP', QQ r , RR' are respectively parallel to QR, RP, PQ : show that the normals at P r , Q! , R' are also concurrent. 49. If (a/3), (a'/3') 5 (a"/3") are the feet of the normals from (hk) to the parabola, prove the relation /3'/3" (a' - a") + /3"/3 (a" - a) + /3/3' (a - a') = o 50. TP, TQ are tangents to a parabola ; TL is a perpendicular to the axis and the perpendicular from T on PQ meets the axis in M. Prove that 2 LM = latus rectum. Exercises on the Parabola 203 51. L, M, N are the feet of the normals from O (hk) to the parabola y 2 = 4 ax : prove that OL 2 . OM 2 . ON 2 = (k 2 - 4 ah) 2 [k 2 + (h - a) 2 ] [Note— Let L, M, N be the points m^ m 2 , m 3 ; let OL = p x , OM = p 2 , ON = p 3 . Then 2 3, Vi + rri! 2 = proj'n of OL on a || to axis of y — k ; •'• p t = Vi + m x 2 ( kj ; similarly for p 2 , p 3 - Also m 1 , m 2 , m 3 are the roots of the cubic in m m 3 k + m 2 (h — 2 a) — a = We have .'. the identity m 3 k + m 2 (h - 2 a) - a = k (m - 1x4) (m - m 2 ) (m — m s ) In this identity substitute successively ± V — I for m, and multiply; /. (h - a) 2 + k 2 = k 2 (1 + nV) (1 + m 2 2 ) (1 + m 3 2 ) A 2 a I / 2 a \ / m i . Agam, _ - k = ( T - m^f T i the value of (- kj (...) (...) is then easily deduced by subs'g 2 a/k for m in preceding identity.] 52. If n lt n 2 , h 3 are the lengths of the normals and t lt t 2 the lengths of the tangents from any point to the parabola y 2 = 4 ax prove that n x n 2 n 3 = at x t 2 53. The tangents at P, Q meet in T and the normals at P, Q meet in T' if the co-ordinates of T are h, k show that those of T' are l2a + h, I V a a/ If T describes a straight line y = mx + c, show that the equation of the locus of T 7 is (a - cm) [acx + a (2 mc — a) y — c (c 2 + 2 a 2 )] = m [a (x + my) — c 2 — 2 a 2 ] 2 [Note — It will be seen hereafter (Chap. XL) that this eq'n represents a para- bola whose axis is _L given line.] 204 Analytical Geometry 54. On a chord PP' inclined at a constant angle to the axis of a parabola a point Q is taken such that PQ . QP' = constant = b 2 ; find the locus of Q. {Note — Let co-ord's of Q be h, k. Take Q as origin of polar co-ord's ; then the equation of the parabola is (§ 208) (k + r sin 0) 2 = 4 a (h + r cos 6) ; .*. 5 2 = - QP. QP' = — product of roots of this quadratic in r = (4 ah - k 2 )/sin 2 0. The req'd locus is .*. the parabola y 2 — 4 ax + b 2 sin 2 = o.] 55. Three tangents to a parabola whose focus is S intersect in A, B, C ; SA, SB, SC meet BC, CA, AB in A', B' ', C/ respectively; show that the perpendiculars from A, B, C to the other tangents from A', B 7 , C respectively are concurrent. CHAPTER IX THE ELLIPSE EQUATION TO ELLIPSE § 237. We have defined the ellipse in § 213; viz. if P is a point on the curve then SP = ePM, S being the focus, PM the ± from P on the directrix KK', and e (the eccentricity) < 1. The equation to the ellipse may be obtained in a simple form thus. k Draw SX _L directrix. Divide SX internally in A' and externally in A in the given ratio e : 1. 206 Analytical Geometry [237. Then SA'= e A'X (1) SA = eAX (2) By def A, A' are points on the curve. Bisect A A' in C ; let A A' = 2 a Adding (1) and (2) we find 2 CA = e . 2 CX i. e. 2 a = 2 e . CX .-. cx = - e Subtracting (1) from (2) we find 2CS = e.2CA .-. CS = ae Through C draw BCB' ± CA; and take CA, CB as axes. Let the co-ord's of P be CN = x, PN = y. The co-ord's of S are (— ae, o) /. SP 2 = (x + ae) 2 + y 2 (§ 10) Also PM = CN + XC a = X+ e Now SP 2 = e 2 PM 2 /. (x + ae) 2 + y 2 = (ex + a) 2 .-. x 2 (i -e 2 ) + y 2 =a 2 (i - e 2 ) X 2 V 2 •. — + ^ =3 I a 2 + a 2 (i-e 2 ) Put a 2 (1 - e 2 ) = b 2 The equation to the ellipse is .*. ,2 y2 a 2+ b 2 x *+r4=i 238.] The Ellipse 207 If we put x = o in this equation we get y = + b .-. if along the axis of y we measure CB = CB' = b, then B, B' are points on the curve. Def's — The point C is called the centre, the line AA' the major axis, and the line BB' the minor axis. FIGURE OF THE CURVE § 238. Let (x'y') be the co-ord's of a point P on the curve. [Fig', § 237.] ... x ' 2 /a 2 + y' 2 /b 2 = 1 ... (_ x ') 2 /a 2 + (- y') 2 /b 2 = 1 i.e. the point (— x', — y') or P', the image of P with respect to the centre is on the curve. .'. all chords through the centre are bisected. Again, x' 2 /a 2 + (- y0 2 /fc> 2 = 1 ; .*. (x', — y') or p, the image of P with respect to the major axis is a point on the curve. Thus the curve is symmetrical with respect to the major axis. Similarly it is symmetrical with respect to the minor axis. We infer that there is another focus S' and a corresponding directrix kX'k', images respectively of S and KXK' with respect to the minor axis. Again, if we write r cos 0, r sin for x, y in the equation of the ellipse, we obtain its polar equation referred to the centre as pole, viz. r 2 cos 2 r 2 sin 2 b 2 = 1 1 cos 2 sin 2 or - = — -7T- + From this we deduce r* =- b 2 a 2 b 2 a 2 b 2 a 2 sin 2 6 + b 2 cos 2 b 2 + (a 2 - b 2 ) sin 2 Now b 2 + (a 2 - b 2 ) sin 2 is least when = o, and then r 2 = a 2 ; thus the greatest value of r is a. The curve is .'. bounded in all directions. 208 Analytical Geometry [239. §239. Def — The double ordinate through the focus is called the fatus rectum. To find its length, substitute — ae for x in the equation of the ellipse. ... y2 = b 2 (i-e 2 )=^ 2 latus rectum = 2 a FOCAL DISTANCES § 24-0. We may express SP, S'P in terms of x, the abscissa of P (fig/, § 237). SP = e PM = e NX = e (CX + CN) = e (? + x\ .-. SP = a + ex Also S'P = ePM , = e(CX , -CN) = e^-x) .-. S'P = a - ex We infer that SP + S'P = 2a Thus the ellipse is the locus of a point P which moves so that the sum of its distances from two fixed points S, S r = a constant 2 a. (Compare § 104.) This gives a method of describing the ellipse mechanically. If the ends of a thread are fastened at two fixed points S, S', then a pencil moved about so as to keep the thread always stretched will describe an ellipse whose foci are S, S' and whose major axis = length of thread. INTERNAL AND EXTERNAL POINTS § 24" I. As in § 218 we can prove that if (x, y) is a point inside the ellipse the function x 2 y 2 + — — 1 a 2 b 2 of its co-ord's is negative, and if (x, y) is outside the ellipse the function is positive. Of course if (x, y) is on the ellipse the function is zero. 2 4 2.] The Ellipse 209 AUXILIARY CIRCLE § 242. Def — The circle whose diameter is the major axis A A' is called the auxiliary circle. From we deduce Let the ordinate PN of the point P (x, y) on the ellipse meet the circle in Q. ^fei- 1 b a -/ a z i.e. PN = ->/CQ 2 -CN 2 a a .-. PN : QN = b : a Thus the ordinates of the ellipse and circle at corresponding points P, Q are in a constant ratio. Cor'— Draw PLD || CQ, cutting the axes in L, D. Then PD = CQ = a Also, by similar As, PL : CQ = PN : QN = b : a .-. PL = b Hence, if we suppose PLD to be a ruler having pins at L and D, and if AA', BB' be grooves in which these pins run, a pencil at the point P will trace an ellipse whose semi-axes are PD, PL. This is the principle of the elliptic compass. P 2io Analytical Geometry [24: Exercises 1. Find the equation of the ellipse whose focus is (i, o), directrix x + y = o, and eccentricity \. Ans. 7X 2 — 2xy + 7y 2 — i6x + 8 = o 2. Show that the point (i|, i|) is inside the ellipse 4X 2 + 9y 2 = 30 3. Find the eccentricity, the length of the latus rectum, and the co-ordinates of the foci of each of the ellipses 4 x 2 + 9 y 2 = 36, 9 x 2 + 4 y 2 = 36 Ans. i V5. 2|> (± V5, o) ; k V5, 2|, (o, ± >/5) [Note — Observe that major axis of second ellipse is along axis of y.] 4. Determine the same particulars for the ellipses 2 (x — i) 2 + 3 (y + 2) 2 = 1, 4X 2 + 5 y 2 — 8 x — 20 y = o Am - VI' *-£-• * 7i' ~ 2 )' Vs' 8 ~f' ( r± ^' 2 ) 5. Show that the line y = x + 2 touches the ellipse x 2 — xy + y 2 + 2X + 4y + 4 = o at the point (—4, — 2) 6. Find the eccentricity of the ellipse 2 _ 2 x L + y* = 3 x also its foci and directrices. T V^2 Ans. -j-; (f, ±f); y =±f 7. The joins of a point P on the ellipse to the extremities of the minor axis meet the major axis in p, q : prove that C p . C q = CA 2 8. If CP, CQ are two semi-diameters at right angles, prove that 1 ill CP 2 CQ 2 [Note — Use polar equation (see § 238).] 243-] The Ellipse 211 9. The major axis of an ellipse is divided into two parts equal to the focal distances of a point P on the ellipse : prove that the distance of the point of division from either end of the minor axis is equal to the distance of P from the centre. 10. Find the co-ordinates of the mid point of the chord which the ellipse x 2 /a 2 + y 2 /b 2 = 1 intercepts on the line Ax + By + C = o Am. - ACa 2 /(a 2 A 2 + b 2 B 2 ), - BC b 2 /(a 2 A 2 + b 2 B 2 ) 11. A line AB of given length moves with its extremities on two rectangular axes OX, OY : find the locus traced by a fixed point P on the line. [Note— Draw PN ± OX so that ON = x, PN = y are the co-ord's of P; let A OAB = 0. Then sin d = PN/PA = y/PA, cos 6 - x/PB Square and add, .-. locus of P is the ellipse x 2 /PB 2 + y 2 /PA 2 = i. Compare Cor' , § 242.] ECCENTRIC ANGLE § 24-3. Def — The angle QCA (fig', § 242) is called the eccen- tric angle of the point P. If the eccentric angle of P is we may express its co-ord's (x, y) in terms of 0. For x = CN = CQ cos = a cos and y= PN = -QN = - (CQ sin 0) =-(asln0) a a a = bsin These values x = a cos 1 y = b sin ) evidently satisfy the equation of the ellipse x 2 y 2 a • b 2 We can thus express the co-ord's of any point of the ellipse in terms of the single parameter 0. p 2 212, Analytical Geometry [244. S 24-4- . To find the equation of the chord joining the points whose eccentric angles are OC, j3. The eq'n to the join of a cos OC 1 a cos /3 } b sin OC ) b sin /3 ) is = o or x y 1 a cos OC b sin OC 1 a cos /3 b sin j3 1 bx (sin a - sin $) — ay (cos OC - cos/3) = ab sin {OC - /3) . oc- 8 oc + 8 :m nr\c oc+ 8 oc- 8 or 2 bx sin cos + 2 ay sin sin = 2 ab sin Divide by 2 ab sin : the required equation is .\ x oc + (3 y oc + 8 oc - 8 - cos (- — sin = cos a 2 b 2 2 This result is often useful. oc- 8 oc- 8 — cos EQUATION OF TANGENT § 245. To find the equation of the tangent at (x'y') If (x' y')> (x" y") are two points on the curve + b^ = i and ^ + V ,/2 IT = I By subtraction (x'- x") (x' + x") , (/-yW + y") / - y" b 2 x' + x" x' - x" a 2 / + y" Hence the equation to the join of the two points, which is y-y' _ / - y // x — x' x' - x" 246.] The Ellipse 213 becomes y — y' b 2 x' + x' x — x' " "a 2 y' + y' In this put x" = x', y" = y' .*. the equation of the tangent at (x' y') is y — y' b 2 x' x — x' ~ a 2 y' or a 2 yy' + b 2 xx' = a 2 y /2 + b 2 x /2 xx' yy' x /2 y /2 a 2 b 2 a 2 b 2 The dexter *of this equation = i, since (x'y') is on the curve. The equation of the tangent at (x'y') is .'. xx' yy' _ a 2 "*" b 2 ~ This equation should be remembered. Cor f (1) — The tangents at the ends of either axis are -L that axis. Thus the tangent at A (a, o) is ax — T = 1, or x = a, &c. a Cor' (2 ) — The normal is the line through (x'y') Its equation is .*. (§71) , xx' yy' _L — + - jL ~ = 1 a 2 b 2 x — x' _ y — y' W (bV S 24*6. In the result of § 244 put j3 — OC : thus the equation to the tangent at the point whose eccentric angle is OC is xy. - cos a + fsinOI = 1 (1) a b v J The normal is the -L to this through (a cos OC, h sin OC) : its equation is .*. x — a cos OC _ y — b sh OC /cos 0C\ /sin OC\ 214 Analytical Geometry [247. ax b y 2 uo fn \ a 2 — b 2 (2) cos OL sin OL These results are often useful. § 24-7. The abscissae of the points where the line y = mx + c cuts the ellipse x 2 y 2 a 2 + b 2 = * are determined by the equation x 2 (mx + c) 2 — 4- = 1 a 2 + b 2 or (a 2 m 2 + b 2 ) x 2 -f- 2 mc a 2 x + a 2 (c 2 — b 2 ) = o As this equation is a quadratic it has two roots, which may be real, coincident, or imaginary. Thus every line meets an ellipse in two points, which may be real, coincident, or imaginary. The roots of the quadratic are equal, i. e. the line y = mx + c touches the ellipse if (a 2 m 2 + b 5 ) a 2 (c 2 - b 2 ) = m 2 c 2 a 4 or c 2 = a 2 m' 2 + b 2 or c = ± Va 2 m 2 + b 2 We can thus draw two tangents || a given line y = mx, viz. y = mx + Va 2 m 2 + b 2 and y = mx — Va 1 m 2 + b 2 Note — That two tangents can be drawn in a given direction may be seen otherwise thus. In fig 7 , § 237, if co-ord's of P are (x'y'), then those of P r are (— x', — /). The tangent at P' is .'. (§ 245) xx / yy r — + =-^- = — 1 a 2 b 2 which is parallel to the tangent at P. H7-] The Ellipse 215 Exercises 1. Find the tangent and normal to the ellipse 3 x 2 + 4 y 2 = 1 2 at the end of the latus rectum in the first quadrant. Ans. x + 2y = 4, 4 x — 2 y = 1 2. Determine c so that y + x = C may be a tangent to the ellipse Ans. c = + % A/13 2 x 2 + 3 y 2 = 1 3. Find the tangent and normal to x 2 /a 2 + y 2 /b 2 - 1 at the extremity of the latus rectum in the first quadrant, Ans. y + ex = a, ey — x + ae 3 = o 4. Find the tangents to the ellipse 3 x 2 + 4 y 2 — 12 which cut off equal intercepts on the axes. Ans. y + x = + V7 5. If the tangent at the end of the latus rectum pass through a point of trisection of the minor axis, prove that the eccentricity of the ellipse is determined by the equation 9 e 4 + e 2 = 1 6. Find co-ord's of inters'n of tangents at two points on the ellipse x 2 /a 2 + y 2 /b 2 - 1 whose eccentric angles are OC, (3. oc + fi/ oc-8 ql + bI a - /3 Ans. a cos / cos , b sin / cos 2 / 2 2 / 2 \Note — Tangents are x y a y . \ - cos OC + {- sin OC = I a b ^ cos/3 + £sin/3 = 1 j x v Determine - : .- : 1 by §6i.l a b J J 2i6 Analytical Geometry [248. 7. P is a point on the ellipse whose eccentric angle is OC ; find equations of AP, A' P. x oc y . oc oc x . a y oc . oc Ans. - cos v r- sin — = cos — , sin f- f- cos — = sin — a2D2 2 a 2 b 2 2 {Note — Ecc' A of A is o and of A' is 77 ; use eq'n of § 244.] 8. P and Q are two points on an ellipse ; AP, A'Q meet at J, and A'P, AQ meet at J'. Prove that JJ' is perpendicular to AA'. 9. P is a point on an ellipse; find locus of inters'n of A'P with perpen- dicular to AP through A. Ans. The straight line x (a 2 - b 2 ) = a (a 2 + b 2 ) 10. P, Q, R are three points on the ellipse; p, q, r the corresponding points on the auxiliary circle. Prove that A PQR : A pqr = b : a [Note — Let OC, /3, y be eccentric As of P, Q, R: then co-ord's of P are (a cos OC, b sin OC) and of p are (a cos OC, a sin OC), &c. ; use § 22.] 1 1 . The vertices of a triangle are three points on the ellipse whose eccentric angles are OC, (3, y. Show that its area . 0-/3 . p-y . y-OC — 2 ab sin sin sin 12. P is any point on the curve ; the perpendiculars through Pto AP, A'P meet A A' in L, M. Prove that LM = latus rectum 13- P is any point on the ellipse, y its ordinate. Prove that cot A PA' oc y GEOMETRICAL PROPERTIES § 248. Let the tangent at P meet the axes in T, t and let the normal meet the major axis in G. Let (x r y') be the co-orcTs of P. The tangent at P is xx^ yy' _ _ a 2 "*" b 2 2 4 8.] The Ellipse 217 To get its intercept on CA, put y = o .-. xx' = a 2 i.e. CN . CT = CA 2 Similarly PN . Ct = CB 2 Again, putting y = o in the normal's equation x — x' y — y' Va 2 / VbV its intercept on CA is determined by X — X = — a' or Hence and x = x'( I -|)=e^x' i.e. CG = e 2 . CN . . . . SG = ae + e 2 x' = eSP (§ 240) S'G = ae- e 2 x' = eS'P .-. SG : S'G = SP : S'P A (I) (II) (HI) ,-. PG bisects SPS' (Euclid VI. 3) Accordingly, The normal bisects the angle between the focal distances, and the tangent bisects the supplementary angle (IV) 2i8 Analytical Geometry [248. ,/2 Again, since the co-orcTs of P are (x'y') and of G are (e 2 x' ; o) /. PG 2 =(x / -e 2 x / ) 2 + y / b 4 = x' 2 |j + y" x' 2 v /2 Also, if C K is central J_ on tangent r^x* r ■ • xx r yy' CK = 1 from origin on —^ + ^ — 1 = o From these we deduce PG . CK = b 2 (V) Further, let SZ, S'Z' be the focal _l_s on tangent ; and let S'Z' meet SP in q. It follows from (IV) and Euclid I. 26, that APS'Z'ee APqZ' Accordingly, S'Z 7 = q Z' and PS' = Pq .-. Sq = SP + PS' = 2a Also S'Z':S'q = i:2 = S'C:S'S .-. CZ'is || Sq, and CZ' = \ S q = a Similarly CZ = a Thus the feet of the focal perpendiculars on any tangent lie on the auxiliary circle (VI) This is proved analytically in § 249. Further, if the eccentric angle of P is 0, the equation to the tangent at P is (§ 246) x ™ rk , y -cos + jr sin = 1 s 4 8.] The Ellipse 219 S'Z' = X on this from (ae, o) = ( I -ecos^)/ /V /^ + ^ and SZ = -L on same from (— ae, o) / , ~ «~^ rk\ I / cos 2 (b sin 2 .-. sZ.S'Z' = (i-e«oo8'^)/(— i? ^ + _ Bi r) q2 _ ^2 Now 1 - e- cos 2 0=i 2 — cos 2 9 b 2 = s\n 2 (j) -\ — 5 cos 2 (f) _ b2 / cos 2 sin»0 \ V a 2 + b 2 / Hence SZ.S'Z'=b 2 . . . . . . (VII) The learner will find it easy to prove this geometrically. Another analytical proof is given in § 250. Again, tangents at the ends of a chord intersect on the diameter which bisects the chord (VIII) Subtracting the equations of the tangents at P (x 1 y 1 ) and Q (* 2 y 2 )> viz - x *t yyi j xx 2 yy* —r + S = * and — T + TZ = 1 a 2 b 2 a? b 2 x (Xj — x 2 ) y (y, — y 2 ) we get . v 1 a — - + J v,yi u 2 ^ = o a 2 b 2 This eq'n represents a line through the inters'n of the tangents. But the centre (o, o) and the mid point of the chord / X + x 2 y, + y 2 \ l~T~' ^~ ) are also obviously points on this line, .\ &c. 22o Analytical Geometry [249. CONDITIONS OF TANGENCY § 249. The condition in § 247 may also be obtained thus. The equation to the pair of lines joining the origin to the inters'ns of X 2 V 2 y=mx+c and - + i = 1 is (§ 119) x 2 y 2 ^ / y- mx V a 2+ b 2 "l c ) If these lines are coincident the points of inters'n coincide .-. y = mx + c is a tangent if ( L H}1\ ( I 1 \ m2 W 2 cV Vb 2 ~ cV " c*" which reduces to c = + \/a 2 m 2 + b 2 , as before. The result may be usefully stated thus — Whatever he the value of m, the line y = mx + -/a 2 m 2 + b 2 is a tangent to the ellipse. Ex. Find locus of foot of _L from focus on tangent. Any tangent is y - mx = Va 2 m 2 + b 2 The equation to the -L on this from S (— ae, o) is my + x = — ae To eliminate m, square and add : this gives (1 + m 2 ) (x 2 + y 2 ) = a 2 m 2 + b 2 + a 2 e 2 = a 2 (1 + m 2 ) /. x 2 + y 2 = a 2 The locus is /. the auxiliary circle. (Compare § 248, VI.) § 250. To find the condition that Ix + my = n (1) may be a tangent. 25 i.] The Ellipse 221 This may be deduced as in § 247, or § 249. For variety we give another method. The tangent at (x'y') is 0) xx' yy' a 2 b 2 If this represent the same line as (1) — = / = 1 a 2 1 b 2 m n . (1UIU al bm n Hence, as (£f + g)' = 1 we obtain a 2 1 2 + b 2 m 2 = n 2 the condition required. This result is often useful. Cor r (1) — X cos OC + y sin OC = p is a tangent if p 2 = a 2 cos 2 OC + b 2 sin 2 OC This important result might of course be proved independently. Cor' (2) — The equation to any tangent may be written x cos OC + y sin OC = Va 2 cos 2 OC + b 2 sin 2 OL Ex. 1. Prove SZ.S'Z' = b 2 (fig' § 248) Let the tangent at P be x cos (X + y sin OC — p =0 Then S' Z' = ± on this from (ae, o) = p - ae cos OC SZ = J. on it from (-ae, o) - p + ae cos a .-. SZ . S' 2! = p 2 - a 2 e 2 cos 2 OC = (a 2 cos 2 OC + b 2 sin 2 OC) - (a 2 - b 2 ) cos 2 OC = b 2 Compare § 248, VII. § 251. Ex. 2. Find locus of point of intersection of tangents at right angles to each other. Substitute — + OC for OC in 2 x cos OC + y sin OC = Va? cos 2 OC + b 2 sin 2 £X . . . . (1) 222 Analytical Geometry [251. The tangent at right angles to the tangent (i) is .' — x sin a + y cos a = Va 2 sin 2 a + b 2 cos 2 a .... (2) By squaring and adding (1) and (2), OL is eliminated. The required locus is .'. the circle x 2 + y 2 = a 2 + b 2 Def — This circle, which is the locus of the inters' n of rectangular tangents, is called the director circle. Exercises 1. CP, CQ are semi- diameters at right angles; prove that PQ touches a fixed circle whose centre is C. {Note — Let JL from C on PQ = p ; let Oi = angle which this _I_ makes with CP. We have to prove that p is constant. cos a 1 sin a 1 _ _, Now = 7^= , = y^r ; square and add p UK p UQ •'• p l2 = cP^ + cc7 = i + b^ (page 2IO > Ex - 8)0 2. Find the polar and rectangular equations of locus of foot of perpendicular from centre on tangent. Ans. r 2 = a 2 cos 2 + b 2 sin 2 0, (x 2 + y 2 ) 2 = a 2 x 2 + b 2 y 2 {Note — Use result of § 250, Coi J (1).] 3. If LSL/ be the latus rectum, and any ordinate PN be produced to meet the tangent at L in Q, prove that SP = QN. 4. Any tangent to an ellipse meets the tangent at A in V and the minor axis in t. Prove that t V = t S. 5. The normal at P meets the axes in G, g. Prove that PG.Pg - SP.S'P. 6. Find the condition that the line x cos a + y sin (X = p may be normal to the ellipse x 2 /a 2 + y 2 /b 2 = 1 Ans. p 2 (a 2 sin 2 a + b 2 cos 2 a) = (a 2 - b 2 ) 2 sin 2 a cos 2 a 252.] The Ellipse 223 7. P is any point on the ellipse. If SP = r and perpendicular from S on tangent at P = p, find the relation between r and p. b 2 2 a Ans. —r = 1 p 2 r 8. A line through C parallel to the tangent at P meets the focal distances of P in d, d'. Prove that Pd = Pd' = a 9. An ellipse slides between two lines at right angles. Find the locus of its centre. {Note— Take given lines as axes; use result of § 251. The req'd locus is the circle x 2 + y 2 = a 2 + b 2 ] 10. Find locus of intersection of perpendicular from focus on any tangent, with join of centre to point of contact. Ans. The corresponding directrix. 11. Points J, J' are taken on the minor axis such that CJ = CJ' = CS; p and p r are the perpendiculars from J, J' on any tangent. Prove that p2 + p ' 2 = 2 a 2 12. The sum of the eccentric angles of two points P, Q on the ellipse is constant ( = 2 y) ; find locus of intersection of tangents at P, Q. Ans. The straight line ay = bx tan y 13. Show that the equation of the tangents from (x'y') to the ellipse x 2 /a 2 + y 2 /b 2 = 1 /x' 2 y /2 \ /x 2 y 2 \ /x'x y'y \ 2 {Note — Proceed as in § 167.] TANGENTS FROM A GIVEN POINT § 252. Any tangent is y = mx + \/a 2 m 2 + b 2 This will pass through a given point (x' y') if y' = mx 7 + Va 2 m 2 + b 2 or (y'~ mx0 2 = a 2 m 2 + b 2 or (a 2 - x ,2 )m 2 + 2xym + b 2 - y ,2 = o . . (i) 224 Analytical Geometry [253. This quadratic gives two values of m, and .-. in general two tangents can be drawn from a given point (x'y'). The roots of (1) are real, coincident, or imaginary, according as x /2 y /2 > = < ( a 2 _ x / 2 ^ (fc>2 _ y /2) i. e. according as x '2 v /2 a 2 b 2 The two tangents are .\ (§ 241) real, coincident, or imaginary, according as the given point (x'y') is outside, on, or inside the ellipse. § 253 . Cor f — If (xy) is any point on either tangent from (x'y r ) (y - y0/( x - x') = m If we substitute this value of m in (i) we get a relation between x and y. The equation to the pair of tangents from (x'y') is .'. (a 2 - x' 2 ) (y - y') 2 + 2 xy (x - x') (y - y') + (b 2 - y' 2 ) (x - x') 2 = o This equation reduces to a 2 Cy - yO 2 + b 2 (x - x') 2 = (xy' - x'y) 2 Note — Another form of this eq'n is given, Ex. 13, page 223. § 254-. If TP, TQ are the tangents from T, then A A STP = S'TQ Let p, q be the _Ls from S on TP, TQ, and p', q' the J_s from S' on TP, TQ. Then pp' = b 2 (§ 248, VII.) = qq' q p' •55-] The Ellipse 225 p sinSTP , q' sinS'TQ q smSTQ p' sin S'TP A .'. ST, S'T divide PTQ into parts having the same ratio of sines; and the proposition is evident. Note — The proposition may also be proved thus. It is evidently true if the line-pairs (TP, TQ) and (TS, TS') have the same bisectors of angles ; and . \ if ||s through the origin to these line-pairs have the same bisectors. Retaining only the terms of the second degree in the equation of § 253, the line-pair through the origin || TP, TQ is (a 2 - x /2 ) y 2 + 2 xyxy + (b 2 - y' 2 ) x 2 = o Again, the ||s through the origin to TS, TS' are y (x r + ae) — y'x = o and y (x r — ae) — y'x = o The product of these is y 2 (x' 2 - a 2 -1- b 2 ) - 2 x'y'xy + y' 2 x 2 = o Each of these line-pairs is bisected (§ 116) by the line-pair x y (x 2 - y 2 ) = (x /2 - a 2 + b 2 - y /2 ) xy Q.E.D. DIAMETERS §255. To find the locus of the mid points of a system of parallel chords. Let the chords be || y = mx. Let (x 1 y 1 ), (x 2 y 2 ) be the extremities of one of the chords; (xy) its mid point. Then 2X = x l + x 2 , 2 y = y 1 + y 2 . . . . (1) Also 5^11^ =m ....... ( a ) Xj X 2 x 2 v 2 x 2 v 2 a 2 b 2 a 2 b J By subtraction — — - — - -f yi y2 . ( Xl - x 2 ) fa + x 2 ) , (y, - y,) (y t + y 2 ) n a 2 + W~ — = ° 226 Analytical Geometry [256. (3) Divide by x 1 — x 2 ; then from (1), (2) we see that x , my — - A — =0 a 2 ^ b 2 This is the equation to the locus. It is .*. a straight line passing through the centre. Cor' (1)— All diameters of the ellipse pass through the centre. (See def, § 229.) Cor' (2) — If we write the equation to the locus (3) in the form y = m'x then m' = - b 2 /(ma 2 ) .-. mm' = - b 2 /a 2 The symmetry of this relation proves that y = mx is the locus of mid points of chords || y = m'x Hence if one diameter of an ellipse bisects chords parallel to a second, the second bisects all chords parallel to the first. Two such diameters are said to be conjugate. Cor' (3) — Two diameters y = mx, y = m'x are conjugate if b 2 mm' = x § 256. If PCP', DCD' are conj" dian/rs there is a simple relation between the eccentric A s 0, (j)' of P, D. The co-ord r s of P are (acos0 } bsin0) .-. if eq'n to CP is y = mx A then m = b sin (t> /(a cos (j)) Similarly, if eq r n to C D is y = m'x then m' = b sin 0'/(a cose/)') But [§ 255, Co/ (3)] mm' = - tf/a 257-] The Ellipse 227 .-. b 2 sin (f) sin 0'/(a 2 cos cos 0') = — b 2 /a 2 .-. cos0 co$(j)' + sin sin 0' = o .*. COS (0' — = dCA-pCA = dCp .*. Cp, Cd are at right angles. Cor r (2) — Since §' = (j) + 90 the equation to the tangent at D, which is ?„~jl# . y a cos + ab = o Again, the eq'n to CP is y/x = b sin $/(a cos (j)) or bx sin (p — ay cos (f) = o .*. tangent at D is parallel to CP. Hence /^^ tangent at either end of any diameter is parallel to the chords which that diameter bisects. This may also be proved geometrically, exactly as in § 230. § 257. If POP' , DCD' are conjugate diameters, then CP 2 + CD 2 is constant ; and the area of the parallelogram whose sides are the tangents at P, P', D, D' is constant. Let the eccentric angle of P be ; then that of D is + 90° [fig 7 § 256.] Let the co-oid's of P be (x'y') and of D (x"y"> Then x' = a cos 0, y' = b sin Also x" = a cos (0 + 90°) = — a sin Q 2 228 Analytical Geometry |_ 2 5 8 - and y" = b sin ((f) + 90 ) = b cos .-. CP 2 = x' 2 + y' 2 = a 2 cos 2 + b 2 sin 2 and CD 2 = x" 2 + y" 2 = a 2 sin 2 + b 2 cos 2 (j) .-. CP 2 + CD 2 =a 2 + b 2 Again, area of O who sides are the tangents at P, P', D, D' = 4 area of □ whose adjacent sides are C P, C D = 8 area of ACPD = 4(x'y"-x"y') [CV,§ 22 ] = 4 ab (cos 2 (j) + sin 2 (j)) = 4ab. Cor' (1)— Denote the lengths of two conjugate semi- diameters CP, CD by a', t/ and the angle between them by to. Since area of A CPD = \ a'b' sin to .*. a'b' sinto = ab Also a' 2 + b' 2 - a 2 If CP, CD are given in position and magnitude, i.e. if a', b' and a) are given, we can determine a and b by solving these equations. Cor' (2) — From the previous equations we get 4 a 2 b 2 cosec 2 o) = 4a 72 b /2 = (a' 2 + b' 2 ) 2 - (a' 2 - b /2 ) 2 .-. 4 a 2 b 2 cosec 2 co = (a 2 + b 2 ) 2 - (a' 2 - b' 2 ) 2 Hence cosec o> is greatest, and .*. sin 0) is least when a' ' = b'. SUPPLEMENTAL CHORDS § 258. Let Q be any point on the ellipse, PCP' any diameter ; then QP, QP' are called supplemental chords. The diameters parallel to supplemental chords QP, QP' are conjugate. For diam' || QP' bisects QP, and diam || QP bisects QP' [Euclid VI. 2]. The diameters are .-. conjugate, since each bisects a chord parallel to the other. ' 1 + b 2 ) j6o.] The Ellipse 229 The proposition may also be proved analytically thus. Let y — mx, y = m'x be eq'ns of diam'rs || QP and QP'. Let a = ecc' A of P and /3 = ecc' A of Q ; then ecc' A of P' is n + OL. The eq'n of PQ is x ol + 8 y .a + /3 a-/3 - cos — + -r sin = cos (§ 244) a 2 b 2 2 b a + /3 .*. m = cot a 2 Again, writing it + OL for (X in this eq r n we get b IT + OC + /3 nrv = cot a 2 b a + (3 = + -tan a 2 b 2 .*. mm' -= - -2 , and /. &c. [§ 255, Cor' (3)] a EQUAL CONJUGATE DIAMETERS § 259. The diameters || AB, A'B (fig' § 237) are equal by symmetry; and by § 258 they are conjugate. The equi-conjugate diameters are .'. the diagonals of the rectangle whose sides are the tangents at A, B, N, B' ; and their equations are b , b y = + — x and y = x J a J a To find their lengths ; since CP 2 + CD 2 = a 2 + b 2 [§ 257] : if CP - CD - a' we find 2 a' 2 = a 2 + b 2 POLES AND POLARS § 260. Def — If P, Q are the points of contact of tangents from T, then PQ is called the polar of T, and T is called the pole of PQ. To find the equation to the polar ofix'y'). Let (hk) be the point of contact of either tangent from (x'y r ). ^3° Analytical Geometry [260. Express that the tangent at (hk), viz. passes through (x'y') hx ky 1 — — = 1 a 2 + b 2 x'h /k a 2 b 2 "" * This equation expresses that (hk) lies on the straight line a 2 + b 2 ~ As foM points of contact lie on this line, it is .-. their join. The equation of the polar of (x'y') is .-. xx' yy' a 2 **" b* Cor' (1)— It is proved exactly as in § 234 that if P lies on the polar of Q then Q Zfey 0?z the polar of P. Cor' (2) — The polar of the focus (ae, o) is a xe = a, or x = - , e i. e. the corresponding directrix ; .\ tangents at the ends of a focal chord intersect on the directrix. Exercises 1. If CP, CD are conjugate semi-diameters, prove that SP.S'P = CD 2 . 2. Find locus of mid point of PD. Am. The ellipse 2 x 2 /a 2 + 2y 2 /b 2 = 1 3. Find locus of intersection of tangents at P, D. Am. The ellipse x 2 /a 2 + y 2 /b 2 = 2 4. CP, CD and CP', CD' are pairs of conjugate semi- diameters. Prove that A PCP' = A DCD'. 26o.] The Ellipse 231 5. CP, CD meet the tangent at A in V, W. Prove that AV.AW is constant. 6. The normal at P meets the major axis in G, and the diameter conjugate to CP in F. Prove that PG . PF = b 2 . 7. SJ, SJ' are perpendiculars from a focus on a pair of conjugate diameters: prove that JJ' meets the major axis at a fixed point. 8. The perpendiculars from S, S' on a pair of conjugate diameters meet at Q. Find the locus of Q. Ans. The ellipse b 2 y 2 + a 2 x 2 = a 2 (a 2 - b 2 ) 9. If the perpendicular from the centre on the tangent at P = p, and CP = r, prove that p 2 (a 2 + b 2 - r 2 ) = a 2 b 2 10. The normals at P, D meet in J. Prove that JC is perpendicular to PD. 11. The perpendicular from P on its polar meets the major axis in y ; n is the foot of the ordinate of P. Prove that C 7 = e 2 C n 12. V is the mid point of a chord PQ; prove that the polar of V is parallel to PQ. [Proceed as in Note, Ex. 9, page 197.] 13. Find the equation of the chord whose mid point is (hk). h k Ans. - ( x - h) + — (y - k) = o [Note — Use result of Ex. 12.] 14. Chords of the ellipse x 2 /a 2 + y 2 /b 2 - 1 are drawn through (x'y') ; find locus of their mid points. Ans. The ellipse x 2 /a 2 + y 2 /b 2 = xx'/a 2 + yy'/b 2 [See Note, Ex. 13, page 197.] 232 Analytical Geometry [261. EQUATION REFERRED TO CONJUGATE DIAMETERS §261. To find the equation of the ellipse referred to any conjugate diameters CP, CD as axes. Let CP = a', CD = b'. Let QVQ' be a chord || CD; so that x = CV, y = QV are the co-ord's of Q. Draw QN _L CA. Put also A PCA 0, A DCA = 0' V ) Project the broken line CVQ on CA and CB. .-. CN = xcos0 + ycos0' QN = x sin 6 + y sin 6' But CN 2 /a 2 + QN 2 /b 2 = 1 .-. (x cos 6 + y cos 0') 2 /a 2 + (x sin 6 + y sin Q'fjb' 1 = 1 _ /cos 2 e sin 2 d\ /cos 6 cos 6' sin 6 sin 0'\ „ /cos 2 6' sin 2 6'\ The coeff ' of 2 xy in this = o v tan tan 6' = - b 2 /a 2 [§ 255, CV' (3)] Coeff' of x' 2 = 1/CP 2 = i/a' 2 [§ 238] Coeff' of y /2 - 1/CD 2 = 1/b' 2 The required equation is .\ t ^t »'2 x^ y z + -=— = 1 a'* b /2 § 262 • As this equation is of the same form as that referred to CA, CB, several of the previous §§ are applicable when any pair of conj' diam / rs are chosen as axes. Thus the tangent at (x'yO is xx r yy' a 2 + b 2 ' 263.J The Ellipse 233 and as in § 248 we deduce that the tangents at Q, Q' meet CP in the same point t such that CV.Ct = CP 2 Again, lx + my = n is a tangent if a' 2 1 2 + b' 2 m 2 = n 2 ; &c. Ex. If the tangent at P meet any two conjugate diameters in t, V, prove that Pt.Pt' is constant. Take the conjugate diameters CP, CD as axes of co-ordinates. Let any two conj' diam'rs be y = nnx, y = m'x ; then .mm' = - b' 2 /a' 2 The eq'n of tangent at P is x = a' The co-ord's of t are determined by x = y .-. y, i.e. Pt = ma' Similarly PV = m'a' .-. Pt.Pt' = mm'a' 2 = - b /2 PARABOLA A LIMITING FORM OF ELLIPSE § 263. Referring to § 237 we may express a, b in terms of e and SX. For SX = CX-CS = a/e-ae = a(i - e 2 )/e .-. a = eSX/(i-e 2 ) Also b 2 = a 2 (1 - e 2 ) = e 2 SX 2 / (i - e 2 ) If now we suppose the focus S and the distance SX to remain fixed while e becomes 1, we see that a and b become infinite. The latus rectum p = 2 eSX and .*. remains finite. Thus the parabola is an ellipse whose axes are infinite, while its latus rectum is finite. = mx 1 234 Analytical Geometry [264. § 264*. The same conclusion may be deduced thus. If we change to || axes through the vertex A' the equation of the ellipse is (x - a) 2 /a 2 + y 2 /b 2 = 1 or y 2 /b 2 = 2 x/a - x 2 /a 2 or- y 2 = p X — px 2 /2 a (1) where p = 2 b 2 /a = latus rectum If a = 00 while p remains finite, (1) becomes y 2 = px § 265. Ex. 1. What does the property of § 248, I., viz. CN . CT = CA 2 become for the parabola ? Put CN = b .-. b {b + AT + NA) = (8 + NA) 2 ... b (AT - NA) = NA 2 .-. AT - NA = NA 2 /g Now put b = 00 ; .-. AT = NA (see § 225) Ex. 2. If in § 254 we suppose the focus S' to move off to infinity; we deduce the following property of the parabola — QTX' = STQ' [see fig', § 231] Thus if tangents are drawn from a point T to a parabola, one of them makes the same angle with the axis that the other does with ST. Exercises on the Ellipse. [Unless otherwise implied, in these questions the ellipse is referred to its axes of figure, its equation being x 2 /a 2 + y 2 /b 2 = 1 The notation of the figures in the chapter is retained throughout ; thus S, S' are the foci, G the foot of the normal at P, P and D extremities of conjugate diameters, &c] 1. Prove that SZ', S'Z meet at the mid point of PG. (See fig', p. 217.) 265,] Exercises on the Ellipse 235 2. If (f) is the angle between the tangents from (x'y'), prove that ( X '2 + y /2 _ a 2 _ fo2) tan ^ = 2 Vb 2 x' 2 + a 2 y' 2 - a 2 b 2 3. If OC, /3 are the eccentric angles of the extremities of a focal chord, then a /3 e - 1 tan — tan — = 2 2 e + 1 4r. Prove that the sum of the squares of the perpendiculars from P, D on a fixed diameter is constant. 5. Find the co-ordinates of the pole of the line x cos OC + y sin OC — p a 2 cos OC b 2 sin OC Ans. — — , P P 6. A chord PQ parallel to AB meets CA, CB in p, q. Prove that Pp = Qq Also, if m and n are the feet of the ordinates at P and Q, prove that 2Am.An = Ap 2 7. Given base of a triangle ABC and also tan \ A tan \ B ; show that locus of vertex C is an ellipse whose foci are A, B. [Note — By Trigonometry- tan \ A tan I B = (s - c) Is ; .'. s is given.] 8. P is any point on the curve; find locus of centre of circle inscribed in triangle SPS r . Ans. The ellipse (i - e) x 2 + (i + e) y 2 = e 2 (i - e) a 2 [Note — Use result of Ex. 62, p. 165.] 9. P is a point on the curve ; find locus of intersection of ordinate of P with perpendicular from centre on tangent at P. Ans. The ellipse a 2 x 2 + b 2 y 2 = a 4 10. Any tangent to the ellipse meets the director circle x 2 + y 2 .= a 2 + b 2 . in p, d ; show that C p, C d are conjugate diameters of the ellipse. ><$6 Analytical Geometry [Note — Let the tangent be [§ 250, Cor' (2)] x cos OC + y sin a = Va? cos 2 a + b 2 sin 2 a ; then the eq'n to the line-pair Cp, C d is (§ 119) (a 2 + b 2 ) (x cos a + y sin a) 2 - (a 2 cos 2 (X + b 2 sin 2 a) (x 2 + y 2 ) = o If the factors of this are y — mx, y — m'x then mm' = - b 2 /a 2 ; .-. &c. (§ 255, Cor' 3).] 11. P is any point on the ellipse ; the perpendiculars to AP, A' P at their mid points meet AA' in J, J'. Prove that JJ' = (a 2 - b 2 )/a [Note — This gives a method of describing the ellipse mechanically. See a paper by Professor Genese in Milne's Companion, p. 210.] 12. The tangents from P make angles 6, 6' with the major axis. Find the locus of P if tan0 tan0' = constant ( = k). Ans. y 2 — b 2 = k (x 2 — a 2 ) ; this equation represents an ellipse if k is nega- tive and an hyperbola if k is positive. 13. Show that normals at the ends of a focal chord intersect on a parallel to the major axis through the mid point of the chord. 14. Find locus of a point whose polars with respect to the ellipse x 2 /a 2 + y 2 /b 2 = 1 touch the circle x 2 + y 2 = r 2 Ans. The ellipse x 2 /a 4 + y 2 /b 4 = 1/r 2 15. An ellipse is cut by a series of parallel chords ; OC, /3 are the eccentric angles of the extremities of one of the chords. Prove that OC + f3 is constant. 16. Three sides of a quadrilateral inscribed in an ellipse are parallel to given lines. Show that the fourth side is parallel to a given line. [Note — Use result of Ex. 15.] 17. If the perpendiculars from the centre on the tangents a't P, D meet these tangents in K, K' and the ellipse in k, k', prove that 1 111 CK 2 .Ck 2 CK' 2 .Ck /2 a 4 b 4 Exercises on the Ellipse 237 18. If PD is inclined to the major axis at an angle (p, and b is the perpen- dicular from C on PD, prove that 2 8 2 = a 2 cos 2 (f) + b 2 sin 2 (f) 19. If - (/>' K /h 2 k 2 tan-?- T_ = AU + _ T 2 v a 2 b a 20. A parallel to PD meets CP, CD in p, d and the ellipse in q. Prove that qp 2 + qd 2 = PD 2 21. Any diameter meets parallels through P, D to any tangent in p, d, and the tangent in r ; prove that C p 2 + C d 2 = C r 2 22. Any line through the vertex A meets the ellipse in q and the minor axis in r ; CP is the semi-diameter parallel to A q. Prove that Aq.Ar = 2 CP 2 [Note — To determine A q use polar eq'n referred to A as origin.] 23. p is any point on the auxiliary circle; A p, A' p meet the ellipse in q, q'. Prove that Ap/Aq + A'p/A'q' = (a 2 + b 2 )/b 2 24. PN, PM are the perpendiculars from a point P on two given oblique axes; if PN 2 + PM 2 is constant (= k 2 ), show that the locus of P is an ellipse, and find its eccentricity. [Eq'n of locus is (x 2 + y 2 ) sin 2 co = k 2 ; this represents an ellipse referred to its equi-conjugates. The semi-axes a, b are determined by k 2 /sin 2 ft> = (a 2 + b 2 )/2, b/a = cot - ; .-. e 2 = 1 - cot 2 -.] 25. PCP r is any diameter of an ellipse. Any line through P cuts the ellipse in q and the tangent at P' in r; CQ is the semi-diameter parallel to P q. Prove that Pq.Pr = 4CQ 2 26. PCP', DCD r are conjugate diameters; O is any point on a circle concentric with the ellipse whose radius is r. Prove that OP 2 + 0P /2 + OD 2 + OD /2 = 4 r 2 + 2 (a 2 + b 2 ) c 238 Analytical Geometry 27. Show that the joins of the vertices of an ellipse to opposite ends of any segment of the directrix which subtends a right angle at the focus meet on the curve. 28. Points Q, Q' are taken on the normal at P such that PQ = PQ' = CD ; show that CQ = a - b, CQ' --= a + b \Note— If ecc' A of P is OC, deduce that co-ord's of Q, Q' are (a + b) cos oc, (b + a) sin a.] 29. TP, TQ are tangents from T. If PQ subtend a right angle at the centre, find the locus of T. [Note— Let (hk) be co-ord's of T. The eq'n to the line-pair CP, CQ is (§ 119) 'xh yk\ 2 _ x 2 y 2 /a 2 + bV ~ a 2 + b 2 These lines are at right angles [§ 114, Cor' (1)] if h 2 /a 4 + k 2 /b 4 = 1/a 2 + 1/b 2 The locus req'd is .*. the ellipse x 2 /a 4 + y 2 /b 4 = 1/a 2 + 1/b 2 .] 30. Find locus of intersection of normals at extremities of a chord parallel to one of the equi-conjugates. [Take equi-conj's as axes ; let eq'n to ellipse be x 2 + y 2 = c 2 . A If the axes are inclined at a) the normal at (x'y') is (x — x') (y' — x' cos to) = (y — y') (x' — y' cos to) This intersects normal at (— y! , y r ) on the line y + x cos to = o ; the locus is .'. the diameter _L other equi-conjugate.] 31. p is any point on the auxiliary circle; the tangent at p meets A A' in t. If pA, pA' meet the ellipse again in d, e, show that the chord de passes through t. 32. p is any point on the auxiliary circle; the perpendiculars from S, S' on the tangent at p meet the ellipse in q, q r . Show that pq. pq' are tangents to the ellipse. Exercises on the Ellipse 239 33. If 6 is the angle between the tangents from T show that 2 ST. S'T cos0 = ST 2 + S'T 2 - 4a 2 34. The tangent at P intersects parallels to the major axis at a distance b/e from it in q, r; prove that P q, P r subtend equal angles at the centre. 35. Deduce the following form of the equation of the tangents from (x'vO viz. a 2 (y - y') 2 + b 2 (x - x') 2 = (xy' - x'y) 2 by expressing the condition that the join of (xy), (x'y') should touch the ellipse. 36. Find the locus of a point such that; tangents being drawn from it to the ellipse, area of triangle which these form with major axis = area of triangle which they form with minor axis. Ans. The equi-conjugate diameters. 37. Two ellipses have the same focus and eccentricity and their major axes coincide in direction. If PN, P'N' are the ordinates of the larger ellipse which are tangents to the smaller, prove that SP - SN = SP' - SN' 38. Show that the normal at P passes through the mid point of the join of the feet of the perpendiculars from P on the equi-conjugate diameters. 39. Prove that the length of the chord joining the points whose eccentric angles are X, /3 h (X - 8 2 b' sin J 2 where b' is the parallel semi-diameter. 40. Find the area of the triangle formed by the tangents at the points whose eccentric angles are (X, /3, y. Ans. ab tan \ (JX - ft) tan j (/3 - y) tan | (y - (X) 41. If P, Q are the points of contact of tangents from T (hk) ; show that 'h 2 k 2 > Gp + p) sp - sq - sts {Note — If abscissa of P is x, SP = p ; then x = (p - a)/e (§ 340) 240 Analytical Geometry Elim' y between x 2 /a 2 + y 2 /b 2 = i, hx/a 2 + yk/b 2 = i ; then subst' preceding value of x. This gives a quadratic in p.] 42. Find locus of mid point of normal PG. Ans. The ellipse 4b 2 x 2 + 4(1 + e 2 ) 2 a 2 y 2 = a 2 b 2 (i + e 2 ) 2 43. Find locus of mid points of chords of the ellipse which touch the concentric circle x 2 + y 2 = r 2 Ans. (x 2 /a 2 + y 2 /b 2 ) 2 = r 2 (x 2 /a 4 + y 2 /b 4 ) 44. Two chords meet the axis major at points equi-distant from the centre ; if OC, (3 and y, b are the eccentric angles of their extremities, prove that tan — tan — tan — tan - = I 2222 45. Any chord PQ is drawn through the fixed point (d, o) on the major axis. If the abscissae of P, Q are x x , X 2 , prove that 2 d ( Xl x 2 + a 2 ) = (a 2 + d 2 ) (x L + x 2 ) Hence show that the product of the perpendiculars from P, Q on a fixed line parallel to the minor axis is constant. [Note—li \ = (a 2 + d 2 )/(2 d) the preceding eq'n gives (x x — A.) (x 2 — A.) = constant The line required is .*. 2 dx = a 2 + d 2 .] 46. P, Q 5 R are three points on the ellipse: the diameter bisecting QR meets PQ, PR and the curve in m, n, d : show that Cm.Cn = Cd 2 47. The three points P, Q, R on an ellipse are the vertices of a triangle whose area is a maximum. Show that QR is parallel to the tangent at P. [Note — By Ex. 10, page 216, A PQR is maximum when A pqr is maxi- mum. It follows by Geometry that A pqr is equilateral. The point P may Exercises on the Ellipse 241 be anywhere on the ellipse ; and if OC is its ecc' A , that of Q is OC + 277/3, and that of R is OC + 4 77/3. The required result follows from §§ 244, 246 ; or Geometrically.] 48. Any two conjugate diameters subtend angles A, A/ at a fixed point on the ellipse; show that cot 2 A + cot 2 A/ is constant. 49. Find the equation of the locus of the pole of a normal chord. Ans. a 6 /x 2 + b 6 /y 2 = a 4 e 4 50. Find locus of intersection of normals at the ends of a chord through the focus (ae, o). Ans. The ellipse a 4 (1 + e 2 ) 2 y 2 + a 2 b 2 (x + ae) 2 = a 3 b 2 e (1 + e 2 ) (x + ae) \_Note — This elimination is worked out in Nixon's Trigonometry, p. 158.] 51. Two vertices A, B of a given triangle ABC move one on each of two given lines OP, OQ ; show that the third vertex C describes an ellipse. [This may be solved geometrically thus. Let join of C to centre of circum circle of A ABO meet this circle in R and S. Then diam' of circle — AB/sin AOB, and is .*. given ; and we see that circle and points R, S are fixed relatively to triangle ABC. A A Hence ROB = RAB = a given angle; .*. line OR is fixed. We have then a line RS of constant length whose ex- tremities move on two fixed red/ axes OR, OS, and .. a fixed point on this line; .*. &c, by Ex. n, page 211. An analytical solution may be obtained; see Nixon's Trigonometry, p. 153.] 52. PCP', QCQ' are any two diameters; if OC, /3 are the eccentric angles of P, Q, prove that the area of the parallelogram formed by the tangents at P, P', Q, Q' is 4 ab/sin (a - /3) 53. P, Q are the points of contact of tangents from T (hk) ; prove that area of triangle CPQ = a 2 b 2 Vb 2 h 2 + a 2 k 2 - a 2 b 2 /(b 2 h 2 + a 2 k 2 ) and area of quadrilateral TPCQ = Vb 2 h 2 + a 2 k 2 - a 2 b 2 R 24-2 Analytical Geometry 54. A chord PR passes through the fixed point (hk) ; if CD, CD' are the semi-diameters conjugate to CP, CP', prove that DD' passes through the fixed / ak bh\ pomt (r"b'"aJ 55. Find locus of mid points of chords of constant length 2 c. x2 y2 c 2 (a 2 y 2 + b 2 x 2 ) i 2 + b 2 + a 4 y 2 + b 4 x 2 - * 56. PGQ, QG' are normals to an ellipse at P and Q, meeting the axis in G and G'. Prove that PG = projection of QG' on QP 57. K is the foot of the perpendicular from the centre on the tangent at P; from K another tangent is drawn to touch the ellipse in Q. Show that the eccentric angles 0, (f) of P and Q are connected by the relation 6 + 6 b 2 tan i = a 2 tan ; 2 and that the other extremity of the diameter through Q lies on the normal at P. 58. If PQ is a focal chord, and if R is the intersection of the tangent at P and the normal at Q, prove that QR is bisected by the minor axis. 59. P and P' are two points on the ellipse whose eccentric angles are + V y . * + - — ) = °> where u = -9 + lb - x V bu au/ a 2 b 2 64. Show that the locus of centres of equilateral triangles whose sides touch the ellipse is 9 (x 2 + y 2 ) 2 - 2 (5 a 2 + 3 b 2 ) x 2 - 2 (3 a 2 + 5 b 2 ) y 2 + (a 2 - b 2 ) 2 = o R 2 CHAPTER X THE HYPERBOLA EQUATION TO THE HYPERBOLA § 266. We have defined the hyperbola in § 213 ; viz. if P is a point on the curve then SP = ePM, S being the focus, PM the _L from P on the directrix KK'; and e (the eccentricity) > 1. To find its equation. Draw SX ± directrix. Take points A ; A' on SX such that SA = eAX (1) SA'^eA'X (2) The Hyperbola 245 Then A, A' are points on the curve. Bisect A A 7 in C ; let A A 7 = 2 a. Adding (1) and (2) we find 2CS = e.2CA .-. CS = ae Subtracting (1) from (2) we find 2 CA = e.2CX /. CX = a/e Through C draw BCB' J. CA ; and take CA, CB as axes. Let the co-ord's of P be CN = x, PN = y. The co-ord's of S are (ae, o). ... SP 2 = (x - ae) 2 + y 2 Also PM = CN - CX = x — a/e Now SP = e 2 PM 2 .-. (x — ae) 2 + y 2 = (ex — a) 2 ... x 2 (e 2 - i)-y 2 = a 2 (e 2 - 1) Divide by a 2 (e 2 — 1), and put a 2 (e 2 - 1) = b 2 The equation to the hyperbola is .\ x* y 2 a 2 "" b 2 " * If we put x = o in this equation, y = + b V — 1 .-. the hyperbola does not meet the axis of y in real points. It is however convenient to measure CB = CB' = b. Def's — The point C is called the centre, the line A A' the trans- verse axis, and the line B B' the conjugate axis. 246 Analytical Geometry [267. § 267. Put x = ae (= CS) in the equation X 2^ a 2 _ y 2^2 = j .-. y 2 = b 2 (e 2 - 1) = b 4 /a 2 .•• latns rectum (the double ordinate through focus) = 2 b 2 /a § 268. If (x'y') is a point on the curve, the points (-x', -/), (x', -y'), (-x>, /) are evidently on the curve. The curve is .*. symmetrical with respect to the centre (so that all chords through the centre are bisected) ; and also with respect to the axes. Since it is symmetrical with respect to the axis of y, there is another focus S' and a corresponding directrix kX r k', images respectively of S and KXK' with respect to the axis of y. FOCAL DISTANCES § 269. In fig', § 266, SP = ePM =e(CN - CX) = ex - a S'P = e PM' = e(CN + CX') = ex + a Co/— S'P-SP = 2a FIGURE OF THE CURVE § 270. Let Us to the axes through A, B meet in J and through A, B' in J'. Now if we write r cos 6, r sin 6 for x, y in the equation of the hyperbola, we obtain its polar equation, C being pole, viz. r 2 cos 2 Q r 2 sin 2 6 a 2 b 2 1 cos 2 sin 2 6 or pa = ~£a b 2 " From (1) we deduce = 1 (1) r 2 = a 2 b 2 a 2 b 2 b 2 cos 2 - a 2 sin 2 b 2 - (a 2 + b 2 ) sin 2 6 It follows from this that r = a when 6 — o; -7I-] The Hyperbola 247 and as 6 increases r increases ; and r = 00 when (a 2 + b 2 ) sin 2 = b 2 or A e = jca After 6 passes this value r 2 becomes negative, and .\ r imaginary. Remembering then the symmetry of the curve (§ 268) we see that it consists of two infinite branches inclosed between CJ, CJ' as in the figure. ASYMPTOTES § 271. Let the ordinate of a point P on the curve meet CJ in Q (fig', § 270). Let CN = x Then x 2 /a 2 — y 2 / b 2 = 1 y or PN = — Vx 2 — a 2 ....(!) Also, the equation to CJ is y = -x J a .-. QN=-x a (») 248 Analytical Geometry [272. Thus PN is always < QN ; and their difference b PQ = - (x - 7x 2 - a 2 ) a = ab/(x + Vx 2 - a 2 ) Thus PQ may be made < any assignable length, by taking x large enough. The curve .'. continually approaches the lines CJ, CJ', but never meets them. The lines CJ, CJ' are the asymptotes of the hyperbola. A general definition of asymptotes will be given later. Cor f — From (i), (2) we deduce QN 2 _ PN 2 = constant = b 2 RECTANGULAR HYPERBOLA § 272. If a = b the hyperbola is called rectangular or equi- lateral. Its equation is X 2 _ y2 _ a 2 Since b 2 = a (e 2 - 1), (§ 266) .-. e = V2 The semi-angle between the asymptotes, being tan -1 b/a is in this case 45 . The asymptotes of a rectangular hyperbola are .'. at right angles. § 273. Most of the work of the preceding chapter is applicable to the hyperbola, if throughout the proofs b 2 is replaced by — b 2 . Some of the results for the hyperbola are — The tangent at (x'y r ) is — _ yy 7 = a 2 b 2 T The polar of (x'y') is xx r yy / a 2 b 2 :73.] The Hyperbola 249 Whatever be the value of m, y = mx + )> (4> 3)« What is its eccentricity ? Ans. 3X 2 — 4y 2 = 12 ; \ V7 6. Two rods whose lengths are a, b slide along two rectangular axes in such a way that their extremities are always concyclic ; find the locus of the centre of the circle. Ans. The hyperbola 4 (x 2 — y 2 ) = a 2 — b 2 25o Analytical Geometry [274. 7. AOB, COD are two lines which bisect one another at right angles ; find the locus of a point P which moves so that PA . PB = PC . PD [Take the given lines as axes ; let AB = 2 a, CD = 2 b.] Ans. The rectangular hyperbola 2 (x 2 — y 2 ) = a 2 — b 2 8. x cos OC + y sin OC = p is a normal to x 2 /a 2 — y 2 /b 2 = 1 if p 2 (a 2 sin 2 OC - b 2 cos 2 OC) = (a 2 + b 2 ) 2 sin 2 OC cos 2 OC 9. P, Q, R are three points on a rectangular hyperbola. If the angle PQR is a right angle, prove that the normal at Q is parallel to PR. 10. A perpendicular from the centre on any tangent to the hyperbola x 2 — y 2 = a 2 meets the tangent in Z and the curve in Q ; prove that CZ . CQ - a 2 § 274*. The subjoined figure corresponds to that in § 248. All the pro- positions of that § are applicable to the hyperbola. (IV.) however is slightly modified : the normal to the hyperbola bisects the exterior A between SP, S'P and the tangent bisects A SPS r . 78.] The Hyperbola 251 CONJUGATE HYPERBOLA § 275. The hyperbola which has BB'for its transverse axis, and AA' for its conjugate axis, is called the conjugate hyperbola. Let p be a point on the new curve (%'§ 279); draw pn _L CB. Then Cn 2 /CB 2 - pn 2 / CA 2 = 1 The equation of the conjugate hyperbola is .\ y 2 /b 2 -x 2 /a 2 = 1 § 276. It should be noticed that the equations to the hyperbola, the conjugate hyperbola, and the asymptotes are x 2 y 2 , . *~h =I • • • • (I) x 2 v 2 z'-t?'- 1 (2) x 2 y 2 a»-F» -0 0) the sinisters being identical. § 277 '. Let a line through C inclined at 6 to the axis of x meet the con- jugate hyperbolas in P, P f . We shall show that if P is real, P r is imaginary and vice versa. The polar equations (see § 270) give 1 _ cos 2 6 sin 2 I sin 2 cos 2 9 CP 2 = ~a? W CP 72 = "b 2 a 2 " .-. CP /2 = - CP 2 .-. CP'=CPV~T7 Note — If the red/ co-ord's of P are (xy), those of P' are (x\/-i,yV-i). DIAMETERS § 278. The locus of mid points of chords parallel to y = mx is (§ 2 55) y = m'x where a 2 mm' = -5 (1) This is the condition that y = mx, y = m'x may be conjugate diameters. %$% Analytical Geometry [279. It follows from (1) that if m < b/a then m' > b/a. But the semi-angle between the asymptotes is tan -1 b/a. It is .*. evident that of two conjugate diameters one only meets the curve in real points. (1) may be written \m) W/ = b* But this is the condition that y = mx, y = m'x should be conjugate with respect to y 2 x- 2 _ b 2 ~ a 2 ~~ * Accordingly two lines which are conjugate diameters of a hyperbola are con- jugate diameters of the conjugate hyperbola. § 279. If the diameter conjugate to a diameter PCP' meet the conjugate hyperbola in D, D', then the tangents at R, D form a parallelogram with CP, CD, of which one of the asymptotes is a diagonal. 79-] The Hyperbola %53 We may assume for the co-ord's of P x = a sec (j) ) y = b tan (p ) For these values satisfy x 2 y 2 a 7 . b 2 == * Then the co-ord's of D are x = a tan (b ) . . , XI ( 2 ) y = b sec

— t: sec = - 1 . . . • (6) (a~b) (seC + tan( ^ = o 254 Analytical Geometry [279. or (7) a"~b .-. the tangents (5) and (6) meet on the asymptote (7). Cor' (1)— CP 2 - CD 2 = a 2 - b 2 For from eq'ns (1), (2) CP 2 = a 2 sec 2 (f> + b 2 tan 2 (p CD 2 = a 2 tan 2 <|> + b 2 sec 2 (J> .-. CP 2 - CD 2 = (a 2 - b 2 ) (sec 2

— a tan (j) . b tan + tan(|>), y = — (secc/) + tan<|)) 2 2 which lies on the asymptote *~ y a b Also, the equation to PD is x y 1=0 a sec (f) b tan <|> 1 a tan (j) b sec (j) 1 or (bx + ay) (sec (f) — tan <£) = ab which is || the asymptote - + £ = o a b Cor' (4) — The portion of a tangent intercepted by the asymptotes is bisected at the point of contact. For P I = CD (Euclid I. 34) = CD' - Pm .'. Im is bisected in P 280.] The Hyperbola Z55 EQUATION REFERRED TO ASYMPTOTES § 280. To find the equation of a hyperbola referred to its asymp- totes as axes. Let Cd = x, Pd = y be the coord's of a point P on the curve referred to the asymptotes CJ', CJ. Draw PN _L CA ; let JCA = )8. Project the broken line CdPonCA, CB. .-. CN = xcos/3 + y cos/3 PN = y sin /3 - x sin /3 But CN 2 /a 2 - PN 2 /b 2 = i .-. (x + y) 2 cos 2 /3/a 2 - (y - x) 2 sin 2 /3/b 2 = i . . (i) But tan ft = b/a and .-. cos 2 /3 = a 2 /(a a + b 2 ), sin 2 /3 = b 2 /(a 2 + b 2 ) .*. (i) becomes 4Xy = a 2 + b 2 This is the required equation. (for' (i)— If the co-ord's of P (fig' § 279) referred to the asymptotes are (x, y), those of D are (- x, y). [§ 279, Cor' (3)]. The equation to the conjugate hyperbola (the locus of D) is .'. 4 xy = - (a 2 + b 2 ) Cor' (2)— If p lf p 2 are the _Ls from P on CJ', CJ (vide fig / ) then p x = P d sin 2 /3 = y sin 2 j3 256 Analytical Geometry [281. Similarly p 2 = x sin 2 (3 .-. p x p 2 - xy sin 2 2 /3 = constant, since xy is constant. Thus the product of the perpendiculars from any point of a hyperbola on its asymptotes is constant. Cor' (3) — Hence the equation to a hyperbola whose asymptotes are Ax+By+C = o, A'x + B'y + C = o Ax + By + C A'x + B'y + C is — — — constant VA 2 + B 2 VA' 2 + B' 2 or (Ax + By + C) (A'x + B'y + CO = k where k is some constant. The equation of the hyperbola conjugate to this is (A x + B y + C) (A'x + B'y + C) = - k The equation of the asymptotes is of course (A x + B y + C) (A'x + B'y + C) = o Thus, whatever the axes of co-ordinates may be, the equations of two conju- gate hyperbolas differ from that of the asymptotes only by constants, whose values are equal and opposite for the two hyperbolas. § 281. To find the equation of the tangent at (x'y') to the hyperbola 4 xy = a 2 + b 2 It is convenient to put (a 2 + b 2 )/4 = k 2 , so that the equation to the hyperbola is __ . 2 If (x'y'), (x"y") are two points on the curve, the equation to their join is x (y' - y") - y (x r - x") + x' y" - x"y' = o Also x'y' = x"y'' = k 2 If we write the equation to the join in the form a + /3 s8i.] The Hyperbola *57 then Oi = _ x'Y - x'y" / - V x" k 2 /x' - x' k 2 /x' k 2 /x'-k 2 /x" x //2 -x /2 x"-x' = x' + x" Similarly j3 = y' + y" The eq'n to the chord joining (x'y'), (x"y") is .*. y x' + x^y' + y""" 1 ' • • * ' (0 Put now x" = x', y" = y' ; the equation to the tangent at (x'y') is .-. to x . y — + —, = 2 x' y' Cor' (i) — 7^" a«y /z«2 cut the hyperbola in Q, Q' and the asymptotes in q, q', then Qq = Q'q'. Let Q be (x'y') and Q' (x"y"). The chord QQ' [eq'n (i)] meets the axes in the points q' (x' + x", o) and q (o, y' + y") Thus QQ' and qq' have the same mid point, viz. / x' + x" y' + y" \ :. &c. C = < (1 — e 2 cos 2 a) (1 — e 2 sin 2 a) i. e. as e 2 > = < 1 Note — It is obvious geometrically that if a and b are very small the form of the hyperbola approximates to that of its asymptotes. Putting a = o, b = o in the equation referred to the asymptotes, viz. 4xy = a 2 + b 2 , it reduces to xy = o Thus a line-pair is a self-asymptotic hyperbola. Exercises on the Hyperbola 1. A circle meets the hyperbola xy = k 2 in four points (x x y x ), (x 2 y 2 ) . . . ; prove that x 1 x 2 x 3 x 4 = k 4 = y iy2 y 3 y 4 2. PQ is a chord of a rectangular hyperbola, normal at P; prove that PQ oc CP 3 3. Any tangent meets the asymptotes in L, \J ; show that L, L/, S, S' are concyclic. 286.] The Hyperbola 263 4:. Show that the polar of any point T is parallel to the joins of the points in which the tangents from T meet the asymptotes. 5. Prove that the circles described on parallel chords of a rectangular hyperbola as diameters are co-axal. 6. Q is any point on a rectangular hyperbola; PCP' is any diameter. Prove that the bisectors of the angle PQP' are parallel to the asymptotes. 7. A chord of a rectangular hyperbola subtends angles at the extremities of any diameter which are either equal or supplementary. 8. Find the equation of the tangents from (x'y') to the hyperbola xy = k 2 Ans. 4 ( X y - k 2 ) (xy - k 2 ) = (x'y + y'x - 2 k 2 ) 2 ; or (x'y — y'x) 2 + 4 k 2 (x — x') (y — y') = o 9. Show that the director circle of the hyperbola xy = k 2 is x 2 + 2 xy cos a) + y 2 = 4 k 2 cos 15. Find locus of poles of normal chords of the hyperbola x 2 - y 2 = a 2 Ans. The curve a 2 (y 2 — x 2 ) = 4 x 2 y 2 16. Show that any tangent to the hyperbola x 2 — y 2 = 2 a 2 is cut harmonically by the two hyperbolas x 2 + xy = a 2 , y 2 + xy = — a 2 17. A line through the centre C of an hyperbola meets the curve in P, and parallels to the asymptotes through the vertex A in T and T 7 . Prove that CP 2 = CT.CT' {Prof Genese). CHAPTER XI GENERAL EQUATION OP THE SECOND DEGREE § 287. The general equation of the second degree ax 2 + 2hxy + by 2 + 2gx + 2fy + c = o . . . (i) represents a conic. We shall suppose the axes rectangular ; for an equation referred to oblique axes may be transformed to rectangular axes (§ 207) and its degree is unaltered (§ 210). The equation to a conic whose focus is (XjyJ, directrix x cos Oi + y sin a — p = o and eccentricity e, is (def § 213) (x — x x ) 2 + (y — yi) 2 = e 2 (x cos oc + y sin a — p) 2 . (2) Comparing coeff's, this will represent the same locus as (1) if 1 — e 2 cos 2 06 — e 2 cos 06 sin 06 _ 1 — e 2 sin 2 06 a h b e 2 p cos 06 — x t e 2 psin 06 — y x _ x t 2 + y? — e 2 p 2 . , - - - f - c ■ (3) We have thus five equations, which are sufficient to determine the five quantities x lt y lt 06, p, e. Better practical methods of determining the particular conic represented by (1) will be given presently. Cor' — Let e = 1 so that the curve is a parabola. %66 Analytical Geometry [288. The first two of equations (3) become sin 2 oc _ — cos a sin a cos 2 at a h b .-. ab = h 2 Thus if the general equation (1) represents a parabola, its highest terms ax 2 + 2 hxy + by 2 form a perfect square. § 288. If the terms of the first degree are absent, the equation is ax 2 + 2 hxy + by 2 + c = o If (x'y') is a point on this curve, (— x', — y') is also evidently a point on the curve. Every chord through the origin is .*. bisected ; or the origin is the centre. DETERMINATION OF CENTRE § 289. If we transform the equation ax 2 + 2 hxy + by 2 + 2gx + 2fy -f c = o to || axes through (x'y') it becomes (§ 204) a(x + x') 2 + 2h (x + x')(y + yO + b(y + y') 2 + 2g(x + x0 + 2f(y + /)+ c = o ... (a) or ax 2 + 2 hxy + by 2 + 2 x (ax 7 + hy' + g) + 2y(hx' + by' + f)+c/=d . . . (/3) where c' = ax' 2 + 2 hx'y' + by' 2 + 2gx' + 2fy' + c . . (y) The new origin is the centre (§ 288), if the coeff's of x and y are zero, i. e. if ax' + hy' + g = o hx' + by' + f = o These equations determine the centre. } (8) 290.] General Equation of the Second Degree 267 Solving them, the co-ord's of the centre are __ hf-bg _ gh -a f * ab - h 2 ' y " ab-h 2 These values are infinite if ab = h 2 ; this case will be con- sidered separately. Again, the equation (/3) becomes now ax 2 + 2 hxy + by 2 + c' = o The value of c' [equation (y)] may be written c r = x'(ax' + h/ + g) +. y'(hx' + by'+ f) + gx' + f/ + c Hence, using (8) we get c' = gx' + fy' + c (e) ab — h 2 ab — h 2 abc 4- 2fgh — af 2 — bg 2 — ch 2 ab- h 2 A [§ "8. ab-h 2 Note — The preceding investigation holds if the axes are oblique. The equations (§) and (e) should be remembered. It is useful to notice that the coeff r s in (8) and (e) are the letters in order which occur in the determinant A = a h g h b f g f c The student who is acquainted with the Differential Calculus will remark that the sinisters of (8) are half the derived functions of 4> (x', /) = ax /2 + 2 hx'y' + &c. with respect to x r and y 7 . POINTS AT INFINITY § 290. If we substitute r cos 6, r sin 6 for x, y, the equation ax 2 + 2 hxy + by 2 + 2gx + 2fy + c = o . . (i) 268 Analytical Geometry [291. becomes r 2 (a cos 2 # + 2 h cos 6 sin 6 + b sin 2 0) + 2 r(g cos # + fsin#)+c = o One root of this quadratic in r is infinite (§ 282) if a cos 2 + 2 h sin 6 cosO + b sin 2 # = o or a + 2 h tan 6 + b tan 2 6 = o This equation determines the lines through the origin which meet the curve at infinity. Substituting y /x for tan 0, we see that the equation to these lines is ax 2 + 2 hxy + by 2 = o (2) This equation .*. represents two lines parallel to the asymptotes. [This also follows evidently from § 280, Co/ (3).] (1) If h 2 — ab > o, the lines (2) are real; the curve (1) is .\ an hyperbola. (2) If h 2 — ab < o, the lines (2) are imaginary. No real line through the origin meets the curve at infinity, or the curve is bounded in all directions. It is .'. an ellipse. (3) If h 2 — ab = o the lines (2) which meet the curve at infinity coincide. The curve is a parabola (§ 287, Cor') S 291. Two roots of the quadratic in r are infinite, if also g = o and f = o The asymptotes of ax 2 + 2 hxy + by 2 + c = o are .\ ax 2 + 2 hxy + by 2 = o This also follows evidently from § 280, Cor' (3). Note — The student will observe that an ellipse has imaginary asymptotes. Thus the asymptotes of x 2 /a 2 + y 2 / b 2 = 1 are x 2 /a 2 + yyb 2 = o, or x/a + yV-i/b = o CENTRAL CONICS § 292. We resume the discussion of the reduced equation of §289, viz. ax 2 + 2hxy + by 2 + c r =o . . . . (i) 292.] General Equation of the Second Degree 269 A Let us turn the axes through 0: we must .-. substitute (§ 205) x cos — y sin for x and x sin + y cos for y The equation becomes a(xcos#— y sin Of +2 h(x cos 0—y sin 0) (xsin # + ycos#) + b(xsin# + ycos#) 2 + c' = o . . (2) The coeff of xy in this is — 2a cos sin + ah (cos 2 # — sin 2 0) + 2b cos 0sin 0* Put this = o ; .-. tan 20= 2h/(a- b) (3) If we give a value which satisfies this, (2) assumes the form ax 2 + |3y 2 + c'=o (4) This represents a conic referred to its axes, unless c' — o, when it represents two straight lines. The values of OC, /3 are easily determined thus. Equate the invariants of (i) and (4), (§ 212) .-. a + /3 = a + b 1 a/3 = ab - h 2 ) .'. OL, /3 are the roots of the quadratic in t t 2 - (a + b) t + ab - h 2 = o Note — This method of determining OC, (3 leaves it uncertain whether the reduced equation is (Xx 2 + /3y 2 + c' = o or /3x 2 + ay 2 + c' = o; this must be settled by other considerations. Another method, which is free from ambiguity, is described in the next §. * This eq'n may be written h (1 - tan 2 6) = (a - b) tan Writing y/x for tan 6, we obtain the equation of the axes h (x 2 - y 2 ) = (a - b) xy This equation is obtained otherwise in § 293. 270 Analytical Geometry [293. § 293. The asymptotes of ax 2 + 2 hxy + by 2 + c' = o . . . . (1) are (§ 291) _ , „ v * y ; ax 2 + 2 hxy + by 2 = o The axes bisect the angles between the asymptotes ; the equa- tion to the axes is .*. (§116) h (x 2 — y 2 ) - (a - b) xy = o . . . . (2) Let the factors of this be y — Ax = 0, y — jjlx = o ; the axes are . . x / ■> y = Ax (3) y = />tx ........ (4) If the equation of the conic referred to its axes is x 2 y 2 • 1 '2 r 2 and r 2 are easily obtained by finding the points where (1) meets (3) and (4). Thus at the points of inters'n of (1), (3) x 2 (a + 2 h A + b A 2 ) + c' = o .-. r* = x 2 + y 2 = x 2 + A 2 x 2 = - & (1 + A 2 )/(a + 2 h A + b A 2 ) § 294". A simpler expression for r x 2 may be found thus : — (2) may be written y (ax + hy) = x (hx + by) /. y 2 (ax + hy) = xy (hx + by) Add to both sides x 2 (ax + hy) ; .-. (x 2 + y 2 ) (ax + hy) = x (ax 2 + 2 hxy + by 2 ) .-. r x 2 (ax + hy) = — c'x, by (1) Substitute in this A x for y ; then divide by x .-. r 2 2 (a + h A) = - c' ... ri 2 = _ c y(a + h A) Similarly r 2 2 = — cy(a + h fi) The equation of the conic referred to its axes is .*. x 2 (a + h A) + y 2 (a + h /a) + c' = o 295-] General Equation of the Second Degree 271 THE PARABOLA § 295. We shall now consider the case when h 2 = ab. As ax 2 + 2 hxy + by 2 is a perfect square, the equation is of the form (ax + /3y) 2 +2gx+2fy + c = o This may be written (ax + /3y+k) 2 = 2 x(ak-g) + 2y(/3k-f) + k 2 -c . (1) Now let Y, X denote the distances of a point from the lines ax + /3y + k = o (2) 2 X (ak-g) + 2 y(jSk-f) + k 2 -c = o . . (3) Then (1) expresses that Y 2 varies as X Thus whatever k is, the line (2) is a diameter and (3) the tan- gent at its vertex. The lines (2), (3) are at right angles if a(ak-g) + /3(/3k-f) = o ... k = (ag + /3f)/(a 2 + /3 2 ) Introducing this value of k, (2) or Y = o is the axis ; and (3) or X = o is the tangent at the vertex. Now Y = (ax + /3y + k)/Va 2 + 8* and X = [ 2X (ak - g) + 2 y (/3k - f) + k 2 -cJ/A/ 4 (ak-g) 2 + 4(/3k-f) a Thus (1) becomes [ Y v / a r +^8 2 ] 2 = 2 X A/(ak-g) 2 + 03k- f) 2 or Y 2 = pX . (4) where p = 2 V( Expanding this, A(ab— h 2 ) + A=o Accordingly, if the conic (i) is denoted by S = o, its asymptotes are S = A/(ab - h 2 ) Cor r — The equation to the conjugate hyperbola is [§ 280, Cor' (3)] S + 2 A = o or S = 2 A/(ab - h 2 ) INTERSECTIONS WITH CONCENTRIC CIRCLE. § 299. The following is another method of determining the axes of the conic ax 2 + 2hxy -f by 2 = i (i) At the intersections of the conic and the circle x* + y* = r 1 (2) we obtain, by dividing (2) by r 2 and subtracting the result from (1) (a-p)x 2 + 2hxy + (b-p)y 2 = o . . (3) We see, as in § 119, that (3) is the equation of the common diameters of the conic and circle. If CP, CQ are two common semi-diameters, then since CP = CQ= r, CP and CQ are equally inclined to the axes of the ellipse. Hence P, Q coincide only at the extremities of an axis. If the lines (3) are coincident (a-i- 2 )(b-i)=h° (4) The squares of the semi-axes are the values of r 2 deduced from (4). As (3) is now a perfect square, if we multiply (3) by a — 1/r 2 and take the square root, we find (a - p) x + hy = o This is the equation of the axis whose length is 2 r. In working numerical examples the learner may either use this method, or that of §§ 293, 294. 274 Analytical Geometry [300. § 300. Ex. 1. Trace the conic 5 x 2 + 6 xy + 5 y 2 — 16 x— i6y + 8 = o Here h 2 — ab = 9 — 25 = — 16 ; the curve is .\ an ellipse (§ 290) The equations to find the centre are (§ 289) 5X + 3y- 8 = o ) 3X + 5y- 8 = o J ... x = 1, y = 1 Next, c' = (-8)(i) -8 (1) + 8 =- 8 The equation referred to || axes through centre is (§ 289) 5x 2 + 6xy + 5y 2 -8 = o The axes are (§ 293) 3X 2 - 3y 2 = o or y = — x and y = + x The equation of the conic referred to its axes is (§ 294) x 2 (a + h A) + y 2 (a + h /x) + c' = o i. e. or 2X 2 + 8y 2 -8=o x 2 + 4y 2 = 4 [§ 289, (€).] 300.] General Equation of the Second Degree 275 Ex. 2. Trace the conic x 2 - 4 xy + 3 y 2 - 4 x + 7 y + ¥ = ° h 2 - ab = 1 ; ::i Here the conic is .*. an hyperbola. For the centre x — 2 y — 2 - 2x + 3y + I ... x = 1, y = - I Next, c / = _ 2 -i + ^=2 The equation referred to || axes through centre is x 2 — 4 xy + 3 y 2 + 2 =0 The axes are — 2 (x 2 — y 2 ) — (— 2) xy = o Solving this quadratic for y/x the axes are y/x = (- 1 ± V5)/2 The equation referred to the axes is (§ 294) x 2 (2 - V5) + y 2 (Vs + 2) + 2 = o x 2 (V5-2)-y 2 (V5 + 2) = 2 or or (2- 9 ) 2 (-6 3 ) 2 = I y \ / 1 / f I / f y i />C"^" X Xl^^'" ^ 1 / T 2 276 Analytical Geometry [300. Ex. 3. Trace the parabola (3X + 4Y) 2 - 8x + i56y - 36 = o Here (3 x + 4y + k) 2 = x (6 k + 8) + y (8 k - 156) + k 2 + 36 Let Y, X be the distances of (x, y) from the lines 3x + 4y+k = o x (6 k + 8) + y (8 k - 156) + k 2 + 36 = o . . . These lines are at right angles if 18 k + 24 + 32 k — 624 = o, .'. k = 12 (1) becomes 3x + 4y+i2=o (2) becomes 80 x — 60 y + 180 = 0, or 4X — 3y + 9 = o . . •■• Y = (3X + 4y + i2)/5, X = ( 4 x - 3y + o)/5 .-• (5Y) 2 = 20 (5 X) .-. Y 2 = 4 X The line (3) is the axis ; (4) is tangent at vertex. (0 (2) (3) (4) 3oo.] General Equation of the Second Degree 277 Exercises 1 . Find the centres of the conies i°, x 2 + xy + y 2 - 3X + 3y 2 , xy — /3 x — OC y = o 3°, 5X 2 + 4xy + 8y 2 — 18 x - 36 y + 9 = o 4°, 4X 2 — 24xy + ny 2 + 40 x + 30 y — 105 = o Ans. (1,1); (0C,/3); (1, 2); (4,3) 2, Find the equations of the preceding conies referred to parallel axes through the centres. Ans. i°, x 2 + xy + y 2 = 3 2 , xy = OK/3 3°, 5X 2 + 4xy + 8y 2 = 36 4°, 4X 2 — 24 xy + 11 y 2 + 20 = o 3* Trace the preceding conies. Determine the inclination of the greater axis of each conic to the axis of x, and the lengths of its semi-axes. Ans. i°, An ellipse; 135 ; \/6, */2 2 , An hyperbola ; 45 ; V2 0CJ3, V2 a/3 3 , An ellipse ; tan" 1 (- £) ; 3, 2 4 , An hyperbola; tan -1 (f) ; 2, 1 4. Find the equation of the ellipse 2x 2 + y 2 — 2xy — 2 x = o when referred to its axes. Ans. (3 - V5) x 2 + (3 + \/5)y 2 = 2 5. Trace the following parabolas; and determine for each the equations of the axis and the tangent at the vertex. Obtain also the equation of each para- bola referred to its axis and tangent at vertex. l0 > (3 x + 4 y) 2 + 22 x + 46 y + 9 = o 2 , 4 (x - y) 2 = 4 (x + y) - 1 3 , (x - 2 y) 2 - 2 (x + 2 y) + 1 =0 Ans. i°, 3X + 4y + 5 = o, 4X - 3y + 8 = o; 5y 2 = 2x 2 , x — y = o, 4 (x + y) — 1 = o ; y 2 V2 = x 3 , 5X - ioy + 3 = o, iox + 5y = 2; 5 Vsy 2 = 8x 278 Analytical Geometry [30c 6. Show that the equation V x/a + vy/b = * represents a parabola ; and find its latus rectum. Ans. 4a 2 b 2 /(a 2 + b 2 )* 7. Find the latus rectum of the parabola (a 2 + b 2 ) (x 2 + y 2 ) = (bx + ay - ab) 2 Ans. 2 ab/Va 2 + b 2 8. Trace the hyperbola y = x + a 2 /x Show that the product of its semi-axes is 2 a 2 . 9. Find the equation to the hyperbola which passes through the point (1, 2) and whose asymptotes are x + 2y— 1=0, 3x — y + i=o Find also the equation of the conjugate hyperbola. Ans. 3X 2 + 5xy — 2y 2 — 2X + 3y — 9 = 0, 3x 2 + 5xy — 2y 2 — 2x + 3y+ 7 = 10. Find the centre of the conic 5X 2 + 11 xy + 2y 2 — 13 x +■ ioy — 28 = o Ans. (- 2, 3) \Note — This conic is a pair of lines whose intersection is the centre.] 1 1 . Find the asymptotes of the hyperbola 6 x 2 — xy — y 2 — x + 3 y + 2 =0 Find also the equation of the conjugate hyperbola. Ans. 2X — y + 1 = o, 3X + y— 2=0; 6 x 2 — xy — y 2 — x + 3 y — 6 = 12. Trace the conies 2x 2 = 3X + 4y+5, y 2 =3X + 4y, (x - y) 2 = a 2 - x 2 13. What curve is represented by the equation (6 x + 2 y — i) 2 + (3 x — 9 y + 2) 2 = 360 ? Ans. An ellipse, the lengths of whose semi-axes are 3, 2 ; and their equations are 3X — 9y + 2 = o, 6x + 2y— 1=0 302.] General Equation of the Second Degree 279 14. Find the equation of the axes of the ellipse 2 x 2 + y 2 — 2 Ans. x 2 + xy — y 2 — 4 x + 3 y — 1 =0 2 x 2 + y 2 = 2 xy + 2y EQUATION OF TANGENT § 301. We shall sometimes use abbreviations for the expression ax 2 + 2 hxy + by 2 + 2gx + 2fy + c It may be denoted by S, or by (p (x, y). Thus S = o means the conic ax 2 + 2 hxy + by 2 + 2gx + 2fy + c = o Of course (f) (x', y f ) means the result of substituting x r , y r for x, y in (x, y) : or ( x '> y') E a*' 2 + 2 hx'y' + by' 2 + 2 gx' + 2f/ + c § 302. To find the equation of the tangent at (x'y') to the conic S = o Let (xy), (x"y") be two points on the curve. Consider the locus represented by a(x-xO(x-x r/ ) + 2h(x-x')(y-y /r ) + b(y-/)(y-y") = ax 2 + 2 hxy + by 2 + 2gx + 2fy + c . . . (1) The terms in x 2 , xy, y 2 cancel ; so that the equation is of the first degree and .*. represents a straight line. If we substitute x', y' for x ; y in the sinister of (1) it vanishes identically. If the same substitution is made in the dexter, it vanishes also, since (x'y') is on the curve. Thus (x'y') is a point on (1). Similarly (x^y'O is a point on (1). Hence (1) is the equation of the chord joining (x'y'), (x"y"). Now put x" = x', y" = Y ; the equation of the tangent at (xy) is .-. a(x-x') 2 + 2h(x-x')(y-/) + b(y-/)' 2 = ax 2 + 2 hxy + by 2 + 2gx + 2fy + c 280 Analytical Geometry [303- or, expanding and cancelling 2 ax'x + 2 h (y'x + x'y) + 2 by'y + 2gx + 2fy + c = ax' 2 + 2hx'y' + by /2 This may be simplified : add 2 gx' + 2 fy' + c to both sides ; then dexter vanishes, since (x'yO is on the curve. The equation of the tangent at (x'y') is .*. ax'x + h (y'x + x'y) + by'y + g (x + x') + f (y + y r ) + c = o In the preceding investigation (due to Prof W. S. Burnside) the axes may be oblique or rectangular. Investigations which are applicable when the axes are oblique will be indicated in future by the sign (Qi). The equation of the tangent at (x'y') may be remembered thus : — Imagine equation of conic written as follows axx + h (xy + xy) + byy + g (x + x) + f (y + y) + c = o ; then dash one letter in each term. DIAMETERS § 303. To find the locus of the mid points of chords of S = o which are parallel to y = mx. We proceed as in § 228. Let (x'y'), (x^y") be the extremities of one of the chords; (xy) its mid point. Then 2X = x'+ x", 2 y = y' + y", m = (/ - y")/(x' - x") . (1) Now since (x'y'), (x"y") are on the curve .-. ax /2 + 2hx'y'+ ••• = o, ax" 2 + 2hx"y" + ... = o Subtract these equations ; .-. a (x /2 - x //2 ) + 2 h (x'y' - x"y'') + ■ • • = ° This may be written a(x' + x")(x'-x") + h[(x'+x") (/-/') + (x'-x") (y'+y")] + b(y , + y /, )(/-y ,/ ) + 2g(x / -x // ) + 2f(y / -y / = o 305.] General Equation of the Second Degree 281 Divide this by x' — x" ; then from (1) we see that ax + h (mx + y) + bmy + g + f m = o or ax + hy + g + m (hx + by + f) = o . . . (2) This is the equation required. This proof is due to Prof J. Purser. Cor' — If (2) is parallel to y = m'x, then m' = — (a + hm)/(h + bm) .-. bmm' + h (m + m') + a = o (12) This is the condition that y = mx, y = m'x may be parallel to conjugate diameters. § 304*. To find the condition that the lines a'x 2 + 2h 7 xy + b'y 2 = o (0 may be conjugate diameters of the conic ax 2 + 2hxy + by 2 + c = o . . . . . . (2) Let a'x 2 + 2 h'xy + b'y 2 = b'(y — mx) (y — m'x) .-. m + m'= — 2h / /b / , mm' = a'/b' But y - mx = o, y — m'x = o are conjugate if bmm' + h (m + m') + a = o [Co/, § 303.] ba' 2hh' •''¥-T + a = ° The required condition is .-. ab'+ a'b- 2 hh'=o (12) Note — This equation expresses (§ 141) that the lines (1) are harmonically conjugate to the asymptotes of the conic (2). We have thus another proof of the proposition of § 281, Cor' (3). § 305. Ex. 1. Find the equation to the axes of the conic ax 2 + 2 hxy + by 2 = 1 (12) Let the required equation be a'x 2 + 2 h'xy + b'y 2 = (1) 282 Analytical Geometry [306. Since the axes are conjugate, .-. ba' — 2 hh' + ab' = o . And since the axes are at right angles .-. a! — 2 h' cos a) + b' = o From (2) and (3), a' — 2 h' b (2) (3) b cosco h — a cos a) a — The required equation is .*. (h — a cos to) x 2 + (b — a) xy + (b cos co — h) y 2 = o Note — This may be expressed otherwise. Eliminate linearly a', — 2 h/, W from (1), (2), (3). The equation to the axes is .*. — xy y 1 h a = o b I COSO) Ex. 2. To find the equation to the common conjugate diameters of the conies ax 2 + 2 hxy + by 2 = 1 , a'x 2 + 2 h'xy + b'y 2 = 1 . . . (X2) Let the required equation be a"x 2 + 2 h"xy + b"y 2 = o (1) .-. ba'' - 2 hh'' + ab" =0 (2) and b r a" - 2 WW + a'b" =0 (3) Eliminate a", — 2 W, b" from (1), (2), (3). This gives the required equation, viz. x 2 — xy y 2 = o b h a b r W a! This proves that any two concentric conies have a common pair of conjugate diameters; and the equations of two such conies can .*. be expressed in the forms ax 2 + by2 = I} a / x 2 + b / y2 = ! POLES AND POLARS § 306. It is proved, exactly as in § 168 or § 233, that the polar of (x' y r ) with respect to S = o is ax'x + h (y r x + x'y) + by r y + g(x + x') + f (y + /) + o = o . . . . (Q) This is of the same form as the equation of the tangent. 3o8.] General Equation of the Second Degree 283 The preceding equation may be written x (ax' + hy' + g) + y (hx' + by' + f ) + gx' + fy' + c = o Cor' (1) — Put yf — o, y' = o : the polar of the origin is .\ gx + fy + c = o (12) Cor' (2) — It is proved as in § 172 that if P lies on the polar of Q, then Q lies on the polar of P. Cor' (3) — If the polars of A and B meet in C, then C is the pole of A3. This is proved as in § 1 74. Cor' (4) — If a line revolve round a fixed point P ; then its pole describes a straight line, viz. the polar of P. This is proved as in § 173. § 307. The tangents at the extremities of a diameter are parallel to the conjugate diameter ; the pole of a diameter is .*. the point at infinity on the conjugate diameter. The polar of the centre is [§ 306, Cor' (3)] the join of the poles of any two lines through the centre. Hence the polar of the centre is the line at infinity. This is otherwise evident : the points of contact of the tangents from the centre, viz. the asymp- totes, are at infinity. Def — Two points such that each lies on the polar of the other are called conjugate points. If O, R are conjugate points it will be shown in § 308 that OR is cut harmonically by the conic. Def — Two lines such that each passes through the pole of the other are called conjugate lines. Two conjugate diameters are conjugate lines ; the pole of each is the point at infinity on the other. § 308. Any line drawn through a point O is cut harmonically by the conic and the polar of O. Let any line through O meet the conic in P, Q, and the polar of O in R. Take O for origin ; let the equation of the conic be ax 2 + 2 hxy + by 2 + 2 gx + 2fy + c = o 284 Analytical Geometry [309. The polar of O is [§ 306, Cor' (1)] gx + fy + c = o Writing r cos 0, r sin 6 for x, y, these become r 2 (a cos 2 6 + 2 h cos s\nO + b si n 2 #) + 2 r(g cos 6 + f sin 6) + c = o . . . . (1) r (gcos# + f sin 6) + c = o .... (2) A Let O PQ make 6 with the axis of x. (1) is a quadratic in r; its roots are OP, OQ. 1 1 2 (g cos# + f sin 6) •'' OP + OQ ~" c^ " Also (2) gives 1 1 gcos# + fsin# OR"" r"" c^ 112 + " OP ' OQ OR i.e. OP, OR, OQ are in harmonic progression. § 309 . If two lines through a point O meet a conic in A, A 7 and B, B', and the direct and transverse joins of these points meet in P, Q; then PQ is the polar ofO. Take the fixed lines for axes ; let OA = 06, OA r = 06', OB = /3, OB / =/3 / , Let the conic be ax 2 + 2 hxy + by 2 + 2gx+2fy+c = o Putting y = o in this, we see that Of, QL f are the roots of the equation ax 2 + 2gx + c = o 309.] General Equation of the Second Degree 285 a a' c w Similarly 11 2f , , + ^=-c (2) The equation to PQ is ^ + ^ + | + ^- 2 = • • • • (3) for (see § 126) this line passes through the inters'n of XV XV and also of x , y xv Using (1) and (2), (3) becomes 2g 2f ^X V — 2 = C C J or gx + fy + c = o ; but this is the polar of O [§306, Cor* (1)]. Cor'— Similarly OQ is the polar of P; and .\ [§ 306, Cor* (3)] OP is the polar of Q. A triangle such as OPQ, in which each vertex is the pole of the opposite side with respect to a conic, is said to be self-conjtigate or self-polar. Note — The geometrical proof of the preceding proposition given in Eiiclid Revised, p. 367, is applicable to conies. Exercises 1 . Find the latus rectum of the parabola Vx + Vy = 3 also the equation of the tangent to it at the point (4, 1 ). Ans. 3 */2 ; x + 2 y = 6 286 Analytical Geometry [309. 2. Deduce the equations of the tangents at (x'y') to the curves y 2 = 4 ax, x 2 /a 2 + y 2 /b 2 = 1, xy = k 2 by the method of § 302. 3. Find the equation of the tangent at the origin to the curve ax 2 + 2 hxy + by 2 + 2gx + 2fy = o Ans. gx + fy = o 4. Find the diameters of S = o which are conjugate to the axes of x and y respectively. Ans. ax + hy + g =• o, hx + by + f = o 5. Find equation to chord of S = o which is bisected at origin. Ans. gx + fy = o 6. Show that the equation to the chord which is bisected at (x'y') is (ax' + hy' + g) (x - x') + (hx' + by' + f) (y - y') = o 7. Prove that one of the asymptotes of S = o will pass through the origin if af 2 + bg 2 = 2 fgh 8. Through a fixed point O are drawn two lines at right angles, meeting a given conic in P, P' and in Q, Q' : prove that 1 1 OP. OP' + OQ.OQ' is constant. [Note — Take O as origin ; let equation of conic be ax 2 + 2 hxy + by 2 +2gx + 2fy+c = o A If OP is inclined at to axis of x, OP and OP' are the roots of the follow- ing equation in r : r 2 (a cos 2 + 2 h cos 6 sin 6 + b sin 2 0) + 2 r (g cos + f sin 6) + c = o 1 a cos 2 + 2 h cos sin + b sin 2 ■"" OP. OP' = c Changing into 7T/2 + 0, 1 a sin 2 — 2 h cos sin + b cos 2 OQ.OQ' = c 1 1 _ a + b , *'• OP. OP' + OQ . OQ' " ~c~~ " J 3io.] General Equation of the Second Degree 287 9. On a line which revolves round a fixed point O and cuts a conic in P, Q, a point R is taken such that OR 2 = OP.OQ: find the locus of R. {Note — Choose O as origin, and equation of conic as in Ex. 8 ; if PQ is A inclined to axis of x at 6, OR 2 = OP. OQ = c/(a cos 2 6 + 2 h cos 6 sin 6 + b sin 2 0) ; and locus required is the conic ax 2 + 2 hxy + by 2 = c] 10. A parallel through any point O to an asymptote meets the conic in P and the polar of O in R : prove that OP = PR {Note — The second point in which OP meets the conic is at infinity ; .'. (§ 3°8) the line OR is cut harmonically in the points P, 00 ; .*. &c. See § 133, VI.] 11. If the equation S = o is transformed to any axes through the same origin, show that g 2 + f 2 is unaltered. [Note — Length of J_ from origin on its polar is unaltered.] 12. Trace the curves y = x — 3X 2 , + y = o J ' I + 2X J Show that they intersect at right angles at the origin ; and elsewhere at the angles tan -1 (7/6) and tan -1 (7/22). SEGMENTS OF CHORD THROUGH GIVEN POINT A § 310. Let a line through T (x'y') inclined at 6 to the axis of x cut the conic S = o in P, Q. We can find the lengths TP,TQ. As in § 208 we transform to polar co-ord's referred to T by substituting x' + r cos 6, y' + r sin 6 for x, y. Thus S = o becomes a (x r + r cos 0f+ 2 h (x r + r cos 6) (y' + r sin 6) + b (y'+rsin#) 2 +2g(x'+r cos#)+ 2f (y'-frsin#) + c = o 288 Analytical Geometry [sir. or r 2 (acos 2 #+2h cos sin 6+b sin 2 0) + 2r[(ax / + hy / +g)cos# + (hx / + by'+f) sin#] + 0(x',y') = o . . . (i) TP, TQ are the roots of this quadratic in r. Hence TPTQ = 0(x^yO/(acos 2 0+ 2 hcos<9sin0+bsin 2 0) . (2) § 311. We can draw some important inferences from the pre- ceding equation (2). A Let another line through T inclined at u cut the conic in R, S. Then TR .TS = (j) (x', y')/(a cos 2 0'+ 2 h cos ff sin &+ b sin 2 0') Hence the ratio TP.TQ:TR.TS is independent of the position of T, if the lines TP, TR are drawn in fixed directions. Thus if lines through another point T f parallel to TP and TR meet the curve in P', Q' and R', S', then TP.TQ :TR.TS = T'P'.T'Q': T'R'T'S'. Cor' (1) — Let T' coincide with the centre ; then we see that the rectangles TP.TQ and TR .TS are in the ratio of the squares of the parallel semi-diameters. Cor' (2) — Let P, Q coincide and also R and S ; and let Cp, Cr be semi- diameters parallel to the tangents TP, TR. We infer that TP : TR = C p : C r Cor r (3) — The joins of two pairs of points of intersection of a circle and :i conic are equally inclined to an axis of the conic. For let the circle cut the conic in P, Q, R, S ; let PQ, RS meet in T. Then TP.TQ = TR .TS (Euclid III. 35, 36) Hence by Cor' (1) the semi-diameters parallel to PQ, RS are equal ; and are .*. equally inclined to an axis. Ex. 1. If a circle intersect an ellipse in four points whose ecc' As arc % /3, y, b, then OC + (3 + y + b = 2 rnr 3i2.] General Equation of the Second Degree 289 For the joins of (X, /3 and of y, 8, viz. x oc + 8 v . will represent any conic through the four given points A, B, A', B'. 294 Analytical Geometry [318. Cor f (1) — If we put OL' = OL, /3 r = /3 we obtain the equation of a conic touching the axes at A, B. Hence (I + 1 - 2 = Xxy (a) is the equation of a conic touching the axes at the points (Oi, o), (o, /3). Cor' (2) — If (2) represents a parabola, then (§ 2S7, Cor') \a/3 2/ a 2 /3 2 .*. A = o or 4/oLfi Neglecting the value A. = o which gives a pair of coincident lines, (2) becomes (1 + y _ A 2 = i^ Take the square root x y /xy Take the square root again ; then, as a square root may be either positive or negative, we may write the result This is .*. the equation to a parabola touching the axes. 8 318. To find the condition that the line * + y h + k " + 1-* CO ?;wj ^7/ o, and an ellipse if q < o. [Note — The names parabola, ellipse, and hyperbola were originally derived from this property.] CONDITIONS DETERMINING A CONIC § 320. Five points determine a conic. If it is required to find the equation of a conic passing through five given points (x^), (x 2 y 2 ), ... (x 5 y 5 ); then assuming that the equation of the conic is ax 2 + 2 hxy + by 2 + 2 gx + 2fy + c = o we have ax x 2 + 2hx 1 y 1 + ... = o, ax 2 2 + ... = o, ... We have then five simple equations which suffice to determine the five ratios a/c, h /c, bfc, g/c, f/c. Thus there is one conic, and only one, which passes through the five given points. [If three of the points lie on a line, then, since a line can only meet a conic in two points, the conic is the line-pair containing the five points.] In practice it is better to proceed as in the following example. Ex. Find the conic through the five points (- 1, 3), (o, 5), (- 2, 4)» (°i IO )> ( x > 2 ) Form eq'ns to joins of two pairs of four of the points ; thus the line-pairs x (x + y - 2) = o and (2 x - y + 5) (3* - y + IO ) = ° are conies passing through the first four points. The eq'n of any conic through these four points is .*. x (x + y - 2) = A (2 x - y + 5) (3 x - y + 10) Expressing now that this eq'n is satisfied by the co-ord's of the fifth point, we find A = ^r. 65 298 Analytical Geometry [3 2 1, § 321 • In general, five-conditions determine a conic. For each condition leads to an equation connecting the five unknown ratios a /c, b /c, . . . There may be more than one conic satisfying the conditions : for if the equa- tion to determine one of the ratios obtained by elimination is of the nth degree, there are n conies. The number of conditions is not necessarily equal to the number of verbal statements : thus to be given the centre (x'y 7 ) is equivalent to two conditions ; for we have two equations ax' + hy' + g = o, hx' + by' + f = o Ex. Two conies can be described to pass through four points and touch a given line. Take the joins of two pairs of the given points as axes ; so that the conic is represented by eq'n (i), § 317. The reader will find it easy to work out the condition that this conic may touch a given line /. /. s x/h + y/k = 1 : the resulting equation is a quadratic in A ; .'. &c. NINE-POINT CONIC S 322. To find the locus of the centre of a conic passing through four fixed points. Such a conic is represented by the equation of § 317, viz. (s^-'X^- 1 )-^ (1) The centre of this conic is determined by the equations !G? + j£- , ) + £(! + ]&- i )- Ax The indeterminate A. is eliminated by multiplying the eq'ns by x and y respectively, and subtracting. Reducing, we find that the locus required is the conic 2 /3/3 r x 2 - 2 aa'y 2 - /3/3' (a + a') x + aa' (/3 + /3') y = o . . (2) Putting y = o we find that the locus passes through the origin [as we might 323.] General Equation of the Second Degree 299 have foreseen, O being the centre of the line-pair conic AA', BB' (fig 7 , /OC + 0i r \ page 284)] ; and through I , o 1, i. e. through the mid point of AA'. As the joins of any two pairs of the points might have been chosen as axes, we see that — Given four points A, B, C, D ; then the six mid points of their joins AB, AC, AD, BC, BD, CD and the three intersections of opposite con- nectors AB and CD, AC and BD, AD and BC lie on a conic. Cor f — If OCOC' = — /3/3' the centre locus (2) is a circle, and the conies (1) are rectangular hyperbolas. The line-pairs consisting of the joins of opposite con- nectors are special conies through the four given points ; these line-pairs are .*. rectangular. Hence the four points are so related that each is the orthocentre of the triangle formed by the other three. Accordingly, we infer that — If D is the orthocentre of a triangle ABC, then the feet of the perpendiculars and the mid points of AB, BC, CA, AD, BD, CD lie on a circle. This circle is called the nine-point circle. [An exhaustive discussion of the nine-point conic will be found in Clifford's Mathematical Papers , pp. 579-583.] TANGENTIAL EQUATION § 323. To find the condition that the line Ax + fiy + v — o (1) may touch the conic ax 2 + 2 hxy + by 2 + 2gx + 2fy + c = o . . (2) The equation of the lines joining the origin to the inters'ns of (1), (2) is (§119) v 1 (ax 2 + 2 hxy +" by 2 ) - 2 v (Ax + /xy) (gx + fy) or + c ( Ax + ^y) 2 = ° x 2 (a^ 2 — 2gz/A + cA 2 ) -f 2xy(hz> 2 — i\v — gixv + cA/x) +y 2 (b v 1 — 2f jxv + Cju, 2 ) = o These lines .-. coincide if (az/ 2 — 2gXp + cA 2 ) (bz/ 2 — 2f/jii/ + c/jl 2 ) = (hi> 2 — fAj/ - g/jLi> + cA/x) 2 This condition reduces to (bc-f 2 ) A 2 + (ca- g 2 )/x 2 + (ab - h 2 ) v 2 + 2 (gh - af)/m/ + 2(hf-bg)j/A + 2(fg-ch)A/x= o . . (12) 3°° Analytical Geometry [324. In future this equation will be written AX. 2 + Bfx 2 + Cv 2 + 2Ffxv + 2G V \ + 2Hkfx = . . (12) The coefficients A, B, C, 2 F, 2 G, 2 H are the derived functions of the discriminant A = abc + 2 fgh - af 2 - bg 2 - ch 2 taken respectively with regard to a, b, c, f, g, h. The co-ordinates of the centre (§ 289) are G/C, F/C PAIR OF TANGENTS § 324*. The points where the join of (x'y'), (x"y") meets the conic S — o are determined by substituting mx" + nx' my" + ny' m + n ' m + n for x, y in the equation ax 2 + 2 hxy + by 2 + 2gx + 2fy+c = o This gives a (mx" + nx') 2 + 2 h (mx" + nx') (my" + ny') + b (my" + ny') 2 + 2 g (m + n) (mx" + nx') + 2 f (m + n) (my" + ny') + c (m + n) 2 = o or m 2 (ax" 2 + 2 hx"y" + by" 2 + 2 gx" + 2 fy" + c) + 2 mn [ax'x" + h (x'y" + x"y') + by'y" + g(x' + x") + f (y' + y") + c] + n 2 (ax' 2 + 2 hx'y' + by' 2 + 2 gx' + 2 fy' + c) — o This quadratic in m /n has equal roots if (ax" 2 + 2 hx"y" + ...) (ax' 2 + 2 hxy + ...) = [ax'x" + h(x'y" + x"y') + ...] 2 Now the quadratic has equal roots if (x"y") is any point on either tangent from (x'y') : writing then x, y instead of x", y" in the preceding equation, we obtain the equation of the pair of tangents from (x'y'). The equation of the pair of tangents from (x'y' ) is .'. (ax' 2 + 2 hxy + by' 2 + 2 gx' + 2 y + c) (ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c) = [axx' + h (xy' + x'y) + byy' + g(x + x') + f (y + y') + c] 2 . . . (12) 8 325 . To find the equation of the director circle o/S = o. The equation of the tangents from (xy ) is (*'> y r ) (x, y) - [axx' + h (xy' + x'y) + ...] 2 = o 325.] General Equation of the Second Degree 301 If (x / y / ) is on the director circle these tangents are at right angles ; the con- dition for this (if we suppose the axes rectangular) is coeff' of x 2 + coeff / of y 2 = o Express this condition and write x, y for x', y' ; reducing, we obtain the equation of the director circle C (x 2 + y 2 ) - 2 G x - 2 Fy + A + B = o Here C = ab — h 2 , &c, as in § 323. Cor f — If the curve is a parabola, C = o. The equation of the directrix (which is locus of inters'n of rectangular tangents) is .'. 2 Gx + 2 Fy = A + B Exercises on Chapter XI 1. Find the equation of the diameters of the conic ax 2 + 2 hxy + by 2 — 1 passing through its intersections with the concentric circle x 2 + 2 xy cosco + y 2 = p 2 . (X2) Ans. (ap 2 — 1) x 2 + 2 (h p 2 — cos co) xy + (b p 2 — 1) y 2 = o 2. Show that the length of a diameter of the same conic bisecting the angle between its axes is 2 p, where (a + b — 2 h cos co) p 2 = 2 sin 2 co [Note — Express cond'n that line-pair in Ex. 1 is rectangular.] 3. If 2 p is the length of an equi-conjugate diameter of the conic ax 2 + 2 hxy + by 2 = 1 then 2 (ab — h 2 ) p 2 = a + b — 2 h cos co ; and the equation to the equi-conjugate diameters is (a + b — 2 h cos co) (ax 2 + 2 hxy + by 2 ) = 2 (ab — h 2 ) (x 2 + 2 xy cos co + y 2 ) 4. Show that the points of intersection of the conies ax 2 + 2 hxy + by 2 = 1, a'x 2 + 2 h'xy + b'y 2 = 1 are on conjugate diameters of the former conic if ab' + a r b - 2 \\W = 2 (ab - h 2 ) 302 Analytical Geometry 5. Show that the following function of the coefficients in the equations of two conies is unaltered by change of axes : ab/ + a'b — 2 hh' sin 2 co 6. The major axes of two conies are parallel : show that their four points of intersection are concyclic. 7. The polars of a point with respect to two given conies intersect at right angles ; prove that the point describes a conic, and that this conic is a circle if the given conies are rectangular hyperbolas. [Note— Choose asymptotes of one conic as axes ; so that eq'ns of given conies are xy = § 2 , ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c = o.] 8. Chords of a rectangular hyperbola at right angles to each other subtend right angles at a fixed point O. Show that they intersect on the polar of O. [Note — Choose O as origin, and parallels to axes of figure of hyperbola as axes of co-ord's ; so that its eq'n is x 2 - y 2 + 2 gx + 2 fy + c = o.] 9. Prove that two parabolas can be drawn through four given points ; and that their axes are parallel to the asymptotes of the conic which is the locus of centres of conies passing through the four points. 10. Find equation of tangents from origin to S = p. Ans. B x 2 — 2 H xy + A y 2 = o {Note — This may be deduced from § 324; or directly thus. Substitute r cos 6, r sin for x, y in S = o ; express condition that the quadratic in r has equal roots. Then replace sin /cos by v/x.] 1 1 . Tangents from P to a conic intercept a given length on a fixed tangent ; prove that the locus of P is a conic. [Note — Take given tangent and corresponding normal as axes.] 12. If tangents from P intercept a given length on a fixed line, the locus of P is in general a curve of the fourth degree. [Take given line as axis of x ; the conic may be represented by the general equation S = o.] Exercises on Chapter XI 303 13. If tangents from P meet a fixed tangent at points equidistant from the point of contact ; prove that the locus of P is a straight line, viz. the diameter conjugate to the given tangent. [Note — Choose axes as in Note, Ex. n. The abscissae of points where tangents from P meet given tangent are determined by a quadratic ; put sum of its roots = o.] 14r. PQ is one of a system of parallel chords of S = o; R is a point on PQ such that p^ + r"q = constant H) If the chords are inclined to the axis of x at an angle 0, prove that the locus of R is the conic S + 2 5 [(ax + hy + g) cos + (hx + by + f ) sin 0] = o [Note— Let co-ord's of R be x', y'; then RP, RQ are the roots of eq'n (0, § 3io.] 15. PQ is one of a system of parallel chords of an ellipse; R is a point on PQ such that PR 2 + RQ 2 + PR . RQ = constant. Show that the locus of R is a conic. 16. Find the equation to the axis of the parabola VxTa + V^//3 = 1 (&) A x y a 2 -/3 2 AnS. 7s H ^ ™ 75 = o a /3 a 2 + /3 2 + 2 a/3 cosa> [Note — Rationalize eq'n ; and proceed as in § ^95, Use cond'n of perpen- dicularity given in § 93, Cor' (2).] 17. Find the equation to the directrix of the same parabola. [Note — Express cond'n that tangents from (x'y') are at right angles.] 18. Find the focus of the same parabola. [Note— The line x/h + y/k = 1 is a tangent if h/(X + k/p =1 (1) 304 Analytical Geometry The circum-circle of the triangle which this tangent forms with the axes is x 2 + 2xy cos co + y 2 = hx + ky (2) Eliminate k from (2) by means of (1) ; eq'n of circle becomes OL (x 2 + 2 xy cos co + y 2 — /3 y) - h (a x - /3 y) This eq'n contains the single parameter h in the first degree ; and whatever be the value of h the circle passes through the fixed point determined by x 2 + 2 xy cos co + y 2 = OC x = (3 y This fixed point is the focus. (§ 330, Ex. 3).] 19. A parabola touches the sides of a given triangle ; show that each of the chords of contact passes through a fixed point. 20. The equation to determine the eccentricity of S = o is / e 2 \ 2 _ (a - b) 2 + 4 h 2 V, 2 - e 2 / ~ (a + b) 2 [Note — Suppose eq'n reduced by successive changes of axes to ax 2 + 2 hxy + by 2 + c' = o (1) cxx 2 + /3y 2 + c' = o (2) Equate the invariants (§ 212) ; .-. a + /3 = a + b (3) «/3 = ab - h 2 (4) Now if (2) is written x 2 y 2 ■ — + — = 1 A 2 B 2 then A 2 = - c'/ot, B 2 = - c'//3 „ A 2 - B 2 „ (3 - OC and e 2 = — ^ 5 •'• e = ~ q~ (5) Eliminate OC, j3 from (3), (4), (5). It will be observed that the equation to determine e 2 is a quadratic ; the existence of a second eccentricity will be accounted for in Chap. XIII.] 21. Prove that the equation of the join of (x'y r ) to the centre of S = o is (ax + hy + g) (hx r + by' + f ) = (hx + by + f) (ax' + hy' + g) [Note — The lines ax + hy + g = o, hx + by + f = o pass through the centre (§ 289).] Exercises on Chapter XI 305 22. Prove that the equation to the axes of S = o is h (u 2 - v 2 ) = (a - b) uv where u = ax + hy + g, v = hx + by + f [Note — If (x'y') is a point on an axis of the conic, the join of (x'y') to centre is _1_ polar of (x'y').] 23. PQ is a chord of a conic ; P q, P r are chords equally inclined to PQ. Show that q r passes through a fixed point, viz. the intersection of the tangents at P and Q. [Note — When P q coincides with PQ so does P r ; the tangent at Q is .*. one position of q r.] 24. Conies are drawn through four fixed points. Prove that the polars of a given point pass through a fixed point. [Let two of the conies be S = ax 2 + 2 hxy + ... = o } S' = a'x 2 + 2 h'xy + ... = o The polar of (x'y') with respect to S + AS'= o is ax'x + h (y'x + x'y) + ... + A [a'x'x + h' (y'x + x'y) + ...] = o This passes through the intersection of ax'x + h (y'x + x'y) + ... = o, a'x'x + h' (y'x + x'y) + ...= o] 25. Find locus of poles of a given straight line with respect to conies passing through four fixed points. [Let the given line be Ix + my + n = o Express that this coincides with polar of (x'y') with respect to S + \& = o: ax' + hy' + g + A. (a'x' + h'y' + g 7 ) _ _ ... ... . = __ = _ _ M> say . Multiply up and eliminate A., \x. Locus is the conic ax + hy + g a'x + h'y + g' I hx + by + f h'x + b'y + f m gx + fy + c g'x + f'y + c' n = o.] 26. A system of conies passes through the angular points of a square. Tangents are drawn from a given point on one diagonal. Prove that the locus of the points of contact is a rectangular hyperbola. X 306 Analytical Geometry {Note — Choose parallels to sides of square through its centre as axes ; so that system of conies is x 2 - a 2 + A. (y 2 - a 2 ) = o] 27. Conies are drawn through four points : prove that the pole of the join of any two of them describes a straight line. 28. The lengths of the tangents from (x'y') to the ellipse x 2 /a 2 + y 2 /b 2 = i are the roots of the equation in I, ay' V |2 _ b 2 S' = bx' Va 2 S'- I 2 + ab V(a 2 - b 2 ) S' where S' - x /2 /a 2 + y' 2 /b 2 - i [Note — Let 8 = length of parallel semi-diameter ; tan -1 m its inclination to axis of x. Then I = b V& [Ex., § 312] and / = mx' + Va 2 m 2 + b 2 [§ 249] Express 8 in terms of m (§ 238) ; then eliminate m.] 29. Find the locus of the Fregier Points (see Ex., § 313) corresponding to points on the ellipse x 2 /a 2 + y 2 /b 2 = 1 Ans. x 2 /a 2 + y 2 /b 2 = (a 2 - b 2 ) 2 /(a 2 + b 2 ) 2 [Note — Let P be a point on the ellipse ; draw chords PQ, PR parallel to the axes. The corresponding Fregier Point is inters'n of diam' QR and normal at P. Express eq'ns of these lines in terms of (X, the ecc/ A °^ P; then eliminate OC.] 30. Find the condition that the line x/h + y/k = 1 may touch the conic (x/a + y//3 - i) 2 = 2 A.xy ^• x=2 (s-fi)(g-iD [Note — Express cond'n that the following eq'n in x/y has equal roots: (x/oc + y//3 - x/h - y/k) 2 = 2 Axy] 31. Prove that the locus of centres of conies inscribed in a quadrilateral is the straight line through the mid points of its diagonals. Exercises on Chapter XI 307 [Take diag's as axes ; let eq'ns to sides be x /p + y/q = r > x /p' + y/q = 1, ■ • • • The cond'ns that these lines touch the general conic are A/p 2 + B/q 2 + C - 2 F/q - 2 G/p + 2 H/(pq) - o . . (1) A/p'2+B/q 2 +...= (2) A/p 2 + B/q' 2 + ...= o . . (3) A/p' 2 + B/q'* + ...= o (4) Subtract (2) from (1) and divide by i/p - i/p' .*. A (i/p + i/p') - 2 G + 2 H/q = o So from (3) and (4) A (i/p + i/p') - 2 G + 2 H/q' = o Whence H = o, A = 2 pp'G/(p + p') Similarly B = 2qq'F/(q + q') Substituting in (1), and simplifying, we get C_2G/(p+p')-2F/(q + q') = o (5) But co-ord's of centre are OC = G/C, /3 = f/c We deduce then from (5) 2«/(p + p') + 2/3/(q + q') = 1 i.e. (JX(3) is a point on /p + p' /q + q' But this is the join of mid points of diagonals. The cond'n = o may be interpreted to show that the diagonals are conju- gate with respect to the conic, i. e. the pole of each lies on the other. This solution is by Prof Genese. It may be seen h priori that the mid points of diagonals are points on the locus ; for a diagonal is the limiting form of a thin inscribed ellipse. The question may also be solved by taking two sides of quad' as axes, and using the cond'n of Ex. 30.] X 2 CHAPTER XII POLAR EQUATION OP A CONIC RE- FERRED TO FOCUS ; CONFOCAL CONICS § 326. To find the polar equation of a conic referred to the focus. By def also From a point P on the curve draw PN _L axis, and PM ± directrix ; let SP=r, PSN=0. Let the semi-latus rectum SL= I SL = eSX SP = ePM = e(XS + SN) = SL + eSN i.e. r = I + ercos# or r (i — e cos 6) — I The required equation is .*. - = i — e cos 6 r Polar Equation of a Conic 309 Cor' (1) — The equation of the directrix is r cos (tt - 6) = SX = - ; or - = _ e cos 6 r Cor' (2) — Let PS p be a focal chord ; then gp = 1 - e cos Q ^- = 1 — e cos (77 + 6) = 1 + e cos fc>p •'• SP + S p 2 112 Thus the semi-latus rectum is a harmonic mean between the segments of any focal chord. A Note — If the initial line make an angle OC with SN, then as PSN is now 6 + OC, the equation of the conic is \fr = 1 -e cos (6 + a) § 327* It is a useful exercise to deduce the figure of the curve from the equation I It — 1 — e cos by giving 6 a series of values increasing from o to 2 77. The cases e = i, e < 1, e > 1, should be separately considered. The last case is very instructive ; the following summary will guide the learner in discussing the question fully. When 6 — o, r = l/(i — e) : this is negative and = SA r (fig', § 266). As 6 increases, r continues negative (the extremity of the radius vector describing the lower portion of the further branch of the hyperbola), until cos 6 = 1 /e, when r = — 00 . The nearer branch is next described in the order PA p (fig', § 270) ; and lastly the upper portion of the further branch. Note — Observe that if a line through S cut two opposite branches of an hyperbola in P and p, then the vectorial A s of P and p are not the same ; they differ by 7T. 3io Analytical Geometry [328. EQUATIONS OF CHORD AND TANGENT § 328. Consider the locus represented by I //, a + j8\ ol — P a t \ - = cos(0 — )sec ecosfl, . (1) r \ 2 / 2 This may be written r /1 a + /3 . . n . a + jSn a — j8 I = r cos a cos ! — h r sin sin sec [_ 2 2 j 2 — er cos or / a + fi . 06+/3\ a-/3 I = ( x cos —- LJ1 + y sin c ) sec - ex \ 2 2 / 2 Thus (1) represents a straight line. Again, if we put 6 = 06 in (1) it becomes I - = 1 — e cos a r .*. (1) passes through the point on the conic whose vectorial angle is 06. Similarly it passes through the point whose vect' A is /3. Hence (t) is the equation of the chord joining the two points on the conic whose vect' A s are 06, 0. Cor' — Put /3 = 06 in (1); the equation of the tangent at the point whose vectorial angle is 06 is .*. - = cos (0 — a) — e cos 6 .... (2) § 329. The equation of the chord may also be obtained thus. Assume that the required equation is - = A cos 6 + B sin (3) r Express that this is satisfied by the co-ord's of the given points : .'•. 1 — e cos OL = A cos OC + B sin OC ) i — e cos/3 = A cos/3 + B sin/3 ) 330.] Polar Equation of a Conic 311 If we solve these two equations for A, B and substitute their values in (3), we obtain after reduction the same equation as before. The student should work this out as an exercise. S 330. Ex. 1. If the tangent at P meet the directrix in Z, then PZ subtends a right angle at S. Let vect' A of P be (X ; the equation of the tangent at P is I It — cos (0 — (X) — e cos The equation of directrix is I It — — e cos At the point Z where these intersect we have .'. cos(0 - a) = o i A 1 .-. - a = + -n/2, i.e. PSZ = -n/2 Ex. 2. If the tangents at P, Q intersect in T, then ST bisects the angle PSQ. Let vect' A of P be (X, and of Q, /3. At the inters'n of the tangents I /r = cos (0 - a) — e cos 6 \It — cos (0 — /3) — e cos we have cos (0 — OL) = cos {0 — ft) (1) ... $ _ a = _ ( _ q) .-. = (a + /3)/a ; /. &c. If, however, P and Q are on opposite branches of an hyperbola, then ST does not lie between SP, SQ. The proper deduction from (1) is now 0-(X = 2TT-(0-l3) .'. = it + (0C + /3)/a In this case ST bisects the angle supplemental to PSQ. Ex. 3. The circum-circle of the triangle formed by three tangents to a para- bola passes through the focus. Let the equation to the parabola be I It = 1 — cos 312 Analytical Geometry [33°- The tangents at the points whose vect' A s are OL, /3, y are l/r = cos (0 - a) - cos 6 (1) l/r = cos (0 - /3) - cos 6 (2) l/r = cos (0 - y) - cos (3) OL /3 At p't of inters'n of (i) and (2), = |(CX + (3), l/r ^ 2 sin sin — (2) „ (3), = W+y)> l/r=2sin^sin^ (3) » W, = %(y + (X), l/r = 2 sin- sin - It is evident that these three points all lie on the circle 1 . /o + /3 + y \ . « . /3 . y - sin I a I = 2 sin — sin — sin - r V 2 / 222 The equation to the circle shows that it passes through the focus. Ex. 4. A chord PQ of a conic subtends a constant angle 2 y at the focus. Find the locus of T the intersection of tangents at P and Q. Find also the envelope of PQ. Let the conic be l/r = 1 — e cos 6 Let the vect' As of P, Q be OL, (3 ; then OL - (3 = 2y The tangents at P, Q are l/r = cos (6 — OL) — e cos 6 l/r = cos {6 — /3) — e cos 6 Where these intersect we have 0L + B . OL - 8 6 = , l/r = cos — e cos 2 / 2 The equation to the locus of T is .*. l/r = cos y — e cos or (I sec y)/r = 1 — e sec y cos 6 The locus of T is .*. a conic having the same focus and directrix as the given conic, and whose eccentricity = e sec y. 33o.] Polar Equation of a Conic 3*3 Again, the equation to PQ is (§ 328) \/r = cos (0 J sec y - e cos 6 or (I cos y)/r = cos ( j — e cos y cos Comparing this with the equation of the tangent [eq'n (2), § 328] we see that PQ touches the conic (I cos y)/f = 1 — e cos y cos (1) at the point on that conic whose vect' A is (a + p)/i The conic (1) is the envelope. It has the same focus and directrix as the given conic, and its eccentricity = e cos y. Ex. 5. Find the equation of the polar of (r ± t ) with respect to l/r = 1 — e cos0 The equation of the conic is I = r — er cos# or I = Vx 2 + y 2 — ex or x 2 + y 2 = (I + ex) 2 or x 2 (1 — e 2 ) + y 2 — 2 lex — I 2 = o The polar of (r x cos y , r y sin X ) is then (§ 306) xrj cos 6 X (1 — e 2 ) + yr x sin O x — le (x + r x cos X ) — I 2 = o Writing r cos 6, r sin 6 for x, y we obtain the req'd eq r n, viz. rr x cos 6 cos O x (1 — e 2 ) + rr x sin sin 6 X = le (r cos + r x cos 0^) + I 2 This may be written n*! cos (6 - 6J = (I + er cos 0) (I + er\ cos 6 X ) or cos (0 - d x ) = (l/r + e cos 6) (l/r a + e cos X ) Exercises [Unless otherwise implied, the equation of the conic in these questions is l/r = 1 — e cos 0] 1. PS p, QS q are focal chords at right angles. Prove that the sum of the reciprocals of P p, Q q is constant. 3 J 4 Analytical Geometry [330- 2. A chord PQ subtends a right angle at the focus. Prove that / 1 A 2 fj_ 1 \ 2 _ e 2 VSP ~ 7/ + Vsq " 7/ I 2 3. Prove that a focal chord varies as the square of the parallel semi- diameter. 4. PS p, QS q are focal chords. Prove that PQ, pq meet on the directrix. 5. Prove that tangents at the ends of a focal chord meet on the directrix. 6. Tangents at P and Q meet in T. If PQ meet the directrix in Z, prove that SZ is at right angles to ST. 7. Find the equation to the normal at (r x X ). Am. r [sin (^ — Q) + e sin 0] = er x sin 6 1 8. Find the equation to the director circle. Ans. r 2 (e 2 — i) + 2 ler cos 6 + 2 I 2 = o 9. A chord PQ passes through a fixed point O. Prove that is constant. tan \ PSO . tan \ QSO 10. P is a point on a hyperbola; the tangent at P meets an asymptote in T. Prove that A A PST = CTS [Note — TP and the asymptote are the tangents from T. Use result of Ex. 2, § 330.] 11. A chord of a rectangular hyperbola subtends a right angle at the focus; prove that it envelopes a parabola. 12. If TP, TQ are tangents to a parabola, then ST 2 = SP.SQ; and the triangles SPT, STQ are similar. [Note — See § 330, Ex. 3 ; use Euclid VI. 6.] 13. Show that the equations I \/r = 1 — e cos 6, l/r = — 1 — e cos 6 represent the same conic. 333-] Confocal Conies 315 \Note — If PSQ is a focal chord whose vect' A is 0, one of these eq'ns deter- mines SP and the other determines SQ.] 14. The equations of two conies having a common focus are S = \Jr - i + e cos 6 =» o, S' = V/r - i + e' cos (0 + /3) = o ; prove that S-S' = o, S + S' + 2=0 are the equations to one pair of common chords. CONFOCALS § 331. The co-ordinates of the foci of x 2 /a 2 + y 2 /b 2 = 1 (1) are (o, + Va* - b 2 ) These are unaltered if we substitute a 2 — A, b 2 — A for a 2 , b 2 . Accordingly, if different values are given to A, the equation x 2 /(a 2 - A) + y 2 /(b 2 - A) = i .... (2) represents a system of conies confocal with (1). If A > b 2 , the curve represented by (2) is an hyperbola. Note — We may briefly refer to the ellipse represented by (i) as 'the ellipse (a, b).' Thus (2) represents 'the ellipse (Va 2 - A, Vb 2 - A).' § 332. Put c 2 = a 2 - b 2 , a x 2 = a 2 - A, bj 2 = b 2 - A ; then aj , b x are the semi-axes of the confocal (2) ; and &1 2 - bi 2 = a 2 - b 2 = e 2 Equation (2) may also be written in the forms x 2 / ai 2 + y 2 /b x 2 =1 (3) x 2 /a! 2 + tjW - c 2 ) = 1 (4) x 2 /(b! 2 + c 2 ) + y 2 /b t 2 = 1 (5) 8 333 . Looking at eq'n (5) we see that — If b x 2 is very small and positive, then a x 2 is nearly — c 2 ; and the confocal is a thin ellipse nearly coincident with the join of the foci SS'. If bj 2 is very small and negative, the confocal is a thin hyperbola nearly 3 l 6 Analytical Geometry [334- coincident with the portions of the axis extending from S, S' to infinity in opposite directions. Thus, corresponding to b x 2 = 0, or A = b 2 , we have as limits of the confocal system both the line-ellipse (the join of the foci), and the line-hyperbola (the complement of this join). § 334. If the confocal pass through a given point (hk), then by eq'n (5) h 2 /^ 2 + c 2 ) + k 2 /b x 2 = 1 or h 2 b x 2 + k 2 (bi 2 + c 2 ) = b x 2 (b x 2 + c 2 ) ... bi 4 - (h 2 + k 2 - c 2 ) b x 2 - k 2 c 2 = o This is a quadratic to determine b x 2 . Its roots are both real ; one is positive and the other negative. Thus two conies of a confocal system can be drawn through a given point ; one is an ellipse and the other an hyperbola. S 335. We deduce similarly from eq'n (4) the quadratic in a^, a x 4 - a x 2 (h 2 + k 2 + c 2 ) + h 2 c 2 = o (6) Let a^, a 2 2 be the roots of this equation ; .-. h 2 c 2 = a! 2 a 2 2 (7) Similarly - k 2 c 2 = b x 2 b 2 2 = (a x 2 - c 2 ) (a 2 2 - c 2 ) . . . . (8) The semi-axes major of the two confocals which pass through a point are called by Lame its elliptic co-ordinates. The preceding eq'ns (7), (8) give expressions for the rect' co-ord's h, k of a point in terms of its elliptic co-ord's Cor' — We see also from (6) that a x 2 + a 2 2 =-- h 2 + k 2 + c 2 (9) § 336. Confocal conies cut at right angles. Suppose that (h k) is a point of intersection of the conies x 2 /a 2 + y 2 /b 2 = i, x 2 /(a 2 - A) + y 2 /(b 2 - A) = i Then h 2 /a 2 + k 2 /b 2 = i, h 2 /(a 2 - A) + k 2 /(b 2 -A ) = i By subtraction h 2 k 2 I O /I o ^ \ w a 2 (a 2 - A) " b 2 (b 2 - A) 339-] Confocal Conies 317 But this is the condition that the tangents at (hk) viz. xh/a 2 + yk/b 2 = 1, xh/(a 2 - A) + yk/(b 2 - A) = 1 should be at right angles. Note — This proposition and that of the last § are obvious geometrically. If P is the given point and S, S' the given foci : then one curve is the ellipse whose foci are S, S' and major axis = SP + S' P ; the other is the hyperbola whose foci are S, S' and transverse axis = SP — S' P*. A The tangents at P bisect SPS' and the supplemental angle; they are .'. at right angles. § 337. The line x cos 06 + y sin Ot = p will touch the conic x 2 /(a 2 - A) + y 2 /(b 2 - A) = 1 (a 2 - A) cos 2 a + (b 2 - A) sin 2 a = p 2 [§25o,CV'(i). .-. A = a 2 cos 2 a + b 2 sin 2 a — p 2 Thus there is one conic, and only one, of a confocal system which touches a given line. § 338 ■ tfp, p' are perpendiculars from the centre on parallel tangents to two confocals ; then p 2 — p' 2 is constant. Let x cos OC + y sin OC — p, x cos OC + y sin OC = p' be parallel tangents to the confocals x 2 /a 2 + y 2 /b 2 = 1, x 2 /(a 2 - A) -1 y 2 /(b 2 - A) = 1 Then p 2 = a 2 cos 2 a + b 2 sin 2 OL p'2 = ( a 2 _ X) cos 2 a + (b 2 - A) sin 2 a .*. p 2 — p r 2 = A = constant. 8 339. To find the locus of the intersection of tangents to two confocals which cut at right angles. The line xcos(X + ysina = \ / a 2 cos 2 (X + b 2 sin 2 a (i) * It follows that the elliptic co-ord's of P are ai - I (SP + S'P), a 2 = \ (SP - S'P). 3*8 Analytical Geometry [340. is a tangent to the conic x 2 /a 2 + y 2 /b 2 = 1 If we change a 2 , b 2 into a 2 — A, b 2 — A and OC into 7T/2 + OC, we obtain the line at right angles to (1) which touches the confocal x 2 /(a 2 - A) + y 2 /(b 2 - A) = 1 The tangent to this confocal at right angles to (1) is .*. - x sin OC + y cos a - V(a 2 - A) sin 2 a + (b 2 - A) cos 2 a . . (2) OC is eliminated from (1), (2) by squaring and adding; the required locus is .: the circle x 2 + y 2 = a 2 + b 2 - A § 34-0. To find the locus of the pole of a given line with respect to a system of confocals. Let the given line be x/h + y/k= 1 (1) If (x'y') is the pole of this line with respect to x 2 /(a 2 - A) + y 2 /(b 2 - A) = 1 we must have 1 x' 1 y' h " a 2 - A ' k^h^A .*. hx'- ky'=a 2 - b 2 The required locus is .*. the line hx — ky = a 2 — b 2 ; which is perpendicular to the given line. Further, the point of contact of the given line with the conic of the system which it touches is a point on the locus. The required locus is .*. the normal to that conic at its point of contact with the given line. Cor f — If the tangent at a point P on a conic intersect a confocal in p, q ; then the tangents at p, q intersect on the normal at P. § 34-1 • If & chord of a conic touch a confocal, its length varies as the square of the parallel semi-diameter. {Prof Burnside.) 342.] Confocal Conies 3*9 Let 8 be the length of a chord of the ellipse x 2 /a 2 + y 2 /b 2 -- i ; OC, /3 the ecc' As of its extremities ; R the length of parallel semi-diam', Q its inclination to the axis of x. The equation of the chord is x oc + p y .a + /3 a-/3 - cos — + f- sin = cos . . . . (i) a 2 b 2 2 v ' tan = — b cos /( a sin — J i cos 2 sin 2 Hence, as ^ = -^- + -^~ oc + 8 oc + 8 we deduce R 2 = a 2 sin 2 — + b 2 cos 2 - (2) 2 2 v ' Again, 8 2 = (a cos OC — a cos /3) 2 + (b sin OC — b sin /3) 2 Reducing this, and using (2), we find 0C- 8 b = 2 R sin " (3) 2 Again, (1) touches the confocal x 2 /(a 2 - A) + y 2 /(b 2 - A) - 1 if s — cos 2 + — — — sin' 2 = cos 2 - a z 2 b 2 2 2 Using (2), this reduces to . a a-j8 AR 2 ,, sm 2 = -tpr- (4) 2 a 2 b 2 ^ Eliminate sin (a — /3)/2 from (3), (4) ; ., 6 = i^ Q . E . D . ab x - S 34*2. Def — Points on two ellipses which have the same eccentric angle are called corresponding points* Thus (a cos OC, b sin OC) and (a' cos OC, b' sin 00 are corresponding points on the ellipses (a, b), (a', b'). Some properties of corresponding points on confocals are given in the Exer- cises which follow. 32° Analytical Geometry Exercises on Chapter XII 1. If tangents are drawn to a confocal system from a point in the major axis, the locus of the points of contact is a circle. 2. Find the confocal hyperbola through the point on the ellipse (a, b) whose eccentric angle is (X. x 2 v 2 Ans. -A, rV = a 2 - b 2 cos 2 a sin 2 a 3. Prove that corresponding points on a system of confocal ellipses lie on an hyperbola. 4. TP, TQ are tangents to two confocals. If TP, TQ are at right angles, prove that the join of T to the centre bisects PQ. 5. Show that the locus of the intersection of rectangular tangents to the confocal parabolas y 2 = 4(X(x + a), y 2 = 4/3 (x + /3) is the line x + OL + (3 = o Prove that the parabolas cut at right angles in two points at a finite distance; and that these points are imaginary if (X and (3 have the same sign. 6. Parallel tangents are drawn to a confocal system ; prove that the locus of the points of contact is a rectangular hyperbola. 7. Tangents from T to an ellipse meet a confocal in R, R' ; S, S'. Prove that i i i i TR " TR' ~ TS ~ TS 7 i i RR' [M>te—z^ ~jR f = TR.TR / ' USe § 34 * and C ° r ' ^' § 3 "^ 8. P and Q are any two points on an ellipse ; p and q are the correspond- ing points on a confocal. Prove that Pq = Qp {Ivory's Theorem?) 9. Prove that the equation in elliptic co-ordinates of the director circle of the ellipse (a, b) is &1 2 + a 2 2 = 2 a 2 Exercises on Chapter XII 321 10. Prove that the square of the semi-diameter of the ellipse (a, b) parallel to the tangent at its intersection with a confocal is A, where A is the parameter of that confocal. 11. Tangents to the ellipse (a, b) from a point whose elliptic co-ordinates are a x , a 2 include an angle (j) : prove that tan %(j) = ^ a x 2 — a 2 [Note — This follows from Ex. 33, page 239, and foot-note, page 317O 12. Prove that the equation of the tangents in the last question referred to the normals to the confocals through the point as axes is x 2 /^ 2 - a 2 ) + y 2 /(a 2 2 - a 2 ) = o 13. The tangents to two confocals from a point P which describes a con- focal ellipse are inclined at angles -ty, \J/ to the tangent at P. Prove that the ratio sinxj/ - : sini!/ is constant. 14. If A, A' are the parameters of the confocals which pass through any point on a directrix of the ellipse (a, b), prove that AA' = a 2 (A + AO 15. If a point describe the director circle of the ellipse (a, b) its polar envelopes a confocal. \_Note — Write eq'n of polar of (x'y 7 ), viz. xx'/a 2 + yy'/b 2 = 1 in the form Ix + my =1 (1) .\ x' -a 2 1, y'=b 2 m Express now that (x f , y') is a point on director circle .-. a 4 1 2 + b 4 m 2 = a 2 + b 2 The line (1) is .*. a tangent (§ 250) to the ellipse / a2 b2 \ 1 VVa 2 + b 2 ' Va 2 + b 2 / 16. P and Q are points on the ellipse (a, b) ; P f , Of are the corresponding points on the ellipse (a r , W). If the tangents at P', Q meet in T, prove that T is the pole of PQ' with respect to the ellipse (Vaa'j Vbb'). Prove also that the tangents at P', Q are at right angles if the tangents at P, Q' are at right angles. {Mr. R. Russell.) Y 322 Analytical Geometry 17. Prove that the join of the points of contact of rectangular tangents to two confocal ellipses (a, b), (a l3 b x ) envelopes the confocal 'bi \ \Va 2 + b^ Va 2 [Note — If the conies in Ex. 16 are now assumed to be confocal ; then if tan- gents at P, Q' are at right angles, T describes the circle x 2 + y 2 = a 2 + b x 2 . Finish solution as in Note, Ex. 15.] 18. The equations to the asymptotes of the hyperbola I /r = 1 — e cos 6 are abe = r (a sin 6 ± b cos 6) 19. Two conies have a common focus S; a variable line through S meets the conies in P and Q. Prove that the intersection of the tangents at P and Q describes a straight line. 20. Two conies have a common focus ; prove that two of their common chords pass through the intersection of their directrices. 21. A, A r ; B, B' ; C, C are three pairs of opposite summits of a quadri- lateral which circumscribes a parabola whose focus is S. Prove that SA . SA' = SB . SB' - SC . SC' 22. Two parabolas have a common focus. Show that the locus of the intersection of two tangents, one to each, cutting at a constant angle, is a parabola. {Prof Purser?) 23. Prove that the polar equation to the normal to the parabola r = a sec 5 - 2 at the point OC is a = r cot — cos — sin I u I 2 2 \ 2/ If 6 be the angular co-ordinate of the point at which the normals at OL, /3, y intersect, prove that 2d = OC + (3 + y 24. A circle is described with the focus S of a conic as centre; a line through S meets the circle in P and the conic in Q. Prove that the tangents at P and Q intersect on a common chord of the circle and conic. Exercises on Chapter XII 323 25. Show that the locus of the extremities of the latera recta of parabolas which have a common focus and a common tangent consists of two circles. 26. Given the focus and directrix of a conic ; show that the polar of a given point with respect to it passes through a fixed point. 27. A circle circumscribes the triangle formed by tangents to a parabola at three points A, B, C. If p is its radius, prove that 2p 2 l = SA.SB.SC where S is the focus of the parabola, and 2 I its latus rectum. 28. Two conies have a common focus ; then axes are inclined at an angle j3. Show that the conies touch if (| _ |')2 = |2 e /2 + |/2 e 2 _ 2ee'H'cos/3 where e, e' are the eccentricities, and I, I' the semi latera recta. [Note — If (X is the vect' A of a point of inters' n of the conies l/r ~ 1 — e cos 6, V J r = 1 — e' cos (0 + /3) the tangents at that point are l/r = cos (0 - a) - e cos 6, l'/r = cos (0 - OC) - e' cos (6 + (3) These equations may be written in the form l/r = A cos 6 + B sin 6 These tangents .\ coincide if (cos a - e) J\ = (cos a - e' cos j3)/ V and sina/l = (sin a + e'sin/3)/!' The result is now obtained by eliminating (X.] 29. Two conies have a common focus, about which one is turned; show that the common chords envelope a conic having a focus at the given focus. (S. Roberts, Educ' Times, xxxix.) 30. Two ellipses have a common focus ; one revolves about this focus while the other remains fixed. Prove that the locus of the point of intersection of their common tangents is a circle. Y 2 CHAPTER XIII ABRIDGED NOTATION; MISCELLANEOUS PROPOSITIONS THE STRAIGHT LINE § 34-3. In §§ 126-130, 139-141, 148, we have given some account of the abridged notation of the straight line ; we shall now give some other propositions. We shall (as agreed on in § 127) use Greek letters (X, (3, &c. as abbrevia- tions for expressions of the form x cos CX + y sin a — p ; and English letters u, v, w, or L, M, N, for expressions such as Ax + By + C § 344-. If U = o, v = o, w = o are the equations to three given straight lines which form a triangle ; then the equation lu + mv + nw = o may, by giving suitable values to the ratios I : m : n, be made to represent any straight line whatever. Let the given lines form a triangle ABC (fig' § 138); and let any other line meet the sides of this triangle in D', E, F. The line BE passes through B, the intersection of u = o, w = o ; it may .*. be represented (§ 126) by the equation lu + nw = o Abridged Notation 525 The line FE passes through E, the intersection of BE and V = o ; it may .-. be represented by the equation (lu + nw) + mv = o, or lu + mv + nw = o Q.E.D. § 345. The preceding general result is of course applicable to the particular case when the equations of the given lines are expressed in the standard form. Thus OC — o, /3 = o, y = o being the equations to the sides of a fixed triangle in the plane of reference, the equation to any other line may be expressed in the form la + m/3 + ny = o (1) This equation expresses a relation between the lengths of the perpendiculars from any point of the line on the sides of the given triangle (which may be called the triangle of reference). Trilinear co-ordinates are thus suggested ; the trilinear co-ordinates of a point are its perpendicular distances from the sides of the triangle of reference. Looking at equation (i) we see that any homogeneous equation of the first degree in trilinear co-ordinates represents a straight line. 8 34-6.* More generally, let the equations to the sides of the triangle of reference be u=Ax+By + C = o, v = A'x + B'y + C - o, w = A"x + B"y + C"= o Let Uj, v 1} w x denote the values of u, v, w respectively when we substitute. x i> y"i f° r x an d y; and u 2 , v 2 , W 2 their values when we substitute x 2 , y 2 for x and y. Then (§ 76) U-l = VA 2 + B 2 x perpendicular from (x x yj on u = o ; &c. Thus we may speak of the u, v, w of a point as its co-ordinates ; these co-ordinates being constant multiples of its perpendicular distances from the sides of the triangle of reference. § 34-7. To find the equation to the join of the points {kx-^]^-^), (u 2 v 2 w 2 ). Let the required equation be lu + mv + nw = (1) * The beginner may omit §§ 346-349, 354, 356, 357 until after he has read the early part of Chap. XIV. 326 Analytical Geometry [34 s - Then we must have lu x + mVi + nw x = o lu 2 + mv 2 + nw 2 = The eq'ns (2) determine the ratios I : m : n ; viz I m :} (2) V X W 2 - V 2 W! w x u 2 - w 2 u x u ± v 2 - U 2 V, If these values are substituted in (1) we obtain the required equation. The result may of course be written w w 9 = o § 34-8. It should be noticed that a homogeneous equation lu + mv + nw = o in reality expresses a relation between two of the ratios u : v : w ; for the equation may be written 1 u v I— + m — + n = o w w This is unaltered if we substitute ku, kv, kw for u, v, w. § 34-9- Ex. 1. We may apply these principles to prove the theorem of § 138. Let the equations to BC, CA, AB and FE be u = o, v ^= o, w = o, lu + mv + nw = o Then mv + nw = (1) is the eq'n to a line through the inters'n of v — - o, w = o, i. e. through A. But (1) may be written (lu + mv + nw) — lu — o The line (1) .-. passes through the inters'n of lu + mv + nw = o and u = o i. e. through D r . Hence (1) is the eq'n to AD r . Similarly the equations to BE, CF are lu + nw = o . . . (2) lu + mv = (3) Subtracting (2) from (3) we see that the equation mv — nw = (4) represents a line through the inters'n of BE, CF; i. e. through O. 349-] Abridged Notation 327 Hence (4) is the equation to AO. But the lines (1), (4) form a harmonic pencil with v = o, W = o [§139, Cor' (1)]. Ex. 2. ABC, A'B'C are two triangles. BC, B'C'meet in P; CA, C'A' in Q and AB, A'B' in R. If P, Q, R are collinear, show that AA', BB', CC are concurrent. Let the equations to BC, CA, AB, and PQR be a = o, /3 = o, y = o, la + m/3 + ny=o By suitably choosing V , the equation VOL + m/3 + ny = o (1) will represent any line through the inters'n of a = o, ICX + m/3 + ny = Hence we may take (1) as the equation to B'C Similarly the eq'ns to CA', A'B' are la + m'/3 + ny = o (2) I a + m /3 + n' y = o (3) Subtracting in pairs the eq r ns (1), (2), (3) we obtain the equations to AA', BB', CC; viz. (m - m') j3 - (n - n') y, (n - n r ) y = (I - I 7 ) a, (I - I') a = (m - m') ^3 The third of these equations is a consequence of the other two ; and .*. the three lines co-intersect. Exercises 1. What is represented by the equation a + C -■= o? Ans. A straight line parallel to OC = o. 2. If u = o, v = o represent parallel straight lines ; show that U + V = o represents a parallel midway between them. 3. If u = o, v = o, w = o are the equations of three parallel straight lines ; then the equation lu + mv + nw = o represents a straight line parallel to them. 328 Analytical Geometry [350. 4. With the notation of Ex. I, § 349, find the equations to DF, DE. Ans. lu + mv — nw = 0, lu — mv + nw = o 5. Prove that the lines v = Iw, w= mu, u = nv co- intersect if Imn = 1 6. The joins of the vertices of a triangle ABC to a point meet the opposite sides in A', B', C : if the equations of BC, CA, AB are u = o, v=o, w = o prove that AO, BO, CO may be represented by the equations mv == nw, nw = lu, lu = mv Prove also that BC, B'C ; CA, CA; and AB, A'B' meet on the line lu + mv + nw = o {Note — Show that equation to B'C is — lu + mv + nw = o; &c] 7. Two triangles are such that the perpendiculars from the vertices of one on the sides of the other are concurrent ; prove that the perpendiculars from the vertices of the second on the sides of the first are concurrent. [Note— Let sides be OC, ft, y, OL', fi f , /. Denote A between QC, (3 by (a/3). Then eq'n of J_ from (a/3) on y / is (§ 128) a cos (/3/) - /3 cos (a/) = o ; &c. The condition of concurrence is found to be cos (a/3') cos (/3/) cos (y a = cos (a'/3) cos (fi'y) cos (/a).] CASES OF S — AS'= O § 350. If S = o, S r = o are the equations of two conies; we have seen (§ 315) that S-AS'=o represents a conic through their intersections. We shall now consider some important cases of this equation. I. Suppose that one of the conies is a line-pair; it follows that — The equation S - ALM = 35i.] Abridged Notation 329 represents a conic passing through the points in which the conic S = o is met by the lines L = o, M = o. Ex. If TP, TQ, T'P', T'Q' are tangents to a conic, the six points T, P, Q, T 7 , P r , Q' lie on a conic. Let the given conic be x 2 /a 2 + y 2 /b 2 = 1 ; let (xV), (x"y") be the co-ord's of T, "P. Then the equation /x'x" y'y" \ /x 2 y 2 \ /x'x y'y \ /x"x y"y \ V1T + ~W ~ V Va~ 2 + b 2 " V = V^ + b 2 " " ' A"^ + "b 5 " " 7 represents a conic which evidently passes through the points (x = x', y = y r ) and (x = x /r , y = y /r ) ; and also (by the preceding principle) through the inters'ns of S — o with their polars. (Wolstenholme, Edutf Times, 6103.) § 351. Again, let P, Q be the points in which the conic S = o is met by the line L = o, and P 7 , Q' the points in which it is met by the line M = o ; and let us now suppose that P, Q move up to coincidence with P 7 , Qf respectively. Then the chords PP', QQ r ultimately become tangents; and M = o becomes ultimately the same line as L = o. We see then that — II. The equation S-AL 2 = o (1) represents a conic touching S = o at the points where it is met by the line L = o. That is, (1) represents a conic having double contact with S — o, "along the line L." Note — Instead of (1) we may write S - L 2 = o ; for the multiplier A. may be supposed to be implicitly included in L. Ex. 1. We may deduce the equation of the tangents from (x'y') to (x, y) = o 33° Analytical Geometry [352. Let P = ax'x + h (x'y + y'x) + ... so that P — o is the eq'n to the polar of (x'y'). Then <£ (x, y) = X P 2 (i) is a conic which has double contact with (f> (x, y) = o at its intersections with the polar of (x'y') ; and the two tangents are such a conic determined by the condition that it is to pass through (x'y'). Expressing then that (i) is satisfied by x = x', y = y f , This determines A. ; and the required equation is 4> M /) $ (x, y) = P 2 (Compare § 324) § 3S2> Ex. 2. If two conies have double contact with a third, then two of their common chords and the two chords of contact meet in one point and form a harmonic pencil. For, subtracting the equations S - L 2 = o, S - M 2 = o we obtain a conic through the intersections of these conies ; viz. L 2 - M 2 = o But this conic is a line-pair. Hence two of the common chords are L+M = o, L — M = o But these lines form a harmonic pencil with L = o, M = o [§ 139, Cor' (1)] Ex. 3. 7$i? two diagonals of an inscribed quadrilateral, and of the quadri- lateral whose sides are the tangents at its vertices, meet in a point and form a harmonic pencil. This is a particular case of Ex. 2 ; the conies S - L 2 = o, S - M 2 = o being now supposed to reduce to line-pairs. § 353. Again, as a particular case of II. it follows that — III. The equation LM = R 2 represents a conic touching the lines L = o, M = o at the points where it is met by the line R = o. 355-] Abridged Notation 33 1 If we suppose the equations to the lines written in the standard form and interpret the equation ^n _ ^2 we deduce the theorem — The product of the perpendiculars from any point of a conic on two tangents varies as the square of the perpendicular from the point on the chord of contact. § 354-. The equation of a conic referred to two tangents and their chord of contact LWj = R 2 is evidently satisfied by the co-ordinates (L/M'R') of a point, if these co- ordinates are in the ratios L' W _ R^ This may be called ' the point ^.' The equation to the join of the points fx, \if on the conic is (§ 347) L M R = o I jix' 2 \xl This reduces to fXfx'L - (fx + /x') R + M = o (1) Putting \x! = fA we get the equation of the tangent at fX, viz. jm 2 L - 2/xR + M = o (2) § 355. Both the conies S = o, S' = o may be line-pairs. We have then the theorem — IV. The equation ay = \l38 (i) represents a conic circumscribing the quadrilateral whose sides are 06 = o, /3 = o, y = o, 8 = In fact the eq'n is evidently satisfied by the co-ord's of the vertices of the quadrilateral, which are the points of inters'n of the lines (06 = o, = o), (ft = o, y = o), (7 = 0, 8 = o), (8 = o, 06 = o). Let these points be A, B, C, D ; and let 7T 13 7T 2 , 7T 3 , 7T 4 be the perpendiculars from any point P of the conic on the lines a = o, & = o, y — o, 8 = 0. S3 2 Analytical Geometry [356. Then (1) expresses that — - — - = constant = A (2) 77\>7T 4 Again, 2 area PAB = 7T 2 . AB = PA . PB sin APB .-. tt 2 = PA . PB sin APB/AB Similar expressions may be obtained for 7T 3 , IT 4, ir 1 . When these values are substituted in (2) it reduces to sinAPD.sinCPB _ sin APB. sin CPD Hence (§ 136) — The joins of a variable point on a conic to four fixed points on the conic form a pencil, whose cross ratio is constant. N 356. We may now interpret the equation XL 2 + jnM 2 + yN 2 = o (1) This may be written - v N 2 - (L VA + M V^jx) (L VX - M V~Z~jl) The lines L VA + M V- /x = 0, L VA - M V- fM = o . . . (2) are .\ tangents, and N = o is their chord of contact (§ 353). The pole of N = o is .*. the inters'n of the lines (2), i. e. it is the inters'n of L = o, M = o. Similarly with reference to the poles of L = o, M = o. The equation (1) .\ represents a conic which is such that the lines L = o, M = o, N = o form a triangle which is self-conjugate with respect to it. § 357. If u = o, v = o, w — o are the equations of the sides of a given triangle, the equation to any conic may be written in the form au 2 + bv 2 + cw 2 + 2 fvw + 2 gwu + 2 huv = o . . . (1) For let (XiYi), (x 2 y 2 ), ... (x 5 y 5 ) be five points on the conic; then expressing that the co-ord's of these points satisfy (1), and using the notation of § 346, we get five simple equations au^ + bv t 2 + ... = o, au 2 2 +...=- o, ... . These equations suffice to determine uniquely the values of the five ratios a/h, b/h, c/h, f/h, g/h. Q.E.D. 357-J Abridged Notation 333 Cor f — Suppose that the conic (i) circumscribes the triangle (u = o, v = o, w = o) If (x x y x ) are the co-ord's of the inters'n of u = o, v = o, the values of u , v, w for this point are o, o, w x . As these values satisfy (i) we must have cWi 2 = o ; .*. c = o Similarly a = o, b = o The general equation of a conic circumscribing the triangle (u = O, V = o, w = o) is .-. fvw + gwu + huv = o Exercises 1 . An ellipse touches the asymptotes of an hyperbola ; prove that two of the common chords of the ellipse and hyperbola are parallel. 2 . Find the equation of an ellipse passing through the centre of the ellipse x 2 /a 2 + y 2 /b 2 = 1 and touching it at two adjacent extremities A, B of its axes. Ans. b 2 x 2 + abxy + a 2 y 2 = ab (bx + ay) 3. Find the circle having double contact with the same ellipse at the ends of its latus rectum. Ans. x 2 + y 2 — 2 ae 3 x = a 2 (1 — e 2 — e 4 ) 4. A circle has double contact with an ellipse ; prove that the chord of con- tact is parallel to one of its axes. 5. A circle has double contact with an ellipse at the extremities of a parallel to the minor axis ; prove that the tangent to the circle from any point on the ellipse is to the distance of the point from the chord of contact in the constant ratio e : I. 6. Two circles have double contact with an ellipse, the chords of contact being parallel ; prove that the sum or difference of the tangents drawn to the circles from any point on the ellipse is constant. 7. A, B, C, D are given points on a conic. On a line which moves parallel to itself and cuts AB, CD in P, Q and the conic in P / , Q! a point O is taken such that p QQ = X OP'. OQ' 334 Analytical Geometry [358. where A is constant; prove that the locus of O is a conic passing through A, B, C, D. [Note — Let S = o, S' = o be eq'ns of line-pair AB, CD and given conic; use § 312.] THE FOCOIDS J RELATIONS OF CONICS TO THE LINE AT INFINITY § 358. Since parallels meet at infinity (§ 132) lines parallel to y = X V - I pass through a fixed point on the line at infinity. The fixed points in which the line at infinity is met by the lines y — x\/ — 1=0, y + x\/-i = o . . . (1) are called the circular points at infinity. The shorter term focoids has been suggested by Dr. C. Taylor. These points are also often referred to as the points I, J (Sal- mon, Higher Plane Curves). The lines (1) have been called the isotropic lines through the origin; they are evidently parallel to the isotropic lines through any other point (x'y'), viz. y — y' = ± (x — x') V — 1 We shall now prove some useful properties of the focoids. § 359. All circles pass through the focoids. For the equation of the circle X 2 + y 2 + 2 gx + 2 fy + c = o may be written (y + x V"^) (y - x V - 1) = (— 2gx — 2fy — c) (o.x + o.y + 1) It follows (§ 355) that the intersections of y ± x V — 1 = o and o.x+o.y+i=o are points on the circle. 363.] Abridged Notation 335 § 360. Every right angle is divided harmonically by the isotropic lines through its vertex. Take the arms OA, OB of a right angle AOB as axes ; then the equations of the line-pairs (OA, OB) and (01, OJ) are xy = o and x 2 + y 2 = o These form a harmonic pencil (§ 141). 8 361.* If the axes of co-ordinates are oblique, the circle x 2 + 2 xy cos a) + y 2 + 2 gx + 2 fy + c = o meets the line at infinity on the lines x 2 + 2 xy cos co + y 2 = o (1) Hence equation (i), or its factors y + x (cos co + V — 1 sin co) = o, y + x (cos co — V — 1 sin co) = o . (2) represent the isotropic lines through the origin. § 362. We can determine the cross ratio (p in which any angle is divided by the isotropic lines through its vertex. Take the arms of the angle as axes ; then the rays of the pencil are x = o, y = o and the lines given by eq'n (2) of the last §. Hence (§ 139) (j> = (cos co + V— 1 sin co)/(cos co — V — 1 sinco) .'. /— 1 into —V—i in(i); the condition that it should be a tangent differs .*. from (2) only in the sign of a/— 1. We may .-. equate to zero the real and imaginary parts of the sinister of (2). The foci are .'. determined as the intersections of the two loci C (x 2 - y 2 ) + 2 Fy - 2 G x + A - B = o, Cxy- Fx- Gy + H = o These equations represent equilateral hyperbolas, unless C = o. In the latter case the conic is a parabola ; and the preceding equa- tions represent straight lines. z 2>$% Analytical Geometry [369. § 369. It may be deduced from the last § that the equations to determine the foci are (C x - G) 2 - (C y - F) 2 = A (a - b) (Cx-G)(Cy-F) = Ah where A is the discriminant. The reader will verify this without difficulty. \ § 370. The process of § 368 is equivalent to this rule — To find the foci of a conic whose equation is given : Determine c so that y = mx + c (1) where m = V — 1 may be a tangent ; then equate real and imaginary parts on both sides of (1). S 371. If the axes are oblique the same rule is applicable; but the value of m is (§ 361) , . — (cos a) + V — 1 sm a)) Ex. Find the focus of the parabola Vx/ol + \/y//3 = 1 Using the condition of § 318, we find that y = — x (cos o> + V — 1 sin to) + c (2) is a tangent if c = 0Lfi/(0C + (3 cos to - (3 V^i sin go) Substitute this value of c in (2) and equate real and imaginary parts. This gives the co-ord's of the focus x = a/3 2 /(a 2 + 2 a^3 cos a) + /3 2 ), y = a 2 /3/(a 2 + 2 ocj3 cosa> + /3 2 ) [Compare the solution, Note, Ex. 18, page 303.] § 372. We shall describe another method of solving the problem of § 368. (1) To find the equation of the tangents to S = o which can be drawn parallel to a given line y = mx. These tangents are drawn from the point at infinity on the given line, viz. (A, m A) where A = 00 . Substitute .-. A, m A for x', y' in the equation of § 324 ; and neglect the terms which do not contain A 2 . We can then divide out by A 2 ; and the required equation is (a + 2 hm + bm 2 ) (x, y) = [ax + hy + g + m (hx + by + f)] 2 . . (1) 372.] Abridged Notation 339 (2) The tangents from the focoids are derived from this by writing + V — i for m ; and as the foci lie on the tangents from both focoids it follows that their co-ordinates satisfy the equations obtained by writing + V — i for m in (1) and then equating real and imaginary parts on both sides of the equation. The equations to determine the foci are .'. (ax + hy + g) 2 — (hx + by + f) 2 = (a — b)

(x, y) Exercises } 1. Find the real foci of the conic 20 x 2 — 32 xy + 20 y 2 = 9 Ans. (1, 1), (- 1, - 1) 2. Find the real foci of the conic 32 x 2 — 24 xy — 20 x + 12 y + 11 =0 Ans. (o, - 1), (i, 2) 3. Prove that the foci of the conic x 2 /a 2 + y 2 /b 2 = 1 are the intersections of the loci xy - o, x 2 - y 2 = a 2 - b 2 Prove also that the latter rectangular hyperbola is the locus of a point such that the tangents drawn from it to the conic are equally inclined to the bisectors of the angles between the axes. 4. Prove that the foci of the conic ax 2 + 2 hxy + by 2 = 1 are determined by the equations x 2 — y 2 xy 1 a - b h h 2 - ab 5. The real foci of an ellipse are S, S' ; its imaginary foci are t 4 -i)-2(ah-bky^7)/i 3 -2(ah + bky^T)/x = o If the roots of this are /Xj, /z 2 , /x 3 , /x 4 then /x 1 /x 2 /i 3 ^ 4 = - 1 (2) and 2/i r /x s = o (3) From (2) cos (a + /3 + y + 8) + v 7 ^ sin (a + /3 + 7 + 5) = — 1 ... a + /3 + y + 5 = (2n + i)?r . . . (4) Again, using (2), (3) becomes /. sin (ql + /3) + sin(/3+ y) + sin (7 + a) = o . . (5) The relations (4), (5) are often useful. § 375. The feet of the normals from (h k) to the parabola y 2 - 4ax = o (1) are (§236) its intersections with the hyperbola 2a(k-y) + y(h -x) = o . . . . (2) Ex. Find the circle through the feet of the normals from (hk). Multiply (1) by 2 y — k, and (2) by 2 x and subtract; the resulting equation evidently represents a locus passing through all the intersections of (1), (2). But this equation breaks up into factors, one of which y = o does not pass through the feet of the normals which are at a finite distance. The other factor 2 (x 2 + y 2 ) — 2 x (h + 2 a) — ky = o is the circle required. Note — The proper multipliers 2 y — k and 2 x could be found by trial ; or by assuming expressions with undetermined coefficients. 342 Analytical Geometry [376. CURVATURE § 376. Two conies intersect (§ 314) in four points. If two of these points coincide the curves touch. If three of them coincide the contact is closer. It is called contact of the second order ; and each of the curves is in this case said to osculate the other. Since three points determine a circle, there is a circle which osculates a conic at any point. This circle is called the circle of curvature at the point. If A is the point of contact the circle will meet the conic in another distinct point D : the chord AD is called the chord of curvature. If all four of the points of intersection of two conies coincide the contact is the closest possible : it is called contact of the third order. § 377. If a circle meets a conic in the four points A, B, C, D; then if we suppose that B, C coincide with A it follows from Cor* (3), § 311, that the chord AD and the tangent at A are equally inclined to an axis. Thus the chord of curvature and the tangent at any point are equally inclined to an axis of the conic. § 378. If OL, /3, y, 8 are the ecc' A s of the intersections of a circle and a conic, then (§ 311, Ex. 1) 0L + fi + y+8= 2MT If the points /3, y coincide with OC we have .*. 3O6 + 8 = 2n7T This equation determines the eccentric angle of the point in which the circle of curvature at 06 meets the conic again. Ex. If 8 is given there are three values of (X, viz. (2 tt - 5)/s, (4W - 8)/3, (6tt - b)/s Thus there are three, points on the conic the circles of curvature at which pass through the point 8 ; and since the sum of 8 and the three values of (X is 4 ir, the three points lie on a circle through 8. 379-] Abridged Notation 343 It is easily verified that the centroid of the triangle formed by the three points coincides with the centre of the conic. § 379. Let L = o be the equation of the tangent at (x'y') to a conic S = o. The intersections of this conic with the conic S - LM =0 (1) are (§ 350, I.) the points in which S is met by the lines L, M ; and as two of these points coincide, (1) is the equation of a conic touching S at (x'y'). Again, if M = Ix + my — Ix' — my' i. e. if M = o is a line through the point of contact (x'y'), then three intersections of the conies coincide with (x'y'). The equation of a conic osculating S at (x'y') is .*. S - L (Ix + my - Ix' - my') = ... (2) If further the line M coincide with L, then all four intersections of the conies coincide with (x'y') ; the equation of a conic having contact of the third order with S at (x'y') is .\ S - A L 2 = o Ex. 1. The equation of a conic osculating the conic ax 2 + 2 hxy + by 2 = 2y (3) at the origin is ax 2 + 2 hxy + by 2 — 2 y = y (Ix + my) .... (4) The conic (4) is a circle if a = b — m and 2 h — I = 2 a cos co The circle of curvature of the conic (3) at the origin is . \ a (x 2 + 2 xy cos go + y 2 ) = 2 y Ex. 2. The conic ax 2 + 2 hxy + by 2 — 2y — Ay 2 = o which has contact of the third order at the origin with (3), is a rectangular hyperbola if a+b-A = 2 hcosco The equation to the rectangular hyperbola having four-pointic contact with the conic (3) at the origin is .*. ax 2 + 2 hxy + (2 h cos a) — a)y 2 = 2y 344 Analytical Geometry [38c Exercises 1. Find the circle of curvature at the point (x'y') on the rectangular hyperbola x 2 _ y 2 = a 2 Ans. a 2 (x 2 + y 2 ) — 4X /3 x + 4y' 3 y + 3 a 2 (x /2 + y /2 ) = o 2. Find the equation of the chord of curvature through the point on an ellipse whose eccentric angle is (X ; the ellipse being referred to its axes. Ans. bx cos OC — ay sin OC = ab cos 2 OC 3. The radius of curvature at any point P on an ellipse is b' 2 /p, where W is the semi-diameter parallel to the tangent at P and'p is the central perpen- dicular on this tangent. 4. Find the co-ordinates of the centre of curvature corresponding to a point (x'y') on an ellipse referred to its axes. Ans. [(a 2 - b 2 )x' 3 /a 4 , - (a 2 - b 2 )y' 3 /b 4 ] 5. At the intersection of two confocals the centre of curvature of either is the pole with respect to the other of the tangent to the former at the inter- section. SIMILAR CONICS § 380. If two lines which are always inclined at the same angle revolve round two fixed points O, O'; and if P, P' are points on the lines such that cyp^ a op where A is constant, the points P, P' describe curves, one of which is a magnified representation of the other. Two such curves are said to be similar. To any point 12 rigidly connected with the first curve there corresponds a point Of similarly connected with the second curve. For taking a point Qf such that POX2 = P'O^ and , I2 / = A012: it follows by Euclid VI. 6, that &'P'= AX2P; 382.] Abridged Notation 345 and the lines SI P, QfP' are inclined at the same angle as OP, O'P'. The curves might .\ be similarly generated by radii vectores revolving round £2, Of. If the revolving lines OP, O'P 7 are parallel, the two curves are said to be similar and similarly placed. They are also said to be nomothetic. If O' coincide with O, and the revolving lines are also coincident, the point O is a centre of similitude. § 381. The polar equations of two parabolas referred to their vertices are rsin 2 = 4 a cos#, rsin 2 # = 4 a' cos If .*. p, p' are radii vectores inclined to the respective axes at the same angle 0, pip' = a /a' = constant. Hence all parabolas are similar. § 382. Let the equations of two central conies referred to their respective axes of figure be x 2 / a 2 - y 2 / b 2 = 1, x 2 /a' 2 - y 2 / b /2 = 1 If then p, p' are radii vectores inclined to the transverse axes at the same angle 6, ... . ,cos 2 sin 2 fl\ // cos 2 6 sin 2 fl \ P /?={-* &-)/{-** b*~) This ratio is independent of 6 if a/b = a'/D 7 Hence two conies are similar if the ratio of the axes is the same for both. Cor' (1) — Similar conies have the same eccentricity. For e 2 = 1 + b 2 /a 2 , e /2 = 1 + b' 2 / a /2 ; .-. e = e' Cor' (2) — The asymptotes of similar conies include the same angle. For this angle = 2 tan -1 b/a = 2 tan -1 b'/a r 346 Analytical Geometry [383- Cor' (3) — The asymptotes of homothetic conies are parallel. For an asymptote is inclined to the transverse axis at an angle tan -1 b/a; and the axes of the conies are parallel and proportional ; .*. &c. § 383. If two conies ax 2 + 2 hxy + by 2 +2gx+2fy + c = o a'x 2 + 2h r xy + ... = o are similar they have the same eccentricity \Cor f (1), § 382]. (a - b) 2 + 4 h 2 (a' - b') 2 + 4 h /2 _ -. ••• (a ; b)2 - V (a ' +b? [Ex. 20, p. 304.] The condition that the two conies may be similar is .*. ab- h 2 a'b'-h' 2 (a + b) 2 ~ (a' + b r ) 2 Note — This condition may be obtained otherwise. The line-pairs through the origin parallel to the asymptotes, viz. ax 2 + 2 hxy + by 2 = o, a'x 2 + 2 IVxy + b'y 2 = o include equal angles ; this gives at once (§ 114) the required condition. § 384-. If two conies are homothetic, the line-pairs through the origin parallel to their asymptotes are coincident. The conditions .*. that two conies ax 2 + 2 hxy + ... = o, a r x 2 + 2h'xy + ... = o may be homothetic are a/a r = h/h r = b/b' Cor* — The equations of two homothetic conies may be so written that the terms in x 2 , xy, y 2 are the same in both. S 385. Corresponding radii vectores of two conjugate hyperbolas are (§ 277) in the constant ratio 1 : 1/ — 1 ; these curves are not of the same shape. That two similar curves may have the same shape, the ratio 1 : A of corre- sponding radii vectores must be real. 388.] Abridged Notation 347 § 386. The equations of two nomothetic conies are S = o, S' = S + Ix + my + n = o {Cor*, § 384-] .*. S' = S + (Ix + my + n) (o . x + o . y + 1) Hence (§ 350, II.) two homothetic conies have the line at infinity for a common chord. Note — This proposition is evident when the curves are hyperbolas. The reader will see this on drawing a figure : the curves continually approach the asymptotes ; and since the asymptotes are parallel they meet at a point on the line at infinity. The result of § 359 is a particular case of this proposition. 8 387. If two conies are homothetic and concentric, then taking the centre as origin their equations are S = ax 2 + 2 hxy + by 2 + c = o, S' = ax 2 + 2 hxy + by 2 + c f = o .-. S' = S + (o . x + o . y + r) 2 (c' - c) Hence (§ 351, II.) two homothetic and concentric conies have double contact along the line at infinity. Note — As the curves have the same asymptotes the proposition is evident when the curves are hyperbolas ; the reader will see this by drawing a figure. Co?' — Concentric circles have double contact along the line at infinity. Exercises 1. A chord of an ellipse which touches a concentric and homothetic ellipse is bisected at the point of contact. 2. A point P describes a conic ; O is a fixed point. Prove that the locus of the mid point of OP is an homothetic conic. 3. Two homothetic and concentric conies intercept equal segments on any line. 4. TP, TQ are tangents to an ellipse whose centre is C ; prove that an homothetic ellipse can be drawn through the points T, P, Q, C. INVARIANTS S 388. Some problems are readily solved by using the invariants. Ex. Find the equation of the equi-conjugate diameters of the conic S = ax 2 + 2 hxy + by 2 +2gx + 2fy + c = o Let (qrf) be the co-ord's of a point referred to the axes of the conic. . 34$ Analytical Geometry [389. Equating two expressions for the distance of a point from the centre, f 2 + ff E (x - G/C) 2 + (y _ F/C) 2 (1) Also (see § 292) S = a£ 2 + /3?? 2 + A/(ab - h 2 ) (2) and a + b = a + /3 (3) ab - h 2 = a/3 (4) Now the equation of the equi-conjugates is 0i£ 2 - /3t? 2 = o This may be written (a + /3) (af 2 + /3r? 2 ) = 2 a/3 (£ 2 + tj 2 ) Substituting from (1), (2), (3), (4) we obtain the req'd eq'n, viz. (a + b) [S - A/(ab - h 2 )] = 2 (ab - h 2 ) [(x - G/C) 2 + (y - F/C) 2 ] ONE-ONE CORRESPONDENCE § 389. If two variables x, x' are so related that to any value of either corresponds one and only one value of the other we may assume that they are connected by a relation of the form xx' + JJLX + vy! + p = O (1) This principle is often useful. Ex. TP, TQ are tangents to a conic ; any tangent meets TP in L and a parallel tangent meets TQ in M. Prove that TL . TM is constant. {Prof Genese.) It is evident that each of the points L, M determines the other uniquely. The lengths TL, TM have .-. a (1, 1) correspondence; and if TL = x, TM =x' we may assume that x and x r are connected by the relation (1). But when L is at T, M is at infinity ; .'. x r = 00 when x = o. Substituting these values in (1), v — o ; similarly fM = o. Thus (1) becomes xx / , _ _ Q Exercises 1. PP' is a given diameter of a conic; Q is any point on the curve. If PP', QP, QP r meet a fixed parallel to the tangent at P in O, p, p', prove that Op . Op' is constant. 39i.] Abridged Notation 349 2. An ellipse is inscribed in a parallelogram ; any tangent meets the sides opposite to an angle L in P, Q. Prove that the area of the triangle LPQ is constant. {Prof Genese.) § 390. If two variable points P, Q on a given line have a (i, i) corre- spondence ; their distances x, x' from a fixed point O on the line are connected by a relation of the form xx' + fJLK + vx' + p = o (i) If further P and Q are interchangeable, then //, = v\ , .: xx' + fJL (x + x f ) + p = o (2) .*. (x + fA) (xf + fX) = constant Hence if M be a point on the line such that OM = — fi, MP . MQ = constant Thus the points P, Q determine an involution whose centre is M. Cor' — Put x = x' in (2) ; the foci of the involution are .\ determined by x 2 + 2/ixx + p = o theeo/n 8 391. A system of conies passing through four fixed points determines an involution on any transversal. {Desarguei Theorem.) For let one of the conies meet the transversal in P, P' . Then P, P' have a (1, 1) correspondence and are interchangeable; .'. &c. The theorem may also be proved thus. Take the transversal as axis of x ; let two of the conies be S = o, S' = o. The points in which the conic S + kS' = o meets the axis of x are determined by the equation ax 2 + 2 gx + c + k (a r x 2 + 2 g'x + c') = o . . . . (1) But the points determined by ax 2 + 2 gx + c = o, a'x 2 + 2 g r x + c' = o are in involution with the points (1). This is proved as in § 148. Cor' (1) — The foci of the involution are the points of contact of the trans- versal with the two conies of the system which touch it. Cor' (2) — As the line-pairs through the four fixed points are conies of the system, we see that — The sides and diagonals of a quadrilateral determine an involution on any transversal. 35° Analytical Geometry [392. PLANE PROJECTION § 392. The principle of § 242 may be extended to points not on the ellipse : thus to any point P (x x y x ) in the plane of the ellipse corresponds a point P', viz. (x 15 Ay 2 ) where A = a/b. It is convenient to call P' the projection of P ; if P describe a straight line or curve, P' describes a straight line or curve which is the projection of the locus of P. Thus the projection of the curve (f) (x, y) = o is the curve (f> (x, y/A) = o This use of the term projection was suggested by Dr. C. Taylor ; its justifi- cation is that the theorems thereby deducible are identical with those obtained by the method of orthogonal projection. (See Taylor's Geometry of Conies, § 90.) § 393. If P', Q', R' are the projections of P, Q, R, then A P'Q'R' = A A PQR. For Xy 1 1 Ay 2 1 Ay 3 1 = A *i yi 1 x 2 y 2 1 x 3 ys * It follows that the proj'ns P r , Of, R' of three collinear points P, Q, R are collinear ; or the projection of a straight line is a straight line. S 394-. Let P, Q be two points on a curve ; P r , O! their projections. Then P' coincides with Q' if P coincides with Q. Hence the projection of the tangent at P is the tangent at P'. § 395. We have seen \Cor f (1), § 256] that conjugate diameters of the ellipse project into rectangular diameters of the auxiliary circle. § 396. If we have four points P, Q, R, S whose co-ord's are (x^), (x 2 y 2 ), ... : then PQ is || RS if (yi-y 2 )/(Xi-x 2 ) = (y 3 -y 4 )/(x 3 -x 4 ) (1) and in that case PQ/RS = ( yi -y 2 )/(y 3 -y 4 ) (2) But the cond'n (1) and the dexter of eq'n (2) are unaltered if we change yx, y 2 > ... into Ay lf Ay 2 , .... 39 6 -] Abridged Notation 351 Hence parallel lines project into parallel lines ; and the ratio of segments measured on the same line or on parallel lines is unaltered by projection. Ex. We shall apply these principles to prove the theorem, Ex. 39, page 239. Let P, Q be the points whose ecc' As are a, j3; Cp the senii-diam' parallel to PQ. Let P' , Q!, p' be the projections of P, Q, p. Then PQ/Cp = P'Q'/Cp' = P'Q'/a But P'Q' = 2 CP' sin \ P'CQ' = 2 a sin $ (a - /3) .-. PQ = 2 b' sin \ (a - /3) Exercises 1. Deduce from properties of the circle the theorems, Ex. 20, 25, page 237 ; Ex. 46, page 240; Ex. 52, 53, page 241. 2. Show that the projection of a parabola is a parabola. 3. A triangle is inscribed in the ellipse (a, b) ; b', b", b ;,/ are the semi- diameters parallel to the sides. If R is the circum-radius of the triangle, prove that . R = b'b"b'"/(ab) Miscellaneous Exercises 1. A circle intercepts segments of given length on two fixed lines : show that the locus of its centre is a rectangular hyperbola. 2. The tangents to a parabola at P, Q meet in T ; any other tangent meets TP, TQ in p, q. Find the locus of the intersection of Pq, Qp. Ans. An ellipse touching TP, TQ at P, Q. 3. P, Q are two points on an hyperbola ; parallels through P, Q to the asymptotes meet in T. If PQ passes through a fixed point, prove that the locus of T is an homothetic hyperbola. $$2 Analytical Geometry 4. A parabola whose latus rectum is 4 a slides between two rectangular axes ; find the curve traced by its focus. Ans. x 2 y 2 = a 2 (x 2 + y 2 ) {Note— Find lengths of -Ls on y = mx + a/m, y= — x/m — am from (a, o) ; call these lengths x and y. Then eliminate m.] 5. Find also the curve traced by the vertex. Ans. x^ys + x» y» = a 2 6. If one of the common chords of a circle and a rectangular hyperbola is a diameter of the circle ; prove that another common chord is a diameter of the hyperbola. 7. Prove that if a parabola be described with a point on an ellipse as focus, and the tangent at the corresponding point on the auxiliary circle as directrix, it passes through the foci of the ellipse. {Prof J. Purser?) 8. A series of parabolas whose axes are parallel have a common tangent at a given point. Prove that if parallel tangents are drawn to the parabolas the points of contact lie on a straight line through the given point. 9. If p, p' are the perpendiculars from the foci of an ellipse upon any chord, R the semi-diameter parallel to the chord, prove that length of chord = ■ a/j _ PP_^ a v b 2 10. CP, CD are conjugate radii of an ellipse, AA', BB' the axes. Prove (1) that PA . PA' = DB . DB'; and (2) the bisectors of the angles APA', BDB' are at right angles. {Prof Genese.) 1 1 . Given a focus, a tangent and the eccentricity of a conic ; prove that the locus of the other focus is a circle. 12. Tangents from any point T meet an ellipse in P and Q, so that, S being a focus, SP. SQ cc ST 2 . Prove that the locus of T is a similar ellipse. Miscellaneous Exercises $53 13. Given a focus and two tangents to a conic; prove that the chord of contact passes through a fixed point. 14. If a parabola always touches a given straight line and has double contact with a fixed circle, then the chord of contact passes through a fixed point. 15. Show that the normals at the points of intersection of the ellipse (a, b) with the polars of the points (x'y'), (x /f y") are concurrent if x'x"/a 2 = yY'/b 2 = - i 16. The normals at the points P, Q, R, S on the ellipse (a, b) are concurrent. Prove that two parabolas can be drawn through P, Q, R, S, and that the angle between their axes is 2 tan~ 1 b/a. 17. Four normals to an ellipse are concurrent. Prove that perpendiculars to these normals from an extremity of the axis major meet the ellipse in four points which lie on a circle. [Mte—The ecc' As of the extremities of the chord A. normal at the point whose ecc' A is OC are o, 2 (X ; use eq'n (4), § 374 and § 311, Ex. 1.] 18. Show that the equation of the normal to the conic xy = k 2 at the point (kA, k/A) is (A 3 - A cos oo) x + (A 3 cos co - A) y = k (A 4 - 1) where a) is the angle between the axes of co-ordinates. If the normals at the points P, Q, R of the above conic meet at a point on the curve, prove that the centroid of the triangle PQR lies upon the conic 9 xy = k 2 cos 2 a) 19. If the normals to a conic at L, M, N, R meet in O, and S is a focus, then e 2 SL.SM.SN.SR = b 2 .S0 2 {Prof Burnside, Educ' Times, 1708.) 20. Show that the product of the three normals that can be drawn from the point (x^) on the ellipse (a, b) is 2 ab(a 2 -e 2 x l 2 )t/(a 2 - b 2 ) 21. The sum of the squares of the four normals from a point to an ellipse is constant ; prove that the locus of the point is a conic. A a 354 Analytical Geometry 22. If the circle of curvature of the ellipse (a, b) at the point whose eccen- tric angle is passes through the centre of the ellipse, prove that tan 2 <£ = (a 2 - 2 b 2 )/(2 a 2 - b 2 ) 23. If p, p' are the radii of curvature at the extremities of two conjugate diameters of the ellipse (a, b), then pi + p'f = (a« + V)/(pb)i {pnf Curtis) 24. Any chord of curvature of a parabola is divided by the axis in the ratio 1 ' 3- {Prof Curtis.) [Note — The sum of the ordinates y l , y 2 , y 3 , y 4 of the intersections of a circle with the parabola is zero ; put y 2 = y 3 = y 4 » &c.] 25. Find the equation of the parabola which meets the ellipse (a, b) in four coincident points (x'y') ; and show that the equation of its axis is x/x' - y/y' = (a 2 - b 2 )/(x' 2 + y /2 ) 26. Prove that the centres of curvature at the points (Xj y^, (x 2 y 2 )> (x 3 y 3 ) on the parabola y 2 = 4 ax are collinear if yiy 2 +y 2 y 3 + y 3 yi = ° 27. From O, the centre of curvature at any point on the ellipse (a, b), the other normals OQ, OR are drawn to the ellipse; prove that the locus of the mid point of QR is (x 2 /a 2 + y 2 /b 2 ) 3 = x 2 y 2 /(a 2 b 2 ) 28. Show that the latus rectum of the parabola which has contact of the third order at P with an ellipse, whose centre is C and axes 2 a, 2 b, is 2 a 2 b 2 /CP 3 29. The locus of the pole of a given straight line with respect to one of a given system of co-axal circles is a hyperbola whose asymptotes cut the line of centres in points equidistant fr.om the radical axis, and which becomes two straight lines, if the given line passes through one of the limiting points. 30. Two circles have double internal contact with an ellipse, and a third circle passes through the four points of contact. If t, t/, T be the tangents from any point on the ellipse to these circles, then T 2 = tt' {Prof Crofton, Educ* Times, 1994.) Miscellaneous Exercises 355 31. Show that a tangent to the conic x 2 /(b + b') + y 2 /(a + a') = i/(ab' + a'b) is cut harmonically by the conies ax 2 + by 2 = i, a'x 2 + b'y 2 = i 32. If the normals to the conic I /r = i — e cos at the points 2 OC, 2 /3, 2 y meet on the curve, then (1 + e) 2 cos (OC + j3 + y) = 2 e cos a cos /3 cos y 33. Prove that all conies through the extremities of the principal axes of the ellipse (a, b) are cut orthogonally by the hyperbola x 2 /a 2 - y 2 /b 2 = (a 2 - b^a 2 + b 2 ) {Prof Crofton, Edud Times, 1602.) 34. A series of parabolas have AB, the hypotenuse of a given right-angled triangle ABC, for a common chord, whilst their axes are parallel to the side AC. Prove that their foci all lie on a hyperbola of which A and B are the foci; and show that if the triangle is isosceles, the hyperbola will be equi- lateral. 35. The centre of a conic and two tangents are given ; prove that the locus of the foci is a rectangular hyperbola. 36. The rectangle under the tangents from (x^) to the ellipse (a, b) is ^^ V(x 2 - y 2 - a 2 + b 2 ) 2 + 4X 2 y 2 Oj + I where Sj = x^/a 2 + y^/b 2 - 1 Hence find the locus of a point such that the rectangle under the tangents from it to a given conic is equal to the rectangle under the tangents to a given confocal conic. {Prof J. Purser.) 37. From a fixed point O a tangent OT is drawn to one of a system of confocal conies, and a point P is taken on the tangent, such that OP. OT is constant ; prove that the locus of P is a rectangular hyperbola. (Prof J. Purser.') A a 2 356 Analytical Geometry 38. PQ is the chord of contact of tangents OP, OQ to a conic whose centre is C, and foci S, S', and OC cuts PQ in V ; prove that OP.OQ :OS.OS' = OV:OC 39. A circle passes through the focus of a parabola and cuts the parabola at angles whose sum is constant. Prove that the locus of its centre is a straight line. 40. Two fixed tangents OP, OQ to a parabola are met by a variable tangent in P and Q ; prove that the locus of the circum-centre of the triangle OPQ is a straight line. 41 . Prove that normals drawn to the ellipse (a, b) from a point on either of the straight lines a 2 x 2 - b 2 y 2 = o meet the curve in points, a pair of whose connectors are parallel. 42. To a rectangular hyperbola with centre C and focus S normals are drawn from a point P. Show that, if these normals make angles 1} 2 , ... with one of the asymptotes, 2cosec2 = CP 2 /CS 2 43. If from any point on a given normal to a conic three other normals be drawn ; prove that the circle through their feet belongs to a fixed co-axal system. {Mr. F. Purser.) 44. Through a fixed point O within an ellipse is drawn any pair of conju- gate lines, and parallel to these are drawn a pair of tangents meeting them in R and R' and one another in T. Prove that the locus of T is an ellipse, and that the locus of R and R' is a similar ellipse which has double contact with the given ellipse at the extremities of the diameter through O. 45. If the line joining P, the intersection of two tangents to an ellipse to the intersection of the corresponding normals is cut by the axes of the ellipse in a constant anharmonic ratio, the locus of P is a concentric conic. {Prof Purser.) 46. If three sides of a quadrilateral inscribed in a conic pass through three fixed points in a straight line, prove that the fourth side passes through a fixed point in the same straight line. [Note — The given line is cut by the conic and sides of quad' in six points in involution (§ 391) ; and as five of these points are fixed, so is the sixth.] Miscellaneous Exercises 357 47. If — _ 3 12 — ; prove that the equation x y 12 = - cos I (OCx + 0L 2 ) + r sin £ (a x + (X 2 ) — cos | (0^ — CX 2 ), &c. £1 D 12.34.56-23.45.61=0 (i) includes as part of its locus the ellipse (a, b). What is the remaining part of the locus? {Prof Purser.) {Note — We find that the sinister of (i) vanishes identically if we substitute a cos 0, b sin for x, y ; we infer that x 2 /a 2 + y 2 /b 2 - i is a factor. The other part of the locus (i) is a straight line through the points ( I2 > 45)> (34> 61), (56, 23) This affords a proof of Pascal's Theorem (see § 431).] 48. Interpret the equation (a + 1 8) (fi + m 8) (y + n 8) = (a - 1 8) (/3 - m 8) (y - n 8) Ans. The equation represents the line 8 = and the conic I /3y + m yOC + n a/3 + Imn 8 2 = o This conic passes through the six points (0t+ 18 = o, p- m8 = o), ... 49. TP, TQ are tangents to a conic. The circle through T, P, Q cuts the conic again in P', Of ; tangents at P r , Q' meet in J'. Show that CT, CT 7 are equally inclined to the axis, and that CT. CT' = CS 2 , C being the centre, and S a focus of the conic. {Leudesdorf, Educ' Times, 6103.) {Note — If T is (x'y'), and T (x"y"), the equation of the conic through T, P, Q, T 7 , P', C^ is given, Ex., § 350. Express cond'ns that this eq'n represents a circle.] 50. TP, TQ are tangents to a parabola ; the circle through T, P, Q meets the parabola again in P', Q r , and the tangents at P', Q' meet in T f . Prove that TT r passes through the focus. 51. T and T' are points external to a parabola ; TP, TQ, T'P', T r Q r are tangents to the parabola. If TF is bisected by the parabola, prove that the six points T, P, Q, T', P' ', Q r lie on another parabola. 35$ Analytical Geometry 52. A conic is drawn through a point P and the feet of the normals from it to an ellipse. Show that its centre is the centroid of the points in which a circle of any radius meets the ellipse, the centre of the circle being at P. 53. PQ, P'Q' are chords of a conic, normals at P and P 7 ; the tangents at P and P' meet in T, and the tangents at Q and Q' meet in T 7 . Prove that if T describes a straight line, so does "P. 54. If the axis of a parabola which touches two fixed lines passes through a fixed point, prove that the focus lies on a rectangular hyperbola. 55. If the chords PQ, P'Q' of a conic are such that each contains the pole of the other, prove that the points P, Q, P', Of subtend a harmonic pencil at every point of the conic. \Note— Let tangents at P, Q meet in T ; let PQ meet P'Q' in "P. Take P for the fifth point on the conic ; then one ray of the pencil is the tangent PT. Then {P.PP'QQ'} = {TP'T'Q'} = - i C§ 308).] 56. If the origin is within the quadrilateral formed by the lines OC, (3, y, h, determine the value of k in order that the equation (X/3 = k yb may represent a circle. Ans. k = — 1 57. A line through the focus of a conic meets the tangents at the ends of the transverse axis in P, Q : prove that the circle whose diameter is PQ has double contact with the conic. / D & r* \ [Prof Genes e.) 58. The directrices are common chords of a conic and its director circle. {Prof Genese.) 59. A circle cuts a rectangular hyperbola in P, Q, R, S. Prove that the join of their centres is bisected by the centroid of P, Q, R, S. {F. D. Thomson, Educ' Times, 5995.) 60. If five points lie on a circle, radius r, prove that the centres of the five rectangular hyperbolas which pass through them, taken four and four together, lie on another circle whose radius is r/2. (J. G?'iffilhs, Educ r Times Reprint, V., page 56.) 61. An ellipse has double contact with each of two concentric circles ; show that the loci of its centre and its foci are circles. Miscellaneous Exercises 359 62. Two supplemental chords of an ellipse PQ, P'Q meet the tangents at P' and P in ~P, T respectively ; prove that PT.P'T' is constant. 63. A conic is drawn passing through two given points and having double contact with a given conic ; prove that the chords of contact pass through one or other of two fixed points. 64. A point describes a straight line ; prove that the locus of the intersec- tion of its polars with respect to two fixed conies is a third conic. 65. If the common tangents of the curves y 2 = 4 ax, x 2 + (y — b) 2 = r 2 make angles OC, /3, y, b with the axis of x, prove that 2 (tan a) = o 66. Show that the equation C tan 2 a — r (sec OC + cos 8) where OC is a variable parameter, represents a system of confocals. 67. A fixed tangent to an ellipse meets the major axis in T; Q and Q' are two points on the tangent equidistant from T. Show that the second tangents which can be drawn from Q, Q' to the ellipse meet on a fixed straight line parallel to the major axis. 68. A variable circle whose diameter is D passes through a fixed point O and intersects a given conic in L, M, N, R : prove that OL.OM.ON.ORoc D 2 69. Two concentric circles are cut by a conic in the points M, N, P, Q and M', N', P', Of respectively: show that a conic can be described through M, N, P, Q whose asymptotes are M'lSK, P'Q'. 70. A line through P cuts a conic in Q, Q'. If N is the foot of the per- pendicular from P on the polar of P, prove that NQ, NQ' make equal angles with NP. 71. An ellipse of semi-axes a, b slides between two rectangular axes. Prove that the equation to the locus of its real foci is y 2 (x 2 - b 2 ) 2 + x 2 (y 2 - b 2 ) 2 + 2 xy (x 2 - b 2 ) (y 2 - b 2 ) = 4 x 2 y 2 (a 2 - b 2 ) 360 Analytical Geometry 72. PP' is a diameter of a conic. If from points on a fixed chord through P / tangents be drawn meeting the tangent at P in R, R' : prove that PR + PR' is constant. 73. A system of conies passes through four fixed points, of which P, Q are two ; PX, QY are two fixed lines meeting any conic of the system in X, Y. Prove that XY passes through a fixed point. 74. The polar of a point P with respect to an ellipse always touches a fixed circle whose centre is on the major axis, and which passes through the centre of the ellipse. Prove that the locus of P is a parabola whose latus rectum is a third proportional to the diameter of the circle and the latus rectum of the ellipse. 75. Prove that the polar of the origin with respect to the general conic S = o is a normal if f 2 (f 2 - be) + 2 fg (fg - he) + g 2 (g 2 - ac) = o 76. Prove that the two conies ax 2 + 2 hxy + by 2 = i, a'x 2 + 2 h/xy + b'y 2 = 1 may be placed so as to be confocal if (a - b) 2 + 4 h 2 ^ (a'- b') 2 + 4 h' 2 (ab - h 2 ) 2 (a'b'-h' 2 ) 2 77. O is a given point on a conic; OP, OQ are any chords through O. Parallels through P, Q to OQ, OP meet the conic in R, S. Prove that RS is parallel to the tangent at O. 78. Two similar conies have a common focus. Prove that a pair of their common chords intersect at right angles. 79. Through a fixed point O on a conic a chord OP is drawn; the circle whose diameter is OP cuts the curve again in Q, Qf. Prove that QQ' passes through a fixed point. 80. With the notation of Ex. 22, page 305, the equation of the joins of the extremities of the axes of S = o is 2 h (u 2 - v 2 ) - 2 (a - b) uv = + (ab - h 2 )z {(a - b) 2 + 4 h 2 }?. S 81. Prove that the locus of the extremities of the principal axes of conies through the four points (+ a, o), (o, + b) is the curve (x 2 /a 2 - y 2 /b 2 ) (x 2 + y 2 ) = x 2 - y 2 Miscellaneous Exercises 361 82. Prove that the product of the latera recta of the two parabolas which can be described through four concyclic points is iW-d^sinco where d x , d 2 are the diagonals of the quadrilateral ; co the angle between them. 83. Prove that an ellipse which has double contact with two fixed confocals has a fixed director circle. 84. Tangents are drawn to the parabola 2 a = r (1 + cos 6) at the points 2 Of, 2/3, 2 y, 2 § : circles are described about the triangles formed by the tangents taken three at a time : and circles are described through the centres of these circles taken four at a time : show that the five centres of these circles lie on the circle 4 r cos (X cos /3 cos y cos 8 = a cos (6 — OC — /3 — y — h) 85. The third diagonal of a quadrilateral inscribed in a circle and circum- scribed to a parabola passes through the focus of the parabola. 86. Four rectangular hyperbolas osculate each other at a certain point. Prove that any other conic osculating these at the same point is cut by them in four other points which have a fixed anharmonic ratio. {Prof Curtis.) 87. P is a point on a rectangular hyperbola whose centre is C. A circle is described with centre P and radius PC. Prove that an infinite number of triangles can be inscribed in the circle whose sides touch the hyperbola. 88. An infinite number of triangles can be inscribed in the ellipse (a, b) and circumscribed to the ellipse (a', b/) provided a'/a + b'/b + 1=0 {Wolstenholme, Educ' Times Reprint, Vol. XV., p. 43.) CHAPTER XIV TRILINEAR CO-ORDINATES § 397. In the plane of reference take a fixed triangle ; this is called the triangle of reference. As in Trigonometry, its sides are denoted by a, b, C and its angles by A, B, C. We shall denote its area by D, and the radius of its circum-circle by R. The trilinear co-ordinates 06, /3, y of a point P are its perpendicular distances from the sides of the triangle of reference : 06 is positive or negative according as P is on the same side of BC as A, or on the opposite side ; and there is a similar agreement with respect to the signs of /3, y. If P is within the triangle ABC, A PBC + A PCA + A PAB = A ABC .-. ao: + b/3 + cy = 2D (1) By examining figures in which P is outside the triangle ABC, it is seen as a consequence of the agreement as to the signs of C*j A y, that the trilinear co-ordinates of any point whatever are connected by the relation (1). The relation (1) may be written in the form a sin A + /3 sin B + 7 sin C = D/R = constant . . (2) Note (1) — Any equation may be rendered homogeneous by means of (1). For example, the equation «/3 2 = ky becomes 4 D 2 a/3 2 = k y (a OL + b /3 + c y) 2 Note (2) — The equations used in trilinears are always homogeneous. We are .-. in general concerned only with the ratios OC : (3 : y. Trilinear Co-ordinates 363 Ex. The equations to the perpendiculars of the triangle ABC are (§ 128) OC cos A = /3 cos B, (3 cos B = y cos C, y cos C = OC cos A These lines obviously co-intersect in the point a/secA = /3/secB = y/secC (3) These equations give the ratios of the co-ordinates of the ortho-centre. Again, by a familiar algebraic theorem, each of the fractions in (3) = (aa+b/3 + c y)/(a sec A + b sec B -1- c sec C) = 2 D /(a sec A + b sec B + c sec C) = I say ; the trilinear co-ord's of the ortho-centre are .'. I sec A, I sec B, I sec C. § 398. We have seen in § 345 that trilinear co-ordinates are suggested by the methods of abridged notation. It is easy to transform from trilinear to Cartesian co-ordinates. Thus, taking the origin of rectangular axes inside the triangle of reference, so that the equations of its sides are OC = xcosa + y sin a — p = o, j3 = ... = o, yE...=o; then since the trilinear co-ordinates of a point inside the triangle are positive, we have the following equations connecting the trilinear co-ordinates OC, /3, y of any point with its Cartesian co-ordinates x, y : OC = p — x cos OC — y sin OC \ £ = p' _ xcos/3 -ysin/3 > (1) y = p" — x cos y — y sin y J Further, since OC, (3, y in the dexters of (1) are the angles made by perpen- diculars on the sides of the triangle ABC with a fixed line, these angles being measured by the amount of rotation in a definite direction from that line ; it will be seen from a figure that we may take (3-0C = 7r -C s y-/3 = TT-A, a-y=-(7T + B) Cor'— cos Q3 - y) = - cos A, cos (y - OC) = - cos B, cos (OC - /3) = - cos C § 399. There is another method of transforming to Cartesians. Let (x, y) be the co-ord's of the point (OC, /3, y) referred to two sides CA, CB of the triangle of reference as axes. We see at once from a figure that a = x sin C, /3 = y sin C It follows then from eq'n (2), § 397, that y = D/(R sin C) - x sin A - y sin B 364 Analytical Geometry [400. § 4-00. The general equation to a straight line in trilinears is(§345) i a+ m/3 + n r = o (.) It is easy to assign the geometric meanings of the ratios I : m : n. Let the line (i) meet the ^ sides of the triangle of refer- ence in P, Q, R ; let the _Ls flXPX/ \ AL ' BM > CN on the line be "n/ \ denoted by p, q, r. We shall assume that Xs VP from points on different sides of the line have opposite signs ; 13 C P^v thus the signs of q, r in our figure are opposite to that of p. The co-ord r s of R are determined by y = o, l(X + m/3+ny = o .-. l/m = - /3/a = - (RA sin A)/(BR sin B) = (AL sin A)/(BM sin B) ; .-. l/m -- ap/(bq) ; and a similar value is found for n/m. The equation to a line in terms of the perpendiculars p, q, r on the line from the vertices of the triangle of reference is .*. apa + bq (3 + cry = o (2) Cor' — The ratios of the J_s from A, B, C on a line very distant from the triangle of reference are nearly = i ; hence for the line at infinity (see page 107) p = q = r The equation to the line at infinity is .\ aa + b/3 + cy = o (3) or Oi sin A + /3 sin B + y sin C = o (4) § 4-01. To find the equation to the join of two points (Otj /3 X y^), (a 2 &y 2 ). Let the required equation be loc+ m/34- ny = o Then I Oj + m ft + n y x = o, la 2 +mft+ny 2 = o 4 y 2 Cor' — The condition that three points (^^y^, (0L 2 {5 2 y 2 \ (#3/3373) may be collinear is «i ft 7i = ° «2 ft 72 « 3 ft y 3 § 4-02. The co-ordinates of the point of intersection of two lines la + m^+ny = o, Vol + m'/3 + n'y = o are determined (see § 61) by the equations a /3 y mn m'n "" nl'- n'l . Im'-I'm These equations give the ratios of the OC, (3, y of the point of intersection ; their actual values may (if required) be deduced as in Ex., § 397. & 4-03. To find the condition that three lines I OC + m ft + n y = o, l'0( + m / /3 + n'y = o, I" a + m" (3 + n" y = o may be concwrent. Eliminating OC, /3, y we obtain the required condition, viz. m m' I" m" n' § 4-04. It follows from eq'n (2), § 397, that the equation of a line parallel to la+m/3 + ny = o (1) is I a + m /3 + ny = k(asin A 4- /3sin B 4- y sin C) Cor' — The equation of the line through (0C x fi x yi) parallel to (1) is I OC + m /3 + n y OC sin A + /3 sin B + y sin C I 0C X + m /3 X + n y x 0^ sin A + & sin B + y t sin C 366 Analytical Geometry [405. § 4-05. Suppose that the third line in § 403 is the line at infinity. This gives (§ 132) the condition that the lines la + m/3 + ny = o, V oi + m' /3 + n' y = o may be parallel ; viz. 1 m n V m' n' sin A sinB sinC = o § 4-06 . To find the length of the perpendicular from ((Xi/3 x yi) on the line I (X + m/3 + ny = o Transforming to Cartesians by the substitutions of § 398, eq'n (1), the equa- tion of the line becomes (I cos OC + m cos /3 + n cos y) x + (I sin a + m sin/3 + n siny)y — Ip — mp' — np" = o The result of substituting the Cartesian co-ordinates x l5 y x of the point {0C 1 ^3j y x ) for x, y in the sinister of the preceding equation is evidently I (X 1 + m /3i + n y x The length of the _L is (§ 76) the quotient of this by the square root of (I cos OC + m cos /3 + n cos y) 2 + (I sin OC + m sin /3 + n sin y) 2 This expression reduces (Cor', § 398) to I 2 + m 2 + n 2 — 2 mn cos A — 2 nl cos B — 2 Im cos C The length of the perpendicular is .*. I OC-l + m /3 X + n y x Vl 2 + m 2 + n 2 — 2 mn cos A — 2 nl cos B — 2 Im cos C This result should be remembered. (0 § 4-07. To find the condition of perpendicularity of two lines la+m/3 + ny = o, I' a + m' /3 + n' y = o Transform to Cartesians as in § 406 ; the lines are at right angles [§ 68, Cor' (3)] if (I cos a + m cos (3 + n cosy) (I' cos OC + m' cos /3 + n' cosy) + (I sin OC + m sin (3 + n sin y) (I' sin OC + m' sin /3 + n' sin y) — o Using the results of Cor f , § 398, this reduces to IK + mm' + nn' - (mn' + m'n) cos A - (nl' + n'l) cos B — (Im' + I'm) cos C = o This is the required condition ; it should be remembered. 409.] Tri linear Co-ordinates 367 § 408. If 8 is the distance between the points (aiftyO, (Ofeftya) it follows from § 14 and § 399 that S 2 sin* C = («! - a 2 ) 2 + (ft - >3 2 )2 + a («! - a 2 ) (ft - ft) cos C We may deduce a symmetrical expression for 8. By § 397, a (a 2 - a 2 ) + b (ft - ft) + c (^ - y 2 ) = o .-. a («! - a 2 ) 2 = - b («! - Oj) (ft - ft) - c (0£ x - a 2 ) (y x - y 2 ) Hence (a x - « 2 ) 2 , and similarly (ft - ft) 2 may be expressed in terms of the binary products of the differences a i - a 2> ft - ft* yi - y 2 We may .*. assume 5 2 = A (ft - ft) ( 7l - y 2 ) + M ( 7l - y 2 ) («! - <%,) , , + V (Oi - Ota) (/Si - ft) . , . (1) where A, /x, l> are constants. Substitute in (1) the co-ord's of the points B, C whose distance is a ; .-. a 2 = A(2D/b)(- 2D/c) ... A = - a 2 bc/(4 D 2 ) ; and jj,, V are similarly obtained. We have .*, 8 2 = - ^52 Ca (ft - ft) (yi - y 2 ) + b ( 7l - y 2 ) (a x - a,,) + c («! - a 2 ) (ft - ft)] ... (2) AREAL CO-ORDINATES S 409. It is sometimes convenient to use the following system of co- ordinates. If ABC is the triangle of reference, the areal co-ordinates x, y, z of a point P are the ratios of the triangles PBC, PCA, PAB to the triangle ABC. They are connected by the relation x + y + z = 1 If x, y, z are the areal co-ordinates of any point, and (X, ft y its trilinear co-ordinates, we have the relations x = a a/(a D), y = b /3/(a D), z = c y/(a D) Ex. Substituting the values of (X, ft y deduced from these relations in the trilinear equation of the circum-circle of the triangle of reference (§ 434), its equation in areal co-ordinates is found to be a 2 yz + b 2 zx + c 2 xy = o 368 Analytical Geometry [410. 8 4*10. It follows from eq'n (2), § 400, that the equation to a straight line in areal co-ordinates is px + qy + rz = o (1) where p, q, r are the perpendiculars from A, B, C on the line. The length of the perpendicular from the point whose areal co-ordinates are y! ', y', z r on the line (1) is px' + qy' + rz! The denominator of the expression (1), § 406, is a constant independent of 0C 1} /3j , y x ; we infer that the length of the perpendicular from (x r y' z r ) on the line (1) must be A (px' + qy' + rz!) where A is a constant independent of x', y', z r . The length of the X from A, (1,0,0) is .'. Ap ; but this length is p ; .*. A = 1, and /. &c. § 41 1. Since the ratios of the J_s from A, B, C on the line at infinity are each = 1 ; the equation of the line at infinity in areal co-ordinates is x + y + z = o § 412. If the areal co-ordinates of the vertices of a triangle are (x 3 y x Z-J, (x 2 y 2 z 2 ), (x 3 y 3 z 3 ) its area is xi yi z x 2 y 2 z x 3 y 3 z The area vanishes only when the three points are collinear ; the condition for this is that the determinant should vanish. We may .*. assume that the area is A times the determinant where A is constant. But taking the vertices of the triangle of reference, (i, o, o), (o, 1, o), (o, o, 1) as the three points, we find A = D ; and .*. &c. § 4-13. To find the relation connecting the perpendiculars p, q, r from A, B, C on any line. The trilinear equation of the line is (§ 400) ap a + bq /3 + cr y = o By § 406 the length of the J_ from A, (2 D /a, o, o) is 2 p D/Va 2 p 2 + b 2 q 2 + &c. Put this = p ; this gives the required relation, viz. a 2 p 2 + b 2 q 2 + c 2 r 2 — 2bcqr cos A — 2 carp cos B — 2 abpq cosC = 4D 2 4i3-] Tri linear Co-ordinates 369 Exercises [The triangle ABC in these questions is the triangle of reference.] 1. Find the trilinear co-ordinates of the mid point L of BC, and of the circum-centre. Ans. (o, D/b, D/c) ; (R cos A, R cos B, R cos C) 2. Find the equation of the median AL. Ans. b/3 = cy 3. Prove that the three medians co-intersect in the point OC sin A = /3 sin B = y sin C 4. The internal bisectors of the angles co- intersect in the point oc = /3 = y = D/s {Note — The bisectors are /3_y = o, y_a = o, a-/3 = o.] 5. The external bisectors of the angles B, C meet on the internal bisector of the angle A. \Note — The bisectors are y + (X = o, OC + (3 = o, /3_y = o.] 6. Find the equation to the join of the in-centre and circum-centre ; and show that it is perpendicular to the line OC + /3 + y = o Ans. OC (cos B — cos C) + /3 (cos C — cos A) + y (cos A — cos B) = o 7. Find the equation of the parallel to BC through A. Ans. b/3 + cy = o 8. Find the equation to the join of the mid points of AB, AC. Ans. b /3 + cy - aOC = o 9. The joins of the vertices of ABC to any point O meet the opposite sides in A', B', C r . BC, B'C' meet in A" ; CA, CW in B" ; and AB, A'B' in C". Prove that A", B", C" are collinear. B b 370 Analytical Geometry [413- 10. What is the equation of the line A"B"C" in the last question, i° if O is the in-centre ; 2° if it is the ortho-centre ? Ans. (X + /3 + y = o ; Oi cos A + /3 cos B + y cos C = o 11. Find the equation of the line bisecting BC at right angles. Ans. /3 sin B — y sin C + OC sin (B — C) = o 12. The in-circle touches BC in D ; find the equation of the join of the mid point of AD to the mid point of BC, and show that it passes through the in-centre. Ans. (b — c)a=b/3 — cy 13. Show that the lines a = k/3, ka = /3 are equally inclined to the internal bisector of the angle C. 14. If three lines drawn through the vertices of a triangle co-intersect; the three lines drawn through the same vertices, equally inclined to the bisectors of the angles, also co-intersect. [Note — If the first point is Oc/\ = fi/m = y/n, the second is I OL = m (3 = n y.] 15. Any line meets the sides BC, CA, AB in P, Q, R respectively; P', Q', R' are points on these sides equidistant with P, Q, R from their mid points. Prove that P f , Q r , R / are collinear. 16. If b is the distance between the points (0^ /3 X y x ), (CX 2 ^2 Y^)* prove that S 2 - ^ [a cos A («! - 0C 2 Y + b cos B (ft - y3 2 ) 2 + c cos C ( 7l - y 2 ) 2 ] 17. Prove that the relation of § 413 may be expressed thus : a 2 (p - q) (p - r) + b 2 (q - r) (q - p) + c 2 (r - p) (r - q) = 4 D 2 18. Find the area of a triangle in terms of the trilinear co-ordinates of its vertices. Oil & yi a 2 & y% «3 /3 S 73 \_Note — See § 412 ; for another method see Nixon's Trigonometry, page 210.] abc Ans - m* 4I5-] Trilinear Co-ordinates 371 GENERAL EQUATION OF THE SECOND DEGREE § 414. The general equation of a conic in trilinears is (§357) (f> (a, A 7) = a06 2 + b /3 2 + cy 2 + 2f/3y + 2gyoc + 2h a/3 = o (1) The equation of the chord joining two points (06' /3' y f ), (OL" $" y") on the conic is a (a - a') (a - a") + b.(j3 - /3') (/3 - 0") + c(7-r ; )(7-7 ,/ )+2f(^-/30(7-/ / ) + 2g( y - /)(a - a") + 2h(a - a')(j8 - /3") = (a, /3, y) For the locus represented by this equation evidently passes through the points {&' fi' y'), (a" /3" y") ; and the equation is really of the first degree in 06, /3, y. Now put 06" = a', fi" = /3', y" = y'; the equation of the tangent at (06 r , j3', y f ) is .*. a 0606'+ bfifi' + cyy' + f (P'y + y'/3) + g(/a + 06 r 7) + h (a'j8 + /3'a) = o (2) As in § 168, if the point (Oi' fi' y') is not restricted to lie on the curve, (2) is the equation of its polar. We shall use the abbreviation P for the sinister of (2). With the notation of Differentials, _ 6d) d(f> dd> 6(b n 6 («', p, y') (f> (a, (3, y) = P* This may also be proved as in Ex. 1, § 351. B b 2 37% Analytical Geometry [416. § 4-16. The condition that the conic may be a line-pair is found as in § 117. It is A = a h g h b f g f c = o § 417. If the line XOC + fx/3 + vy = o coincide with the line represented by eq'n (2), § 414, then aa'+h/3' + g/ ha'+b/3' + f/ ga' + f/3' + c/ (1) (2) A \k v The pole ((X'fi'y) of the line (1) is .*. determined by the eq'ns (2). We may thus determine the centre, which is the pole of the line at infinity OS. sin A + /3 sin B + y sin C =0 Again, if the line (1) is a tangent, its pole (CX / /3 / y / ) is on the line. The condition of tangency is .'. obtained by eliminating Off, /3 / , y / between equa- tions (2) and \ (a, P, y) = k (Xoc + ix(3 + vyY (i) represents (§ 351) a conic having double contact with the given conic at its intersections with the line \oc + /u/3 + vy = o (2) The conic (1) is a line-pair (§ 416), if a — k A 2 h — k A.//, g — kkv =0 h — kkfji b — k ju, 2 f — kij,v g — kkv f — kjjLV c-kr 2 If this equation is expanded the terms in k 2 and k 3 vanish identically ; and we shall obtain the following simple equation to determine k : A - k 2 = o, where 2 has the meaning assigned, Note, §417. The equation of tangents to the given conic at its intersections with the line (2) is .". 2 (a,/3, y) = A (Aa + /u/3 + z/y) 2 (3) Cor' — Since the asymptotes are the tangents to the conic at its intersec- tions with the line at infinity, the equation to the asymptotes is derived by substituting sin A, sin B, sin C for A, fJL, v in (3). § 421. Let IX, (3, y be the co-ordinates of the extremity of a semi- diameter p a parallel to BC ; (OC, (3, y) the centre. Then projecting p a on perpendiculars to the sides of ABC, we see that OC = OC, ft = /3 + p a sin C, y = y - p a sin B 5 .*. (a, j3 + P a sin C, y - p a sinB) = o 374 Analytical Geometry [422. This equation is a quadratic in p a ; and since the sum of its roots is zero, the equation when expanded becomes (p (a, J3, y) + p a 2 (f> (o, sin C, - sin B) = o or (p (a, /3, y) + p a 2 (b sin 2 C + c sin 2 B - 2 f sin B sin C) =0 This determines p ; and p~ , p are similarly expressed. If the conic is a circle, p a = p^ — p c ; the conditions for a circle are .*. (j) (o, sin C, — sin B) = <£ (sin C, o, — sin A) =

2 = (1^ + nA)(mi/+ n/x) This condition reduces to l/A + m//x + x\jv = (3) Hence (3) is the condition that the line (2) may touch the in- scribed conic (1). Note — This condition may also be deduced as a particular case of eq'n (6) ?j § 417 ; the rationalized form of (i) being used. 378 Analytical Geometry [428. S 428. To find the equation of the inscribed circle. Let A', B', C be its points of contact with the sides. Put OC — o in the equation V\OC + Vm /3 + Vn y = o ; the co-ordinates of A' .*. satisfy the equation n/m = /3/y .-. n/m = A'C sin C/(A'B sin B) = r cot i C sin C/(r cot | B sin B) = cos 2 \ C/cos 2 \ B A similar value is found for I /m ; the equation of the inscribed circle is .\ cos § A V(X + cos \ B Vj3 + cos| C Vy = o . . . . (i) AWtf — The equations of the escribed circles are similarly obtained. The escribed circle touching BC is cos \ A V^~OL + sin \ B Vfi + sin | C Vy = o . . . (2) Exercises [The conic in these questions is the inscribed conic V\OL + Vm (3 + Vn y = o The points of contact are denoted by A', B', C] 1. Prove that AA', BB' ; CC are concurrent. 2. If AA', BB', CC meet the conic again in A", B", C", prove that the tangents at A", B", C" meet the opposite sides in three points lying on theline la + m/3 + ny = [Note — The equation of the conic may be written ny(ny-2la-2m/3) + (ICX-m/3) 2 = o The tangent at C" is .*. (§ 353) x\y — 2 I a — 2 m /3 = o.] 3. If the centre is (OLfiy), prove that a/(n sin B + m sin C) = /3/(l sin C + n sin A) = y/(m sin A + I sin B) 429.] Trilinear Co-ordinates 379 {Note — If

/. A dd> / dd> / d^/ smA= d|/ sinB = d7/ sinC Otherwise thus : It is known that the joins of A to the mid point of B^' and of B to the mid point of CW pass through the centre ; form the eq'ns of these joins and determine OC : f3 : y.] 4. Prove that the centre lies on the join of the mid points of AA r , BC. {Note — Use Ex. 3. The theorem also follows from Ex. 31, page 306, by supposing two sides of the quadrilateral to coincide.] 5. Prove that the conic is a parabola if I /sin A + m /sin B + n /sin C = o 6. If one focus describes the straight line Xoc + jtx/3 + vy = o the other describes the conic X/oc + fi/fl + v/y = o {Note— If the foci are (a'/3'/), {OL"fi"y") then (§ 248, VII.) Oi'OL ff = square of semi-minor axis = fi' 'ft" = y'y n '.] 7. Prove that the equation of the chord joining ((xp'y'), {3! f fi' f y") is VI (a - oo/ya' + V«") +... + ...=* o ; and the tangent at (JX'fi'y') is OC \f\/a' + (3 v/m//3 r + y \/n// = o 8. If A', B', C are the mid points of the sides ; the equation of the conic in areal co-ordinates is /- /- /- vx + Vy + Vz = o NINE-POINT CIRCLE 8 4-29. If the circle represented by eq'n (1), § 425, pass through the mid point of BC (a = o, j3 sin B = y sin C), we must have sin A sin B sin C + 2 sin B sin C (m sin C + n sin Bj = o .'. 2 m /sin B + 2 n /sin C = — sin A/(sin B sin C) 380 Analytical Geometry [430- If the circle pass also through the mid points of CA, AB we have two similar equations. Solving these, we find — 2 I /sin A = (sin 2 B + sin 2 C — sin 2 A) /(2 sin A sin B sin C) .*. 2 I = — cos A ; and 2 m = — cos B, 2 n = — cos C The equation of the nine-point circle is .\ 2 (j3y sin A + y(X sin B + a/3 sin C) — (a cos A + /3 cosB + y cosC) (a sin A + /3 sin B + y sin C) =0 . . (1) or a 2 sin 2 A + /3 2 sin 2 B + y 2 sin 2 C — 2 /3y sin A — 2 yOC sin B - 2 a/3 sin C = o . . . (2) § 4-30. The nine-point circle of a triangle touches the inscribed and escribed circles. (Feuerbach's Theorem.) The following proof is by Prof r Genese. Let A', B', C f be the mid points of the sides of the given triangle ABC ; take A / B / C / as triangle of reference. The equation to the nine-point circle is thus a/3y + byOL + cCX/3 = o (1) The eq'ns to BC, CA, AB are b/3 + cy = o, c y + a a = o, a a + b /3 = o Let the equation to a circle touching these be a/3y + byOC + ca/3 - (a a + b/3 + cy) (la + m/3 + ny) = o . . (2) Putting b/3 + cy = o, (2) becomes (a/3 + ba)(-^/3) + ca/3-(la + m/3- £b/3)aa = o This is to be a complete square / ab\ / b 2 anb\ 2 or ±2 Vabc Val = c 2 — b 2 + abn - cam Similarly that (2) may touch CA, AB, + 2 A^abc Vbm = a 2 — c 2 + bcl — abn + 2 Vabc Vcn = b 2 — a 2 + cam — bcl Adding we find + Val + Vbm + Vcn = o 43i.] Trilinear Co-ordinates 381 But this is the condition that the line I (X + m/3 + ny = o should touch (i), and this line is the radical axis of (i) and (2) ; .*. (2) touches (1). pascal's theorem; other geometrical theorems § 4-31. Pascal's Theorem. If any hexagon is inscribed in a conic, the intersections of its three pairs of opposite sides are collinear. Let the hexagon be AC / BA / CB / ; the reader can draw the figure. Using abridged notation, let the equations of AC, C'B, BA', A'C, CB', B / A, and AA' be respectively U = O, V = o, w = o, u' = o, v' = o, v/ = O, (f) = o Since the conic circumscribes AC'BA', AB'CA', it's equation may (§ 355) be written in the forms uw — X v (f) = o, u'w' — A/ v' <£> — o ,-. uw-Av=A"(uV-W(|)) (1) A, A/, A" being numerical constants. From (1) we infer uw - A" u'w' = 4> (A v - a'a" vO This identity shows that the conic uw — X" u'w' = o (2) is a line-pair, one of the lines being (j), and the other passing through the inter- section of v and v'. But the conic (2) passes through the intersections of u, u f and of w, w' ; these points are .*. collinear with (v, v'). Q.E.D. Note (1) — Another proof is indicated in Note, Ex. 47, page 357 ; and a proof of the converse theorem in Ex. 48, page 357. We add an outline of a proof by trilinears. Take ABC as triangle of reference ; let the other points be A' ((Xj /3j y^), B'(a 2 /3 2 y 2 ), C'(a 3 /3 3 y 3 ). Express that A', B', C are on the circumscribing conic I /a + m//3 + x\Jy = o and eliminate I, m, n. This gives i/cxx i/a i/ri i/a 2 i//3 2 i/y 2 i/a* J /^3 J /yz = (3) 3 8 2 Analytical Geometry [432. Next, form the equations of AC, A'C, &c. ; and find the co-ord's of the intersections of the three pairs of opposite sides. The condition that these points should be collinear reduces to (3). The reader should work this out. Note (2) — The six points A, C, ... may be taken in sixty different orders, thus forming sixty ' hexagons.' The theorem is applicable to each of these hexagons ; so that there are sixty Pascal Lines corresponding to six points on a conic. Note (3) — Many particular results are deducible from Pascal's Theorem. For example, suppose that A coincides with C, and A' with C ; then AC, A'C are the tangents at A, N. We infer that — The tangents at two opposite vertices of an inscribed quadrilateral meet on its third diagonal. § 4-32. Brianchon's Theorem. If a hexagon circumscribe a conic, its three diagonals co-intersect. Take the triangle formed by producing three alternate sides as triangle of reference; then the equation of the conic is (§ 426) VToc + Vm jtf + Vn y = o Let the sides of the hexagon respectively opposite to OC, 8, y be X 1 OC + fafi + v x y = o (1) A 2 OC + \x 2 /3 + v 2 y = o (2) A. 3 a + n s p + v s y = o (3) We have then (§ 427) three equations l/Xi + m/ju,! + r\/v ± = o, ..., ... The elimination of I, m, n gives = o (4) i/Ai \j\K x i/ Vl 1/A-2 1/M2 t/v* 1/A3 1///3 i/v s This is the condition that the six lines should touch a conic. Again, one diagonal is the join of the inters'n of the line (2) with y to the inters'n of the line (3) with /3. Its equation is .*. = 0, or a + /x 2 fi/k 2 + v 5 y/X 3 = o oc (3 y - M2 ^2 v 3 -A, 434-] Tri linear Co-ordinates 383 Similarly the other diagonals are X t , ^ Aa + /a/3 + vy = o (1) touches the conic la 2 + m/3 2 + ny 2 = o (2) if X 2 /l + /x 2 / m + ^ 2 /n = ° (3) Note — In questions relating to four lines it is convenient to assume that their equations are XOC ± nj3 ± vy = o (4) Thus each side of the quadrilateral (4) touches the conic if the single condition (3) is satisfied. The reader will find no difficulty in proving that the diagonal triangle of this quadrilateral is the triangle of reference. 437-] Trilinear Co-ordinates 385 In questions relating to four points it is convenient to assume that the points OC (3 y - «= ± - = ± -; these points are on the conic (2) if lp 2 + mq 2 + nr 2 = o § 4-37. In questions relating to two conies it is convenient to choose the harmonic* triangle of the quadrangle formed by their four intersections (see Cor* > § 314) as triangle of reference ; so that the equations of the two conies are I CX 2 + m /3 2 + n y 2 = o, V OL 2 + m'/3 2 + n' y 2 = o Exercises 1. Find the locus of the pole of a given line X(X + ju/3 + vy = o with respect to conies inscribed in the quadrilateral p0£ + q/3±ry = o Ans. The straight line p 2 a/A. + q 2 /3/ju, + r 2 y/z> — o 2. Find locus of centres of conies inscribed in this quadrilateral. Ans. The straight line p 2 a/sin A + q 2 /3/sin B + r 2 y/sin C = o [Compare the solution, page 307.] 3. Find the locus of the pole of a given line ACX + ju/3 + vy = o with respect to conies through the four fixed points (p, + q, + r). Ans. The conic A p 2 /(X + /x q 2 //3 + V r 2 /y = o 4. Show that the triangle of reference is the harmonic triangle of the quad- rangle whose vertices are (p, + q, + r). 5. Find locus of centres of conies circumscribing this quadrangle. An s. The conic p 2 sin A /(X + q 2 sin B//3 + r 2 sin C/y = o [Compare § 322.] * Let A, A', B r , B be the vertices of a quadrangle (fig r , page 284) ; the triangle OPQ whose vertices are the intersections of the three pairs of lines joining the points is called the harmonic triangle of the quadrangle v Casey) ; the triangle whose sides are the three diagonals AB r , A'B, OP is called the diagonal triangle (Cremona). c c 386 Analytical Geometry [438. 6. Find the conditions that the conic la 2 + mj3 2 +ny 2 = o may be (i), a parabola ; (2), a rectangular hyperbola. Ans. sin 2 A/l + sin 2 B /m + sin 2 C /n =o;l+m+n = o 7. Prove that a rectangular hyperbola passes through the centres of the inscribed and escribed circles of any triangle which is self-polar with respect to the hyperbola. 8. The polars of A, B, C with respect to a conic form a triangle A / B / C / ; prove that AA', BB', CC co-intersect. 9. Prove that the locus of centres of rectangular hyperbolas passing through the vertices of a triangle is the nine-point circle. 10. Find the director circle of the conic I OL 2 + m /3 2 + n y 2 = o Ans. I (m +. n)a 2 +... + ...+ 2 mn /3y cos A +... + ...= o {Note — The eq'n of the tangents from (OL'fi'y 1 ) is (§ 41 }) (l(X /2 + m/3' 2 + n/ 2 )(la 2 + m/3 2 + ny 2 ) = (I OtOt' + m /3/3' + nyy') 2 If these are at right angles the coeff's satisfy the cond'n for a rectangular hyperbola (§ 419).] 11. If the curve is a parabola, find the equation of its directrix and the co-ordinates of its focus. I Cm + n) m (n + I) n n (I + m) sin A sin B ^ sin C ' (X sin A _ )3 sin B _ y sin C m + n n + I I + m {Note — Use Ex. 10. The locus of intersection of rectangular tangents breaks up into factors, one of which is the directrix, and the other the line at infinity a sin A + /3 sin B + y sin C =0 The focus is the pole of the directrix.] CONIC REFERRED TO TWO TANGENTS AND THEIR CHORD OF CONTACT § 438. The equation of a conic touching the sides CA, CB of the triangle of reference at A, B is (§ 353) a/3 = k 2 y 2 (1) The results of §354 are applicable if we write a, /3, ky for L, M, R. 44°.] Trilinear Co-ordinates 387 Ex. If the bisector of the angle C meet AB in R, and PQ is any chord through R ; prove that PR, RQ subtend equal angles at C. The eq'ns to CR, PQ are OC - ft = o . . (2) a - ft + Ay = o (3) Eliminate y from (i), (3) ; this gives the equation to the pair of lines CP, CQ, viz. k2 (a _ ^2 _\*ot(3 = o The factors of this are of the form given, Ex. 13, page 370 ; and .'. &c. § 4-39. If fa, |U 2 , fi 3 , /x 4 (see § 354) are four points on a conic, the cross ratio of the pencil passing through these points, and whose vertex is any fifth point on the conic, is constant (§ 355) ; this cross ratio is briefly designated ' the cross ratio of four points on a conic;' Take A for the vertex of the pencil ; its rays are (§ 354) (3 = faky, ft - fcky, The cross ratio of the four points is .*. (§ 140) (Mi ~ Ms) (M* ~ M2) (I) (Ms - M2) (Mi ~ M 4 ) Four fixed tangents cut any fifth in a range whose cross ratio is constant, and equal to that of the four points of contact. Let in> M2> M3' M4 De * ne f° ur P omts 5 M * ne P oul t °f contact of the variable tangent. Let the variable tangent meet the other tangents in P l5 P 2 , P 3 , P 4 . The elimination of y from the equations of the tangents at jtx and [X\ > fjL 2 (X — 2[xky + ft = o, jUjl 2 (X — 2/oi 1 ky + /3 = o gives fAjJLiOC = ft This is the equation of CP X . Hence {^P^P*} = {CP 1 P 2 P 3 P 4 } = cross ratio of pencil fxu-i OC — ft, /uju 2 OC = ft, . . . . ; the latter cross ratio, which is given by the formula of § 140, reduces to the preceding expression (1). Q.E.D. 8 44-0. In working with homogeneous equations we may replace the co-ordinates by any quantities proportional to them. Thus we may take (1, /u. 2 , ju,) as the co-ord's of a point on the conic LM = R 2 c c 2 388 Analytical Geometry [441. Ex. 1. Prove that the tangents at fi, \i' meet at the point [1, \i\i', (n + \i')J2\ These co-ord's are obtained by solving for L : M : R the eq'ns of the tangents /u 2 L-2/xR + M=o, /x /2 L- 2ju'R + M = o § 4-4-1. Ex. 2. If the three sides of a triangle touch a conic S and two of its vertices move on another conic 2, the locus of the third vertex is a conic inscribed in the quadrilateral formed by the common tangents of S and 2. Take two of the (real or imaginary) common tangents as two sides of the triangle of reference ; so that the equations of S and 2 are LM = R 2 , LM = (I L + m M + 2 n R) 2 Let fi 1} jJL 2 , fi 3 be the points of contact ; then expressing that the inters'ns of the tangents at /x x , fi 3 and at fi 2 , [X 3 ^ e on S> we see tnat Mi anc * M2 are tne roots of the quadratic in \l \i\i z = (I + mjujUcj + r\\i + n/x 3 ) 2 (i) We can .*. write down the values of ju^ + \i 2 , \i x \i 2 • But, by Ex. 1 these are 2 R/L and M/L, where (L, M, N) is the inters'n of the tangents at /x 1? /x 2 ; hence 2 R/L and M/L are expressed in terms of fi 3 : /ut 3 is easily eliminated ; and we find for the locus an equation of the form LM = (I' L + m' M + 2 n' R} 2 The locus is .'. a conic touching L, M ; but L, M are any two common tangents ; .'. &c. Note — The reader should work out this solution in full, supplying the sup- pressed details. It will be seen that the locus becomes two coincident lines if Im = n 2 ; i.e. if the chord of contact of 2 with L, M touches S.. INVARIANTS § 4-42. The following convenient notation is due to Prof Cayley. (a, b, c, d) (x, y) 3 means ax 3 + 3 bx 2 y + 3 cxy 2 + dy 3 Thus (a, b, c, f, g, h) (a, /3, y) 2 = a a 2 + b/3 2 + c y 2 + 2 f (3y + 2 gyoc + 2 h a/3 and (a, b, c, f, g, h) (x, y, i) 2 E ax 2 + 2 hxy + by 2 + 2 gx + 2 fy + c § 44-3. Let the result of substituting the values of OC, /3, y, given in § 398 in (j> (OC, fi, y), be (a, b, c, f, g, h) (a, /3, y) 2 E (a lf b 1} c l7 f 1} g x , h x ) (x, y, i) 2 444-] Trilinear Co-ordinates 389 Let the discriminant of cf) (a, j3, y) be A. Also, as in § 323, A = be - f 2 , &c. ; i. e. A, B, ... are the minors of the determinant A. We shall also put = a , b x - h x 2 , 0' = a x + b x The conic is a parabola if = o, and a rectangular hyperbola if 6 f = o ; and 0' are the invariants of § 212. Then by substituting the actual values of a x , b 1? h x we shall obtain the following relations : 6 E a x b x - h x 2 = (A, B, C, F, G, H) (sin A, sin B, sin C) 2 . . (1) 0' E a x + b x E a + b + c — 2 f cos A — 2 g cos B — 2 h cos C . . (2) Ex. 1. Suppose that the conic is a circle; and let its centre be the origin of rectangular axes, so that (a, b, c, f, g, h) (a, p, y) 2 E q (x 2 + y 2 - p 2 ) .-. 6 = q 2 , 6' = 2 q The square of the tangent from any point (OL, /3, y), or (x, y) is .'. 2 (a, ^, y)/^ Ex. 2. 7%^ circum-cirde of any triangle which is self -polar zvith respect to an ellipse cuts its director circle orthogo?ially. (M. Faure.) Take the given triangle as triangle of reference ; let the equation of the ellipse be I a 2 + m /3 2 + n y 2 = o The equation of the director circle is given, Ex. 10, page 386 ; the square of the tangent from the centre of the circle (R cos A, R cos B, R cosC) is .*. (by Ex. 1) 2 R 2 [I (m + n) cos 2 A +... + ... + 2 mn cos A cos B cos C + ... + ...] I (m + n) +... + ...— 2mn cos 2 A —...—... We find by easy Trigonometrical reductions that this expression = R ; this proves the theorem. § 44-4-. Suppose that the axes of x and y are the axes of figure of the conic ; so that . . (a, /3, y) - (A/0) (D 2 /R 2 ) = q (x 2 /p 2 + y 2 /p' 2 ) Subtract this from (1), .-. - q = (A/0) (D 2 /r 2 ) (4) The relations (2), (3), (4) suffice to determine all constants connected with the conic. Thus we have at once expressions for p 2 + p' 2 and p 2 p' 2 ; and we find that the semi-axes are the roots of the following equation in p : R 4 3 p 4 + R 2 D 2 A00'p 2 + D 4 A 2 = o § 4-45. If S, S' are two conies, the conic through their intersections, S + kS^o (1) is a line-pair (§§ 118, 416) if its discriminant vanishes; i. e. if a + ka' h + kh' g + kg' =0 . . . . (2) h + kh' b + kb' f + kf g -t- kg' f + kf c + kc' This cubic in k is usually written A + k + 0' k 2 + A' k 3 = o (3) The substitution of the roots of this cubic in (1) gives the line-pairs through the intersections of S = o, S' = o. § 4-46. If by a change of axes S and S' become S and S', then S + k S' becomes S + k S' ; and if k is such that S + k S' breaks up into factors, so does S + k S 7 - Hence the roots of the preceding cubic (3), and .*. the ratios of its coefficients, are independent of the axes of co-ordinates to which the two conies are referred. This conclusion leads to many interesting properties of a system of two conies ; for these the student is referred to Salmon's Conies. Exercises on Chapter XIV [In these questions ABC is the triangle of reference.] 1. Find the trilinear co-ordinates of the centre of the nine-point circle. Ans. \ R cos (B - C), \ R cos (C - A), |R cos (A - B) 2. Prove that the in-centre, circum-centre, and ortho-centre of the triangle ABC are collinear if cos 2 A (cos B — cos C) + cos 2 B (cos C — cos A) + cos 2 C (cos A — cos B) — o 446.] Exercises on Chapter XIV 39 1 3. If p, q, r are the perpendiculars from A, B, C on the join of the centroid and in-centre of ABC, prove that p cot | A + q cot | B + r cot | C = o 4. Prove that the equation of the circle through the three centres of the escribed circles is a a 2 + b/3 2 + cy 2 + (a + b + c) (/3y + yOL + a/3) = o 5. Prove that the equation of the nine-point circle may be written a 2 /(D - a a) + b 2 /(D - b (3) + c 2 /(D -cy) = o {Note — Let A', B', C be mid points of sides. Prove that length of ± from (a'/3'/) on B'C is a'- D/a; then form eq'n of circum-circle of A'B'C] 6. Find the radius of the circle j3y sin A + yOL sin B + a/3 sin C = 2 k Ans. R Vi - 4 k/D 7. Through A', B', C , the mid points of the sides of the triangle ABC, lines A r L, B'M, C'N are drawn perpendicular to the sides and proportional to them ; prove that AL, BM, CN co-intersect in a point whose locus is /3y sin (B - C) + yOL sin (C - A) + a/3 sin (A - B) - o 8. Find the equation of the circle whose diameter is BC. Ans. /3y sin A + yOL sin B + a/3 sin C = (D/R) a cos A 9. If S = o is the equation to the circum-cirilc of the triangle ABC, the equation of the circle through the three points (^(3^^, (a 2 /3 2 y2)> (^3/3 3 y 3 ) is S a j3 y = o Si Oiftyi 5 2 a 2 /3 2 y 2 5 3 oc B (3 3 y 3 10. The equation of the circum-circle of the triangle whose sides are a = o and the bisectors of the angle A is sin (B - C) {/3y sin A +... + ...} + (J3 sin C — y sin B) (a sin A + /3 sin B + y sin C) - o This circle and its two analogues are co-axal, the radical axis being a sin (B - C) + (3 sin (C - A) + y sin (A - B) = o {Prof Purser?) 39 2 Analytical Geometry 11. The radical axis of the in-circle and nine-point circle of the triangle of reference is aa/(b - c) + b /3/(c - a) + c y/(a - b) = o 12. Prove that the intersections of the lines la + m/3 + ny = o, l / a+m / /3 + n / y = o with the circum-circle and self-conjugate circle respectively, lie on the same circle if cos A cos B cos C m n = o r nr n' [Note — The radical axes of the three pairs of circles are concurrent.] 13. The equation 0C(3 = y 2 represents a circle if the two sides CA, CB of the triangle of reference are equal. 14. Show that the conic V\0C + Vm /3 + Vn y = o is an ellipse, parabola, or hyperbola, according as Imn (l/a + m/b + n/c) > = < o 15. Show that the theorem, Ex. 3, page 311, is a particular case of Ex. 6, page 379- 16. Determine A. so that the equation a 2 = A/3y may represent a parabola ; find the equation to its directrix. Ans. A = 4 bc/a 2 ;a(XcosA = c/3 + by 17. If p, q, r are the perpendiculars from A, B, C on any tangent to a parabola touching AB, AC at B, C, prove that p 2 = qr 18. Prove that the normals to the conic /3y cot I A + yOL cot | B + 0C(3 cot | C =0 at A, B, C co-intersect. 19. The locus of centres of rectangular hyperbolas touching the sides of a given triangle is the self- conjugate circle. 20. The locus of centres of rectangular hyperbolas having ABC for a self- conjugate triangle is the circum-circle. Exercises on Chapter XIV 393 21. The circum-centre of any triangle, self-conjugate with respect to a parabola, lies on the directrix. 22. A point describes a straight line ; prove that the locus of the intersection of its polars with respect to two conies is a conic circumscribing their common self-conjugate triangle. 23. Find the relation between the perpendiculars p, q, r from A, B, C on any tangent ; i°, to the inscribed circle ; 2°, to the ellipse touching the sides at their mid points. Ans. i°, (s - a) qr + (s - b) rp + (s - c) pq -= o 2°, qr + rp + pq = o [Note — Use eq'n (2), § 400, and eq'n (3), § 427.] 24. Through each vertex of the triangle of reference a parallel is drawn to the opposite side; these parallels form a triangle A'B'C. If p, q, r are the perpendiculars from A, B, C on any tangent to the circum-circle of A'B'C, prove that sin A Vq + r — p + sin B Vr + p — q + sin C Vp + q — r = o 25. The major axis of a conic inscribed in ABC passes through the point in which the external bisector of the angle A meets BC ; prove that the locus of its foci is the conic a 2 = (3y [Note — See Note, Ex. 6, page 379, If (CX/3y) is on? focus, ( — , -= , — 1 is the other; express cond'n that these are collinear with (o, 1, — 1).] 26. The minor axis of a conic inscribed in the triangle of reference is given ( — P) > P rove that the locus of its foci is the curve Otfiy (aa + b/3 + cy) = p 2 (a /3y + b yOC + c Otfi) 27. Prove that the locus of the foci of conies touching the four lines a = o, (3 = o, y = o, l(X-t-m/3 + ny--=o is the cubic I m n I 2 + m 2 + n 2 — 2 mn cos A — 2 nl cos B — 2 Im cosC _ -j- — .f. _ _ CX. (3 y la + m/3 + ny 28. Prove that the six points in which tangents to any conic from the vertices of the triangle of reference meet the opposite sides lie on a conic. 394 Analytical Geometry \_Note — Let the conic be represented by the general equation in areal* co-ordinates + ^ y, z) = o The equation of the tangents from A (i, o, o) is (§ 415) a (ax 2 + by 2 + cz 2 + 2 fyz + 2 gzx + 2 hxy) = (ax + hy + gz) 2 or Cy 2 - 2 Fyz + Bz 2 = o (1) The roots of this equation in y/z are the ratios in which BC is divided by the lines (1) ; the product of these ratios is .*. B/C. The result follows by Carnot's Theorem (Ex. 2, § 311).] 29. Two points are joined to the vertices of the triangle of reference. Prove that the six points in which the joins meet the sides lie on a conic. If the areal co-ordinates of the points are (x'y'z') and (x // y // z // ), find the equation of the conic. Am -£p*"*"~(&! + F?)y+~ 30. A parabola drawn through the mid points of the sides of a triangle ABC meets the sides again in A", B" '., C" ; prove that AA", BB", CC" are parallel. 31. If lines are drawn through a point O parallel to the sides of a triangle, the six points in which the parallels meet the sides lie on a conic. 32. If a triangle be self- conjugate with respect to a parabola, prove that its nine-point circle passes through the focus. 33. CA, CB are tangents to a conic at A and B ; P is any point on the curve. Any line through C meets AP, BP in Q, R. Prove that BQ, AR meet on the curve. 34. Having given five tangents to a conic, show how to determine their points of contact. {Note — Let ABCDE be a circumscribed pentagon. Then AD, BE meet on the join of C to the point of contact of AE. This follows from Brianchon's Theorem, by supposing two sides of the hexagon to coincide.] * Areal co-ordinates possess some advantages over trilinears. If we put x = o in the homogeneous equation of a curve in areal co-ordinates, the roots of the resulting equation in y/z are the ratios in which the curve divides BC. It is worth noticing that the point (xyzj is the centre of gravity of masses x, y, z placed at A, B, C. Exercises on Chapter XIV 395 35. A parabola inscribed in the triangle of reference touches the line X.OC + fx(3 + vy = o; prove that the equation to the directrix is «ootA0-0 + ^cctB(£-0 +y cotc(I-i) = o 36. The four common tangents to two conies intersect two and two on the sides of their common conjugate triangle. 37. Two triangles are self-polar with respect to a conic; prove that their six vertices lie on a conic. 38. If tangents are drawn to a conic from any point of a straight line whose pole is P, the sum of their distances from a fixed point Q divided respectively by the distances of P from the same tangents is constant. 39. A transversal drawn through a fixed point O meets a conic in P, Q. Prove that the algebraic sum of the distances of P, Q from a fixed straight line divided respectively by the distances of P, Q from the polar of O is constant. 40. If the summits of three angles are collinear, their arms are the opposite sides of a hexagon whose vertices are on a conic. [See Ex. 48, page 357. This is the converse of Pascal's Theorem.] 41. The three diagonals of the quadrilateral whose sides are (X = o, /3 = o, y = o, Aa + ju,/3 + vy = o are divided harmonically by the conic I (X 2 + m /3 2 + n y 2 + r (XOC + jj.fi + vy) 2 = o {Hesse.) 42. The equation to the isotropic lines through A is /3 2 + y 2 + 2 j3y cos A = o 43. The trilinear co-ordinates of the focoids are proportional to cos B + V — 1 sin B, cos A — V — 1 sin A, — 1 {Note — Solve for //3//3' + sin C s/y/Y = ° 60. Prove that the equation to the Sim son's Line of a point (CX / '/3 / 'y / ) on the circum-circle of the triangle of reference is / . A R sin 2 A\ (X[smA ^r— j +... + ...= 398 Analytical Geometry 61. The locus of centres of conies inscribed in a triangle, and such that the normals at the points of contact co-intersect, is a cubic which passes through the vertices of the triangle, the centroid, the ortho-centre, the in- and ex-centres, and the mid points of the sides and perpendiculars. 62. Tangents are drawn from a fixed point P to parabolas inscribed in a given triangle ; prove that the locus of the point of contact is a cubic. 63. If the triangle of reference is equilateral, find the equation to the axes of the conic I a 2 + m/3 2 + n y 2 = o Ans. I 2 (m — n) a 2 + ... + ... + 2 mn (m — n)/3y +... + ...= o \_A T ote — The axes are the loci of points whose polars with respect to the conic and its director circle are parallel.] 64. Obtain equations to determine the foci of the conic in the last question. Ans. I (m + n) a 2 + mn (/3 + y) 2 = m (n + I) /3 2 + nl (y + a) 2 = n (I + m)y 2 + Im (a + /3) 2 [Note — The equation of the tangents from a focus satisfies the analytical conditions for a circle. The foci may also be obtained by the methods of §§ 368, 372, using the result of Ex. 43, page 395 ; the equations thus obtained are however unsymmetrical.] 65. If the general trilinear equation represent a circle whose centre is (Oi, ft, y) and radius r, prove that r 2 = - 2 $ (a, J3, y)/6' [Note — See § 443. The square of the tangent from the centre is — r 2 .] 66. If the general trilinear equation represent a pair of lines, prove that the angle between them is . - — tan-i^y-4 0/0' 2 67. The co-ordinates of the focoids are proportional to e e v ~*, e^^, e^ v:r * where 0, (f), \jf are the angles which the sides of the triangle of reference make with any line ; these angles being measured round from the line in the same direction. {Prof Cenese.) 68. Two concentric and similar conies are, one inscribed and the other circumscribed to a triangle. Prove that the locus of the common centre con- sists of two circles. (Faure.) Exercises on Chapter XIV 399 69. If 6, //" are the angles which the sides of the triangle of reference make with an axis of the conic represented by the general trilinear equation, prove that a sin 2 6 + b sin 2 (f) + c sin 2 \j/- + 2 f sin (<^> + \j/) + 2 g sin (\J/- + Q) + 2 h sin (0 + (y2 _ b2) = x2 y2 or x 2 /(b 2 + c 2 ) + y 2 /b 2 = 1 § 4-4-8 . If (h is indeterminate, the envelope of Lcos<^> + M sin = R (i) is L 2 + M 2 = R 2 (2) Substituting fx = tan | (j), (1) becomes L (1 - fx 2 ) + 2 Mju = R (i + m 2 ) The envelope is .-. (L + R) (R - L) = M 2 ; and .-. &c. Ex. If CP, CD are conjugate semi-diameters of an ellipse, find the envelope of PD. If 6 is the eccentric angle of P, the equation of PD is found to be (ay — bx) sin + (ay + bx) cos = ab The envelope is .". the ellipse 2a 2 y 2 + 2 b 2 x 2 = a 2 b 2 § 4-4-9. To find the envelope of the line Xoc + fjLj3 + vy = o (x) where (a, b, c, f, g, h) (A, jlc, v) 2 = o (2) D d 4 02 Analytical Geometry [450. Eliminating v from (i), (2), we see that the lines of the system which pass through a given point (OL'fi'y) are determined by the equation (a, b, c, f, g, h) IX, /x, - r -^- ) = o Expressing the condition that this quadratic in X/fJ. has equal roots, and reducing (as in § 323), we obtain for the equation of the envelope Aa 2 + B/3 2 + Cy 2 + 2F/3y + 2 G y(X + 2 H a/3 = o . . (3) where A, B, &c. have the usual meanings. § 4-50. By writing x, y, 1 instead of OC, {3, y, the preceding investigation becomes applicable to Cartesians. It is convenient to interchange the large and small letters A, a, &c. Hence the envelope of the line Xx + fj.y + v = o (1) where (A, B, C, F, G, H) (X, [i, v)* = o (2) is the conic (a, b, c, f, g, h) (x, y, i) 2 = o (3) where a = BC - F 2 , f = GH - AF, &c. Since eq'n (2) is the condition that the line (1) should touch the conic (3), this result should be equivalent to that of § 323. This is verified by the identities BC - F 2 = A a, GH - AF = A f, &c. where the letters A, A, a, &c. have the same meanings as in § 323. Cor' (1)— The centre of the envelope is (G/C, F/C). [§ 323.] Cor' (2) — The equation to the director circle of the envelope is given (§ 325). The envelope is a rectangular hyperbola if the radius of this circle is zero, i.e. if G 2 + F 2 = C (A + B) Cor' (3) — If the axis of x is a tangent to the envelope, the tangential equation (2) is satisfied by A = o, v — o; hence B = o. The envelope .'. touches the axes if A — o B = o Cor' (4) — The envelope is a parabola (§287)ifC=o; this condition may also be deduced by noticing that if the line at infinity is a tangent the tangential equation is satisfied by X = o, n = o Cor' (5) — If the origin is a focus, the tangential equation is satisfied by A == 1, fJL = ± V- 1, V = o; .-. A - B - o, H = o 452.] Envelopes 4°3 § 451. Ex. i. The locus of centres of conies inscribed in a quadrilateral is a straight line. Take the asymptotes of one inscribed conic as axes ; so that its tangential equation is [§ 450, Cor' s (1), (3)] 2 E C z/ 2 + 2 H a/x = o Let the tangential equation to another inscribed conic be 2' = (A', B', C, F, G', HO (A, p, z>) 2 = o Then if the line Ax + juty + z; = o touches both conies, we have 2 = o and 2' = o ; .-. S + kS'-o (1) Hence (1) is the tangential equation of conies inscribed in the quadrilateral formed by the common tangents of 2 and 2'. The centre of the conic (1) is [§ 450, Cor' (1)] x = kG'/(C + kC), y = k F/(C + kC) Eliminate k ; the locus of centres is .'. the line x/y = G'/F [Compare solution on page 307.] Ex. 2. The director circles of conies inscribed in a quadrilateral form a co-axal system. For the director circle of the conic whose tangential eq'n is (1) is (§325) C (x 2 + y 2 ) + k [C (x 2 + y 2 ) - 2 G' x - 2 F y + A' + B'] = o This is the eq'n of a circle co-axal with the director circles of 2 and 2' ; and the radical axis is the line 2 G' x + 2 F y = A' + B' It follows from the result of Ex. 1 that the radical axis is perpendicular to the locus of centres. TANGENTIAL CO-ORDINATES § 452. The Boothian tangential co-ordinates of a straight line are the reciprocals of its intercepts on the axes. Thus (I, m) are the tangential co-ordinates of the line lx + my = 1 The equation to a point (x'y') in this system is lx' + my' =1 . (1) i. e. the co-ordinates of every line passing through the point are connected by the relation (1). D d 2 404 Analytical Geometry [453- The envelope of a line is a conic if its co-ordinates I, m are connected by a relation of the second degree ; for the results of § 450 become applicable to this system if we write I, m, — 1 for A, jjl, y. Ex. Let the tangent to the ellipse (a, b) at the point whose eccentric angle is (p» q> "") = ( a > b > c > f > g> h ) (p. q> r ) 2 = ° Let (p'q'r'), (p /, q // r // ) be two tangents to the conic. The equation of the chord joining (OL'fi'y), (a"/3' /, y // ) is given in § 414. p'r q'r r* v In this equation replace OL, (3, y by p, q, r; * 0L f , /3', •/ by ^—j , ~~ , — T ; n"r a"r r ff r and ^, ^,/' by L, !_,___. Then the resulting equation is satisfied if we substitute p', q', r' or p", q", r" for p, q, r; and .*. the given lines are tangents to the envelope represented by the equation. Further, when the equation is expanded the result is divisible by r; and the quotient is the equation of the point of intersection of the tangents (p'qV), ^p"q"r"). Hence, as in § 414, we deduce the equation of the point of contact of (p'q'r') ; this may be written in either of the forms d(f> dd> d(f> dd>' d' v dp ^ dq dr r dp' ^ dq' dr' If (p'q'r') is not a tangent, either of these equations represents its pole. * It might seem simpler to replace OL' by p', &c. ; the result, however, when expanded is in that case of the first form given in IV; and#is the equation to a circle. 406 Analytical Geometry VI. The conic is a parabola if its equation is satisfied by the co-ordinates of the line at infinity (i, i, i) ; i. e. if a + b + c + 2f +2g + 2h= o VII. The centre being the pole of the line at infinity, its equation is d d (b d d> — - 4. — L- + — L = o dp dq dr Exercises on Chapter X"V 1 . P is any point on the parabola y 2 = 4 ax ; PM, PN are perpendiculars on the axes. Find the envelope of MN. Ans. The parabola y 2 = — 16 ax 2. Find the envelope of a line which forms with the axes a triangle of con- stant area k 2 . Ans. The hyperbola 2 xy sin co = k 2 3. Two fixed lines meet in O ; if P, Q are points on the lines such that m OP + n OQ is constant ; the envelope of PQ is a parabola touching the lines. 4. If circles are described on double ordinates to the axis of a parabola as diameters, their envelope is an equal parabola. 5« Find the envelope of a line such that the sum of the squares of the perpendiculars on it from two fixed points (c, o) and (— c, o) is constant ( = 2 k 2 ). Ans. The conic x 2 /(k 2 - c 2 ) + y 2 /k 2 = i 6. Find the envelope if the difference of the squares of the perpendiculars = 2 k 2 . Ans. The parabola c 2 y 2 = k 2 (k 2 — 2 ex) 7. Two fixed lines intersect at O ; a circle passing through O and another fixed point P meets the given lines in L, M. Show that the envelope of LM is a parabola touching the given lines. Exercises on Chapter XV 407 8. The axes of an ellipse are given in position. If the product of the lengths of its semi-axes = k 2 , find its envelope. Ans. The hyperbolas 2 xy = + k 2 9. The normals at four points P, Q, R, S on an ellipse co-intersect : if PQ passes through a fixed point, prove that RS envelopes a parabola touching the axes of the ellipse. [Note— Use the eq'ns of PQ, RS given at end of § 373.] 10. Being given the radius of the director circle of an ellipse, and two conjugate diameters in position : show that the ellipse touches four fixed straight lines. 1 1 . Prove that the envelope of the polar of a given point P with respect to a system of confocals whose centre is C is a parabola whose directrix is CP. 12. A transversal meets the sides of a given triangle in P, Q, R. If PQ : QR is a given ratio, the envelope of the transversal is a parabola touching the sides of the triangle. 13. A line which revolves round a fixed point O meets a given line AB in P. Prove that the envelope of the bisector of the angle OPA is a parabola whose focus is O and directrix AB. Prove also that the envelope of a line through P inclined at a constant angle to OP is a parabola which touches AB and has its focus at O. 14. The diagoDals of a quadrilateral inscribed in a circle intersect at right angles in a fixed point. Prove that the sides of the quadrilateral touch a fixed conic, whose foci are the fixed point and the centre of the circle. 15. The envelope of chords of a conic which subtend a right angle at a fixed point O is a conic, whose focus is O and directrix the polar of O. 16. If a straight line is cut harmonically by two circles, its envelope is a conic whose foci are the centres of the circles. 17. A transversal is cut harmonically by two conies. Show that its envelope is a conic. [Note — Refer the conies to their common self-conjugate triangle. Adopt the notation of Ex. 51, page 396 ; let the transversal be koc + /x/3 + vy = o (j) Eliminating y between (1) and the equation of each conic we obtain the 408 Analytical Geometry equations of the line-pairs joining the vertex C of the triangle of reference to the intersections of the transversal (i) with the conies. Using the condition of § 141 we find that the transversal (1) is cut harmonically if (mn'+ m'n)A 2 + (nl' + n'\) (j? + (lm / + I'm) z; 2 = o . . (2) The envelope of the line (1) subject to the condition (2) is (§ 449) the conic = a 2 /(mn'+ m'n) + /3 2 /(nl' + n'l) + y 2 /(lm' + Km) = o.] 18. Prove that the eight tangents to two conies at their points of intersection touch the conic . 19. The vertices A', B', C of a triangle move along fixed lines BC, CA, AB; two sides A'C, B'C pass through fixed points Q, P. Prove that the envelope of the third side A'B' is a conic touching AQ, BP. {Note— Take ABC as triangle of reference; let P, Q be (OL'fi'y), (a"/3"y")-] 20. Given three points on a conic; if one asymptote pass through a fixed point, the other will envelope a conic touching the sides of the given triangle. 21. Two sides of a triangle inscribed in a conic pass through fixed points P, Q. Prove that the envelope of the third side is a conic touching the given conic at its intersections with PQ. {Note — Let the given conic be a/3 = Ay 2 ; PQ being the line y = o.] 22. Two tangents making a constant angle are drawn to two given circles ; prove that the join of the points of contact touches a fixed conic. 23. The envelope of a transversal on which two given circles intercept equal chords is a paraboia. 24. PQ is a variable diameter of a conic, and the chords PR and QR make equal angles with the tangent at R ; prove that all the lines PR and QR touch the same conic. 25. Show that the foci of a conic touching the sides of a parallelogram lie on a rectangular hyperbola circumscribing the para] lei ogram. [Note— See §§ 369, 451.] 26. The locus of foci of conies inscribed in a quadrilateral is in general a cubic. Exercises on Chapter XV 4°9 27. If p? q> r are the perpendiculars on a line from three given points A, B, C : find the envelope of the line if p2 = q 2 + r 2 Ans. The conic x 2 = y 2 + z 2 28. The envelope of chords of the parabola y 2 = 4 a x which subtend a constant angle (X at the vertex is the conic (x — 4 a) 2 + 4y 2 + 4 cot 2 a (y 2 — 4 ax) = o 29. Conies are drawn with a fixed point within a given circle as one of their foci, and touching two fixed tangents to the circle ; show that their chords of intersection with the circle envelope a second circle. 30. The envelope of a circle on a chord of a conic fixed in direction as diameter is a conic. 31. A straight line moves so that the sum of the squares of the perpen- diculars on it from any number of fixed points is constant : prove that its envelope is a conic. 32. If the normals drawn to the ellipse (a, b) from any point on the normal at (hk) meet the ellipse in P, Q, R; prove that the sides of the triangle PQR touch the parabola (xh/a 2 + yk/b 2 + i) 2 = 4 hkxy/(a 2 b 2 ) 33. A triangle is inscribed in a conic S, and two sides touch another conic S / ; prove that the envelope of the third side is a conic passing through the intersections of S, S r . 34. Interpret the tangential equations pq + qr + rp - o, p 2 = qr, p 2 = 4qr Ans. (1) The ellipse touching the sides of the triangle of reference at their mid points. (2) The parabola touching AB, AC at B, C. (3) The ellipse touching AB, AC at B, C, and passing through the centroid of ABC. 35. Find the tangential equations to the nine-point circle and the self- conjugate circle of the triangle of reference. Ans. a Vq + r + b Vr + p + c Vp + q = o, p 2 tan A + q 2 tan B + r 2 tan C =0 410 Analytical Geometry 36. Interpret the tangential equation p (a 2 q + a x r) - a 3 (p - q) (p - r) Arts. The triangle of reference being ABC, the form of the equation shows that it represents a conic inscribed in the quadrilateral whose vertices are A, the mid points of AB, AC, and the point dividing BC in the ratio a x : a 2 . 37. The tangential equation of a conic whose foci are (x'y'z'), (x^y^z") is (x'p + y'q + z'r) (x"p + y"q + z"r) = A £1 where 12 is the sinister of the equation of § 413, and A is a constant. 38. If the equation to the normal at either extremity of a chord PQ of an ellipse be Ax + /uy + 1/ = o prove that (a 2 - b 2 ) A/x + OCfJLV — /3i>A = (1) the ellipse being referred to its axes, and (CX, /3) the pole of PQ. {Prof Burnside, Educ' Times Reprint, Vol. VI., p. 108.) 39. Hence prove that a parabola can be described touching the two normals, the chord PQ and the axes of the conic ; the diameter conjugate to the chord being the directrix. [Note — The preceding equation (1) is satisfied by (A = o, jjl = o), Qut = o, v = o), (A - o, v = o) ; hence the curve of the second degree represented by this equation touches the line at infinity and the axes of the conic. It is also satisfied by (A = a/a 2 , /oc = /3/b 2 , i>--i); hence this curve touches the chord PQ. If Ax + /ay + v — o is the eq'n of one of the tangents from (CX, (3), then Aa + /x/3 + v = o (2) Eliminating v between (1) and (2) we find that the sum of the coefF of A 2 , fx 2 is zero, or the tangents to the parabola from (CX, /3) are at right angles. Hence (CX, /3) is a point on the directrix ; also the centre is another point on the directrix, since the parabola touches the axes.] CHAPTER XVI METHODS OP TRANSFORMATION RECIPROCAL POLARS § 4-55. By taking the polars of all the points and the poles of all the lines in a plane figure A with respect to a fixed conic S, a new figure B is constructed ; if B is treated in the same manner, the figure A is reproduced. Two such figures are said to be polar reciprocals. If p, q * are the lines in one figure corresponding to two points P, Q in the other, the intersection (p, q) corresponds to the line PQ \Cor' (3), § 306J. To a series of collinear points P, Q, R, ... in one figure cor- responds a series of concurrent lines p, q, r, ... in the other \Cor' (4), § 306]. Thus from any theorem relating to the position of points and lines we can deduce another relating to lines and points. Ex. The following theorem is the reciprocal of Ex. 7, p. 174. If the three vertices of a triangle move one on each of three concurrent lines, and two of its sides pass through fixed points ; then the third side passes through a fixed point. § 4-56. Let P, Q, R, ... be a series of points on a curve 2 ; their polars p, q, r, ... with respect to the fixed conic S envelope another curve 2'. * It is sometimes convenient to denote lines by single letters p, q, r, ... ; the point of intersection of two lines p, q is denoted by (p, q). 4 T 2 Analytical Geometry [457. Also the points (p, q), (q, r), ... are the poles of the lines PQ, QR, ... • If now we suppose that the points P, Q are consecutive, then PQ becomes the tangent to 2 at P, and (p, q) becomes the point of contact of p with its envelope 2'. Hence 2 is the envelope of the polars of points on 2' with respect to S ; and the relation between the curves 2, 2' is reciprocal. The relation between the reciprocal curves 2, 2' may also be stated thus : each curve is the locus of poles of tangents to the other with respect to the auxiliary conic S. Cor* (i) — To a point on any curve and its tangent correspond a tangent to the reciprocal curve and its point of contact. Cor' (2) — To a point of intersection of two curves corresponds a common tangent to the reciprocal curves. Cor' (3) — If two curves touch, their reciprocals also touch. For the original curves have a point and the tangent at that point common ; the reciprocal curves have .*. a tangent and its point of contact common. § 4-57. The polar reciprocal of a conic is a conic. Refer the given conic 2 and the auxiliary conic S to their common self-conjugate triangle ; let their equations be, to S, La 2 + M/3 2 + Ny = o and to 2, I a 2 + m/3 2 + ny 2 = o The reciprocal polar of 2 is the locus of poles of tangents to 2 with respect to S. Expressing (§ 436) the condition that the polar of (a! 'fi'y') with respect to S, viz. La'a+ M/3'/3 + Hy'y = o touches 2, and writing 06, /3, 7 for a', /3', y' \ we find that the reciprocal polar of the conic 2 is the conic 2 r , L 2 oi 2 /l + M 2 /3 2 /m + N 2 7 2 /n = o 460.] Methods of Transformation 413 § 4*58. Let P, Q be two points on 2 ; PT, QT the tangents at these points. Also, let p, q be the points on 2' corresponding to the tangents PT, QT ; let the tangents at p, q meet in t. Then T is the pole of pq and t is the pole of PQ with respect to the auxiliary conic S {Cor 1 (3), § 306.] Hence — To a line and its pole with respect to a conic correspond a point and its polar with respect to the reciprocal conic. S 4*59. Ex. 1. The following theorem is the reciprocal of Ex. 24, p. 305. The locus of the pole of a given line with respect to conies inscribed in a given quadrilateral is a straight line. Ex. 2. Reciprocating Ex. 25, p. 305, we find — The envelope of the polars of a given point with respect to conies inscribed in a given quadrilateral is a conic. Ex. 3. The theorem, Ex. 33, p. 409, is the reciprocal of Ex. 2, § 441. Ex. 4. Reciprocate Pascal's Theorem (§ 431). Let A, B, C, N, B f , C be the vertices of the hexagon. Their polars with respect to the auxiliary conic S are six tangents a, b, c, a', b/, c' to the reciprocal conic 2'. To the opposite sides AB, A / B / of the first hexagon correspond opposite vertices (a, b), (a', b') of the reciprocal hexagon ; and to the intersection of AB, A' W corresponds the join of the intersections (a, b), (a', W). Similarly with reference to the other pairs of opposite sides of the given hexagon; and as the intersections of its three pairs of opposite sides are collinear, the joins of opposite vertices of the reciprocal hexagon cointersect. This is Brianchon's Theorem (§ 432). Note — The learner will remark that in many cases the process of reciproca- tion reduces to a mechanical interchange of the words ' point ' and ' line, ' ' locus ' and ' envelope,' ' inscribed ' and ' circumscribed,' &c. RECIPROCATION WITH RESPECT TO A CIRCLE § 4-60. If the auxiliary conic S is a circle, the construction for the reciprocal polar of a given curve 2 may be stated thus. Let O be the centre and h the radius of the circle S. [See fig', page 415.] Let P be any point on the curve 2 ; draw OM perpendicular to the tangent at P. On OM take a point p such that OM . O p = P . (1) Then 2' is the locus of p» 4H Analytical Geometry [461. If b changes, the curve 2' will evidently remain similar to itself. As we are usually concerned only with the shape of 2', it is unnecessary to mention the radius b ; and we may simply speak of 2' as the reciprocal of 2 with respect to the point O. The point O may be called the origin. Cor' — We see from the relation (i) that — The distance of any point from the origin varies inversely as the distance of its reciprocal therefrom. § 4*6 1. Let the lines a, b be the reciprocals of the points A, B ; then a is perpendicular to OA and b to OB. Hence the angle between the two lines (a, b) is equal to the angle AOB, which the join of their poles subtends at the origin. This principle is often useful. 8 4-62. Returning to the construction of § 458, let the point T coincide with the origin O. Then pq, which is the reciprocal of the origin, is the line at infinity ; t is the centre of the reciprocal conic 2', and tp, tq are its asymptotes. The centre and asymptotes of the reciprocal of a given conic 2 with respect to any origin O are .*. determined by this construction — Draw tangents OP, OQ to the conic 2; the centre of the reciprocal conic is the reciprocal of PQ, and its asymptotes are the reciprocals of the points of contact P, Q. Cor? (1) — The angle between the asymptotes of 2' is the supplement of the angle POQ (§ 461). Cor* (2) — Hence 2' is a rectangular hyperbola if POQ is a right angle ; i. e. if the origin O is on the director circle of the conic 2. Cor 1 (3) — As either 2 or 2' may be regarded (§ 456) as the original conic, we see that — The reciprocal of a rectangular hyperbola with respect to any point O is a conic whose director circle passes through O. Cor* (4) — To find the condition that the reciprocal conic 2' may be an hyperbola. The points p, q in which it is met by the line at infinity must be real and different ; the tangents OP, OQ must .*. be real and different, i. e. the point O must be outside the conic 2. Similarly the reciprocal curve 2' is an ellipse if the origin O is inside the conic 2; and it is a parabola if the origin O is on the conic 2. 4 6 5-] Methods of Transformation 4i5 § 4-63. To find the polar reciprocal of a circle whose centre is C and radius r with respect to an origin O. Make the construction of § 460 ; let the polar of the point C with respect to O meet OC in X. Draw pn perpendicular to the polar of C Then PM is the polar of p, and nX is the polar of C. .-. Op : OC = pn : CP [§ 202.] .-. Op : pn = OC : r The reciprocal polar is .'. a conic whose focus is O, directrix the polar of C, and eccentricity OC/r. Cor 1 (1) — The latus rectum of the conic varies inversely as r. For it = 2 e OX = 2 (OC/r) (5 2 /0C) = 2 p/r Cor' (2) — The centre of the reciprocal conic is the reciprocal of the chord of contact of tangents from O to the circle (C). This is proved in § 462. § 4-64- . Ex. 1. 'The product of the segments of chords of a circle drawn through a fixed point O is constant.' (Euclid III. 35, 36.) The reciprocal of this is {Cor', § 460) — The product of the perpendiculars from a focus of a conic on two parallel tangents is constant. This is equivalent to the Theorem of § 248, VII. Ex. 2. ' The vertices of a rectangle circumscribing a conic lie on the director circle.' If this is reciprocated with respect to a focus we obtain the theorem, Ex. 14, page 407. § 465. It follows from § 463 that concentric circles reciprocate into conies having the same focus and directrix. Ex. Two conies have the same focus and directrix ; their eccentricities are are e and e'. If e' = 2 e, prove that an infinite number of triangles can be inscribed in one conic and circumscribed to the other. 416 Analytical Geometry [466. The reciprocal of this is — If the radii of two concentric circles are r and r', then if r - 2 K an infinite number &c. ; this is proved at once by Elementary Geometry. § 4-66. The reciprocal of a system of co-axal circles with respect to either limiting point is a sj/stem of confocal conies. For the conies have one focus common, viz., the limiting point ; also, the limiting point having the same polar with respect to every circle of the system (Ex., § 189), the conies are concentric [Cor' (2), § 463]. The second focus is .". the same for all the conies. Ex. The theorem of § 340 is the reciprocal of Ex. 31, page 161. sjj 4^7. If we are given the trilinear equation of a curve, (a, (3, y) = o (1) it is easy to obtain the tangential equation of its reciprocal with respect to any point OCfciftyO. Let b be the radius of the reciprocating circle, A'B'C the triangle reciprocal to the triangle of reference ABC ; then A' is the pole of BC, and A is the pole of B'C, &c. ; and «! . A'O = p 1 ■ B'O = y x . CO = 5 2 (2) Let p, q, r be the perpendiculars from A', B', C on the reciprocal of a point P (Otfiy) on the curve (1) ; this reciprocal is a tangent to the reciprocal of the curve (1). Then, by § 202, A'O : PO = _L from A' on polar of P : J_ from P on polar of A' = p : a .-. gybfj : PO = p : OC .-. Oi'.pOC, = PO:52 = /3:q/3l==y:ryi (3) The tangential equation to the reciprocal curve is .*. 4>(p&i> q&» r yi) = ° (4) Again, from (3), p : q : r = (x/(X x : &ffi x : y/y x If .*. the tangential equation of a curve referred to A'B'C is +■ (P» q, r) = o the trilinear equation of its reciprocal referred to ABC is yfr (a/ot!, pfa, y/y,) - o Ex. The tangential equation of the circum-circle of A'B'C is sin A' Vp + sin B' Vq + sin C Yr = o [§§ 4 2 4, 4-3; 4°°-l 467-] Methods of Transformation 4 J 7 Now A' = 180 - BOC, &c. ; and the reciprocal of the circle is a conic inscribed in ABC, and having its focus at O. The equation of a conic inscribed in ABC and having its focus at (tti/3 x yi) is .-. sin BOC x/a/ax + sinCOA \/ P/fc + sinAOB yj f yfy x = o Exercises 1. Prove that the ellipses (a, b) and (a', b') are polar reciprocals with respect to the ellipse (Vaa', Vbb'). 2. Find the equation of the reciprocal of the ellipse (a, b) with respect to a circle whose centre is (x'y'), and radius b. Am. (xx'+ yy' + 6 2 ) 2 = a 2 x 2 + b 2 y 2 3. The reciprocal of a parabola with respect to a point on the directrix is a rectangular hyperbola. 4. Find the polar reciprocal of the conic 2 I 0C(3 = -f with respect to the conic 2 m QC/3 = y 2 Ans. I y 2 — 2 m 2 a/3 = o 5. Prove by reciprocation the results of Exs. i ; 2. 4. § 330; Exs. 19, 20, 24, page 322 ; Ex. 26, page 323. 6. Two conies are described with given directrices and a given common focus S ; the sum of the squares of the reciprocals of their latera recta is given. Prove that their common tangents envelope a conic having one focus at S. 7. Prove that Ex. 29, page 323, reduces by reciprocation to the theorem : — If on a line which revolves round a fixed point and meets a given circle in P a point Q is taken so that OQ is to OP in a constant ratio, the locus of Q is a circle. 8. A conic touches three given lines, and its director circle passes through a given point ; prove that the conic touches another fixed straight line. [Note — This is the reciprocal of Ex. 2, § 315.] E e 4i 8 Analytical Geometry [468. 9. Reciprocate the theorem, Ex. i, § 315. [See Cor' (3), § 462. The reciprocal is equivalent to Ex. 55, page 397.] 10. Reciprocate the theorems of § 183 ; (3), § 190 ; also Ex. 40, p. 162. 11. A circle is reciprocated with respect to a point O. The second focus of the conic is the reciprocal of the radical axis of the circle and the point O. 12. 'Rectangular tangents to a conic meet on a concentric circle.' The reciprocal of this is Ex. 15, page 407. 13. ' The sum of the focal distances of the extremities of a variable diameter is constant.' What is the reciprocal theorem ? 14. A straight line is drawn across a rhombus so that its segments included between the arms of opposite angles subtend supplementary angles at the centre. Prove that it envelopes an inscribed conic. \Note — Reciprocate with respect to centre of rhombus.] 15. Deduce the equation of Ex. 59, page 397, from that of Ex., § 467. 16. A triangle is inscribed in a given ellipse (a, b), so that the focus is the centre of its inscribed circle. Prove that the radius of this circle is b 2 /(a + V 2 a 2 - b 2 ) PROJECTION § 4-68. The lines joining the points of a figure in a plane IT to a point V in space form a cone ; the section of this cone by a plane 7T is a figure which is called the projection of the given figure. The point V is called the vertex ; the plane ir is called the plane of projection. The projection of a point P is the point p where the join VP is cut by the plane 77 ; we shall denote points in the plane IT by capital letters, and their projections by the corresponding small letters. § 4-69. The joins of the points in a straight line to V form a plane ; this plane intersects the plane ir in a straight line. Hence the projection of a straight line is a straight line. 47i-] Methods of Transformation 419 Cor' (1) — If a system of lines co-intersect in a point P, their projections co-intersect in p. Cor' (2) — To a chord PQ of a curve corresponds a chord pq of its projec- tion ; and if P, Q are consecutive, so are p, q. Hence curves, tangents, and points of contact project into curves, tangents, and points of contact. § 4-70. The projection of a conic is a conic. Let A, B, C, D be four fixed points, and P a moveable point on a conic ; let the lines PA, PB, PC, PD be intersected by any line in the points A', B', C, D'. Let the projections of these points be a, b, c, d, p, a', b', c', d'. Then (§ 137) {P. ABCD} = {A'B'C'D'} = {V. A'B'C'D} = {a'b'c'd'} = {p. abed} But {P. ABCD} is constant (§ 355); .*. {p. abed} is constant. The locus of p is .*. a conic passing through a, b, C, d. Q.E.D. Cor r (1) — The statement in § 213 that any section of a right circular cone by a plane is a conic is a particular case of the proposition just proved. Cor' (2) — It is evident from the preceding proof that the cross ratios of ranges and pencils are unaltered by projection. ~v Cor r (3) — The properties of poles and polars with respect to a conic are projective : this follows evidently from the Cor's to § 469, and the preceding Cor' (2). § 4-71. Let LM (see fig', next §) be the line of intersection of the primitive plane IT with a plane through the vertex V parallel to the plane of projection it. Then points on LM are projected to infinity. Hence lines which cointersect in a point on LM are projected into lines which cointersect on the line at infinity in the plane 77-, i. e. into parallel lines. Similarly, systems of parallels in the primitive plane are projected into systems of lines which co-intersect in points situated on the line of intersection of the plane of projection with a plane through V parallel to the primitive plane. e e 2 420 Analytical Geometry [472. § 4-72. Let POQ be any angle in the primitive plane ; LM the line which is projected to infinity, i. e. the line of intersection of the plane IT with a plane through V parallel to 77". Let OV, OP, OQ meet the plane 77" in O', P', Q'; then O'P', O'Q' are the projections of OP, OQ. Now O'P', VP are parallel ; for these lines are the intersections of the parallel planes 77" and VLM by the plane VOP. Similarly O'Q', VQ are parallel. .-. P'6'Q' = PVQ [Euclid XL 10.] A A i. e. POQ is projected into an angle = PVQ. § 4-73- Any line and any two angles being chosen in a plane ; the line can be projected to infinity and the angles at the same time into given angles. Let LM be the line ; let the arms of the angles meet LM in p, q and p', q' respectively. Draw any plane through LM ; and let segments of circles described, in this plane on pq and p'q', and containing angles respectively equal to the given angles, meet in V. Take V as vertex, and any plane parallel to VLM as plane of projection; it follows from §472 that the given figure will be projected in the manner described. 475-] Methods of Transformation 421 Cor' — The locus of V is a circle in a plane perpendicular to LM. If this locus meet the primitive plane in (f), then any angle is projected into the angle which the intercept of its arms on LM subtends at (j). Thus by a plane construction we can determine the angle into which any given angle is projected. § 4-74. A conic and a line in its plane being given ; the line may be projected to infinity \ and the conic at the same time into a circle. Let LM be the line, C its pole with respect to the conic. Project (§ 473) LM to infinity, and the angles between two pairs of conjugate lines through C into right angles. Then c, the projection of C, is the centre of the projected conic ; and this conic has two pairs of conjugate diameters at right angles. The projected conic is .*. a circle. Cor' — As the pole of the line which is sent to infinity becomes the centre, the theorem may be stated thus — Given a conic and a point in its plane ; the conic may be projected into a circle whose centre is the projection of that point. § 4-75. By projecting the figure, the truth of a general theorem of position can be inferred from that of a simpler particular case. Ex. I, Prove Pascal's theorem (§ 431). Project the conic into a circle, and the join of the intersections of two pairs of opposite sides to infinity. The theorem is then — If a hexagon inscribed in a circle has two pairs of opposite sides parallel, the remaining sides are paralieL This is proved at once by Elementary Geometry. Ex. 2. LOL', MOM', NON', ROR' are four concurrent chords of a conic. If P is any point on the conic, then {P.LMNR} = {P. L'M'N'R'} Project the conic into a circle whose centre is the projection of O. Ex. 3. Any quadrilateral can be projected into a square. Project the third diagonal to infinity : then the quadrilateral becomes a parallelogram. Project the angles included by two adjacent sides and by the diagonals into right angles : the projected figure is then a square. Ex. 4. Two conies can be projected into concentric conies. Project a side of their common self-conjugate triangle to infinity. e e 3 422 Analytical Geometry [476. § 4-76. Any two points can be projected into thefocoids. Let the points be L, M ; draw any conic through LM. Project the conic into a circle, and LM to infinity (§ 474). Cor f (1) — Conies passing through four fixed points can be projected into coaxal circles. Let L, M, N, R be the four points. Project LM to infinity and one of the conies into a circle, i. e. project L, M into the focoids. Then all the conies are projected into conies passing through the focoids, i. e. into circles ; and these circles all pass through n, r, the projections of N, R. Cor f (2) — Conies having double contact can be projected into concentric circles. This follows from Cor' , § 387. Let L, M be the points of contact ; project LM to infinity and one of the conies into a circle. Cor' (3) — Conies inscribed in a quadrilateral can be projected into confocal conies. This follows from Cor' ', § 367 ; project the extremities of the third diagonal into the focoids. Ex. I. Two sides of a triangle inscribed in a conic pass through fixed points P, Q ; prove that the envelope of the third side is a coitic having double contact with the given conic. Project the given conic into a circle, and PQ to infinity : then the theorem becomes — If two sides of a triangle inscribed in a circle are parallel to fixed lines, the third side envelopes a concentric circle. Ex. 2. To prove Desargues' Theorem (§ 391). Project the conies into co-axal circles whose common chord is CD. Let any line meet CD in O and the circles in P, P'; Q, Q' ; R, R' ; &c. Then (Euclid III. 35, 36) OP. OP 7 = OQ . OQ' = OR . OR' = &c. Hence O is the centre of an involution determined on the line by the circles ; and the theorem is proved. Ex. 3. If the six sides of two triangles ABC, A'B'C touch a conic, their six vertices lie on a conic. Project B', C into the focoids, then A' becomes a focus ; and the theorem is reduced to that of Ex. 3, § 330. 479-] Methods of Transformation 423 § 4-77 • The polar s of four collinear points with respect to a conic form a pencil whose cross ratio is eqtial to thai of the four points. Project the conic into a circle whose centre is the projection of a point on the line of collinearity ; then the four polars become parallel. The four points in this case, and the four intersections of the polars with the line of collinearity, form an involution, whose focus is the centre. [§§ 143, 145.] PROJECTION OF ANGLE PROPERTIES § 478. Let i, j be the projections of the focoids I, J. Then (§ 360) two lines at right angles are projected into lines cutting the fixed segment ij harmonically. Also, (§ 362) two lines including an angle 6 are projected into lines cutting the fixed segment ij in the cross ratio cos 26 + V — 1 sin 2 Ex. 1. 'If A, B are fixed points and P a variable point on a circle, the angle APB is constant.' From this we infer — If a, b, i, j are fixed points, and p a variable point on a conic, the cross ratio {p. aibj} is constant. [Compare § 355.] Ex. 2. ' Rectangular tangents to a parabola meet on the directrix. The generalization of this by projection is — The locus of the intersection of tangents to a conic catting the segment determined by two fixed points i, j on a given tangent harmonically is a straight line ; this line is the polar of the intersection of the other tangents from i, j to the conic. PROJECTION OF METRICAL PROPERTIES § 4-79. Lemma. If\,J are fixed points in a plane, and A, B, C, D any other points in the plane, then if (A, B) - <£ (C, D) (1) is unaltered by projection. Let a, b, &c. be the projections of A, B, &c. ; let P, p be the perpendiculars from the vertex V on the planes IT and tt. 424 Analytical Geometry [4 8 °- Then, by known theorems in Solid Geometry, the volumes of the pyramids subtended by AIB and aib at V are in the ratios P.areaAlB : p.areaaib, and VA .VI .VB : Va .V i .Vb Al _ pVA.VI.VB, ... .*. area AIB = ^ —. — ... ... (areaaib) P Va.Vi .Vb Similar values are obtained for the other areas AJB, &c. ; and substituting these in (i), it reduces to (j> (a, b) -^ (p (c, d) § 4*80. Let us ascertain the value of (A, B) = - AB/(2 A3) We may .". in any equation connecting the distances of points A, B, &c. replace each distance AB by I Willi COIAEEEU. 9 -' BOSTON COLLEGE 9031 01550368 3 I 5~0 Si MATH. C BOSTON COLLEGE LIBRARY UNIVERSITY HEIGHTS CHESTNUT HILL, MASS. 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