t:7 ^ls! s~ THE ELEMENTS OF ANALYTICAL GEOMETRY; COMPREHENDING THE DOCTRINE OF THE CONIC SECTIONS, AND THE GENERAL THEORY OF CURVES AND SURFACES OF THE SECOND ORDER. INTENDED FOR THE USE OF MATHEMATICAL STUDENTS IN SCHOOLS AND UNIVERSITIES. BY J. R YOUNG,..uthor of "eAn ElementaTry Treatise on.Algebra," "Elements of Geometry," &c REVISED AND CORRECTED BY JOHN D. WILLIAMS, AUTHOR OF " KEY TO HUTTON'S MATHEMATICS." PHILADELPHIA: BUBLISHED BY E. H. BUTLER & CO. 1850. ADVERTISEMENT TO TH E AMERICAN EDITION. THE work now submitted to the notice of the American public, will, it is hoped, supply in part the want that has lmng been felt by the heads of instruction in this country, of a good elementary treatise, in our own language, upon that all important branch of mathematics-the application of analysis to the solution of Geometrical Problems. The Professors of several of our public institutions, convinced of the absolute necessity, to the student, of a thorough knowledge of this subject, prior to his entering upon the study of the Calculus, and its varied applications, have been induced to place in the hands of their pupils the works of the French writers in their native tongue. Among others. the Essai de Geometrie Anatlytiq/ue of Blot. the T'heorie des Courbes of Boucharlat, and the Application de l'Analyse a la Geometrie of Bourdon. have been used to much advantage. Indeed it may be questioned if the use of the French authors as models be not almost absolutely necessary to the writer of a work on this subject; for nowhere else can we find that simple, and, at the same time, elegant and highly finished analysis, for which they are so justly distinguished in the scientific world. Mr. Young, as will be seen by his preface, has drawn largely from these sources; and the eminent superiority of his elementary treatises on the mathematical sciences, is mainly to be attributed to the liberaity of spirit with which-casting off the trammels imposed upon themselves by the countrymen of Newton-he has freely availed himself of every discovery and improvement in analysis, though such have been chiefly made on the French side of the channel. In the present edition few alterations or additions could have been made which would improve the original, with the exception of a careful correction of the typographical errors-and whether or not the Editor has faithfully executed his task, the work itself will show. JOHN D. WILLIAMS. NEW YORK, August, 1833. PRE F AC E. THE application of algebra to the theory of curves and surfaces may be regarded as the fundamental branch of modern analytical science, and as the principal instrument, in conjunction with the differential and integral calculus, with which the continental mathematicians have worked such wonders in almost every department of. the mathematics. The remarkable contrivance, first introduced by Descarites, of representing lines and surfaces by algebraical equations, enables us to embody in such an equation every property and peculiarity belonging to any curve or surface, when we know the law of its description, or any of its distinguishing characteristics; and then, to develope these several properties of the curve or surface, we have only to perform so many easy, and generally obvious, transformations on the equation which represents it. The superiority of this method over the geometrical, both in ease and fertility, immediately led to its general adoption among the French mathematicians; and the method of co-ordinates, which the Cartesian geometry involved, was afterwards applied to mechanics, and, indeed, to every other:part of mathematical physics, each of which has been improved and extended by its introduction.?t English mathematicians have, however, been singularly slow in appreciating these decided advantages; so slow, indeed, that, till the year 1823, when Dr. Lardner published the first part of his Algebraic Geometry, the English language possessed not a single book on this subject. Besides this work, a treatise on analytical geometry has also emanated from the University of Cambridge, which, although a work of much originality and ability, the ingenuous author has since publicly acknowledged to be unsuited to the purposes of elementary instruction. Dr. Lardner's book will, no doubt, when completed, present a valuable body of analytical science, accessible, however, to those only who have a knowledge of the differential and integral calculus. * Maclaurin, in his "Treatise on Fluxions," first suggested this happy idea. which threw a new light on the entire theory of mechanics. But, unfortunately, this simple principle has been neglected by later English authors, and much of what our mathematicians at present know and practise of this method, we owe chiefly to the re-importation of it through the medium of modern French works; and man), perhaps, who are admiring the facility which is thus thrown into mechanical investigations of the greatest difficulty, are unconscious that this thought had its origin in our own land. Encyclopsedia MIdetropolitana, art. AgMechanics, p. 5. 6 PREFACE. The present little volume is then an attempt to fill up a chasm which seems still to exist in our mathematical courses of instruction, and to supply the connecting link between elementary geometry and trigonometry, and some of the most interesting applications of the transcendental analysis. For such an undertaking, the French language offers copious materials; and I have, accordingly, carefully examined and freely used the performances of Bliot, Lacroix, Boucharlat, Boutrdon, &c.; and I shall consider myself particularly fortunate, if it be found that I have in any degree imbibed the spirit of these elegant writers. As regards arrangement, however, I have differed from most other authors, adopting that which appeared to me most likely to facilitate the progress of the student, without waiting to consider whether a more strictly methodical disposition of parts might not be devised. Conformably, too, to this determination, I have, in one or two cases, not hesitated to introduce a geometrical property to supply the place of analysis, where such introduction appeared of unquestionable advantage in shortening the process; instances of this occur at pages 162 and 163. The total rejection of all geometrical aid in most of Frenchbooks is perhaps carried to an injudicious extent, and seems to be, in some cases, the result of caprice or affectation, for such aid is obviously allowable, and even adviseable, where simplicity may be attained by it. As to the arrangement here adopted, it may be briefly stated as follows: The volume consists of two principal parts, Analytical Geomnetry of Two Dimensions, and Analytical Geometry of Fhree Dimensions. The first part contains an introductory section on the algebraical solution of geometrical problems, and on the geometrical construction of algebraical equations; then follows, in three sections, an examination of the various. properties of the lines of the second order, deduced from the most simple forms of their several equations; these three sections are, therefore, complete in themselves, comprehending, in the compass of one hundred and thirtyeight pages, a pretty copious treatise on the Conic Sections. The fourth section enters more at large into the theory of these curves, by discussing very fully the most general forms of their equations, their, positions in reference to any assumed axes, the determi. nation of their varieties, &c. and the use of -these researches is illustrated in Chapter iii. by their application to a variety of interesting problems on geometric loci. This first part terminates with a supplementary chapter containing some very useful theorems, such, for instance, are those at pages 211 and 214, the former of which is necessary in one very elegant mode of establishing the fundamental problem of physical astronomy, viz. that the planets move in elliptic orbits, having the sun in one of their foci; and the other problem is the foundation of the method of interpolations, so useful in the construction of tables, and ini practical astronomy. PRE FACE 7 The second part is devoted to the consideration of lines and surfaces in space, the developement of their properties, and the general discussion of their equations. As to the first part, so here, a supplementary chapter is appended, containing many curious and interesting applications of the preceding theory. Most of the problems in this chapter have appeared before some in the Annales Math6matiques, others in Leybourn's Repository, &c. but the solutions here given are for the most part new, and I think improved. By way of index to the various topics embraced in the work, I have prefixed to the volume a very copious table of contents, which indeed precludes the necessity of extending further these prefatory remarks. I therefore conclude with the hope that the little volume now submitted to notice, though its pretensions be as humble as the form which it has assumed, may yet prove of some service to the mathematical student, in the earlier stages of his progress. J. R. YOUNG. CONTENTS. P ART I. ANALYTICAL GEOMETRY OF TWO DIMENSIONS. SECTION I. Application of Algebra to Geometry. tArticle Page I Introductory Remarks 1...... 19 2 Knowing the base and altitude of a triangle, to find the side of the inscribed square....... 19 3 To divide a straight line in extreme and mean proportion. 20 3 Geometrical signification of the signs + and- 2.. 20 4 From a given point without a circle, to draw a secant such that the intercepted chord may have a given length... 21 5 To divide a straight line so that the rectangle of the two parts may be equal to a given square.... 21 6 Given the perimeter of a right-angled triangle, and the radius of the inscribed circle, to determine the triangle'.. 22 7 Given the chords of two arcs to find the chord of their surrm 22 8 Given the three sides of a triangle to find the radius of the circumscribed circle... 22 9 Given the sides of a triangle to find its surface.. 23 10 Another expression for the surface. 23 11 Expression for the radius of the circumscribed circle.. 24 12 Miscellaneous problems ~.. 25 13 Construction of some simple algebraical expressions. 81 14 More complicated expressions.... 81 15 Of rendering algebraical expressions homogeneous. 82 16 Construction of irrational expressions... o 83 SECTION 1I. On the Point, Straight Line and Circle. 1 On Analytical Geometry... 85 2 On the equation of a point... 85 3 Situation of a point fixed by the signs of its coordinates 86 4 On the equation of a straight line... 86 5 When the axes of coordinates are rectangular.. 87 6 When they are oblique... 88 7 When the line does not pass through the origin.. 88 8 Equation of a straight line passing through a given point 90 9 Equation of a straight line through two given points. 91 10 Equation of a straight line through a given point and parallel to a given straight line... 91 10 To find the point where two straight lines intersect.. 91 2 10 CONTENTS. Article Page 11 Io find the expression for the angle of intersection of two straight lines.......... 92 12 Expressions for the sine and cosine.... 93 13 Equation of a straight line through a given point and having a given inclination to another straight line... 93 14 Expression for the distance between two points... 94 15 Expression for the distance of a point from a line.. 94 16 Sine and cosine of the inclinations of two straight lines determined by last article....... 95 17 To determine whether perpendiculars from the vertices to the sides of a triangle meet in a point... 96 18 To determine whether perpenwdiculars from the middle of each side meet in a point..... 97 19 To determine whether lines from the vertices bisecting the opposite sides meet in a point....... 98 20 To determine whether lines bisecting the angles meet in a point 98 21 To express the area of a triangle in terms of the coordinates of two angular points...... 99 22 The locus of -a simple indeterminate equation of two variables is always a straight line....... 100 23 To construct the locus...... 100 24 Various equations of the circle..... 101 25 Equation of a tangent to a circle... 103 26 To draw a tangent from a point without the circle. 104 27 From a given point to draw a straight line through a circle, so that the intercepted chord may have a given length.. 105 28 To find the coordinates of the intersections of two circumferences. Deduction of geometrical properties.... 107 29 Remarks on the foregoing investigations.. 109 30 The locus of the equation x2 + y2 + Ax +- By + C =0 is always a circle, a point, or an imaginary curve. -109 31 The base of a triangle and the sum of the squares of the sides given to determine the locus of the vertex... 110 32 To determine the locus when the base and vertical angle are given..... 111 33 The base and vertical angle being given to determine the locus of the intersection of perpendiculars from the vertices to the sides 112 34 To find the locus of the centre of the inscribed circle, when the base and vertical angle are given.. 112 35 The same being given to find the locus of the intersection of straight lines from the angles to the middle of each side.. 113 SECTION III, On Lines of the Second Order. 36 Preliminarv remarks...... 114 37 On the transformatian of coordinates.... 114 38 Investigation and determination of formulas 115 39 Modifications to be introduced when the axis of A/ is below the axis of x........ 11 CONTENTS. 11 Article Page 40 Other modes of expressing the formulas of transformation. 118 41 Equation of the ellipse... 119 42 Determination of the limits of the curve. 120 43 Ex)pression for the distance of any point from the centre 120 43 Different forms of the equation of the ellipse. 121 44 Squares of the ordinates as the products of the parts into which they divide the major diameter; parameter a third proportional to the major and minor diameters.. 122 45 Expression for the angle inscribed in a semi-ellipse.' 23 45 Inferences from this expression.... 124 46 Pioduct of the tangent of the angles which lines from the ends of a principal diameter and meeting in the curve form with that diameter shown to be constant.. 124 47 Expression for the radius vector... 124 48 Transformation of the equation of the ellipse from rectangular to oblique conjugate diameters; inferences from the transformed equation. 125 49 and 50 Peculiarities of the principal conjugates; properties of supplemental chords. 126 51 The property in (46) true for oblique conjugates as well as for rectangular... 127 52 Transformlation of the equation from the oblique to the original conjugate axes.... 128 52 Sum of the squares of the principal diameters equal to the sum of the squares of any system of conjugates 128 53 Circumscribing parallelogram with sides parallel to any system of conjugates always equal to the rectangle of the principal diameters. 128 53 Of all circumscribing parallelograms those about conjugate diameters are least; the sum of the equal conjugate diameters is the greatest, and of the principal diameters the least 129 54 The axes of an ellipse and the vertex of any diameter being given to find the length of that diameter and of its conjugate 130 55 The axes and inclination of a system of conjugates given to determine them in length and direction 13C 56 Equation of the tangent to the ellipse. 131 57 Equations andlengths of the tangent, normal, &c.. 132 58 Preceding equations collected together for reference.. 133 59 Method of drawing a tangent to the ellipse from a i point in the curve - 133 60 Commodious method of drawing a tangent from a point without the curve.. 134 61 Properties derived from the expression for the normal.. 135 62 Properties derived from the -equation of the tangent 135 63 Other properties unfolded by the same equation. 136 64 Determination of the shortest tangent that can be included between the prolonged principal diameters.... 137 65 Properties of the tangent at the extremity of the latus rectum 137 66 Radii vectores equally inclined to the tangent. 138 67 Another method furnished by last article for drawing a tangent to an ellipse.. 140 12 CONTENTS. Article Page 68 If pairs of tangents intersect at right angles the locus of the intersection will be a circle...... 140 69 On the hyperbola and mode of describing it. 141 70 Determination of its equation... o. 142 71 Determination of its vertices and asymptotes... 142 72 Expression for the distance of any point from the centre o 143 73 Different forms of the equation of the hyperbola.. 1.4 75 Equation given as a function of the excentricity... 144 76 Squares of the ordinates as the products of the parts into which they divide the transverse diameter; parameter a third proportional to the transverse and second axes.... 145 78 Expression for the angle included by supplemental chords. 145 79 Expression for the radius vector..... 146 80 Transformation of the equation of the hyperbola from rectangular to oblique conjugates... 147 81 Properties deduced analogous to those in the ellipse.. 148 82 Diameters parallel to a system of supplemental chords from the principal transverse are conjugate.. 148 83 The same property true for any system of supplemental chords 149 84 Transformation of the equation from the oblique to the original conjugates; properties unfolded by this transformation. 149 85 Absolute length of an imaginary diameter determined. 149 86 The axis and vertex of a diameter being given to find the length of that diameter and of its conjugate.. 150 87 Il, "Tu es and the inclination of a system of conjugates given to determine them in length and direction... 150 88 On tangents, normals, &c., to the hyperbola.. 151 89 The equations and lengths of these lines... 152 90 91 93 Properties analogousto those in the ellipse 153 154 94 Equation of the hyperbola when referred to its asymptotes. 154 95 Equation of the tangentwhen referred to the asymptotes. 155 96 Properties of lines drawn between the asymptotes.. 156 97 To construct the curve when a point in it and the asymptotes are given........ 157 98 On the parabola and method of describing it 157 99 Its equation and vertex determined.. 158 100 Equation in terms of the parameter.... 159 101 Transformation of the equation.... 159 102 The diameters all parallel to each other.... 160 103 General equation of the curve in reference to any system of conjugate axes...... 160 104 Expression for the parameter of any diameter 160 105 106 Equations of the tangent, normal, &c.; the subtangent double the abscissa, and the subnormal constant.. 161 107 Properties of the focal tangent.. 162 108 The radius vector and diameter through the point of contact equally inclined to the tangent, &c..... 162 109 To find the locus of the intersection of pairs of rectangular tangents..... 163 110 On polar coordinates.... 164 111 Polar equation of the ellipse, when the focus is pole o 1 65 CONTENTS. 13,Articge Page 112 Polar equation, when the centre is the pole.. 166 113 Polar equation of the hyperbola, the focus being the pole. 166 114 Polar equation, when the centre is the pole... 166 115 Polar equation of the parabola..... 166 116 Determination of the polar subtangent in each curve. 167 117~ Properties of focal chords.... 168 SECTION IV. General examination of indeterminate equations of the Second Degree. 118 Preliminary observations... 169 119 determination of the locus of the equation My2 i:'N x P 170 119 - y__ -Q. 171 120 On the varieties of the three curves.... 171 121 Mode of obtaining the properties of the parabola from those of the ellipse..... 172 122 Locus of any equation of the second degree containing two variables must be one or other of the three curves.. 172 123 Examination of the equation when some of its terms are absent 174 124 Criteria for determining the nature of the curve represented by any equation of the second degree.. 174 125 Meansof constructing the equation.. 176 126 Construction of central curves... e. 176 127 Table of formulas to be employed in constructing central curves 177 128 Examples of their application..... 178 129 Means of judging when the equation represents a system of parallels....... 180 130 Examples ofsuch equations. 181 131 Determination of formulas for constructing parabolas.. 182 132 Table of these formulas, and examples of their application. 182 133 The preceding formulas apply only when the locus is referred to rectangular axes.... 183 134 Discussion of the general equation by the separation of the variables......... 183 135 Examination of the irrational part of the expression for either variable...... ~. 185 136 Discussion of the equation when B -4AC'0.. \. 186 137 B2 —4AC 0... 187 137* - 2, y2, are absent.. 188 138 Construction of asymptotes... 189 139 Discussion of the equation when B2 —4AC0. 190 140 Examples on the preceding discussion.. 191 141 Conditions which exist when the locus meets the axes of coordinates...... X. 195 142 Given the base of a triangle and the sum of the tangents of the base angles to find the locus of the vertex.., 195 143 Given the base and difference of the tangents to find the locus of the vertex.... 196 144 Given the base and difference of the angles at the base to find the locus of the vertex...~.. 196 B 14 CONTENTS., Article Page 145 To find the locus of a given point in a straight line of given length of which the extremities move along the sides of a given angle 197 146 Two straight lines are given in position, fromt any point in one of which a perpendicular is drawn to the other, and from agiven point in this latter, with a radius equal to the perpendicular, an arc cutting the perpendicular in P is described; required the locus of P........ 197 147 To find the locus of the vertex of a parabola touching a given straight line and having a given focus... 198 148 To find the locus of the focus of a parabola which shall touch a given line and have a given vertex 199 149 Given the base and altitude of a triangle to find the locus of the intersection of perpendiculars from the angles to the opposite sides...199 150 Given the base and sum of the sides to find the locus of the point of intersection of lines from the vertices bisecting the opposite sides......... 200 151 Given the base and sum of the sides to find the locus of the centre of the inscribed circle 200 152 Given the base and sum of the sides to find the locus of the centre of the circle touching the base and prolongation of the other two sides.... 201 152 Given thebase and difference of the sides to find the locus of the centre of the circle touching one side and the prolongation of the base and of the other side.... 201 154 Two straight lines are perpendicular to each other, and, through two given points in one, straight lines are drawn, forming with the other angles, the product of whose tangents is constant; what is the locus of their intersection... 202 155 From two given points two straight lines are drawn so as to intercept a given portion of a straight line given in position;, what is the locus of their intersection.... 203 156 Tangents to a parabola form a given angle with each other; what is the locus of their intersection? ~. 203 157 Tangents to a parabola form angles with the principal diameter, the product of whose tangents is given; what is the locus of. their intersection?,... 205 58 To find the locus of the intersections of pairs of tangents to any line of the second order when they make angles with the principal diameter, such that the product of their tangents may be given 205 159 Problems without solutions.. 207 160 If to any line of the second order two secants parallel to the sides of a given angle be drawn, the two rectangles contained by the parts intercepted between their point of intersection and the curve, will have a constant ratio..... 208 161 To determine the general equation of the tangent to any line of the second order'..... 08 162 If through any point chords are drawn to a line of the second order and tangents be applied to their extremities, these tangents will intersect on a straight line... 211 163 If a curve of the second order be referred to a system of conjugate CONTENTS. 15 Article Page axes, and a point in its plane be found, such that its distance from any point in the curve be a rational function of the abscissa, the point thus found will be the focus.. 211 164 To find a cube that shall be double a given cube.. 212 165 To trisect an angle...,.. 213 166 Five points being given on a plane of which no three are in the same straight line, it is possible to describe a line of the second order passing through them all... 214 167 To determine a curve which shall pass through any proposed number of points..... 14 168 General Scholium, with remarks on the higher curves. 218 PART II. ANALYTICAL GEOMETRY OF THREE DIMENSIONS. SECTION I. On the Point, Straight Lne,'and Plane, in Space. 169 Introductory remarks...... 221 170 Equations of a point..... 222 171 Modifications of these equations.. 223 1_72 Equations of the projections of a point.. 223 173 Equations of a straight line in space... 224 174 Equation of either projection determined from the other two. 224 175 Determination of the points where the coordinate planes are pierced by a given straight line.. 225 176 Equations of a straight line passing through a given point. 225 177 two points. 226 178 Equations of a straight line through a given point and parallel to a given line....... 226 179 Conditions of intersection of two straight lines.. 227 180 Expression for the distance of two points in space. 227 181 Sum of the squares of the cosines of the angles which any straight line makes with three others mutually at right angles equal to unity...... 228 182 Expression for the angle of intersection of two straight lines in space......2 s. 28 183 On the generation of a plane surface... 229 184 Equation of the plane... 230 185 From the equation of the plane to determine the equations of its traces ~.... 230 186 Equation of a plane passing through three points.. 231 187 Conditions of parallelism of a straight line and plane. 232 188 two planes.. 232 189 Through a given point to draw a. plane parallel to a given plane 233 190 Conditions of perpendicularity of a straight line and plane. 233 16 CONTENTS. drt icle Page 191 To draw a perpendicular from a given point to a plane, and to determine its length...... 234 192 To determine the inclination of a straight line to a plane. 235 193 To determine the inclination of two planes... 235 194 The locus of every equation of the first degree containing three variables must be a plane.... 236 SECTION II. On Surfaces of the Second Order. 195 Surface of the second order defined... 237 196 To determine the equation of the sphere.... 237 197 To determine the intersection of a sphere with a plane. 237 198 Equation of a tangent plane..... e 238 199 Cylindrical surfaces described...... 238 200 Equation of a cylindrical surface determined... 238 201 Conical surfaces described.... 238 202 Equation determined..... 240 203 Equation of the right cone.... 240 204 Equation of the oblique cone 2. 40 205 Surface of revolution defined... o 241 206 General equation determined.. 241 207 Equation when the generating line is straight a.. 241 208 an ellipse.. 242 209 Equation when the generating line is an hyperbola.. 242 210 a parabola.. 242 211 Surfaces of the second order in general... 242 212 Division of these surfaces into two classes.. 243 213 Diametral planes of central surfaces ~... 243 214 Diametral planes of surfaces without a centre.. 244 215 Discussion of central surfaces; the ellipsoid... 244 216 The hyperboloid of a single sheet... 245 217 The hyperboloid of two sheets... @ 248 218 Discussion of surfaces which have not a centre; the elliptic. paraboloid.... e 248 219 The hyperbolic paraboloid.... 249 220 Tangent planes to surfaces of the second order. 251 221 On conjugate diametral planes... 253 SECTION III. On Orthogonal Projection, Transformation of Coordinates, Discussion of the General Equation, 4c 222 Projection defined.... 253 223 The projection of a plane surface equal to the area of that surface multiplied by the cosine of its inclination to the plane of projection ~....... 253 224 The square of a surface equal to the sum of the squares of its projections on three rectangular planes... 255 CONTENTS. 17 Article Page 224 Sum of the squares the same for every system of rectangular planes...... 255 225 If a surface be projected on three rectangular planes, and then these projections be projected on a given plane, the sum of these last projections equal to the projection of the surface on this given plane...... 255 226 If any number of areas be projected on different systems of rectangular planes, and those on each plane be collected into one sum, the squares of the three sums furnished by each system have a constant amount..... 257 227 Direction of the principal plane determined. 257 228 Sum of the projections on any plane equally inclined to the principal plane constant. 256 229 Transformation of coordinates... 258 230 Formulas to be employed when the primitive axes are rectangular and new ones oblique.. 258 231 Formulas to be used when both systems are rectangular. 260 232 To pass from rectangular coordinates to polar. 260 233 Investigation of the formulas employed by Lagrange and Laplace for this purpose.. 261 234 Exhibition of these formulas..262 235 Case in which the formulas become simplified. 263 236 Of intersecting planes... 263 237 Sections of the cone shown to be lines of the second order. 263 238 Sections of the other central surfaces determined 264 239 Sections of surfaces that have not a centre 265 240 The centres of parallel sections all on a diameter 265 241 Discussion of the general equation.. 266 242 Final reduction of the equation. 269 243 Cases where the equation represents a cylindrical surface 270 244 Identity of curves of the second order with the conic sections 270 245 Criteria for ascertaining when the equation represents a central surface... 270 246 Criteria for ascertaining whether central surfaces are limited or not...... 272 247 A central surface of the second order has an infinite number of systems of conjugates..... 273 248 In a central surface the longest principal diameter is a maximuni, and. the shortest a minimum... 273 249 Area of a triangle in terms of the coordinates of its vertices 274 251 Equation of a plane in terms of the perpendicular upon it from the origin..... 274 252, 253, 254, 255. Monge's property of the pyramid, and remarkable inferences from it. ~.. 275-6 256 Position of a plane determined, so that if a given triangle be projected upon it the projection may be similar to another given trianole......... 276 257, 258. Remarkable algebraical properties.... 277 259 In a central surface of the second order the sum of the squares of any system of conjugates is equal to the sum of the squares of the principal diameters....278 3 B2 18 CONTENTS. Article Page 260 The sum of the squares of the faces of the parallelopiped whose edges are any system of semi-conjugates is equal to the sum of the squares of the faces of the rectangular parallelopiped whose edges are the semi-principal diameters; also the volume of the former is equal to that of the latter... 2.79 261 The squares of the reciprocals of. any system of, rectangular diameters are together equal to the squares of the reciprocals of the principal diameters...... 279 062 Three rectangular planes touch a central surface of the second order; required the locus of their point of, intersection. 280 263 Chords are drawn to a surface of the second order so as all to pass through a fixed point; to find the locus of their middle points 281 264 Planes passing through a fixed point cut a surface of the second order; to find the locus of the centres of all the sections. 282 265 Three straight lines mutually at right angles meet in a point and constantly touch a surface of the second order.; to find the -locus of the point...... 82 267 Miscellaneous propositions...... 283 267 On curves of double curvature... 284 PROBLEM XXIV. Within the sides of a given angle is inscribed a straight line of a given length; what is the locus of the- point which divides this line in a given ratio?.Ins. An ellipse. PROBLEM XXV. To determine the curve of which each ordinate is a mean between the corresponding ordinates of two given straight lines. ANALYTICAL GEOMETRYo SECTION I. INTRODUCTION. APPLICATION OF ALGEBRA TO GEOMETRY. CHAPTER I. Jrticle (1.) ALGEBRA has been properly defined as that branch of mathematics in which calculations are performed by symbols. The signification given to these symbols is quite arbitrary, so that, in the practical application of this science to our inquiries about real quantity, it matters not whether Qhe subject relate to time or space, number or motion. Whatever in nature can be submitted to calculation may always, and generally most commodiously, be treated algebraically; and hence Sir Isaac Newton and others have, with great propriety, called algebra zziversal arithmetic. In the present introductory chapter we propose to show how alge. bra may be applied to the solution of a geometrical problem; in the next chapter will be explained how Geometry may be applied to the construction of an algebraical expression. PROBLEM I. (2.) Knowing the base and altitude of a plane triangle, to find the side of the square inscribed in it. Let ABC be the triangle, and put the altitude AI =a A = 78, the base BC = b = 42, and the side of the inscribed square DG = x. D i Then, because in similar triangles the base. s are as the altitudes, (see Young's Geometry, p. 94,) we have B BC: DE:: AI: AH, or b: x:: a: a-x. G ab that is ax=ab-bx, or ax + bx=ab, that is x — = 27.3 a+b Hence the side of the inscribed square will be a fourth proportional to the three lines a + b, a and b. 20 ANALYTICAL GEOMETRY. PROBLEM II. (3.) To divide a given straight line in extreme and mean proportion. (Geom. p. 113.) Let AB be the given straight line and call F A F B it a. Suppose F to be the point of division, and put' -' AF-=x, then x is to be determined, so that a:x::x: a-x......(I) hence we must have x2 =a - ax.'. x2 + ax=a2..... (2), a J2~a2.x= 2 + a + (3). Ofthese two values of x, one we perceive is negative, viz. 2 -t a2 and is in absolute magnitude greater than a; the whole line: this value, therefore, although it does fulfil the algebraical condition, (1,) cannot answer the geometrical conditions of the question, for the point of division, F, roust necessarily fall between the extremities, A, B, of the proposed line, that is, AF, or Ix, must be less than AB, or a. The other value of x, viz. -x- a + a2- - is less than a, and therefore properly determines the point; F, required by the question. The reason why we have been furnished with two values of x instead of one, is that, having had to determine x, so that the condition (1) might exist, the algebraical process very properly led us to not only the one value sought by the- question, but to every value that could fulfil that condition: and it afterwards remained for us to select that value as a solution to the question which involved no geometrical absurdity. However, although the negative value of x does not come within the geometrical restrictions of the question, yet it admits of a geometrical representation. In order to explain this, we may remark that a negative quantity in algebra may always be considerecl as resulting from the subtraction of a greater quantity from a less, thus the negative quantity - q may be conceived to result from subtracting the greater quantity (p + q) from the less, p, for it may always be supplied by the expression p —(p' + q,) whatever quantities the symbols p and q represent. Applying these remarks to the case in -which the symbols -denote lines, and taking the present problem as an example, we have AF the positive value of x equal to BA - BF, and for the negative value, BF must exceed BA, that is, F must be on the other side of A, as at F', hence making AF' equal to the absolute value of the negative root of the equation (2); the two roots of that equation will be geometrically represented by AF and AF'. By taking the negative value of x, the condition (1) becomes ANALYTICAL GEOMETRY. 21 AB: AF':: AF: BF' so that the equation (3) is the complete solution to the question thus modified, viz.: Given two points A, B, to find on the line AB, or on its prolongation, a point such that its distance from the point A inlay be a mean proportional between its distance from B and the distance between A and B. The one point F answering these conditions is given by the positive value of x, and the other point F', on the opposite side of A, is given by, the negative value; and in like manner, whenever it is required to determine the distance of a sought point from a given point, measured along a fixed straight line, and the solution furnishes both positive and negative values, if the positive values be taken in one direction from the fixed point, and the negative values in the opposite direction, every- point so determined will solve the problem, and every possible solution will be obtained. PROBLEM III. (4.) From a given point without a circle, to draw a secant such that the intercepted chord may have a given length. Let ACDB be the given circle, and P D the given point. Draw PAB through the centre, and let PCD represent the re- C quirled line. Put PB = a, PA b, PB CD- c, and PC = x, then (Geom. p. A 0 106.) or Euclid III. 36 Cor. PDPC = PB'PA, that is (x + c) x ab..... (1) and, solving this quadratic, we get x - ~ ab + The positive value expresses the length of PC, the negative value gives no geometrical solution, although it fulfils the algebraical condition (1). PROBLEM IV. (5.) To divide a given straight line so that the-rectangle of the two parts may be equivalent to a given square. Let AB be the given line, which call a; put x for one A B of the required parts, and c for the side of the given F' F square, then we have (a x) x = c2 or x = - - -: - ca2 Both these values being positive, the line may be divided in two points, F, F', as the problem requires. These points are obviously' evalquidistat from the extremities impof the linble. If c exceeds the value of x is impossible. 22 ANALYTICAL GEOMETRY. PROBLEM V. (6.) Given the perimeter of a right-angled triangle and the radius of the inscribed circle, to determine the triangle. Let ABC be the triangle, and D, E, F, the points A where the inscribed circle touches its sides, then (Geom. p. 106,) AF =AE BF = BD, CD = EC - radius OD of the circle; hence, putting the perimeter =p, OE -=r, AF = x FB = y, wehave 2 x +2y+2r=p, hence y +sr = - -x. (1.) E-... Now (Georn. p.. 22.) pr is twice the area of the triangle, but (x + r) (y + r) is also equal to twice C the area therefore (x + r). (y + r) -= pi.... (2) D that is, by substituting for y + r its value in equation (1.) (x + 2). -x p or (p r ~D zx-. ( p -r) i/ V -' ( p-r)2 2 prt. y=( p-r)-x=1 (-p-r) i-V 1 (~p-r)2 -1-PrL Adding r to each of these expressions, we get AC-I ('p + r)~/ -pr2 p B 2 2C('p+r)~/ V 2i-)2-Pp The double sign showing merely that if AC be made equal to any one of these values, BC will be equal to the other. PROBLEM VI. (7.) Given the chords of two arcs to find the chord of their sum. Let AB, BC, be the given chords, then it is B required to determine the chord AC. Draw the diameter BD and join AD, CD, then A (Geom. p. 21 1,) AB-CD + AD'BC = AC'BD, that is, putting BC a, AB = b- AC - c, BD = 2r, and recollecting that AD — = / (BD=AB2,) and CD = v (BD - BC2,) we have b v (4r2 - a2) +- t b a (4r2 - b2-)=2cr, or V'(4r - a2) t 2/(4r2 - b2)c..() 2r 2r the expression sought. If the given chords are equal, then the expression for the chord of the sum is _a (4r2 - a2) From equation (1) we may determine a when b and c are given; that is, when the chords of two arcs are known, we may find the expression for the chord of their difference. PROBLEM VII. (8.) Given the three sides of a triangle to determine the radius of the circumscribing circle. ANALYTICAL GEOMETRY. 23 Let us represent the three sides by a, b, c, and call the radius sought r, then we shall have to determine r from equation (1,) last problem. By transposing, we have 2 / (4r2 -2) c- 2 (4r2 b2 2r 2rr and by squaring and transposing a2- _/ (4r2- b2) b2 — c.. r (a2 + c2 - b2) ac v 4r2- b2 Squaring again r2 (a2 c -- b)2 2 4a2C2 r_ a2b22 abc Hence r 4 (2 c2 b2)2I the expression sought. PROBLEM VIII. (9.) Having given the three sides of a triangle to determine the expression for its surface. A A Put B C -a, AC = b, AB c, perp. AD= y, BD = x, then DC =a - x, or —a-+x,according as the perpendicular fallsB B within or without the triangle. by geometry we have y2 + x2 = c2....(1.) y2 + (a. —x)2 b2.. (2.) The second equation is not altered by substituting (x —a for (a -x.) Subtracting equation (2) from equation (1), we have - a2 + 2ax = c2_b2.. x a2 -+ c2 b2 2a Putting this value of x in equation (I) we get 2 a- c2 - b1 Y=c =I.(. Y v4a2 c2 — (a? + c2 b2)2 BC-AD ay Now, calling the surface of the triangle S, we have S -- = therefore, substituting for y the value just found, we have S -= a / 4a2 C2 - (a2 + c2 2)2 (3.) for the expression required.(10.) Since the quantity under the radical is the difference of two squares, we may substitute for it the product of the sum, and differ) ence of their roots. This sum and difference is (2ac + a2 + c2 - bh and (2ac - a2 - c2 + b2) which is the same as (a + c)2 - b2 and b2 - (a - c)2 and since each of these is also the difference of two squares, they may, in like manner, be replaced by the products. (a + c +b) a + c - b) and (b + a — c) (b -- a c) Hence the expression (3) is the same as s =: V (a -+ c) - ) (b -c e a — c)(b — c -a) 24 ANALYTICAL GEOMETRY. or putting, for shortness, 2p for the perimeter, S- /Ip (p-a) (p —b) (p —c).} (11.) Cor. The expression for the radius of a circle circumscribing the triangle has been found (Prob. 7) to be abc v1 4 aS c —(a2 " r C2 b2 )2 } therefore, putting for the denominator of this fraction its value 4S, as abc given by equation (3,) we have r = -- 4S' PROBLEM IX. The three sides of a triangle being given, to find the segments formed by letting fall a perpendicular from a vertical angle upon the base, the perpendicular itself, the area of the triangle, and radii of inscribed and circumscribed circles? Let ABC be the given triangle, call BC, a; A AC, b; AB, c. If AD is a perpendicular from the vertical angle on the base, we have by Prop. 13 Book 2, Euclid, AC2 =AB2 + BC2 — 2BC X BD; therefore c a2 +C2a- b2 BD- a = one of the segments formed by the perpendi. 2a cular. The other segment DC=BC- BD-= a2 + 2_ Th 2a. above value of BD gives AB —BD2 or AD2-=2 (a2 +c2-b~ 2+ 2a 4a2c2-(a2+c2 —b2)2 AD- 4_a2C2-(a2+-C2_b2)2 4a2 A 2a Let S represent the area of the triangle, and we shall have S = ~BC X AD, Consequently S = _ /4a2c2 - (a+ c2 — b)21 = 4 -V (2a2b2-+ 2a&c2 + 2bc2 - a4 —b4 — c4). But this formula may be exhibited in a shape better adapted for logarithmic computation; to this end we may observe that the quantity 4a2c — (a2 -+ c2- b2)2 is the product of the two factors 2ac+(a2+c2 —b2) and 2ac-(a2+c2-b2); the first (a+ c)2_b2-(a+c+b) (a+c-b); the second=b2 —(a-c)2 =(b +a-c) (b-a+-c);' therefore we shall have S=- 4 v(aqb+c) (ci+b-c) (a+c-b) (b+c — a.): Now if we makeab-cp, we find a-b+c —-2p, a+b-c=2p-2c, a-c-b-2p-2b, be+-c —a =2p-2a and by substituting these values in the above formula for S, and reducing, we finally obtain S- V/ (p.p-a.p-b.p —c). From this we see that to find the area of a triangle whose three sides are given, we must find the half-sum of the three sides, subtract from the half-sum successively each of the three sides, which will give ANALYTICAL GEOMETRY. 25 three remainders, multiply continually these three remainders and the half-sum of the three sides, and lastly extract the square root of the product: this root will be the area of the triangle. Let now z represent the radius of the circumscribed circle, and u the radius of the inscribed circle, then by Prop. C, Euclid, Book vi. BA X AC = z X AD, consequently BA X AC BC X BAXAC - abc 2z= -- AD 2 AD X BC S' 1 abe and therefore z, or the radius of the cirumscribed circle = 1 ac4 S ~ abe Or by writing for S its value z = ab 4 Ip(p —a) (p —b) (p-.-C) Lastly it appears from Prop. 14, Euclid, Book iv. that if from the 3 angles of any triangle we draw 3 straight lines to the centre of' the inscribed circle, we shall divide the triangle into three right-angled triangles, the area of which will be respectively represented by au bu cu- a+b+e -,-, -, and therefore S- =pu, and consequently u, z' z' z t or the radius of the inscribed circled = -. Or by substituting for S its value u= ((Pc) (p) ( ) PROBLEM X. Having given two contiguous sides a, b, of a parallelogram, and one of its diagonals, m, to find the other diagonal. Or the adjacent sides, a, h, and the diagonal, d, of a parallelogram, to find the other diagonal,:. Let ABCD be a parallelogram A ID whose sides AB, AD-a, b, respectively. Suppose BD=nm, and AC-x. Draw AE, DF perp. to BC. Because ABCD is a parallelogram,n kg/ AD=BC, but AD-=EF, B C F. BC- EF and BE=CF. Now AC2=AB2+BC2-2CB'BE by Prop. 13. B. 2. Euclid. And BD2=BC2+CD2+2CB'CF by Prop. 12.. by addition AC2+BD2-2AB2+2BC2 i. e. 2+=n-2a +2b2.. x~= Ix2a2+2b2 —m2 = AC. 4 C 26 ANALYTICAL GEOMETRY. PROBLEM XI. Given the altitude (a); the base (b), and (s) the sum of the sides of a plane triangle, to find the sides. Let ABC be a triangle whose A base BC=b, and altitude a. Let DB = x, then AB= Va2+x2; also DC=b —x...AC= V a2+ (b — )2 B Now AB+AC=s..a2+x2+a2+(b )22=sB.D C 2. a —(hb )2 2 — s_ =Sa2a2 l 2 Square both sides, And a2+(b-x)2=s2 2s. +a2+z2' 2+a2+x2 Or b2 — 2bx=s —2s { c2+x2 2 Transpose and b2 2bx — s2= 2s a-+-x21 Square both sides, And (b- s2)2 — 4bx (b2 - s2) + 4b2- 42a% + 4s22 Arrange the quantities and divide by 4(s2 - b2) s2 b2 a2s2 Then x2 - bx = And x= 2 1J- s2b2 =BD. Hence AB which = -/ia2 + x2l is found, and AC which /v a2 + (b _-x)21 can be determined. If a=4, b-8, and s=12 Otherwise, Let S-= 6 - ~ sum of th'e sides C and let = / difference. Then AC = S.I+ x CB S -x. Put AB= b = 8'and CD = 4-p. Then AD = / (S + x)2 / o — p__ AndDB-= V'I(S-x)2 —p2 By B the quest. 4/[(S - x)2 -p2 -_ -f (2 (S_ —)2 - P22 b. Or (S x)2-p23 b - V/(S _ x)2-p2- by squaring (S + x)2 - = b2 — 2V(S 2_)2 _p2} + (S oX)2 p2. 4Sx b2- 2bV f(S - -2 p2 that is 2bV(/ S - x)2 - P2 b = b2- 4Sx; squaring again we have 4b2(S - X)2 -- 4b2p = b4 - 8 b2Sx + 16S2x2; Or 4b2S2 + 4b2x _ 4b2p-2 b4 +16S2X2 b2 (b2 —4 2 +4p ) (1+ 4=(b - 4 1)- -b = 4V (1- - ) - 4 = 1 x 5, and cnsequently AC = — 6 + V/5 andU BC- 6 _-4 - 5. ANALYTICAL GEOMETRY. 27 PROBLEM XII. Suppose the town A to be (a) miles from B, and B (b) miles from C, and C (c) miles from A, to find where a house, 0, must be erected 3qually distant. from A, B, and C. Construction.-Make a triangle ABC whose a lides AB, BC, and AC, shall be equal to 30, 25, ind 20, by (Euc. I. 22.) and about this triangle lescribe a circle by Euc. iv. 5. then OC, OA /o r OB are'an equal distance from each other. A In the triangle ABC, all the sides are iven to find the angle BCD, as AB 30: BC +-AC 45::BC — AC5: BD AD= F 7.5. Whence ~ (7.5 + 30) = 18.75 = BD. Againas 25:18.75 radius: sin. BCD = 48~ 35' 25". Produce CO to the circumference in F and join BF, then in the right angled triangle BCF are given the angle BCF = ACD = 48~ 35' 25//, and AC = 20; Asrad.: sec. 48~ 35' 25/,:: AC 20: CF = 30.2371; then -CF = 15.1185 = CO distance required; or thus: Let AB = a=30 BO =- b 25, CA - = 20; and put BD =x and AD = y; then + y = a, andx - y2 = c2 _ b2 by division, x y -- c~a C2 qc. b~2 2 2 2 by addition 2x- or x = -BD, and by a 2a 4a2 ). 4 by Euc. vi. C. Diam. X CD = AC X CB; whence Diam. 500 80 80.V7.2 _5 -- 7 - - 30. 2371578; whence AO or CO or 2 >7 47 7 BO = 15 1185789 distance sought. Draw AD perpendicular to the C base, and let AE = x. then a: b ~c:: b c:BD DA..BD -DA_ — BD + DA=a E a b' c2 _ a2-bi+c2 A 2 2a 2a NowCD='~ cIl(U b2 c) } { 4a2c2-('c2 —b2-2)2 } a c2a 44 be be W4& _(O- _ b2 + C2)2 But CD by Prob. 13. -_ - c 2x 2 _ 2a abc =x ~ ~-(I-2 ~+ ~~)2~- the distance of the house E ~' x 4ec.2 - (e 2 t- b 22 — 28 ANALYTICAL GEOMETRY. rom each of the angles A, B, C. If a = 30 miles, b 25, c 20 30 X 25 X 20 40 30X25X20 —-- -- - = 15.12 miles. V (1440000 - 455626)-, V7 The above is nothing more than having the three sides of a triangle given to find the radius of the circumscribing circle. See prob. 7. PROBLEM XIII. If a, b, c be the three sides of the plane triangle; R, r the radii of abc circumscribed and inscribed circles: show that Rr Let AB = a, BC -b,- AC = c, R = NA in figure 2, and r = IF in figure 1. Then, since NF, NE, NG are perpendicular to AB, AC, CB respectively, the area of zN ANC =- re The area of aL ANB = kra The area of Ai BNC _= Irb Area of A ABC= (a+-b- cc).- - G Now, if AD be drawn perpendicular to BC in fig. 2; since -the angle BNF = - an A. gle BNA — angle BCA, and the right angle BFN = the right angle ADC,.. the triangles -BNF, ADC are similar; hence F, AC:AD:: BN:BFOrc:AD:: R: a ca -.'. AD= —- Now the area of triangle 2 2R x ABC= A-D XBC \ a+ b+ c. r abe And Rr ab 2 4R 2(a + b +t c) PROBLEM XIV. The diagonal of a rectangle ABCD, and the perimeter, or sum of all its four sides, being given, to find the sides. Let the diagonal AC = d, half the perimeter AB + BC = a, and the base BC = x; then will the altitude AB = a - x. And by Euc. I. 47. AB2 - BC2- AC2, we shall have a2 - 2ax -+ x + xa2 =d, or x- - ax = -l- - a. Which last equation, being resolved, gives x = ah V/(2d" a 2) Where a must be taken greater than d and less than d V/2 *The reader must recollect that the triangle presented by ABC in each figure, is -supposed to i'' of the same magnitude. AN.LYT'CAL G.EOMETRY. 29 PROBLEM XV,. Given the three sides of a triangle, to determine its area. Let ABC be any triangle whose A sides AB, AC, BC =, b, c respectively. Draw AE perpendicular to BC. Then b2 = a2 - c - 2c. BE. (Prop. 13. b. 2. Euclid.) - B [g C-- { Also AE = a2 2c 2c ~a24-c2-b~2 ~ + C2 ~h2P J b -(a — c)2 (ac)2- b2 \/ { 2c - } { -2 2c = ( -f a —c) ( _,______________ (b+a-c) (b -a+c) (a +c b)b) (a c-.b) } =2j — { (a+b+) -(a-+bc ) (a+a-) 2cb) a-(-ab c)(a+ )( a + b+ c\ AE.c(a + b + c ) 2 AR ] { (a-+ b +-c) (a-l b-}-c ) (a + h +b c _b) EWe find the aea= { ( ( ) } Cor. by help of this problem we can determine the radius of a circle inscribed in a triangle in terms of the sides. If the figure were con* structed it would readily appear that the area of the triangle=(a+b+-c),x see Prob. 13. ~ (a+ + -' —a -+ -b 2 2 41 2(2 ) 2. 2 We r= 4 { s(2s.)( b)(- c) } C2 30 ANALYTICAL GEOMETRY. PROBLEM XVI. To find the side of a regular octagon inscribed in a circle whose radius is known. Let AB be the side of a square inscribed in A the circle AFB, whose centre is E. Draw: EG perpendicular to AB, then AG = GB, and the arc AF = arc FB. Join AF which is the side of a regular octagon. Let r=radius EA, / andy=AF. Then AB2=2r. AG=2'/2r2 Now, since the LAGE is a right angle, and AEG half a right angle,.. GAE is half a right angle, and AG —GE=-' /2r Hence FG = r A V2r, But y2 —AG2 - GF r= - r2 + V r - 2r2 2 =r r 2- 2 y. 2 2 =r. 2'- 2 the value required. PROBLEM XVII. To find the side of a regular decagon inscribed in a circle. Suppose AB to be the side of a regular deca- D gon. Join CA, CB and produce BA to D, making AD=AC. Join DC. Then since the ZACB- -th of 4 right angles or ~ of two right A angles.'. each of the angles CAB, CBA is equal to -} of two right angles. Also the ZL CAD -- C of two right angles; now AD=AC,.'. each ofB the angles ADC, ACD is equal to I of two right angles, and.. the triangles BDC, BAC are equiangular. Let BC=r, AB=y. Hence BD: BC::BC: BA i. e, r y:- r yrr'y ~. y2+ -ry And y=~r V/ (5) —1, as required. PROBLEM XVIII. Having given the side of a regular decagon inscribed in a circle whose radius is known, to find the side of a regular pentagon inscribed in the same circle. Let AB (see figure to Prob. XVI.) be the side of a regular pentagon, and AF the-side of a regular decagon inscribed in the circle AFB. Suppose AF-a, AE=r, AB=y. Now GE= /(r2 —-y2). FG —r — (r2 —_ y2) *~ a2 — 2+ r2 2r(r2- y2) + r'2 — } y2 =2r -2r 2r v (r,-.. 2 2-y2) —2r2- a2, 4r4-r2y2= 4r4 4a2r2- a4'.a y2 4av2 - a4 2 } Now a = 1 r ( 5- 1 ) by the last problem;.. by substitution, y2=- + 4r2 (6- 2V5) - r2 -+- a2., y - v/ (r -a2. ) Hence the square of the side of a regular pentagon inscribed in a circle, is equal to the square of the side of a regular decagon, together with the square of the radius. ANALYTICAL GEOMETRY. 31 PROBLEM XIX. To find the side of an equilateral triangle inscribed in a circle, whose radius is (a), and that of another circumscribed about the same circle. Suppose ABC an equilateral triangle in- A scribed in the circle ABC. Find the centre / G, and from G draw GH perpendicular to AC, then AH = HC, Prop. 3. B..IIi Euclid. Produce GH to meet the circumference in / O K; and join AK. Then because AG, GIH G are equal to CG, GH, and AH = CHI;. the ZAGH = /_CGH. Now the LAGC Bt -' - is one third of four right angles, ~.. AGK one third of two right angles, and each of the angles GAK, GKA is one-third of two right angles; hence the triangle AGK is equilateral; and since A-I is perpendicular to GK,. GH = HKi; i. e. GH =- I GA. Now by Prop. 47. I. Euclid,.GA2 = GiH2 + HAA2 AG +HA2. Hence HA = AG v /.. 2HA or AC = AG. V/3 = a /3 =a side of the inscribed triangle. By describing an equilateral triangle about the circle, if one of its sides touch the circle in the point K, it may be easily shown that AC: a side of the circumscribed triangle:: GH;:GK 1:: 2,.'. a side of the equilateral triangle described about the circle= 2a'4 3 PROBLEM XX. From the given point C, to draw the straight line CF, which, together with'two other right lines AE, AF given in position shall constitute a triangle AEF of a given magnitude. Through C draw CD parallel - to EA, meeting FA produced in / D. Draw CB, EG perpendicu- lar to DF. Now, because C is given in position with respect to D B GA EA, AF,.'. CB may be considered as given; as also AD. Let CB = b, AD = a, AF = x, and A = the area of the triangle EAF. Then by similar triangles DF: AF:: DC: AE:: BC: EG bx bx2 Or a +- x: x: b: = EG --- = 2A, and bx2 = 2Aa a + x a + x 2A 2Aa AO2 2A+A2 A2+ 2Aab +,2Ax, -.. —F.X = b Or. 2A 2= A V A2 + 2Aab. A t V A2+ 2Aab whic h.b, - t - w.hich derie bt tr det ermines the point F; join CF, and the area of~he triangleAEF=A. 32 ANALYTICAL GEOMETRY. PROBLEM XXI. In a plain triangle, having given the perpendicular (p) and the radii (r, R) of its inscribed and circumscribed circles, to determine the triangle. Let. ABC be a triangle whose sides AB, BC, AC are respectively equal to x, y, z. Also let p = AD, the perpendicular fiom the vertex. Then by Problem / Y y XIII. page 28, we have B D C X +- +- PY also Rr + )..xz =2pn Now, X2 =y2+_z2 2y. CD.~. CD, Andp z 2y j 2y 2y = ~ ~ I(z+y X _ + x ). X + y _Z) t 2 2y _ 2 r 2r) 2r 2r... 1 (py- 2r x )( PY- 2 rz )( p - 2r) 4pr = — p2y —- 2pry. ( x + z ) + 4rxz p-p2r But x + z — PY y --.y., and xz = 2pR p-2 Or - 8Rr2 2r p.y p-2r 4r4 8pr2R + 16r3 -. y p —2r. 2 - Hence y ( 2) 2PR - 4rR 2r 2' —. = 2-r 2pR-4rR-r2 2 - the base The other sides are easily found from the equations x + 2 p -- r = ——.y, and xz __pR. r ANALYTICAL GEOMETRY. 33 PROBLEM XXII. Given the area of an equilateral triangle, CEF, whose base, EF, falls on the diameter, and its vertex C in the middle of a semicircular are: required the diameter of the circle. Let ACB be a semicircle whose diameter is AB and centre D. Draw DC perpendicular to AB and let ECF be an equilateral triangle whose area- a. Suppose DC = x. Now a -x X EF.-.EF 2a x Or EC -2a x;butEC -V/ ED'2+DC2! A E 1_) BF B. 2a - x = v I(a2+ x2) + X2 Or 4a2 + x2- (a'2 X2)+ X2 and X4 3a2.-. x = (3a2)-.'. 2x the diameter = 2 (3a2)~ If the area be represented by 100, the diameter of the circle = 20 i / V3a= 20(3)PROBLEM XXIII. Through a given point P, in a given circle ACBD, to draw a chord CD, of a given length. Draw the diameter APB; and put CD - a, A AP = b, PB = c, and CP-x; then will D PD = a - x. But, by the property of the circle (Euc. III. 35.) CI X PD = AP X PB; whence x (a - x) bc, or x2 - ax = bc. which equation, being resolved in the usual way, gives x a / (~I a 2 — bc); Where z has two values, both of which are positive. B PROBLEM XXIV. The base BC, of any plane triangle ABC,:the sum of the sides AB, A C, and the line AD, drawn from the vertex to the middle of the base, being given, to determine the triangle. Put BD or DC a, AD = b, AB + AC =s, and AB- z; then will AC = s - x. A (Geometry B II. Prop. 13,) AB2 + AC2 = 2BD2 + 2AD`4; whenrce x2 + (s - )2 = 2a2 + 2b2, or z2 _ sx = 2 +-. b2 - 1s2. Which last equation, being resolved as in the former instances, gives- s a/( 1 s2), for the values of the two sides AB and B D C AC of the triangle; taking the sign +- for one of them, and - for the other, and observing that a2 + b2 must be greater than I P. PROBLEM XXV. The two sides AB, AC, and the line AD, bisecting the vertical angle of any plane triangle ABC, being given, to find the base BC. 5 34 ANALYTICAL GEOMETRY. Put AB a, AC - b, AD = c, and BC = x; then, bv Euc. vi. 3, (See last Problem,) we shall have AB (a): AC (h) BD: DC. And consequently, by the composition of ratios (Euc. v. 18,) a + b: a:: x' BD = aax -- (a + b)V, and a +-b: b:: x: DC =lbx + (a+ b. )}, But, by Euc. vi. 13, DC X BD+ AD2 =AB X AC; therefore, also, {abx2 I (a /)22 + c2=ab, or or')X2 = (ab + b)2 X (ab —c). From which last equation we have x = (a+b)V 4(ab - c) + ab = BC. PROBLEM XXVI. Determine a triangle; having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle. Let ABC be the triangle, F and BD the line bisecting the vertical angle. Draw the diameter FG at H right angles to the base. If / BD be produced it will meet the circumference in G, because equal angles stand on equal arcs. Put a for BD, b for AE or EC, c for FG, and x for EG. Then (Geom. A D 90) x (c- x) =b2, hence x- + c'/( ~ c2 - 2); for this value of x put e;, and join BF, also let y represent DG. The angle GBF being (Ge- G om. 52) a right angle, the triangles GDE, GFB, are [Geom. 96] similar;.. y:e:: c: a+y, that is, ay -- y2 = ec, hence y= ~_ ec + I-a2- A'a. Put J for this value of y; and z = DC. Then [Geom. 108] af = 2bz - z2, that is, z = b 6 / b2-af. Put this value of xz = g, and from B draw BH at right angles to FG. As GD. GB:: DE: B, that is, As J: (a+f):: (b-g): (b- fg~ _- ) = BH the distance of the perpendicular from the middle of the base,- EP. Join BP GE: EH (=BP)::GD: DB, that is, As e: BP::f: a, or BP = ae — f- the perpendicular of the triangle ABC. But AB= AP2+PB2J1( b~(bBg+ j-a++ ag) 2 +42) [where p represents PB]; And BC = v PC2 - BP2 = (bIt( {= b(b.-g+cc g) } +p2) as reqluired. ~c~___~~_~_~ ~ 9 P ANALYTICAL GEOMETRY. 3b5 PROBLEM XXVII. In a right angled triangle having given the side of its inscribed llqiare (12) and the radius of its inscribed circle (7) to determine tJhe triangle. Put DF = A B =12 = S. OP = OG = O!=7=r. BCD -; X BA = y; Then'Y =S G (see Prob. 1. P.19). AndAG =X -- r=AP. CI = y - r= AP. Hence C E I B CP + PA=AC= - y - 2r, and by (47. e. 1.) X2 + y2=X2 + y2 +4r2~2xy - 4'r; Or r 2yr=xy + -4y; Orr= xy = 2r (x + y) - 2r2 = Sx + Sy; Whence 2rx + 2ry. — 2ra = S x,2ra 2r2S + S y; And x Y= 2r S xy=S(x Y)= 2rS By substituting the values of r and s, we have x- y=49 and xy=588, and then we easily find x = 21. and y = 28. Let ABC be the proposed triangle, BFDE the inscribed square, OG and OP radii of the inscribed circle at the right angles to AB and AC, and BD a diagonal of the inscribed square; also let BQO be perpendicular to AC from the right angle. Put a for the side of the given square, b for the radius of the given circle, and x for the segment AOQ of the base AC by the perpend. BQ, Then FG=a-b sinceGB=OG; anda- b: a:: b:'ab (a-b). I B(Q (because GF: BF:: OD: BD:: OP: BQ). Therefore (since BD = 2a2) DQ = z/2a2 b) Let this value (a - b L s ab of DOQ to be recognized in c, and put d for _ — It is as x: d:: d: - =C [Geom.p. ]. And x+c: -c::x: d.? HIIence dx - dc = d2 -c. That is x = AQ, thered -Fc.c fore the triangle is determined. * For the trihgles AFD, DEC are similar; also the triangle AQB is similar to AFD, and consequently to DEC. Therefore AD: DC:: AF. DE (- DF) A.: CQMB. 36 ANALYTICAL GEOMETRY. PROBLEM XXVIII. To determine the radii of three equal circles, described in a given circle, to touch: each other and also the circumference of the given circle. Let ABC, be the given circle, whereof O is the centre. Inscribe in it the equilateral triangle PQR (Eucl. iv. 15); join OQ; OR, OP; and produce OP till PS equal the half of PR, or of PQO. Draw SR, aid parallel to SR through the point P draw PM meeting OR, in iI; and through M draw ML parallel to RP; through L, LN parallel to PQ;'and join MN. CUTS L, M, N are the centres of' three circles that shall touch one another, and the circumference of the given circle ABC. For bisect PQ. in H, and join SH; and parallel to SH draw PF. Because POQ is an isoceles triangle, and that, LN is parallel to the base, PL is equal to QN: And because SH, PF are parallel, (PH, LF being also parallel,) and that, SP, PH are equal, PL, LF are equal. In the same manner it may be proved that QN is equal to NF, to NE, to EM, to MR, &c. Moreover it is evident that, NF, FL are in the same straight line. Putting, therefore,-a = radius of the given circle, and x - radius of one of the inscribed circles; It is, (because NF = M IN, and, by a -- x'~ similar triangles OE - OM), (a - x)2 - (- = x2. Whence x -3 a:l P12a2 = 2a v3- 3a. Which IVas re, quired. ANALYTICAL GEOMETRY. 37 PROBLEM XXIX. Given the base and difference of the sides to determine the triangle, when the rectangle of the longest side and difference of the segments of the base is equal to the square of the shortest side. Let b = the base, x = the shorter side, d = the difference of the sides;then (byGeom.p. 36.) as b:2 2 + d:: d: (a + 2x) = the dilfference of the segments of the base, and by the question, this X by the longer side, or (d + 2x). (d — x) = x2; and reduced we have 3Cd2x dS d3 36d4 6ci2 b - 2d b-2d' ~ - -2v ( - 2d)2 - 2d PROBLEM XXX. When a parish was inclosed, the allotment of one of the proprietors consisted of two pieces of ground; one of which was in the form of a right-angled triangle; the other was a rectangle, one of the sides of which was equal to the hypothenuse of the triangle, the other to half the greater side: but, wishing to have his lagd in one piece, he exchanged his allotments for a square piece of ground of equal area, one side of which equalled the greater of the sides of the triangle which contained the right angle. By this exchange, he found that he-had saved 10 poles of railing. What are the respective areas of the triangle and rectangle; and what is the length of each of their sides; Let 2x = the greater side of the triangle, and y = the less; that is V 4x2 + y2,- =the hypothenuse; and also the greater side of the rectangle, and x = the less side of the rectangle;.'. xy - the area of the triangle, and x V/(4x2 - y2) = the area of the rectangle;.. 4x2= xy + xV (4x2 + y2), or 4x - y = (4x +- y2); also 8x+- 10 2x + y +V-(4x2 + y2) + 2x + 2V(4x2 + y2), or 4x + 10 -y + 3 / (4x2 + y2); in which equation substituting the value of,/ (4x2 + y2) found above;.~. 4x~- 10 = y + 3(4x - y)= 12x- 2y, that is by transposition, 2y = 8x- 10, and y = 4x - 5;.- from the first equation,.5 = v 4x2 + (4x 5)2, and 25 = 4x2 + 16n2 - 40x + 25; by transposition 40x -2x2; that is 2 = XC and y _ 4x- 5 -- 3; the sides of the triangle are 3, 4, and 5; the sides of the rectangle are 2 and 5;.and the area of the triangle'and rectangle are 6, and 10 respectively. PROBLEM XXXI. In an oblique-angled plane triangle; there is given the difference of the sides which includes- the angle of 71~ 10,' equal to 11, and the irne that bisects the said angle is equal to 24; from whence is,reD 38 ANALYTICAL GEOMETRY. quired a theorem that will determine the base and sides of the said triangle. The figure can be supplied by the reader. Let ABC be the triangle, and make CH _ CA the shorter side, draw AH. and it will cut the bisecting line CD at right angles in P; make PE = PD and EH = DH = AD, then will the triangles CEH and CDB be similar. Put CD = 24 = b, HB -11 = d, the triangle BCD = ACD = 35~ 35', whose cosine call q, and let x = CH = CA. Then as rad. ( = 1): x:: q: qx = PC; Hence 2qx - b = EC; then by similar triangles 2qx-b. b x:: x: + d=CB; therefore bx = 2q -bx + 2dq -- bd, this equation being solved will give x -Q- ~ 4= 2q - - =25.00218; and hence we get AB = 36.60737, and BC = 36.00218. PROBLEM XXXIi. Given the base, the perpendicular, and the ratio, of the two sides of a triangle; to find the sides. Call the base AB = a; the perpendicular Ci CD,=b; and AD=x; and let the givenratio' of AC toCB be that ofm ton. Then BD=a-x; AC2 = CD2 + AD2= b2 + x2, and by (Euclid I. 47). BC2 = CD2 -- BD2=b2 + a2 —2ax + X2 and therefore m2: n2 b:: b2 + a2 - 2ax +- x2. Multiplying extremes and A GD.B means, we obtain m2b2 + m2a2 - 2m2ax - m2x2 - n2b2 + vn2. 2am2 a2m2 That is, x2 --. -2 = - b2, which equation resolved, am'2 am:m' a%2 2 2 2m a2in S 2 ns aml a22m4 a2 m2 Sx=,2 2+ _ 2~-2 —- _ -- 22 r m2 (m- n2) 2m -2 ) Let a represent the base, x one of the segments of the base by the perpendicular, r the side of the triangle adjacent to, s the other side of the triangle, b the perpendicular; and m: n the ratio of r: s. Then a - x is the other segment of the base. Also x2 +- b2 r2, and x2 + b2 +a 2 =ax S. But m2: n2::'(x2 + b2):(x2 + b2 + a2 _ 2ax) Therefore (n2 - n2)2- 2am2x = (n2 - n2) b2- an2m as before. PROBLEM XXXIII. To find the area of a plane triangle, when two of its sides and the included angle are given. ANALYTICAL GEOMETRY. 39 Let ABC be the triangle of which the A. area is required; BC, AC, the given sides and C the given angle; from A draw AD perpendicular to BC, or BC produced; then by trigonometry p. 681. AC sin. C. as R: sin. C:: AC: AD, and therefore AD + R AC.BC. sin. C and I AD X BC -R = the area of the triangle. PROBLEM XXXIV. In a right-angled plane triangular field, the legs are 3xx and x3z; and the line that bisects the right angle = -x chains: what is the content in acres? y2 Put y z-f; then 3y = 3x =BC,y2 = x2- = CD, andy3 = x3X A=. Let fall the perpendicular DE and DF; then it is evident (per question) that the / DCA = the LZ F DCB, and the sides FC-CE=ED = DE = 7(~- - ) Y=. Put the /2=m; then by siy2 y2 milar triangles, as AF: DF::AC: BC; that is as y3:: y3. - n m 3y3 y 3 y2 3y; therefore 3y4 — _, or 3y -- -; whence by comm, on n s i s pleting the square, &c. we find y = 3.346065, 3y = 10.038195 = BC, y2 = 11.196152 = DC and y3 = 37.463053 =- and the content = - BC X AC = (.4y4 = 2y4 = 188.030715 chains. PROBLEM XXXV. The area of a right-angled triangle, whose sides are in arithmetical progression, being given equal to 216; to determine the triangle. Call the least side x - y, and let the common difference be y, then the three sides will be x - y, x, and x + y. Now by the nature of a right-angled triangle we have (x - y)2 + (~ = -(X + y)2, and by question the area — x (x- y) = a2 = 216. From the former equation we get 2 2 - 2y + y2 = x2 + 2xy - y2 and by reduction 4y = x, substituting this value for x in the second equation we obtain 2y X> 3y = a2, from which y = v a2, and x= 4v a2, consequently the 3 sides are 3v/(6 a2) -18; 4V/(-as) = 24; and 5 V (_a2) 30. -, 40 ANALYTICAL GEOMETRY. PROBLEM XXXVI. To produce a given straight line (a) so that tile rectangle under t1he ajiv-n line and the whole line produced, may be equal to the square of the part produced. fLet x = the part produced, then per question (x +- a) a X 2 or ax = -2 or by completing the square x2 - a + - a2 4a2 and by extracting the root x -a = 1 /(5), orx a- a ( 5 -+ 1) = the part required. PROBLEM XXXVII. Find the side of an equilateral and equiangular dodecagon in a circle whose radius = r. Let EBF denote the given circle, and apply the radius from A to C then AC = the side of a regular hexagon inscribed in, the circle (Legendre's Geometry Art. 271.) bisect the arc AC in B, E - F and join BC, and BC is the side of the regular dodecagon sought; Join BO and it bisects AC A D perpendicularly, (LegendreArt. 106.) Join OC, then since DC = 2OC and OC2 = OD2 + DC2 B (Lecrendre 186.). OC2= OC2 OD2 or OD = -(OCV 3) and BD O -OD OC (2 -V 3) (since OC = r,) DC -r, DB -='r (2 - V/3) and BC2 DC2 + BD2 -- 2 (2- /3.)'. BC = r4 V (2- V3) = the line sought. PROBLEM XXXVIII. There is a triangular piece of ground whose area = 525 square yards, and two of the sides measure 30 and 42 yards respectively find the remaining side. Let p = the perpendicular to the side 42 from its opposite angle, then l!p X 42 -- the area = 525 (Legendre art. 176.).. p = 25 yards, and 302 252 = 52 X 11 or 5 /11 _ the segment of the side 42 adljacent to the angle formed by the given sides, and 42- (5 / 11 ) = the remaining segment of the side 42, this segment and p are the legs of a right-`-angled triangle, of which the side sought is the hypothenuse, hence if x = the side sought, there results the equation, 2 _(42-5 / 11)2-+ pa2= (42 -5 / 11)2+252=22(666105 V/11) and x = 2 V/(666 - 105 V/11) = 35.65 yards nearly. NoTE. —.It is evident that there is another triangle which answers this question, whose remaining side being denoted by y will be found by the equation y = 2 V/ 666 + 105 Vi, in this case the perpendicular (p) falls without the triangle. See Prob. vi:i.:p. 23. ANALYTICAL GEOMETRY. 41 PROBLEM XXXIX. The four sides of a field, whose diagonals are equal to each other, V.re 25, 35, 31, and 19 poles respectively: what is the area. Since the sums of the squares of the opposite. sides of the trapezium ABCO are equal (by the question), the diagonals will cut each other at right angles in D; so that putting BC = 25=a, BA=35 =b OA = 31 = c, and CO 19 -a, and AC -- BO= x, we shall have:x b +et:: b " a' AD CD -a A) — x; and: b c:: b-cBD OD - (b2 2 2xa c2)x; whenceAD - 2 b- CandBD =x — &' and;. q( x -'i)2- ( 2 - _ =; 2 + 2x 2 22 2x from which equation 2x4 - +a2 +- c2= X 2x2 62 - b a22 -4{b2 - = 0 this solved gives x* - (a2 + c2):i b2d2 - X (c2 - a2)2 = e37.9.'and the area = ~z2 = (37.9)2 as required. PROBLEM XL. In a right-angled triangle, there are given the ratio of the sides as 3 to 4, and the difference between the area of its inscribed circle and inscribed square = 20.2825. Required the area and sides of the triangle. Let 20.2825 = a, 3x = the perpendicular, 4x = the base, and 5x = the hypothenuse; also 2x-= circle's diameter; Put y - the side of the inscribed square; therefore 4x - y = the base minus the side of the inscribed square, and by similar triangle we have as 4x: 3x:: 4.- y: y;.-. 4xy = 12xz2 3xy, or 7xy = 12x2. therefore y = Ix; and by the. (2x)2 2 X; andl by the question ( = a; and x = 10, and the sides T49 are 30, 40 and 50, respectively. PROBLEM XLI. There is a triangular field, whose content is known to he = 15 acres, 2 roods, and 16 perches; the perimeter 78 chains; and one of the anigles 126~ 52' 12". It is required to find the sides of the field separately, by a general theorem, that may be of use to the practical surveyor. Put a - area of the triangle = 156 square chains, r = the sum of'all the sides = 78 chains, and s and q for the sine and cosine of ha If the sum of the unknown angles respectively; also let _ t; then q is the cosine of half the difference of the said unknown angles 6 D2 42 ANALYT1CAL GEOMETRY. =.9911223 = 7~ 38' 25"//, which being added to and z2 — 4aI subtracted from the half sum, gives 34 12' 19//, and 180 55' 29" for z5 -4at the two unknown angles. Now the longest side AC = 2 — 2z = 37 chains; consequently AB -- 26, andBC — 15. Or if it be put z 2a1 equal to the cotangent of half the given /_ ABC, then will --- 2 z -HC -- 37 chains as before. This question may likewise be solved by finding the radius of the inscribed circle, which is equal to the area divided by the perimeter. PROBLEM XLII. There is a triangular garden, the length of whose-sides are 200, 198, and 178 yards. Now there is a dial so placed in the garden, that, -if walks be made from each of the three angles to the dial, they will exactly divide the said garden into three equal parts: from whence is required. the length of each walk. Let ABD be the proposed angular garden, ini which having the sides given I find (by plane trig. p. ), the angle BAD= 530 7' 48" (whose sine call s and cosine c), the perpendicular Cd = 52.8, and Ch = 533. Now b - cd c 52.8, Ch = 531-, x and y = sine and cosine of /_ CAd; then sy - x- sine L CAh, and as b b bsy — bcx:b::1: -— AC; also as: -::sy —cax: Ch-d; I bsx therefore bcx + dx = bsy; that is -- - C y ba-t-d.496875, the natural tangent of the angle CAD, -.26 25' 18//; fere; from hence is found CD= E\P F 107.5838; AC = 118.6587, and BC = 106.3425, the length of each walk as required, A ] PROBLEM XLIII. In any right-angled triangle, the area (= 294), and the difference between the hypothenuse and perpendicular (- 14), being given to find all the other parts of a triangle by a simple equation. Given AB- AC = 14-a, and BC X A AC = 588 = b, put x and y for the sine and cosine of the L at B; then (By trigon. p. axI have 1- - xa: a: -x= BC and 1 -::: -ay -- AC ay -- b; con-..C`1 — X (1 — r.2 -1 ANALYTICAL GEOMETRY. 43 sequently ( ) - the tangent of half the angle BAC; ~. the side AC = 21, BC = 28, and BA = 35. PROBLEM XLIV. In a plane triangle ABC, there is given the sides AC and BC equal to 24 and 30 poles respectively; and supposing a circle inscribed in the same, so as to touch all its sides, a line drawn from the angle C, to the centre thereof is found to measure 12 poles. Find, the base of the triangle by a simple equation. Let AC = q 24, BC= d = 30, CO = m 12, AC + BC =- n= 54 and AB = y; then C n:y:: q:AD= and:: d: DB dy m:: q + y: CD = m -; Hence we have dq = (m + my )2,whichreducedgives Y _ (dq - )n 36, the base as required. dq " - m2 PROBLEM XLV. Having the perpendiculars let fall from each angle of a triangle to the opposite sides given, to find the opposite sides and area. Here is given the perpendicular BE=100-=m, C AF —98-,=n CD=95-a; let x=AB, then ax =twice the triangle's area, therefore ax a ax AC —;and BC-; per (trig. p. ) xP xx=DB-AD; hence DB= ( 22 —A X x, butn a an: 1:: DB: cosine /. B = 2- + — 2 2- =.5349834, the cosine of 57Q 39' 27"// - B; hence AB= 115.9951, BC-= 112.4443, AC =-110.1953, and the area = 5509.765. Otherwise algebraically. Put BE =b =21, AF c = 20, CD = d= 19, AB=, AD z, then by similar triangle (Euclid I. 47.) equating the different processes, completing the square, &c. we get =( 44 - =2Cd _ 24.34. 2b4C2 24 2b2c2d4 Y, + 2 b2cld- 2 b4d,4d /t 44 ANALYTICAL GEOMETRY. PROBLEM XLVI. From the vertex A of a triangle, draw a straight line meeting the base produced in D, so that the rectangle BD.DC = AD2. Let ABC be the given triangle, whose base is BC, A and let AB be greater than AC, then make the angle CAD -ABC and produce BC.to intersect AD in D, then is AD the line which was to be drawn. For since the angle D is common to the triangle ADB, B ADC, and because the angle B-CAD the remaining C angles of those triangles are equal, (Legendre 74.) Hence the proportion CD: AD: AD: BD (Legendre 202.) Or BD.CD=AD2 at required.. CD is easily calculated, for put a -- AB, c = AC, C B3, then because the triangle ABD, ADC are similar there results the proportion AB: AC2:: BD2: AD2= BD.CD, or AB2: AC2:: DB: CD and by division of proportion AB2 - AC2: AC2:: BC: CD bc2 ol in symbols we have a2 -c2 c2:: b:CD,.. CD= a2_ A2ns. bC2 ba2 Note, that BD = BC +'-,CD = b + a2 cc22 = a c — the whole side produced. THEOREM XLVII. In any quadrilateral figure whose diagonals intersect at right an-'les, if S.D be respectively the sum and difference of two opposite sides, and S, d the sum and difference of the other sides, then S + d:s- -D:: s-D: S-d. Required the proof. Let ABCD denote any quadrilateral whose diagonals AC, BD intersect at right angles in F and suppose that AD is greater than AB then A De = AE2+DE' and AB2= AE2+BE2 and by subtraction AD2 - AB2= DE2 - BE2 in the same manner DC2 - CB2 = DE2- BE2,.. AD2- AB2 = DC2-BC2. Put AD + BC=S, AD -BC-D, then AD= (S-D) -2 and BC=(S -D)' 2; also put DC+AB=s, DC - AB —d, * then as before DC =(s + d) - 2, AB = (s - d) + 2, by substituting these values in the equation AD2- AB2 - =DC2 CB2 it becomes (S t+- D)2 - ( -d)2=(s + d)2- (S- D)2, or by reduction S2-+D2 - 2 d+2 or S2-d=s2- D2, or (S + d).(S - d)=(s + D).(s-D) which gives S +t d: s D: s- D: S - d as required. PROBLEM XLVIII. Having given the perimeter (p) of a rhombus, and the sum (s) of its two diagonals, to find the diagonals Since a rhombus is a parallelogram whose sides are all equal, we havep =pto one of the sides, put x -one of the diagonals then s - x ANALYTICAL GEOMETRY. 45 =the other; hence (Legendre 195.) (s x)2 + X2=4 (p)2 or by reduction 2x2 — 2sx - pa - S and x - s = s ( p2_ 4s2) -. by quadS i 1/(p2 - s) 1 ratics X — and x =2v2 (s 2~ (p2-s2) 22 212 2 hence s —,x 2 /2 q V(p2- 22) )as required. PROBLEM XLIX. Given the area (a2) of a right angled triangle, whose sides are in geometrical progression to find the sides. Let x, y be the-two legs y 7 x, then 1 y2 + x2 = the hypotheo nuse and (per question.) 1' 2+: X:: y x or x y2+xU l y and 2y2 + x4 = y4 also xy =, 2a, 16c8 16cr2 that is x2ya = 4a4, and y4 -= - hence 4ac + x4 = or 0 - 4a4x4 = i16as and by quadratics x4 + 2a4 = 20ac8 = 2a(5)4 oi 2a( 2a. t=al /(2 A15' 2)i then y — = ( 2 / 2 )-4,-wy\-r~ VY4 X (2,V5 —2)4 as required. 2a2 Otherwise, call the perpendicular x, then the base will be -, and the hypothenuse iV (2 ~ —-2 - ) And by the question x:: x:',/ 4a4a )x —- 22 ( 4a4' 2 (x2~ ). Hence x = 2 x2 + 2) and squaring we have the final equation, x4 -- x2 + 4a or -x8- 4a4 4 16a; completing the square and extracting the square root we have,:4 = a4 1 2 +- 20 I, and x a ( 2 - /20)~ as before. PROBLEM L. Having given the base (b) of a plane triangle, its area (as) and the ratio of the two sides as m: n. Required the values of these sides. Draw a perpendicular from the vertical angle to the base then 2cr2 - = the perpendicular, let x = the lesser segment of the base, then 2 +- } = the lesser of the remaining sides, and{ b2 b the greater hence the proportion n 4 ~- ( b - x)2l -=the greater; hence the proportion in2:' n'' -2- " 46 ANALYTICAL GEOMETRY. 4ac4 2 bin2 4a4(n2 - in2) 6- b-2 - t- (b —x)2 which gives x2- 2 2- ( -2 - -- this quadratic solved gives x, and thence the sides become known. PROBLEM LI. Having given the base (b), the area (a2), and (c) the difference of the two sides, to find the sides and the perpendicular altitude of the triangle. By using the same notationl &c. as in the last question the perpendicular- - the lesser side =+ 2 and the greaterrJ{ b2 ( ) }.(per question.) +x + -- (b _X) } or 2c { 4a 2 -b2-2bx, or =,I i 4a4 t — %-r q-x} =B — 2b, oa 2cJ 4a+x2 V -bc2- 2bx or 6c + 4C22 b4 + c4 + 4b62x2-4b3x- + 4c2 bx- 2b262 or 4 (b2- c2) X2- 4b (b2 — c2) x = 6C42 16a4e% _ (b2- C2)2 62 _ (b2-_2)2 1 and 4x2-4bx = 16ace —b ) (b2;2)2 C 16a4 2(b2 c2) - b2 (b2 _c) C hen e -x' =- b —b -3C2 4'a- 4a4 hen J /+ — lb +x2= the lesser side and 4 (- 2)2 } =I/-(4~ ~+b2) +c=the greater side as required. PROBLEM LII. Having the lengths of three perpendiculars,, EF, EG, EH, drawn from a certain point E, within an equilateral triangle ABC, to its three sides, to determine the sides. Draw the perpendicular AD, and having A joined EA, EB, and EC, put EF a, EG= b, EH-c, and BD (which is ~BC)_ x. Then, since AB, BC, or CA, are each = 2x, we shall have, by Euc. I, 47, AD= V (AB2 —BDD2) = (4x2 -- 2)= 3x2 =x / 3. And because the area of any plane triangle is equal to half the rectangle of its base and ~ D? 0 perpendicular, it follows that L ABC =CX AD- =x X xV3 x23 A BEC = BC X EF = x X a =a, ANALYTICAL GEOMETRY. 47 A AEC = -AC X EG x X b = bx L AEB = AB X EH x'X c = ex. But the last three triangles BFC, AEC, AEB, are together equal to the whole triangle ABC; whence x2 V 3 = ax + 6bx - x. And consequently, if each side of this equation be divided by x, we shall have x V/3 = a + b + c, or x (a a+ b + c). V 3. Which is, therefore half the length of either of the three equal sides of the triangle. Con. Since, from what is above shown, AD is = x /3, it follows) that the sum of all the perpendiculars, drawn from any point in an equilateral triangle to each of its sides, is equal to the whole perpendicular of the triangle. PROBLEM LIII. Having given the lengths of two lines AE, DC, drawn from the acute angles of a right-angled triangle ABC, to the middle of the opposite sides, it is required to determine the triangle. Put CD = a, AE=b, BD or LAB-=x, C and BE or I BC =y; then, since (Euc. I.47,) BD2 + BC2 = CD2 and BE2 +AB2 AE2, we shall have x2 +-4y2 = a2, and y2 b +x2 - 2. Whence, taking the second of these equations from four times the first, there will arise A no 15y = 4a2- b2 or y = v (4ca2 b2)* — 15. And., in like manner, taking the first of the same equations from four times the second, there will arise 15x2' 42 - a2, or -- (4 b2 -- a2) 156. Which values of x and y are half the lengths of the base and perpendicular of the triangle, observing that b must be less than 2a and greater than - a. PROBLEM LIV. Having givern the ratio of the two sides of a plane triangle ABC, and the segments of the base, made by a perpendicular falling from the vertical angle, to determine the triangle. PutBD=aDA=b, BC=x C AC = y, and the ratio of the sides as in to n. Then, since by the question, BC:AC::: n, and by (Geom. p. 36) CB2 - AC2 =BD2 —DA2, we shall. have x: y::rn:n, and x2-y2a2 —b2. A. D But, by the first of these expressions, nx = my, or y = nx -min whence, if this be substituted for y in the second, there will arise, 2 - (n2 -- -2) = - b2, or (in2 n2) = m2 (a2 — 2). 48 ANALYTICAL GEOMETRY. And consequently, by division and extracting the square root, we a2 b2 a2 - b2 shall have x = n V 2 2 and y = n V 22); which are the mrnn.m -- values of the two sides BC, AC, of the triangle, as was required. PROBLEM LV. Given the hypothenuse of a right-angled triangle ABC, and the'sides of its inscribed square DE, to find the other two sides of the triangle. Put AB = h, DF or DE = s, AC = x, and CB - y; then, by similar triangles, we shall have AC (x): CB (y)::AF (x-s):FD (s). And consequently, by multiplying the means D F and extremes, xy - sy = sx, or xy = s (x-+- - y),... (1). But since, by Euc. i. 47, AC -+ CB2=AB2 we shall likewise have, x2 + y2 = h2. (2). B E C Whence, if twice equation (1) be added to the equation (2), there will arise x2 2xy - y=h2+2s ( x+y), or (x+y)2 -2s(x+y) h2. Which equation, being resolved after the manner of a quadratic, gives x +- y = s~t V (h2 s2), or y = s- x+ V (h2 s2). Hence, if this value be substituted for y in equation (1), there will arise Xls — (1h2 + 2)= 8 IS+:' v (h2 + s2) }, or — 2 -s- (s + sx) x= — s s+ V (h- +s2). And consequently, by resolving this last equation, we shall have x - 2 s~ V (h2 + s2) V 1h2 - s 1 2 2 ~ M I - S S I 2 12 Is V(h2 +s-2)}; and y =:s/V(h2 s2) ~ q: V h — s: -s g (h2 + s2) l; which are the values of the perpendicular AC and base BC, as was required. PROBLEM LVI. Find the radius of that circle in which the side of a regular pentagon is 1. Let r = the radius, then V (r2 — ) = the perpendicular from the centre of the circumscribed circle to the side of the pentagon. Hence V 4 + (r-/ (r2 _ ~))2} the side of the inscribed decagon, but (Legendre 273.) r: V I 4 + (r V ( r2 ~ ) )2 1 + (rV( ~))2' —~- r ] (r ( r ))2 iputV} 4 —+ (r(r2 _) )2 =x, then r;:x: r —-x or x2 r2 - rx and:2 +rx:~. x=~:r(/5__l) —VI~. —~(r —-g Ir2 3) )~12 r'2-(3 —-5 ) r2 X Z x- ( /L -5 +1 )= + (r-) )2r 2(3 V o ) -'+ (r V (r2 4 ) 4 )2 + r2 - 2r r + r-+ and by reduction r ( -+ V 5) = 4 V(r - -) o r (3+- V 5)=8r2- 2 2 5+:V5 -- o 5) as required. ani... 5~A/5 1 or as required. ANALYTICAL GEOMETRY. 49 PROBLEM LVII. Given the lengths of two chords, cutting each other at right angles in a circle, and the distance of the point of their intersection from the centre, to determine the diameter of the circle. Let AB, CD be two chords within the cir- C cle ABD, cutting each other at right angles A B in F. Take the centre E, and draw EG, EH respectively perpendicular to CD, AB. Then CD, AB are bisected in the points G, H. Join EB, ED, EF. Let AB= a, CD = b, EF =, and x = EB or ED. Then 2 EBD2 = EH2 ~ HB2 or HB = EH'+ -- - Also ED2 = EG2 + GIYD, or x2 = HF2 - + * 2x = EH2 + HF2 + (a2b) =m2~+ (as+ b2)., X 2+a2 2.. 2x, or the diameter = V l 2m2 +- (a2-+ b2)}. PROBLEM LVIII. Given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle; to determine the sides of the triangle. Put a and b for the two segments of the base, x for the side adjacent to a, and y for the other side of the triangle; and m: n, the ratio x has to y. Then, (since in: n: x: y) it follows that y =- X. But Mx x2 _ 2 b2 = ya a -- 2 a2 y2 _ b2 = 2 2. That is, nn22 2- m 22 —m2 ba2 m2. m2a2s m2b2 n m262a mb2 Hence x = z - I'2, Andy= - / 5 2 2 — which were to be determined. PROBLEM LIX. Given the difference of the segments of the base made. by a line bisecting the vertical angle, the difference of the sides, and the difference of their squares, to construct the triangle. Let a, b, and c, be the three differences as per quest., then c — b = the sum of the sides, and l a2 + c} -2' 2b, and!c —b2; + 25 are the two sides; b: a:: the greater side: the greater segment = b2 — C c —b2 c 2b2 X a, and X a = the less, hence the base — 2and thus we have theorems for all the sides. 7 E 50 ANA'LYTICAL GEOMETRY. PROBLEM LX. Given the area, or measure of the space, of a rectangle, inscribed in a given triangle; to determine the sides of the rectangle If a be put for the perpendicular, and b for the base of the triangle; c for the area of the rectangle; and x for the length of the rectangle parallel to the base of the triangle, it will be as b: a: x ax + b the difference of the altitudes of the rectangle and triangle a Therefore the altitude of the rectangle is a —b x. But c = x(a —— b ) = ax-+b x2 - i X. b= b -- c C the length of the rectangle; consequently bi b = the breadth; which were to be determined. PROBLEM LXI. The base of a right-angled triangle being given = 24, and the difference between the perpendicular and hypothenuse = 12; required the other sides. Put 24 - a, 12 = d, and the perpendicular = x; then the hypothenuse --- x + d, and by 47 Euclid, Book i. we have a2 q — x= a2 - d' (x -+ d)2, that is a +. x2 = x2 + 2dx + d2 and - 2d x + d = d Substituting for a and d their values we find the 2d perpendicular = 18, and hypothenuse = 30. PROBLEM LXII. The area of a certain isosceles triangular field is eight acres and a half; find the sides and angles, when the area of the inscribed circle is equal half the field. Put 1360 = 2ac, = the area of the triangle A in perches; 29.424 -= b, = the dia. of the inscribed circle; draw GF parallel to BC, and 4a 4a-bx put x = BC; then- AE; X X 4a = AO; and by similar triangles, as -: a: X E C 4a — bx 4ax - bx2 x- = %=-FG 4ax2 - bc oxo 4a c —e b2; wx= hich brought out of fractions, and reduced, gives Ie3 92.44., $80033.28. ANALYTICAL GEOMETRY. 51 By converging sines x is found = 38.5296 = BC, and 70.566 = AE Then (by trig. p. ) A 15~0 16', and / B= Z C = 74~ 44/. PROBri'M LXIII. To determine a right angled triangle; having given the hypothenuse, and the radius of the inscribed circle. Construct a right angled C triangle, and inscribe in it a circle; also draw right lines from the centre to the points of contact, and to the two acute angles, as in the figure ABCDEFG. E Put x for the base (AB), y for the perpendicular (BC) and let a represent the given/ radius (DG, DE, or DF), b the hypothenuse (AC.) Then, because GB and DF are equal, y- a is the expressionB for CG; and xz-aAF. F A Also CE and CG are equal [Euclid, I. 26]; and AF is equal to AE. But AE+ CE = AC; that is, (y- a) + (x- a) = b, Orxy -- = 2a — + b. Now x2 + y2 -= b2; comparing, therefore, the double of this equation, with the square of the preceding, x2 - 2xy + y2 b2 4ab -4a2. Hence x- y = 2 — 4ab - 4a2, consequentlyx = a + b:: 2 / (b2 —4ab -4a2), and y=a+- -b -T (b2 —4ab-4a2). PROBLEM LXIV. To find the side of an equilateral triangle, inscribed in a circle, whose diameter is d and that of another circumscribed about the same circle. Let AFGC be the given circle, and ABC K the required inscribed equilateral triangle.- / IJoin A and the centre E; also join CE, and produce it to D. Then by (Euc. IV. 2.) the angle D is a right angle, and the triangles ADE and ADC are similar. But AD DB = AC, therefore also DE - ~ AE. Let then AE = radius = Id; and consequently ED — AE = ~d; also put AB = x, or =AD — x. Then by (Euc. I. 47.) -x2 _ —d2 —d' = 2 6d2;- Whence= V 3-d'= - d - 3, the side of the inscribed triangle. Again produce CD to F, and AE both ways to G and H; and draw HI, IK, perpendicular to EF and EG. Then it is obvious from the proposition above referred to, and (Euc. 6~2 ANALYTICAL GEOMETRY. iv. 3.) that EF = IHE, and HF ='HI. Let now Hi (the side of the circumscribed triangle) = y and we shall have HE2 = —EF'E -+ HFP; and since EF - d, HE = d, and the above becomes ds -I dl d _+ ~y2, or y2 3d2, or y = d / 3, the side of the circumscribing triangle. (See Prob. xix. p. 31.) PROBLEMIV LXV. rTo describe a circle through two given points A, B, that shall touch a right line CD given in position. Join AB; and through 0, the assumed centre of the required circle, draw FE perpendicular to AB; which will bisect it in E E (Euc. III. 3). Also join OB; and draw EH, B OG, perpendicular to CD; the latter of which will fall on the point of contact G (Euc. III. 18). Hence, since, A, E, B, H, F, are given points, put EB= -a, EF - b, C F G H D EH = c, and; EO =x; which will give OF b - x. Then, because the triangle OEB is right angled at E, we shall have OB2 = EO'2+ EB2, or OB = (x +a2). But, by similar triangles, FE: EH::FO: OG or OB, or b: c: b - -x: OB; whence, also, OB- = - (b - x). And, consequently, if these two values of OB be put equal to each other, there will arise, /V (x2 + aC) = b-(b — ). Or, by squaring each side of this equation, and simplifying the result, (b2-c2) x~ + 2bc2x = b2 (c2-a). Which last equation, when resolved in the usual manner, gives, bc2 C4 (2 -a. - b2_ 2 ab v (b2 c2)2- b2- c2 for the distance of the centre O from the chord AB; where b must evidently be greater than c, and c greater than a. PROBLEIMY LXVI. The three lines AO, BO, CO, drawn from the angular points of a plane triangle ABC, to the centre of its inscribed circle, being given, to find the radius of the circle, and the sides of the triangle. Let 0 be the centre of the circle, and on A AO produced, let fall the perpendicular CD; and draw OE, OF, OG, to the points of contact E, F, G. Then, because the three an- F gles of the triangle ABC are, together, equal to two right angles, (Euc. I. 32.) the sum of 3v their halves OAC + OCA + OBE will be B B C ANALYTICAL GEOMETRY. 53 equal to one right angle. But the sum of the two former of these, OAC +- OCA, is equal to the external angle DOC; whence the sum of DOC + OBE, as also of DOC +-OCD, is equal to a right angle; and consequently, OBE- OCD. Let therefore, AO- a, BO = b, CO = c, and the radius OE, OF, or OG = x. Then, since the triangles BOE, COD, are similar; BO: OE:: CO: O D,, or ex:~~ ~~2c2x2 b: x:: c: OD; which gives OD -, and CD = V/ (c2 ), or 2b2 - (2 -x2). Also, because the triangle AOC, is obtuse angled at 0, we shall have (Euc. II. 12.) AC2= AO2+ C02 — +2AO X Do C2acx b(a2 + c 2) + 2aCCX XOD; or AC- = (a2- - ), or / - b 2b But the triangles ACD, AOF, being similar, AC: CD::AO: OF b(a~2 + c0) q- 2ac) C or v ( 6 + ) +- ) (b2 x2) a:::. Whence, multiplying the means and extremes, and squaring the result, there will arise bx2! b(a2 + c2) + 2acI, = a2c2(b2 - x2). Or, by collecting the terms together, and dividing by the coefficient of the highest power of x, we have the equation, X3 -' - -a! = 2 From which last equation x may be determined, and thence the side of the triangle. PROBLEM LXVII. From one of the extremities A, of the diameter of a given semicircle ADB, to draw a right line AE, so that the part DE, intercepted by the circumference and a perpendicular. drawn from the other extremity, shall be of a given length. Let the diameter AB = d, DE = a, and AE = x; and join BD. Then, because the angle E ADB is a right angle, (Euc. III. 31.) the trian- D gles ABE, ABD, are similar. And consequently, by comparing their like sides, we shall have AE: AB:: AB: AD, or x: d::d: x - a. Whence, multiplying the means and ex- tremes of these proportionals, there will arise x2- ax d2. Which equation, being resolved after the usual man ner, gives x — = a- (a2+ d2). PROBLEM LXVIII. A gentleman has a triangular piece of land, whose sides are in the proportion of 3, 4, and 5, and the area of it is equal to the cube of one-fifth part of the base; required the sides and area in numbers. E2 64 ANALYTICAL GEOMETRY. The given proportion denotes a right-angled triangle; let therefore 3x be the base, then 4x and 5x will be the hypothenuse and perpendicular, and, by the question, the cube of three-fifths of the base =- 3 z = 2 13x X 4xl, whence x X 3=831 = the base, w11e-_ the perp. and 139A -the hypoth. the area being 4a629s. If 4; be called the base, then the three sides will be, 35~~, 46, and 581 area 82349; but 18, 24, and 30, area 216, if the hypoth. be base. And had the triangle been oblique, an equation might have been formed and the sides found, with the same ease. PROBLEM LIX. To find the area and the sides of a rectangle of which the perinieter and the diagonal are given. Let ABCD be the proposed rectangle call D C the perimeter p, and the diagonal BD, d. Let AB = x, AD =y, and we shall have A B y = p and X2 + y2 = a2. Squaring the first equation, we obtain x2 + 2xy + y2 = ~p2, and therefore (x2 + 2xy +y2)-(x2+y2) 2xy 4p2 a2. Consequently xy, or the area -= -pp — 2, and x AB = 1 +a (a a-p2), and y D = AD - / ( 2 —p2). PROBLEM LX. Having given the segments of the base, made by a perpendicular falling from the vertical angle, and likewise the ratio of the two sides, to determine the triangle; i. e. to determine the actual value of the sides of the triangle. i C In this problem, BD, DC are given;.*. let ED = a, )DC- = b. Likewise the ratio of BA: AC is given: but since the actual values of AB, AC are unknown, let BA = x, and AC= y, andlet the ratio of BA: AC be that of:. NowBA:AC::x: y::m: n nx = my and y =. Also AD2= AB —BD=2-a, But AD2=AC2 — CD2 = y2 3n k.._2 a2 = _ a2_ y — b2, And by substitution x2-a2 = -i — 262 X by m2 and nmx'-x'-m22 = 2 x2 - 262, By trans. mll - n 2 -2 -' /b2.' = { - } = the value of AB. By a similar process y = n.j {2 -- the value of AC ANALYTICAL GEOMETRY. 5b PROBLEM LXXI. Given all the three sides of the triangle, to find the radius of the inscribed circle. Conceiving the figure constructed (in the accom- G panying diagram) draw lines from the centre to each e of the angles, and to the points of contact. Produce either of the lines joining 0 the angular points and centre (as BO) indefinitely through the opposite side, B cd A and on it produced let fall a perpendicular (as AG) from that angular point from which the perpendicular falls without the triangle. Put a for Ad, b for dB, and c for Cf. Also x for Od= Oe Of ax + b] Then / b2 + x2 [BO]: z [Od]:: a + b [AB]:,/ +{ _ [AG]. V b2 + X2J ab +- b2 But BO: Bd:: AB: BG. that is, v b2 +2} 2/b2 + X2 - ab —. x'2 av X; 2+ -And AG: OG::Cf: Of, whence ax + bx: ab - x2 6:x. Therefore x = ( V a- b + -.) Which was required. PROBLEM LXXII. To determine a right-angled triangle; having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. In the annexed fig. let CD be the greater, and AD the less of the twoe lines of which the difference is given, and let DH be a production of CD, and AH perpendicular to CH, AH is A equal to HD, be- gA cause the angle ADH is equal to the sum of the angles ACD, CAD, together equal to half a right angle, and the angle at H a right-angle. If a be put fir AC the hypothenuse, x for CD, y for AD, b for the, 56 ANALYTICAL GEOMETRY. difference of x and y, r for DH, and s for AH. Then (Geom. p. 34) r = s y -V2. But 2 +- Y2 -2x = a2 x2- y2 xy/2. Now substituting x - b for y, and c for the V2, it will be Y2 - 2bx +b2+ x2 C- cbx a2, that is, (2+c) 2 —(2 + c) bx+b- -a'2. Whence x / 2Ic 2 ~ -,+and b2 2 + 4 Y V 42+ + ~-b }- -b. Consequently the radius of the inscribed circle is known, and the triangle determined. For put i a2 + b2 - =CD n- a2 b2+ 2+c + b-CD. +2~V — b = AD. And let DE, Df, Dg, be three radii at right angles to the sides of the triangle; likewise put w for Ag, and z for Cg. z'2-W 2 n = i2 -n2. Also z + w = a; and by division z w-w a a2 2 b2 n2 a5 2 q2 + that is, z= a +;w= 2a-r 2a; 2a (a2'-jn2 - n2)2 And Dg.= 2 -- = the radius of the inscribed circle. PROBLE[ LXXIII. Given the perpendicular, base, and sum of the sides of an obtuse angled plane triangle ABC, to determine the two sides of the triangle. Let the perpendicular AD = p, the base BC = b, the sum of AB and AC = s, and A their difference = x. Then, since half the difference of any two quantities added to half their sum, gives the greater, and, when subtracted, the less, we shall have, AB -(s + x), and AC = — ). / /- / But, by (Euc. I. 47,) CD2=AC2 -AD2, or CD= I(s —x)2 _p2;..by Geom. E C D p. 37) AB2 = BC(2 + AC2 + 2BC X CD; whence ~ ( s +- x)2 _ 21 Ty, -- 2.SX _ b2!i2+ (S-x)2~- 2bV(s x)2p2, or s-b2 = 2bV 4(s-x)2 -p21. And if each of the sides of this last equation be squared, there will arise, by transposition and simplifying the result, (s2 - b2).,2 =b2 (S2- b2) -4b2p2, or x -b i (1- 42). Whence, by addition and subtraction, we shall have, AB -=- as + b 3 (1 — s b 4p2 and AC = -s —b V (1- -2-_-) as required. 2 2 82- -b ANALYTICAL GEOMETRY. 57 THEOREMlI LXX1V. If two chords in a circle intersect each other at right angles, the sum of the squares described upon the four segments, is equal to the square described upon the diameter. Let ADC be a circle; and let the two chords C AB, CD cut each other at right angles in E. Find the centre G; join GC, GB; and draw GK, GH perpendicular to AB, CD. Then AK = KB and CH =HD. (Euc. III. 3.) Now because AB is divided equally in K andunequally in E,.. AE2 + EB2 = 2AK 2+ 2KEE2 (Euc. II. 9.) Also DE2+EC2- D 2DH2 +- 2HE2... by addition AE2 + EB2 + DE2 + EC2 =-2 AK2 + 2KE2 + 2DH2 X 2HE2 =2(BK' +'KG2) + 2(GH2 + Cfam) -a BG2+ 2GC2 =4 GC2 = (diameter)2. PROBLEM LXXV. Upon a given straight line as an hypothenuse, describe a right-angled triangle which shall have its three sides in continued proportion. Let AD be the given straight line; upon it describe a semicircle ABD. Let AD = a, AC / = x. Then AD: BD: BD: DC. (Euc. vI. 8.) But by the question, AD: BD:BD: AB,.. AB=DC. Hence V/ ax= a-x-..ax a2 2ax + X2, or X2I- 3oax= - a2, Complete A C D the square, and x2 3ax-+ 9-a2 = a2, Or x- a -= ( a V/ 5 ):'.x — ~3arkaV 5 - =-a. 3 + V/5 1 which determines the point C. Draw BC perpendicular to AD, join BA, AD, and the triangle ABD will have its sides in geometric progression PROBLEM LXXVI. Having given the perimeter of a right-angled triangle ABC, and the perpendicular CD, falling from the right angle on the hypothenuse, to determine the triangle. Put p = perimeter, CD = a, AC = x, and A BC - y; then AB p- (x +-y). But, by right-angled triangles (Euc. I. 47,) AC2 + BC_ —AB2; whence x2 J y2=p2 2p (X + y) -4 — }2-2xy + y2. Or, by transposing the terms and dividing by 2, p(x+-y) - p2=-y.. (1). And since, by similar triangles, AB: BC:: AC: CD, we shall also have, by multiplying the nimeans and extremes, AB X CD = BC X AC, or ap a (x+-y) 8 58 ANALYTICAL GEOMETRY. = xy... (2). Whence, by comparing equation (1) with equation (2), there will arise (a + p) X (x +- y) = ap + -?p. Whence, x + y p(a+-) ory p(4 - ) x. And if these values be now substituted for x + y and y in equatiol (2), the result, when simplified and reduced, will give (a +-p)2. — p(a --- i-) = ap2. From which last equation, and the value of y above found, we shall have, xor ACp(a~+ - P) -P (a- p)2-2a2 2(a+p) 2(a +p) and y or BC P(a+p)a 2 2 2(Andie 2(a) -— p) And if the sum of these two sides be taken from p, the result will give AB p - (x + — y) = Which expressions are, there2(a +p) fore, respectively equal to the values of the three sides of the triangle. PROBLEMY LXXVII. To find the side of a regular pentagon, inscribed in a circle, whose diameter is d. It appears, from (Euc. iv. 10,) that the side of F an equilateraland equiangular decagon inscribed in a circle, is found by dividing the;radius of the circle into extreme and mean ratio, the greater part of which is the side of the decagon. 0 \ Hence calling the radius OB =r, and the greater part OD x, we must have r(r - x) = A X2, or x2 +- r x = r2; whence x — r +- V' 5r, B or x — r (-1 5), That is, BC or AB in the above figure -r (-1 + 5) Produce BO to E, and join EX; then by (Euc. i. 47,) EC2 —EB2- BC=r ( + - / 5), or E -r\ (+5 1, + 5), Again, as EB: EC" EC: ED-=r(- --- 4 /5), and EB: BC:: BC: BD r (4 - - ~/5), whence DC (ED X DB) -- =Vr2 Q - 5) ( - I \5)-rv(10-2 /5,) or AC= 4d n 0I l -2 /: -. the side of the pentagon required. PROBLEM LXXVIII. To find the side of a square, inscribed in a given semicircle, whose diameter is d.' et A13CD be the given semicircle: AB, its diameter: G, its cc'tree:;:and.CDFE, the te:iuired squae. Then, since DF CE, we hav: FG -G GE.;~ Let therefore AB d or CGi',d; also E-x, and consequent- A -- B'F G l5 GE!~',th; t!ben by (Euc. i. 47,) CG2= ANALYTICAL GEOMETRY. 59 GLE2 + CE2, or x2 +- x2 = 2d2. Whence 5x2 z d2, or x = d v = IdV5. Otherwise, let ADCB be a semicircle whose diameter AB- a, Suppose CDFE to be a square inscribed in the semicircle. Find G, the centre of the circle, and join GC. Let GE = x. Then AE.BE -- CE2 by (Euc. iII. 35.) Or (AG + GE).(GB - GE) = CE2 i. e. -l:a+:{) - ai - =CE2. Or 1 a2 -x2 = CE. Now since FECD is a square, CD-DF=FE = FG + GE=2x,,, - x_=4x2 x+ 4x2 I a2 or 5x2 I a2, x- = a,\;-I i.~ 2x2a -- =, V- = = a side of the square. PROBLEM LXXIX The lengths of three lines drawn from the three angles of a plane triangle to the middle of the opposite sides, being 18, 24, and 30, re spectively: it is required to find the sides. Let ABC be the required triangle, and AF, BE, C and CD, the three given lines bisecting the three sides CB, AC, and AB. Make AF a, BE = b, (D = c, also CB -, AC =- y, and AB - z. E pP Now it is a well known property of triangles, that "double the square of a line drawn from any angle of A B a triangle to the opposite side, together with double D the square of half that side, is equal to the sum of the squares of the other two sides;" that is 2a2 - I - y2 _+ z2,2b2+ 2-y2- 2+z2 and 2c2 + Iz2 x2 + y2, Or y2 2 z2_ a_ 2 = 2a2, x2 y2 + Z2 262, and:X- y2 - 22 2c2. From whence by taking the former of these equations from twice the sum of the two latter, there comes out 4 x- zI2 = 2 (2b2+2c 2 a2); 2 -- V (2b2+2C2 _-2) 1n like manner y -2 V (2a2 -{ 2c2 - b2), and z 2= V /(2C q- 2b2 _ C2);'Where, by substituting the given values of a, b, and c, viz. a 18 b = 24, c = 30, we have x = 34.176, y = 28.844, and z=20, which are the sides required. PROBLEM LXXX. Given the base (194) of a plane triangle, the line that bisects the vertical angle (66), and the diameter (200) of the circumscribing circle, to find the other two sides. Let ABC be the proposed triangle, AB its base = 194 = b; IK the diameter of the cir- C cuinscribing circle = 200 = d, drawn parallel to AB: and DC the bisecting line -- 66 = a. Then we shall have HII = GK = J (IK - AB) I/ L - =3... AH = GB = V(IG X GI,) = V197 X>3 = /591 = c. Let now CD be produced to meet the circle in. E; Then, because CD bisects the angle ACB, it will bisect the are 60 ANALYTICAL GEOMETRY. AEB, and therefore the perpendicular ELM will pass through the centre L;. EM = 100 + / 519=-e is also known, as is also MN - 100 — 519.=f Now let DE - x; then, since the two triangles ENC and MDE are similar, we have ME: DE CiE: N-E, or e: x:: +x: d; whence x2-ax - de, or x -- -a+ V( 4-a2 + de), which thus becomes known; and consequently the rectangle CD X DE, or a X -- -la- / ('a2+ de) - r is also known. But this latter rectangle CD X DE- AD X DB; therefore AD X DB, and AD + DB, are both known. Assume now, AD --- y, and BD = z, then we have y + z = b, yz -r, whence is readily found, y = -I-b + I (b2 -4r2) =, and z= b - I/V (b2- 4r2) —n, Again, calling AC = v, and BC u, and we have v u: mn: n, or v un um~ n; also vu = 2 + mrn. Substituting urm + n for v in the second equation, gives mu2 + n a2 - mn.,t + a. - - ac2n +r mme a2m- + nm2n Whence =,/ ( ), e-V = / - the m U fl sides required. The numerical values of which may be found by substituting those of a, b, and d, in the original equations. PROBLEM LXXXI. It is required to draw a right line BFE from one of the angles B of a given square BD, so that the part FE, intercepted by DE and DC, shall be of a given length. Bisect FE in G, and put AB or BC = a, A_ FG or GE = b, and BG = x; then will BE A _ x + b and BF = x - b. But since, by right-angled triangles, AE2 = BE2- AB2, F we shal have AE = / ( x + b )2- a2. And because the triangles BCF, EAB, are similar, BF: BC: BE: AE, or a(x + b ) - - (x - b) / I (x -+ b)2 a'2! By squaring lit C each side of this equation, and arranging the terms in order, there will arise x4 — 2(a2 + b 2) x2 = b2(2a2- b2). Which equation, being resolved after the manner of a quadratic, will give x -= I2 + b2 + a / (a2 + 4b2)} And consequently, by adding b to, or subtracting it from, this last expression, we shall have BE = / a2-+ b2 + V/(a2+ 4b2)I + b, or BF = a2+ b2+aV(a2+4 41,2) - -b. Which values, by determining the point E, or F, will satisfy the problem. Where it may be observed, that the point G lies in the circumference of a circle, described from the centre D, with the radius FG, or half the given line. PROBLEM LXXXII. In a plane triangle, there is given a base (50), the area (796), and the difference of the sides (10) to find the sides and the perpendicular. ANALYTICAL GEOMETRY. 61 Let ABC be the proposed triangle, of which the base AB is given, - 50 =2b, Then since the area is also given - 796, the per- C pendicular = 7- p 5- is also known. Make now BE = half base = b, and CD - p, and ED = x; then BD = b-+x, and AD= b -x: BC= V!p' + (b+ x)2, and AC = /p2 + (b — X)2'. Whence, calling the given difference = 10- = D E E d, we have V /p' + (b + ) -- d- = d 4p + (b - x)2I, which equation, squared, gives p -2 (b + x)2 - 2d V' p2 4- (b + x)2} + d2 = p2 + (b — x)2; and this, by reduction, becomes 4bx ~+ 2 = 2d V IP2 + (b 4- x)2. And this, again squared, produces 16b'2. -8bd' + c4 -d, 4d' X (p2 + b2+2b x+ x): Or 16b'L2 h nc4 - 4d2 X/ (b2)4d2 X ( + p2C ) + 42 Whence x _ g/( X16b P - -). Where, by substituting the numeral values for b, p, and d, the answers for the sides will be found. PROBLEM LXXXIII. In a right-angled triangle given the hypothenuse 30, and the difference between the base and perpendicular = 6; required those two sides. Let the hypothenuse = 30 = a, and the given difference = b; and call the base x, then the perpendicular = - b. Now by Euclid, i. 47.,) x2 + (x -- b)2 - a2, whence x2 - bz -a, cornpleting the square and extracting the root x = V/ (-a _b2) + lb, Substituting for a and b their values, x = 24, = the base, and the perpendicular x- b 24 - 6 = 18. PROBLEM LXXXIV. Given the three sides AB, BC, CD, of a trapezium ABCD, inscribed in a semicircle, to find the diameter, or remainuing side AD. Let AB - a, BC - b, CD = c, and AD x; then, by (Euc. VI. Prop. D). AC X BD -_- AD X BC + AB X CD = bx+ ac. But ABD, ACD, being right angles, (Euc. III. 31 ) we shall have AC = V (AD2 — DC2), or (x2 -c), and BD- (AD -.AB2), or & I V (x2 - a'). Whence, by substituting these two values in the former expressions, there will arise V/ (' - c2) X V/ (x - a2) bx + ac. Or, by squaring each side, and reducing the result, x — (a2 -- b2 + C2) = 2abc. From which last equation the value of x may be found, as in the last problem. (p. 52.) F 62 ANALYTICAL GEOMETRY. PROBLEM LXXXV. Given the perpendicular (24), the line bisecting the base (40), and the line bisecting the vertical angle (25), to determine the triangle. Let ABC be the proposed triangle, and C make the perpendicular CD = 24 = p, CE the line bisecting the angle ACB = 25 = b and CF, the line bisecting the base, = 40 = c. Then (Euc. I. 47.) ED = / (CE2 -CD2) / = 7 = n, Also FD = V (FC2 - CD2) = 32 =n; And in order to simplify, let EF -q. Also let half the base AF = FB — =x; then AE = A F ED3 q -- q, EB = X-q; AD =x -- +n, DB - - n Hence AC = (x + n)2 + p2 SBC= V (x - n)2 + p25 And from (Euc. vi. 3.), we have AC:BC AE:EB, or /-( - n )+ +P': 1V (x-n)2+ p2:: x+q: x -q; Whence t(xx+r) -tp2 X (n —q)2 -- (x -n)2a + p2 X (x + q)2, Which by multiplying, cancelling, &c. becomes nx(x2 + q2) qx(x2 + 2 + p2). Where $x q2 ~ _+ qP - nq2 _2 q2 2 2p2 -W'-qre- -,or 2x:2 %/q n-q I — q the base of the triangle; wlhich, by substituting the proper numeral values of q, n, and p, gives 2 ~V/ 14; from which and the given lines the other two sides are readily obtained. PROBLEM LXXXVI. Given the hypothenuse (10) of a right angled triangle, and the difference of two lines drawn from its extremities to the centre of the inscribed circle (2), to determine the base and perpendicular. Let ABC be the proposed right angled trian- C gle, and 0 the centre of its inscribed circle; and let CO - AO = 2 = d, and AC = 10 =h. Produce CO to D, and let fall upon it the perpendicular AD; which put- x. Then, since CO and AO, bisect the two angles, C A and A, and these two angles together are equal to a right angle, it follows that the two angles D OAC and OCA = half a right angle. But the outward / of any triangle, being equal to the two inward opposite Ls, z AOB Z OAC + LOCA. Whence also AOD = half a right angle, and since D is a right angle, DAO is also = half a right angle. Therefore DO = AD = x, and AO = V/2x2 - x -V2; and consequently CO = x /2-+ d, and CD = x +x V/2 + d _ (1 + — v2)x: +- d. Now AD2+-DC2 — AC2, or x12 +- (1 +/2) x-+-d a =h, or1 I + (1 + ~ 2)2 x2 + 2d(1 +-V2) x=h2 or (4-+ 2 V2):C2+ 2d (1 _+') x =-2 -2, or ANALYTICAL GEOMETRY. 63 21 - V/2 h2- _dl +2 + 2 ~d - V- + Then by quadratics we have 4 x= -4+2/2 2 )22 +2 By reducing these surds to rational denominations, we have x --- dV2~t vd2 + (2- v2) (h2 d2)=- d-V2~ 2 1Vd2 2+ (2-v2) (h + d) (h- d) = 3.10850255 = AD = DO. Hence OA = xV2 = 4.39608645 = -m. O -6.39608645 = n. That is AO = mn, and OC = n; now to find the segments AG and GC, (by Geometry, p, 360.) we have As h: n + rn:: -m: C — GA = (n+ ) X (n-m),but CG+ GA = h. lence CG =-h (n + mn) X (n-) d - (n + rn) X (n - m) ~+ 2h, — - and GA = rn; now 2h' 2h = 1.98822509, CG = h + (n + m) X (n-n) = 6.07921729, & 2h GA - 3.92078271; consequently, AB = AG + OG = 5.9090078, BC = CG + OG = 8.06744238. NOTE. The answers given in Bonnycastle's Algebra appear to be wrong, (from which this is taken.) Proof. The sum of the squares of those numbers above is 99.9999997 which should be just 100, viz. the square of the hypothenuse 10, the error bejngTv:wvwF; Otherwise, Let x and y denote the base and perpendicular, then as is well known ~(x + y - a) = the radius of the inscribed circle, (Simpson's Algebra, Lemma, Page 345, Published by Carey & Sons, Philadelphia,) or (Young's Geometry, just published by Carey & Lea,) put (x + y — a) - r then g I (x-r)2 + r" - =the line drawn from the extremity of the base to the centre of the inscribed circle; and / (y - r)2 +- tr = the line from the centre to the extremity of the perpendicular. Hence supposing x7y (per ques.) V/ x2- 2rx+2r'2 = b~+ V y2 -2ry+-2iV or x2 —2rx =2 + 2b v -2iy -+ 2r2 +-y2 - 2ry, or a - y+2r (y - x) —b2=2b / y2- 2ry+-2r2, or restoring the value of r, a(x - y) - b2 = 2b V/ -l (a - X)2+ y2a =2ba/,a -axl, (since x2 - y2 = a2),.,' =a a a2-z2 a2 _ z- -z2- 2 bz Put/V / ax- =z, orx = —-, then y = __ a- (z+b)2. 2a4-2a+z —-2a(z+-b )2 +(zz+b)4 64 ANALYTICAL GEOMETRY. or by reduction, (a2 - z)2 - 2a2(z + -b)2 + (z + b)4 - 0, or (a2 - z2)2 - 2(a2 - z2)(z + b)2 + (z + b)4 —2z2(z + b)2, or by extracting the square root we have a2-z2_-(z+b)'= 2(z+b) / 2, or 2- b2 = 2z2 + 2bz + 2 (z + b) V2 = (2 + V2) X (z2 + b2); z.. z+bz a2 —2 2- X a2 — b2; or by quadratics, we have z = V/(4 - 2,/2 ).( a2- b2) + b2 - b 2 a2 _ a 2 and(z+b)2 hence x-= - and y = become known. a a Otherwise, Geometrically. Let CH denote the given difference of the lines drawn from the acute angles to the centre of the inscribed circle, then draw the indefinite line AH so as to make the angle AHL-O= -P, (where P = two right-angles) then with C as a centre and the given hypotenuse as radius, describe an arc cutting AH in A, and join AC, then draw BC through C so as to make the angle BCO=ACO and let fall the perpendicular AB from A to BC, then ABC is the triangle sought. For, draw AO bisecting the angle BAC and meeting CH produced in 0, then evidently 0 is the centre of the inscribed circle; but since B = a right-angle=-P,? A+C is also= —P.~ OAC + OCA= 2(A+C) =P, and AOC=P —P= —I P.. in the triangle AOH we have AHO+HAO='P, but AHO=-P.. HAO —-P and the triangle AOH is isosceles, and CH —the given difference of the line CO and AO as required.-See calculation, p. 286. PROBLEM LXVI. Having given the sides of a quadrilateral which has two of its opposite angles equal; determine its area. Let ABCD denote the quadrilateral whose four sides are given, and which has the angle B = D; join AC and draw the perpendiculars AF, AF' to DC, BC. The triangles ADF, ABF' are evidently similar, hence AD: AB:: DF: BF':: AE: AF.. B.F= AB.DF —AD, and AF'=AF.AB -- AD, hence if A=the area we AF AD.DC + AB.BC have A- X -D.DCAB A. Put AB=a, BC=b,CD-=c, ~AD 4)~~~p cd -+ ab AD=d, AF=p, DF=-x; then we have X 2 also AC2 = AD2 + DC2 - 2DC.DF, and AC2 =AB2 + BC2 - 2BC.BF/. a2 + b2 — 2b. BF = d2 + c2 2cx, or (since BF= ), a2+-h2 _ = d2+2-2 —2cx. hence we have (d + c -.. p=(a2 +/2)d d-x2 becomes known, and 2(cd- ab) thence A becomes known by (1). ANALYTICAL GEOMETRY. 65 PROBLEM LXXXVIII. Having given the base of a plane triangle = 2a, the perpendicular - a, and the sum of the cubes of its other two sides equal to three times the cube of the base; to determine the sides.@ Let x + d, x- d, denote the sides, then (x + d)3 (x - d)3= 3 X (2a)3 = 24a3, or by reduction x3 + 3xd2= 12 (1), also by the common rule for the area of a triangle when the three sides are given, (Hutton's ilathematics vol. 1. p. 405), we have V (2- a2)' (a2 - d2)- = the area of the triangle, but ~(a X 2a) =a2 (Hutton p. 403), = the area, hence results the equation (x2 — a2) (a2 - d2) = a4, (2); by (2) 2a4 22 __-2 which substituted in (1) gives x5 + 2a2X — 12a3xa26a4x -+ 12a5 = 0, this equation is satisfied by putting x = 2a, hence 2a4-a2x2 2a2 6a2 a: 22 3 9 ord= /V6,..x+d=a(2+-~V'6), a2-x2 3' 3 x- d - a(2- -V/6) are the sides sought. Remark, that the solution of question 51 page 46, might have been much simplified by the method used in the solution of this question. For by denoting the sides by x + —~c, x- ~c we have the area of the I 2 I 2 __ triangle=-/t (z 2_xb6) (~ 42 — c2 ) -a2 ( per question) and x24a~4 b22 + 4 2, or x-= b2C-2+, hence the sides come out the same as in the solution cited. Questions~ of this kind may be made as follows: Let the base of a triangle be = b; the sum of the other two sides = mb, and the sum of their cubes = nb3; Then the difference of the sides will be = b V/ ( (4n - m3)' 3m); where (4rt —. m)' — 3m may be any positive number with the following limitations: Since, from the nature of the question, in must always be greater than unity, it follows that n must be greater than 4, and because b3 X ( m1 + 4ms) is the greatest limit of the sum of the cubes of the sides, it is evident that n must be less than 4m3 + 3ml. The two least whole numbers for m and n (unity excepted) that will answer the question: Let m-2, and n=3; then the difference of the sides xwill be 2b +- v 6, and the sides themselves b + b.- 6 and - b -./ 6, and the perpendicular — b as given by the question. For by (Trig. p. ) b: 2b:: h. / 6: 4b. v/6 = twice the distance of the perpendicular from the middle of the base: hence the greater segment = b + 2b V 6; therefore (b + b + 2 6)2 (~b + 2b +-' 6)2 = 4b2 the square of the perpendicular. In this manner, when the values of in and n are chosen within the above limits, the perpendicular, and thence the area may be determined. If m = 3, then the least whole number for n will be 7; and the three sides of the triangle will be b and lb, and 1ib, and the triangle is right angled. 9 F* 66 ANALYTICAL GEOMETRY. Let BC be the given base = 2a; take a line M I a mean proportional between BC and I BC; on BC describe a triangle such, that the side AB = BC 0/ + M, and AC = BC - M, and the thing is done. For let fall the perpendicular. It is well known that AB3 ~ AC3 = 2BC3 -- 6W2 X BC; and byA construction, 6MI2 BC2,'therefore AB3 AC3 A G K - 3B'(3, one of the conditions. Again, BC: AB + AC (2BC):: AB -AC (2M): 4M, hence N BK I -BC +- 2M; and (by Euc. i. 47,) AK2 = AB2- BK2 (BC + M)2 - (~hBC + 2M)2= ~BC2- 3M2_ -BC2 or AK = ~BC = a, the given perpendicular. t Let the given base BC be bisected, in G by the perpendicular NM; upon which take GE = ~GB, GHE = GC, HL = 2GB = BC; upon the diameter LE describe a circle cutting HA parallel to BC in A, the vertex of the required triangle ABC. For it is evident'that GL = 3BC, GE = IBC, HE = -BC, and HA or GK=BC,/ 3 (by Euc. vI. 13.). Also BK=BC X ( + V 2), and CK = BC X (2/ — -). Therefore BA2 = BC2 X (74- + ) (Euc. I. 47,) and BA = }LBC X (2 4- / 2). In like manner, AC= ~BC X (2- 2 /-). Whence the whole is manifest, for (2 + V/3)3 + (9- V2)3 8=3. Cor. 1. GK = BA —AC. Cor. 2. BA + AC = 2BC. Cor. 3. If CO be perpendicular to BC, the cubes on BOQ and QC are' together equal to seven times the cube on BC. Also BQ + QC = 3BC, and B - QC = ~BC. tDraw BC = the given base, which bisect in G; make GK BC V2, and at K erect the perpendicular KA = BG; join AB, AC, and ABC will be the required triangle. For (by Euc. I. 47,) AB2 = BK2 + AK2 = BGf2 + 2BG X GK + GK2 +- BG2 = BC2 X (7 + 2 -3), or AB = BC V (I +/ 2)= BC X (1 + V1/). Also, AC2 = CK2 -KA2 = BG-2BG X GKt-GK2- BG2= BC2 X (~7- -), or AC =BC X(1- ) Now it is obvious that if AB and AC be each cubed and added together, all the terms except the first and third in each will destroy each other, viz. AB3 -+ AC3 = BC3 X (2 + 1) = 3BC3. Let ABC be the proposed triangle, in which BC = 2a, and AD = a, ED=x.. A Then BD - a + x, and ED = a-x, or - (x - a) we have AB = / (2a2+-2ax + x2), AE = V (2a2- 2ax -x 2). And by the question (2a2+2ax4x 2)2+ / (2a2- 2a + x2) = 24a3. Let 2a2+ - 2axt +x2= n, and 2a2 —2ax-+x2=n, 11 E ANALYTICAL GEOMETRY. 67 3 3 3 Then mn + n2= 24ac. And by squaring, m3 + ns + 2m2 n2 — 576a6 Or 4rA3n3= (576a6 - m- n3)2. Where, substituting the above values of m and n, the equation reduces to this; viz. x =aV/. Whence BD (1 + 38)a, and ED = (1 — / )a. Which, being negative, shows that the perpendicular falls on the base produced. Therefore AB = a V.( 2 - 34 +V6 ), and AE - aV 2A4 4 6). And this, by extracting the roots, gives AB - a( 2 +- 6 ), AE = a ( 2.~ V6), which are the two sides required. See page 40. PROBLEM LXXXIX. Having given the base of a plane triangle (15), its area, (45), and the ratio of its other two sides as 2 to 3, it is required to determine the lengths of these two sides. Let ABC be the proposed triangle. Let AB c a5 = a, CD =- 7,_ b, and AD - x, therefore BD = a —x; also the ratio AC: CB:: 3: 2, or m: n; then AC2 = b2+ x2, and BC2= b2 + (a- X)2.'. b2 +: +(a -x)2:: m2I: m 2:. WThence, we have n2b2 + m2a2 - 2ne2ax + m2x2 A D ]g = n2b2 + n2X2, or (in2 - n2) x2 - 2m2 ax = (n2 - n2) b2 - 2a2 therefore, by solving this quadratic and substituting the values of a, b, mn, and n, the numeral value of x may be determined, and hence those AC and BC. PROBLEM XC. Given one angle, a side adjacent to the saict angle, and the difference of the other two sides, to determine the triangle. Put a = AB; d = the difference of the sides C AC, BC; c = cosine / A; and x = AC; then BC = x ~t d, and (by Geom. ) AB+-AC2 - 2c.AB.AC — BC2, that is a2+ - x - 2acx — x2~ 2dx + d2; therefore x (a2 —d) - (2ac A B d) = —AC. PROBLEM XCI. Given one side, the difference between the square of the other side, and the square of the base, and the difference of the segments of the base, made by a perpendicular from the vertical angle, of a plain triangle, to determine the triangle. Put a = the given sides b2 = the difference of the squares of the other two, d = the difference of the segnments, x - the base, and y = the other side; then y2 -a2 = ~ dx (by Euc. 2II. 36,) and x2- y2= b2, from which equations we get xa- dx = a2~ b2, an equation which exhibits the four cases of this problem. 68 ANALYTICAL GEOMETRY. Observation I.-Since by the conditions of the question, d can never be greater than x, it is evident that the least value of x2 - dx=a'-2_/2 is 0; or that, in this case, b must not be given greater than a. Observation I. —Since x2 + dx can never be less than 2dc, this is the least value of a2- b2 in that case. PROBLEM XCII. A gentleman has a garden in the form of an equilateral triangle, the sides whereof are each 50 feet: at each corner of the garden stands a tower; the height of A is 30 feet, that of B 34 feet, and that of C 28 feet. At what distance from the bottom of each of these towers must a ladder be placed that it may just reach the top of each tower, and what will be the length of the ladder, the ground of the garden being horizontal? OBSERVATION.-Had the height of A been 38, B 42, and C 45, and the distance from A to B = 501 B to C = 40, and from C to A = 47 feet, the operation would have been more difficult-The leng;th of a ladder in this case would have been 49'552, H and hence, the distances would have been found as in my key to Hutton's Mathematics. Admit P to be the point sought, from which C let fall the perpendiculars PG and PF, and produce PF to K; let AD BE, and CH represent D E the three towers, then the angles DAP, EBP, \: and HCP will each be right angles, and lines drawn from D to P, from E to P, or from H to a P will be equal to each other, and equal to the length of the ladder. A l L AD = a, BE = b, CH = c, AB = d AC = e, BC=f, AF -- x, AG = y, PF = z, GP - v. First, we have AE2 +- FP2 (AP), and BF2 + FP2 = BP2; But AD2 + AP2 - BE2 + BP2; hence AD2 + AF2 + FP2 = BE2+BF2 ~+ FP2, that is AD2+AF2 - B. 2;' + BF2; let AF = x, BF = d — x; hence a2- x2 = b2 + -- 2dx: b2 + d2 M a -x2, then b -AF. 2d - Proceeding exactly by the same process, we shall find AG. For AG2 4 — GP2=AP2, and GC2-GP2=PC2; But AD'2 4- AP2= -ClH2 + PC2, hence AD2 + AG2 + GP2 - CH2 + GC2 +'Gp', that is AD2 + AG2 - CH2 + GC2; Let AG = y, GC = e - y, hence nc.2 --- y2 = cC2 + y2 2ey + e2, from which y - (c2 - e2- a2) 2e.AG Again, because the 3 sides of the A ABC are given, the segments, AI and IB are found thus, AB; AC + BC:: AC - BC: A - IB; Then — AB + 1 (AI — IB) =AI= d +- e- ~ + e — f this quantity r, then(A2- - 2 (e2 _ r2) 2d this quantity = r, then V (AC2 - A I,) = ~ /(ea — r~ =IC. ANALYTICAL GEOMETRY. 69 Now because FK is parallel to IC, the angle AKF = angle ACI; and the angles AFK, AIC and KGP are J angles: these triangles are similar AL IC:: AF:FK, r / ( e2 — r2)::( r ) X g (e2 —r2); put this quantity =s. Again AI: AC:: AF: AK; or r: e:: x: (ex -r). Now AG = y, therefore AK - AG = (ex r) -- y - GK, and KP -FK -PF = s- z. But AF2+ EP2 AG2 + GP2, Viz. x2 + 2 = 2 y v2, and GP2 + GK2 = KP2, viz. v2 -+- [(ex - r) - y]2 = S2 2sz + z2, from the first of these equations 2 =x2 + z2 y; and from the second v2 = s2- 2sz+z2 [(e -r) _ y ]2. Hence 2 + z2- y2= S2 2sz- 2 - [ (ex r) -y]2; and z = (s2+y2 2- [(ex. r) -y]2) 2s. And lastly, because AF2 + FP2 AP2, and that V/ (AP2 + AD2) = the length of the ladder, it follows, that V (AF2 + FP2 + AD2)the length sought. Now AF = x, FP = z, and AD = a, all of which quantities are known. If d = 50, e = 47, f- 40; a = 38, b=42, and c = 45, then will AF = x = 28,2; AG = y= 29,68085; AI-r= 31.09; FK s= 31.971365, GK = (ex- r) - y= 12.95022180, FP = z = 14.7033573: And V (AF2 + FP2 + AD2) = / (x2 -+ z2 + a2) = ~ 2455.4287158914 = 49.55228 the length of the ladder. PROBLEM XCIII. Given the four sides of a quadrilateral figure, two of which are parallel; determine its area. (See Diagram to Problem 97, p. 73.) Let ABCB denote the quadrilateral having AB parallel to CD and its four sides given; suppose DC \AB then through A draw AE parallel to BC meeting DC in E, then AE =BC, AB =EC (Geom. p. 84.).. the three sides AD, DE=DC-AB, AE = BC of the triangle ADE are all known and thence its altitude AF=-the latitude of its quadrilateral is easily found by known methods, and consequently we find the area of the quadrilateral -IAF X (AB + DC) becomes known as required. PROBLEM XCIV. A quadrilateral figure has two of its sides parallel; determine its area, when its altitude, one of the parallel sides and the two adjacent angles are given. (See fig. last question.) Let ABCD be the quadrilateral, in which the side AB, CD are parallel, and CD together with the angles at C, D and the altitude AF are known; then by making the same construction as in the last question we have the triangle ADE whose angles are all known, for the angle D is given, and the angle AED=BCE (Geom. p..) ~. the remaining angle is easily found; also the perpendicular is known, hence the sides of the triangle are easily found by (trig. p..) Hence the sides AB = CD - DE becomes known, and then the area =I AF. (AB + CD) becomes known also. 70 ANAL kTICAL GEOMETRY. PROBLEM XCV. The lengths of two lines that bisect the acute angles of a right angled plane triangle being 40 and 50 respectively, it is required to determine the three sides of the triangle.t Let ABC be the proposed triangle, AE = a, C and DC = b, the two given lines. Also let x and y represent the sine and cosine ABC re- / spectively: then by (trig. p..) we have F E V ~- Y_ cos. BAE, andV/ + c = cos. 2 2A B 1q-x b 1q-x BCD. Also x: b:: = C, and y: 2 X 2 -l.y a l +y b 1-+ x a I-+y 2,/- - ---— =AC: Whence ---- or x ax 2tan-A I/1 +-=b-. Again by trig. p., sin. A -- itadA=' and 1- tan2~A cos. A-1 + tan; Putting, therefore, tan BAE = t, and substi2t I -_t2 1 q- t 2at tuting x = - T, and y = 1 We have -2 O2(1 t) or 1 + P 1~2 t2(22)' b(1 +- t) (1 -- t2) = 2atV 2, or t3 t2 - (1 + 2a /2) t 1. b Which is a cubic equation, whence the value of t may be determined: viz. the tangent. of the angle BAE: and hence, also, the ant gles BDC and BEA become known, and consequently the sides AB = 35.80737, BC - 47.40728 and AC = 59.41143, as required. Mt Make AE_=a, CD=b, s = sine of LAFD O CFE = 45~, y= tang. L_ EAB or EAC, 1 -x = its cosine, and y x of course = its sine; also (1 - y2) 2y = tang. ACB, and (s +-sy) -- = sin. D= cos. DCB or DCA. Now, (by trig. p,) as 1: a:; -ABa ndlb sy bs- bsy _ AB, and b1 bs: — + b s = CB. Hence tang. X X x X a I 1 - y2 LACE a B X1 =' from whence by a cubic equation Ts' +y 2y the angles are found and consequently the sides as before. Otherwise, let ABC be the right angled triangle, AE, CD the two given lines bisecting the two acute angles and intersecting one another in F. Draw LFG perpendicular to AE and KFH perpendicular to CD; and join EG, KD. (By Geom. p..) That each of the angles at F is equal to half a right angle, and consequently that FL ~ A denotes any angle; but in this example it is put for the angle BAC See. trig. by Dr. Day, President of Yale College. ANALYTICAL GEOMETRY. 71 -FG= FE, andFK = FH=FD. Put AE = a, CD = b, FL =FG = FE= x, and FK = FH =FD-y; then AF = a-x CF = b-y, EG = xV2, and DK = y V/2. By similar triangles, AF. FD:: AE: EG, and CF: FL:: CD: DK, which give these two equations, viz. ay = (a -- x). x /2, and bx (b - y) X y V 2; from the former y = (a- x). x /2 - a, which substituted in the latter, gives, (a — x)2. 2xV2 = (a - 2x). ab, a cubic equation by which x will be found as before. PROBLEM XCVI. Given the triangle ABC, AB=24, and BC=-37.44; together with the segment of the base, DE= 16.80; made by the two lines BD and BE drawn from the vertical angle to the base; the angle ABE = CBD = 90~. To determine the triangle Let ABC represent the triangle, BE and B) DR the perpendicular to BA and BC, and ED the given distance, instead of the number on the question it will be more convenient to take the c numbers 50, 78, and 35, or 5, 7.8 and 3.5, which are in the same ratio with the others. D - Let AB - 5 = a, BC = 7.8 =b, ED = 3.5 =c. and BF = x. Then AF V Ib2 —x, CF = V/ - a 2 x; by similar triangles CF: FB B FB2 x2:: FB: FE, or, EF = —- / —. and c CF Vja -x3 DF E FB2 _ _x2 FD =AF but EF +- FD = ED or=c, consequently, - =. + 1 —— 1 -( _y2)2 t h Which is made rational byputting x2-a2(1+2)2_b2(1-y) the 4y2 segments are +(b2-a2) + and (b2 2)1 andby re ) 2y 2yreduction we get the final equation a2.(1 + y2)2 b2(1 y2)2= c /(b2 - a2) y (1 -_ y4. ) " In numbers, a being equal to 5, b = 7.8 and c = 3.5. then we have x2 x 2 X/ _68_____ = 3.5, by a few trials we easily V160.84 -x2 / V 25 —x2 I find x -3. Hence 5: 24:: 3: 14.4= BF. Whence AF =-/ Ia2 -x23 = 19.2, FC = V/ bh2 -x2 = 34.56; consequently-, 19.2 + 34.56 = 53.76 = the base AC; and EF = xV2 X2 2/.=.2 —6; andFD = -b2 10.8. HenceBD15 6 and BE = 18, and the area = AC X ~BF = 387.072 feet. 72 ANALYTICAL GEOMETRY. Let AB = c'; BC = a; DE - b; the angle A = %, C=cp'. Draw the perpendicular BF to the base AC; then it is evident that it will fall within the triangle DBE, and that the angle EBF = A - qp; also, that the angle DBF = C = cp'; but BF = AB X sin. A = c/ sin. p; and EF= BF X tan. FBE = c' sin. p tan I; and DF = c' sin. p tan..'. c' sin. g X (tan. 9 +- tan. -') = b (1); also c/ sin. p =-a sin. Ip (2); it is easy by (2) to eliminate tan. (p' = from (1) whence.there would be had an equation in terms of qp and known quantities; but it appears to me better to use (1) and (2) as they stand, by assuming in (2) a certain value for qp and then by (2) calculate p' substitute these values of %,' in (1 )and if they satisfy it, 9 was rightly assumed, if not then by the usual methods of trial and error qp, can be found to any degree of accuracy desired; and thence every thing else becomes known also; remembering always that the sines and tangents are to be taken to the radius (1); also that each of the angles, 9, qp/ must be less than 900, and indeed it is evident that their sum cannot exceed 90~ supposing the point D. E not to lie in AC produced. Again put DF = x, BF = y; the right-angled triangle DBC gives BF44 -F - CB2 BF2, ory = 2_2 y4 X2y2 = a2x2 (3) y2 DF2 xIIr= =-y,..y4+Ay=2 (3);... - x V/ 4a2 +t x) - xa2; similarly by the right-angled triangle ABE, y4 + (b- )2y2 = c(b — x)2 (4);. y2 = I (b — x) v/4c2 ~ (b- x)2 - (b- )2I; put these values of y2 equal to each other and there results the equation x V/ S4a2 + x2 - _ = (b- x) X /4c'2+ (b -)2 -(b —x)2 (5); x can be found by (5) by the usual methods and the problem will be solved as before. By constructing the curve whose equation is (3) and the curve whose equation is (4) we shall find the vertex B of the triangle ABC at their intersection; remembering that in these equations x is to be reckoned on DE from D towards E, and that y is to be drawn through the extre mity of x at right angles to DE above it. The point B can also be found by polar equations; for put DB = r, BE r'; then r tan BDC = rcot. p' = a, or r=a tan. P', (6); r'/ c' tan. p (7); construct then (6) and (7), and the vortex B will be given by their intersection as before. It appears to me, however, that the solution by I) and (2) will be more easy in practice than any of the other methods which I have mentioned. Otherwise. Let AB — 50 = a, BC = 78 = b, ED = 35 = c and the base AC = x. Then AC: AB -+BC:: BC - AB: FCAF, or FC- AF=, and AF + FC -- x. Hence FC = x2+ b2 - a - n nBF2AB2AF2 or, —, and AF= BF2 = AB2 2x 2x 2x 2x ANALYTICAL GEOMETRY. 73 a2 - n2 4x22-(x2- )2 ag — a —4 —. 4 ) Again CF:B::BF: FE 4x2 4x2 4 BF2 BF 4x2a2 - x - _ n2 CF 2 X x2 + n' AF= 22 X lx2 — n but EF + FD = c, whence 4x2 - (2 - n) + 2 x (x2+ n) 4x2a2 - (x2- n)2 1 1 -c, Or 1 27 X i- 2x X ~ qx- n+ 22 X ~x x2 n C a x4x ~ }Or _ 2=4a2a2 -2ne2. By clearing it of fractions 4x3d2 - x5 +- 2x3n2 - f2'2 = Cx4- cn2. Or zc5 -+ cx4 - (2n- +4a2) x3+ n2 x - c2 0; in number x5 + 35 x4 — 17168 X3 + 12845056 x- 449576960 = 0. By rule given in Young's Algebra, p. 212, just published by Carey and Lea, Philadelphia, we find x = 112; Therefore 50: 24:: 112 "2._T n:53.76 -- AC and AF = -- 40, and CF = 72, BF 30. 2x Whence EF = 12.5 and FD = 22.5; consequently EB =32.5, 24 + 32.5 24 X 37.5 BD = 37.5 and =15.6 -EB and 50 =18=BD, 50 50 and the area=387.072 feet as before. PROBLEM XCVII. A triangular field ABC whose sides are given, is to be divided into two parts in the ratio of 2: 1, by a fence passing across from a given point D, in AC to BC. Determine its length. Let ABC denote the triangular field whose sides A B are all given, D the given point in AC, and suppose that AD is not greater th n DC it is evident that the division line will meet e BC at some point between E C' F' B3 and C. Join BD, and through A draw AE parallel to BD meeting BC produced in E, join D F then the triangle DEB=-the triangle ADB, for they have the same base DB and the same altitude (By Geom. p. ) hence the triangle ABC the triangle DEC, then take EF: CF:: 2: 1; join DF, which represents the fence both in length and direction; for the trialngle DEF, DFC having the same altitude are as EF: FC:: 2: 1, but the triangle EDF=the quadrilateral ABFD; hence the quadri. lateral ABFD: triangle DFC:: 2: 1 as was required. The calculation is easy, for since the three sides of the triangle 10 74 ANALYTICAL GEOMETRY. ABC are given, the angle C is easily found. Also, since DB is parallel to AE; we have CD: AD:: CB EB (Legendre 196.) Hence EB is found.. EF -— 3EC is found also, or FC:-EC is found; then in the triangle DFC we have the two sides CD and CF and their contained angle C, whence BF is easily found by (trig. p. 120,) also the angle FDC becomes known, whence DF is found both in length and direction. REMARK. If AB is greater than DC the same construction will hold provided the point F does not fall further from C than the point B; but should F lie beyond B, then draw the division line from D to AB in a similar manner to what has been done above, and the position and length of the fence become known as before. It may also be observed that the above process will serve if two parts of the piece are in the general proportion of m: n, where m and n are any two given numbers. PROBLEM XCVIII. Having given the sides, a = 6, b =) 4, c = 5, and d = 3, of a trapezilm, inscribed in a circle, to determine the diameter of the circle.Let ACE denote the circle, ABDC A the inscribed trapezium; draw the, diameter AE, join EC, EB, the angles ACE, ABE, being each in a semicircle are right angles, (Geon. p. 52.). CE / AE2- AC2, EB =/ AE2 AB2 I, (Geoi p. 35.) Join CB, then AB' CE + AC EB _ AE' C B (Geomn. p. 233) put AB = a, AC b, CD = c, DB F = d, AE = 2r, then the equation becomes,aV 4.-2 —bI+ b5,/i4r2 —Cl 2r ~ CB, in like manner we have, c 1/14r~ —I + d v/l 4r' — c22r CB,.. a 4r2- I -4- b b- a } v 4r-2 d2 - + d / f4r2 _c2 (1), in the same manner we find bv 4r2 — c2 + c, 64r2 —b2 =-d v4r2-c2 t aV4r2 -d2, or b V /4r2_-c2 -a V 4a — i2-dV/I42-a2 —c V I4r2- b2t, (2,). By squaring (1) and reducing we have 2(a2 + b2 -c2 _ d2) r2 (a2b2 -_C2)-c) v 14r2 - d2 X / 4r -2 c2 -- h V I412 b2l x V t4r2 a2a (3), an.l by squaring we have (2), 4b2 b2-22ab 4r2 - c2 X -/ 4r2 - cl2 + 4a2r2 - a2d 2 4r2d- a2d - 2dc V 4r2 - a2 X V 4r 4rc2 b2-c2b2, and redLucig 2 ( a2+b2 c2 ) r2 - ab 4/ 4r —c2 X V 4r2 C dc V 4r2 - &a X -V 4r 2 b2 (4), eli minating / 4r2 —d'2 X V/ 4r2- c2, by (3), & (4), we have 2(ab —cd) X (-2+b2c2 —-()2 (aWb _c9d2) ab = - (2b- C2d) X V 6j-2 t X V 8 _ b4 ANALYTICAL GEOMETRY. 75 or by rejecting the factor ab- cd, 2(a2 + b2- c2-d2)r2-(ab+-cd)ab = — ab+cd)XV/4r2-a2 X V ] 4r2_b2j (5) by squaring (5) 4( a2 + b2-c2- d2 )2r4- 4ab( ab+cd )' ( a2 + b2- c2 - d2)r2 - e,2b(ab + cd)2 - (ab + cd)2 X (4r2'- a2) X (4r2 - b2) = (ab + cd)2 X (16r4-4r2(a22b2) +a2 b2), or 4(a2+b2 -c2- d2)2r2-4ab(ab+cd) (a2 + b2 -_ c2 -d2) (ab + cd)2 X (16r2- 4(ca2 4 + b2) ) Or ( a2- + b2 -c2 -d2)2 -4(ab + c d)2Ir2-= ab(a2 + b2 -c2 — d2) - (ab + cd) X (a2 + b2)X (ab-+cd) - ab(c2 +- ) + cd(a2+b2) l X Iab + cd l or there results 4(ab + cd)2- (a2 + b-c2-2)2- ad - la2b(c+c2)+cd(a2-+b2) X (ab+cd) consequently we have' - ab (C2 + d2) +cd (a2 - b 2) X (ab +- d) O t4 (ab ~ d)2 ( +-a — b2 -- 2 + O 2) e h:*ave rJ- } (6). Again we have 4 X I ab+-cd 2 _- a 2 - c+ -- d2'2 -2 t 2 + b2 + 2abC2 —d2+ 2c] X - C2 d+a' 2ca —-- b + 2ab l = I(a + b)2(c - d)2 X { (c+d)2- (a- b)2f 2 aa-b-b+c - d? X (a+b+d —c) X(a+ c - d-b)X(b+c +d- d c); puta+ b+c+d=2s, then a+b+c-d=2(s- d),a+b+d -c=2(s-c), a+c+d- b -2(s - b), and b + c +- d - a= 2 (s — a), consequently we have 1 (ab + cd) X (ac + b) X (ad + bc) } required. 4- (s - a) X (s - b) X(s-c) s —d) (7) as required. Cor. If d=0 the trapezium becomes a triangle and (7) reduces to 1 abc r — X. ( —- -- = the radius of the circle 4 V((s-a)X(s- b) X (s-c)Xs) which will circumscribe the triangle. " Put the angle CDB = x by (trig. p. 100), a2 + b2 — 2ab. cos. x - BC2 - c2 + d2 —2cd. cos. (180~ - x) - c2 + d2 + 2cd. cos. x a2 + b2 - c2-d2 Hence cos. x = -- The area of DCB = ab. sin. x - 2(ab - cd) CB. DG But BC = 4 (a2 + b - 2ab. cos. x) = m. Hence DG ab. sin. x --— = * By (Euc. vI. C) ab = D X DG (the diameter of the cir- m ab. sin. x 2Th cumscribing circle, ab a sin- X D,.-. D=. =m. cos. c. x. 1n si 8. X PROBLEM XCIX. In a given circle inscribe an equilateral triangle; and within this triangle describe a circle, &c.; then if r = radius of the first circle, find the sum of the areas of all the circles and all the triangles ad infinitum. It is easy to see that the radius of the first circle is twice that of the second, the second twice that of the third, and so on indefinitely,. the radii are r, r, -r, ~r, and so on. Put p=3.14159265 &co. 76 ANALYTICAL GEOMETRY. then the areas of the circles are pr2, ~pr2, 9-lopr2,'14pr, &c., (Geom. p. 291.) hence pr2 +.pr2+ &c. pr2(1 + I + - + &c. ad infinitum) = the sum of the circles required. Put this sum =s, now 1 +~ — + &c. is a decreasing geometrical progression whose ratio of decrease=', hence by the common rule for finding the sum of such a series we have 1 +- &c.- 1 1 1 1 3 -.'. s' 4p-"=the sum of all the circles, as required. It is also evident that the sides of the first triangle are twice those of the second, and so on as before. By putting a-b = c and s =3a in the cor. to the solution of problem 98, there results r=a- V/3 and a =r V/ 3 =the expression for the side of the equilateral triangle inscribed in the circle whose radius=-r; by the common rule for the area of the equilateral triangle we have (a2 / 3) _4 — the area of the first triangle, and (a)2I X 33- that of the second and (~a)2 X V 3 - that of the third, and so on. Let s' denote the sum of these areas and we have s' I +/ 3 C 1 =3. s-16 If a= 10 feet, then s=57.735 square feet very nearly. REMARK. If within any triangle we inscribe another by joining the middle of its sides, and within this second triangle we inscribe another by similar means, and so on, the sum of the triangles so formed together with the first is easily found in an analagous manner to the methods used in the above solution. For let s denote the area of the first triangle, then it is evident that ~s is the area of the second, rs that of the third and so on, let s denote the sum of these areas continued ad infinitum; then s'=s(1 - 4 + -1 + &c.) ad infinitum, =4s —. 3 as required. PROBLEM C. If from any point within an equilateral triangle perpendiculars be drawn to the three sides, their sum is equal to a perpendicular drawn from one of the angles on the opposite side. Required proof. From the point within the triangle draw straight lines to all the angles of the triangle, and they will evidently divide it into three triangles, whose bases are all equal to each other, being each one of the sides of the equilateral triangle. Let a=one of the sides of the equilateral triangle, p=the altitude of the triangle, then (Geom. p. 176.) if A-its area we have A= —ap; also let x, y, z denote the perpendiculars from the point within to the sides of the triangle, then I (ax -4aLy+az) = the sum of the three triangles into which the triangle was divided=A.. la(x + + 2z)l ap, and x + y +- = p as was to be proved. PROBLEM CI. Given the four sides of a quadrilateral inscribed in a circle, to find the diagonals. ANALYTICAL GEOMETRY. 77 Let ABDC (see fig. to prob. 98.) be the inscribed quadrilateral, having AD, CB for its diagonals, put AB=a, AC=b, CD=c, BD -d, CB-=x, AD=y. Now (Geom. p. ) xy=ac+bd, (1), also (Ge:-n. p. ) ab + ed: ad- c:: y: x, (2), or ab + cd: ad+ bc xy:x2, by(1) x=J (ad + bc) X(ac + bd) hence ab + cd hence I b (ab+cd) X (ac+b) ) Y =- - {/ i(a + d + which are the diagonals. PROBLEM CII. If a, b, c, d, be the four sides of a quadrilateral, inscribed in a circle and s = a + b + c + d, it is required to prove that the area -,V(!-s-a) ( s -b) (Q-s-c) ( s-d)I Let ABDC (see fig. to prob. 98.) be the quadrilateral inscribed in the circle ACE, from the angle A draw the perpendiculars AF, AG to the sides CD, BD, respectively, let A denote the area of the trapezium, then 2 (AF'CD) = the area of the triangle ACD and ~(AG'BD) that of the triangle AB D, (Geom. p. 176.) but these triangles make up the trapezium.~. A=-(AF CD + AG' BD.) Now the two angles ACF, ABD when added make two right angles (Geom. p. 130) also ABD + ABG = two right angles (Geom. p. 28.)... the angle ACF=ABG, and ACE being acute ABD is obtuse, and the perpendicular AG falls without the triangle ABD; hence (Geom. p. 191, 192.) AD2 = AC2+ CD2- 2CD. CF, and AD2= AB2 +BD2+ 2BD ~ BG, whence AB2+ BD2 + 2BD BG =- AC2 + CD2 - 2CD CF, or BD' BG- + CD CF = ~(AC2 +- CD2- AB2- B2) (i), put AB a, AC = b, CD _ c, BD=d, and (1) becomes BG X d+-CF X c= -(b2 - + C2 -a2- d2), (2). Now since the angles at F and G are right they are equal (Geom. p. 27.), and since ACF - ABG the triangles ACF, ABG are equiangular, (Geom. p. 74.),. similar and AC: CF:: AB: BG, AC: AB: AF: AG (Geom. p. 202.), or BG - CF * AB.'. A.C CF X & AG AF Xa hence AF CF +-a2-b-ed2 A= (cb~+ad) (3), also (2) givesCF =b +2 - d) -2b ()v b - 2(cb +- ad) AF2 but CF = ~/(AC2 - AF2)= (b2 - AF2), hence V/ (1 b2 ) = b2 —+- a2d2 or AF2 b2l1 { +c 2 = =4(bc + ad)2- (b+ c2-a2- d)2 AF Or4a(bc +- ad)2 b A/ 54(be + ad)2- (b2+ c_2 _ a2 d2)2?/~4(bc2q- ) a), this value when oubstituted 2(be - ad) 7p8 ANALYTICAL GEOMETRY. 4(bc +ad)2- (b2 + C2_ a2 _d2)l in (3) gives aV -- (b cad) (4) Now 4(bc + ad)2 — (b2 -+- c —2 - d2)2- (b2+c2 + 2bc a2 _ - +2ad) (aC2+d2+ 2+ad- b2-c2+2bc)=(f b + c}2_- a- d2) (za d-c ab - C2) + b+c - d b + c +- d- a ae d + b -c' a + d + c- b =(since s= a+ b+c+d)= Is —2a Is-2bl's- 2c' s —2dl, hence (4) becomes by substitution A -V/( (-s-a) (Is — b)( s — c)(s — d) as required. Otherwise. It has been proved (Geore. p. ) that the product of the three sides of any plane triangle = its surface multiplied by twice the diameter of its circumscribed circle; hence (supposing the same notation as in. problems 98, 101, and the present problem,) AC' CD A.D=the area of the triangle ACD X 4r and AB-BD'AD -the area of the triangle ABD X4r or (since the area of the two triangles the area of the trapezium=A), by addition (AC CD + AB BD) X AD (bc + ad) X AD=4Ar, or substituting the value ofAD~= W ( aab+ d) X abc + bd) as found in prob. 101o we have A= / ((ab + cd) X (ac-}-bd) X (adbc)) -4r = (by (7) found in the solution of problem 98.) ( (S —a) (S —b) (S —c) (S —d) ) which agrees with the result found above, for S=Is. Cor. If one of'the sides (as (d) for example = 0,) the trapezium becomes a triangle and the area = x/ iS(S-a) (S - b) (S c) which agrees with the common rule for the area of a triangle when the three sides are given. PROBLEM CIII. Find the side of a square inscribed in a circular segment, which is contained by a chordl and one third part of the whole circumference, Let PAU denote the given circle, apply the radius from A to B, then from B to C, and join AC, B and the are ABC =one third of the circumference (Geomn. p. ). Let EDFG be the inscribed > square having DE parallel to AC, draw the radi- ( ~' us OB perpendicular to AC and it will be perpen- J 0 dicular to DE (Geoer. p. 65.) hence (supposing that OB meets DE in z, and AC in y,) y and z are the middle points of AC, and DE; again because the chord AB= the radius AO and that Ay is perpendicular to BO, BO is bisected in y, (Geore. p. 47) or Oy. OB't2. Join OE and let r=the radius OB and EG=yz=x, then Ez= -x, Oy-= r, Oz =-~r+x, then since by the right angled triangle OzE we have OE2=0z2+Ez2.. by substitution (-r+x)2+4x2= 2 or by reduction {5xq+rx= —r2, or X2+~rxz-r2, which by quadratics we have x= r( 19 —2) -.47178r nearly. If r- = 00, then x=47.178 feet as required. ANALYTICAL GEOMETRY. 79 (104.) Having given the area and its four sides respectively of a trapezium, to determine the length of the greatest diagonal. Put AB =a, BC=.b CD=C, AD =d, angle BAD - x, angle BCD -y, and the double area = e. Then by Young's trigonometry, page 47, Amer. Edition. a2 + d2 - 2ad cos x = BD = b c2 - 2be cos y, therefore 2ad cos x -2be cos y = ca + d2 - b2 - c. Again, double the area of the triangle BAD = ad sin x, and double the area of the triangle BCD = bc sin y; therefore, putting 2r =a'2+ tl2 - b2 -_c2, we have ad sin z + bc sin y = e and ad cos x -bc cos y = r. Hence a2d2 sin2 x = (e - bc sin y)2, and a2d2 cos2x = (r + bc cos y)2, therefore by addition, and remarking that Sin2x2 + cosx2Z 1, and sin 2y + cos 2y 1, ad2 =- (e -bc sin y)2+ (r + b cos y)2 = e2 -2ebc sin y + ri + 2rbc cos y + b2c2, __b%2 - e2 __r' a2d2 b2c2 e2 A or, rcosy - esin y -- bc Whence r cos y - m = e sin y; or; r2 c - 2 cos 2y 2rn cos y +sin y = - e2 cos 2y, therefore (r'- + e2% cos 2y - 2r in cos y = e - na2, and, dividing by r2 + e2 and completing the square, CoSy- _ 2 X e / (r2 + e2-2) - + rm and the diagonal BD — =/ (b2 + c' -- 2be cos y). Or since (be sin y)2 = (e - ad sin x)2 and (bc cos y)2 = (r - ad cos x)2, we have as before Ca2d2 _ b'22 + e2 + -12 e sin x + r cos x — = n, 2ad therefore cos x 2 X r v (r-2 + e2 —n2) rnt7 Ar -J+ e2 and the diagonal BD is then / (a2 + d2- 2ad cos x). Here it may be remarked, that when the value of e is such as to make either in2 or n2 greater than r +- e2, the part under the radical becomes negative, and consequently the problem does not then admit of a solution. Therefore the limit of possibility, or the case in which the area is the greatest possible, will be when mn2 and n2 are each equal to r2 + e2. The values of mn and n are then equal, but have a different sign, and the above expressions for cos y and cos x give rmi r mrn -- r cos - -= 2 I e2 or (r2 _-); cos x = r + / e; therefore cos y is = - cos x, and consequently the one angle is the supplement of the other, and the trapezium is inscribed in a circle. Again a2d2 -_ b22 2 e2 _ r2 a2d2_ b2%2 +t- 62e1'2; because mn = -- n 2a 2bc 2ad therefore by reduction e2=(ad+Ebc)2 —r2, or e= V (ad + bc)2- — 2 Hence, when the four sides are given, if the double area be greater than V lad q+ be)2 -- TrA the problem is impossible. 80 ANALYTICAL GEOMETRY. The common rule for finding the area of a trapezium capable of being inscribed in a circle, when the four sides are known, may be deduced immediately from the above expression for e, by considering that (ad+bc)2 —2 is - (ad + bc +r) (ad +- be -r)= - (2ad + 2bc + a +d2- b2 — c2) (2ad + 21)c - a -- d'2 + b2 c2) 4(a d + cl - b)(a -+ d -- c+ b)(b + c+a- d)(b+ c+ d - a). (105.) A dodecaedron is a solid composed of twelve regular pentagonal pyramids, whose vertices meet in the centre of tile circumscribing sphere, and the bases of the pyramids form the superfices of the dodecaedron. Now suppose a dodecaedron having the side of each pentagon composing the superficies thereof 8 inches, and supposing every two of its composing pyramnids to be hollowed out in the form of the greatest hemnisphere, cylinder, cube, cone, triangular pyramid, and square pyramid: What will the remainder of the dodecaedron weigh after having been hollowed orl scooped out as above described, supposing each cubic foot of the matter of which it is composed to weigh 601b? The dodeceadron weighs, after all the twelve cavities are cut out, 1746.6646016 inches = 1.0108013 feet solidity remaining; which at 601b. a foot, weighs 60.6480771b. (106.) In gauging a spheroidical ale cask, I found the diameter of one head to measure 18.1 inches, that of the other 16, the bung diameter 20, and the distance between the two heads 20.6 inches; also, by the cask lying a little obliquely, I observed that the liquor just rose to, or touched the upper extremities of the two' heads. Having noted these dimensions, I was informed that there were in the cask a ball of iron weighing 601b. another ball of lead weighing 901b. and a cube box, a foot square. What quantity of liquor was in the cask? INCHES. The spheroid's greatest distance from bung to head. 12.054 The lesser distance. 8.547 The content of the cask in ale gallons,.. 20.763 The iron ball equal to cubic inches.. 217.048 The leaden ball....... 219.717 The cask's vacuity.... 117.814 The box emerged....... 1723.000 The sum, cubic inches. 2277.579 which are equal to ale gallons 8.076 which deducted from the whole content leaves 12.687 ale gallons, the true quantity of liquor remaining in the cask. 81 CHAPTER II. CONSTRUCTION OF ALGEBRAICAL EXPRESSIONS. (13.) Having, in the preceding chapter, given several examples of the algebraical method of solving problems of geometry, it will be proper now to show how the algebraical may be converted into geometrical solutions. We shall commence with the construction of rational expressions. The simplest of these are such as denote lines; they are necessarily of one dimension, and are called linear expressions; they may always be reduced to one or other of the forms ab 2 x = -- b + c- d &c. x =-, x = —, in which a, b, c, &c. C C represent lines of known length, or rather they express the number of linear units contained in these known lengths. The construction of the first of these expressions, when put under the form x = a + c +, &c. - (b +- d +, &c.) is obvious. All that is necessary is to draw a line equal to the sum of the lines, a, c, &c and to take from it another, equal to the sum of the lines, b, d, &c the remainder being the line represented by x. ab The construction of x - is reduced to the finding, geometrically, a fourth proportional to the three given lines, c, a, b, for the above expression reduces to the proportion c: a:: b: x. The expression x2 = - requires us to find a third proportional to two given lines, c, a, since c: a:: a': x. (14.) Let us now proceed to more complicated expressions. 2abc 1. Suppose we had to construct the expression X 3d; then, decomposing it into factors, in order to apply the foregoing element2ab c ary constructions, we have x =-3 X-. The first factor represents a fourth proportional to the three lines 3d, 2a, and A; hence, constructing this line and calling it m, the proposed expression beme comes x = -, which represents a fourth proportional to the three known lines, e, m, and c. 11 82 ANALYTICAL GEOMETRY. 2as b2 c 2. Let x = 32a'2g be the expression proposed, then, putting it 2a2 a b b c nder the form, x= 3-' X X - X fX - we shall have first to con2a2 struct the fourth proportional 3d to the three lines, 3d, 2a, and a. Calling the line thus found nm, the proposed expression becomes, mna b b c - -d X;.X - X -, and we have now to construct the fourth prod f f g portional, d-, to the three lines d, r, and a. Calling it in, we have fni b b c -— z X -X- Constructing in like manner the fourth proporf f g m' b mn//b c tional,,and calling it in", x becomes, x = X - and this is I f g constructed, as in the last example, so that the line x will be constructed after finding five fourth proportionals. And it is obvious that in every such expression the construction will require the aid of as many fourth proportionals as are equal to the sum of the exponents of the letters in the denominator. After these examples the construction of such compound expresa2 be e3 f2 g2 n3 sions as x k=2~2 2 2 p2 &c. can present no difficulty. (15.).Before proceeding further, it should be remarked, that every algebraical expression, admitting of geometrical construction, must have its terms all of the same dimension, that is, each term must be either of one dimension, and thus represent a line; or, secondly, each must be of two dimensions, and so represent a surface; or, lastly, each must have three dimensions, and denote a solid. It is plain that if this uniformity of dimension does not belong to all the component terms of an algebraical expression, that such an expression involves a geometrical absurdity, for we can in nowise combine a line with a surface, or a surface with a solid. Nevertheless, it often happens that an expression really admitting of construction does appear under this unsuitable form, but such a result can arise only from the linear unit having been represented in the calculation by the numeral unit, 1, thus causing every term into which it entered as a factor to appear of lower dimensions than the other terms. Whenever, therefore, for convenience of calculation, the linear unit is so represented, the result should be made homogeneous, by introducing it and its powers into the defective terms. Thus, if we happened to have such a result-as ANALYTICAL GEOMETRY. 83 x = ab, then calling 1 the linear unit, we should change it into the homogeneous equation Ix = ab.'. x = —, showing that the line x is a fourth proportional to the lines 1, a, and b. In like manner, if the result were x = abc, we should change it into abc ab c l1x = abc, whence x = -X, which expression we have 12 T X already seen how to construct. 1. Let now the expression to be constructed be a compound frac a3 +-3be — a tion, such as x a —- 3b to admit of geometrical representab + 2e + 3 tion, both numerator and denominator of this fraction must be homogeneous; and to represent a line, x, the terms in the numerator must be one dimension higher than those in the denominator; so that intro. ducing the linear unit, 1, the expression to be constructed must be a + 31bc - 12 a a + 31bc- 1a a' 3be la XZb= b++3that isxl, Z = I b23lh )k+kk lb +- 21c +- 3/ t (b-q- 2e - 31) -' +'k where k is put for the sum of the lines b + 2c + 31; hence the problem is reduced to the construction of simple fractional expressions, such as have been considered in art. (14.) 2. As another example of this kind, let there be proposed the ex2a3 — 3a2 b-+ b2c pression x = 2 -2ab - b 2 This may be constructed as the preceding, if we can represent the denominator as a single product, and this we may do, by putting b2 = va, or by determining v so that b2 2a3 3a2 b +- b2 c v = -, for then the expression becomes x = a a(a- 2b + v) 2a2 3ab yc k -3-b -vk where k is put for the line a -2b - +v. 3. Again, let x = abd gh be taken. To reduce the denomlimnp + qrs nator to a single product, put qr = vm.~ v == is a known line, and the denominator becomes m (np + vs), it remains then to reduce np -+ vs to a single product. Put, then, vs = wn.. w = - is a known line, and the proposed expression becomes finally abed + efgh abed efgh m~ = = - = ra+ where k is put for the p +- I. mn (p + w) mnrk mnk (16.) We now proceed to consider irrational expressions. These mayalways be reduced to one or other of the following simple forms, viz, x =/ ab, x = V/ aa2 + b2, x-= V i a2 - b21, we shall, therefore, begin by constructing these elementary expressions. E2 84 ANALYTICAL GEOMETRY. From the first, x = /ab, we deduce x2- ab a.. ax: x::: ther&e fore x is determined by finding, geometrically, a mean proportional between the given lines, a, b, (Geom. p. 136.) From the second expression, x = a2 + b2 t, we have x2= a2 - b2, so that x is the hypothenuse of a right angled triangle, of which the sides are a and b. (Geom. p. 58.) The last expression, x = -/ a2- b2t, when put under the form x = v/ (a + b) (a - b) t, represents a mean proportional between the two lines a + b and a - b. (1.) As a first example, letx -= V/l a2 - b2 + ca - d2 + e2 - &c. I be proposed. Put m = Va2 -- b2, and construct this line; then in2 = a2 - b2, and x = em2 + c2 d2. + e2 _ &c. t Put now n = v/t m2 + c2t, and construct the line n; then, since n2=r - mn2+ - 2, and x =i/ -n2 d2 + e2 - &c. This series of constructions being continued we shall at length have but two squares under the radical, and the construction of this last expression will be the line sought. In the same way may any numerical surd be accurately represented by a line first assuming some fixed length for unity, for any number may be decomposed into square numbers, thus V7/ V2222 - 1 2; 2 1 1 = v 32+-1 + I and /13 - / v32 + 22 I; V43= - 62 + 32 1 1 2. Let the expression to be constructed be x - 1/ a + 3bc. Put 3be = V2.'. v = V'3bc is a known line, and the expression is reduced to x = v I a2 + v2. Or the same expression may be constructed by putting 3bec = au.'. x = 4/ a (a + -u)}.'. x is a mean proportional between a and a + u. ab2 b2 3. Let x ~+ — deI; then, putting - = m and de = an, we ~~~~C C have x = x!a (m —- ) }.-. x is a mean between a and n — n. 4. As a last example, let x then, put ting it under the form x = / { ( — b + we can first con a-b a3 b2 struct the line - 2c + 3b, then 42 and, calling these in and n, 42 a-b' we shall have, lastly, to construct x - Vnmn. We shall leave the student to point out the-constructions of the following expressions, viz. = a~+ Ja4a +bc% x_ = ac -fg + eh + mn. i a' bxc + -.}a- - U3d+2 d3F 85 SECTION II. CHAPTER I. ON THE POINT AND THE STRAIGHT LINE. (1.) In the preceding section we have endeavoured to show the use of algebra, when combined with geometry, in the solution of problems. in the remaining part of the present treatise we shall proceed in a manner more strictly analytical, dispensing with the truths of geometry, except a few of the simplest kind, and depending upon analytical expressions, as well for the establishment of theorems, as for the solution of problems; that is, as well for the determination of the form and properties, as of the magnitude of geometrical quantity. It is this extended application of the principles of analysis that, strictly speaking, constitutes the science of ANALYTICAL GEOMETRY. On the Equation of a Point. (2.) Let AX, AY, be two assumed straight lines, intersecting, in any angle, at A; and let P be a point in the same plane, whose position it is required to deternine relatively to these assumed lines. Let the lines PC, PB, be drawn respectively y parallel to the lines AX, AY; then, if the lengths R of the former be known, it is obvious that the position of the proposed point will be easily determined. It will be situated at the intersection of C two lines, CQ, BR, drawn, the one parallel to AX, from a point, C, in AY, the distance of which XI B X from A is the given length, BP, and the other parallel to AY, from a point, B, in AX, the distance of which from A is the given length, CP. The two lines AX, AY, in reference to which the position of the point is to be determined, are called axes, and their point of intersection, A, is called their origin. The distance, AB, is denominated the abscissa of the point; P, and BP, or its equal, AC, is called the ordinate of the same point; hence the axis AX is distinguished from the axis AY by the name axis of abscissas, the latter being called axis of ordinates. The abscissa and ordinate of a point, when spoken of together, are, for the sake of brevity, called the coordinates of the point, and, for a like reason, the two axes are referred to as axes ofcoordinates. An abscissa is generally denoted by the letter x, and an ordinate by the letter y; and often, for shortness, the axis of abscissas is called the axis of, and the axis of ordinates the axis of y. 86 ANALYTICAL GEOMETRY. We have just seen that a point becomes determinable when its coordinates, x and y, are known; it follows, therefore, that, in order to this determination, we need only have the two equations x=a, y=b, in which a and b are given. These equations are, therefore, called the equations of a point. (3.) It is of importance to remark, that not only the absolute values of a and b must be given, in order to fix the position of a point, but also the signs of these quantities. If the axes are produced through the origin to Xi and Y/',it is obvious that the abscissas reckoned in the direction AX' ought not to have the same sign as those taken in the opposite direction, AX, nor should the ordinates taken in the direction AY' have the same sign as those taken in the opposite direction, AY; for, if there were no distinction in this respect, the position of a point, as determined by its equations, would be ambiguous. Thus the equations of the point P would equally' belong to the points P', P", P'", provided the absolute lengths, of the coordinates of each were respectively equal to those of P. All / P this ambiguity is, however, avoided, by denoting the axes in one direction +) and in A the opposite direction —. Hence then, regarding the abscissas to the right of the origin, A, as positive, those to the left will be negative. In like manner, considering the ordinates above the origin as positive, those below it will be negative. We thus have for the point P the equations x=a, y=b, for the point P', x - a, y= b,' for the point P x- =-a, y =- b and for the point P"', x = a, y =- b. If the point be situated on the axe AX, the equation y = b becomes y = o, so that the equations x = a, y-= o, characterise a point on the axis of abscissas, at the distance a from the origin. If the point be on the axe AY, then x ='a becomes x = o, hence the equations x - o, y = b, characterise a point on the axis of ordinates, at the distance, b, from the origin. And lastly, if the point be common to both axes, that is, if it be at the origin, its position will be expressed by the equations. X=(2, y=o. A point is said to be given when its coordinates are given, and, instead of referring to it as the point whose coordinates are x y, it is more briefly and more usually designated as the point (x, y.) On the Equation of the Straight Line. (4.) Let it now be required to determine the equation of a straight ANALYTICAL GEOMETRY. 87 line, or, m other words, to find an analytical expression by which it may be characterized. Let MN be any straight line, and, in the same plane with it, let two straight lines AX, AY, be taken for axes of coordinates, and, for greater simplicity, let their origin, A, be upon the proposed line. From any two points, D, B, in AB, let DC, BE, parallel to AY, be drawn, then will AC, C D, be the coordinates of the point D, and AE, EB, the coordinates of the point B, and they form this proportion, viz. AC: CD:: AE: EB CD EB A AC-AE hence each abscissas is to its ordinate in the same constant ratio for every point taken in the proposed line. If, therefore, we represent this ratio by a, which will obviously be an abstract number, there will always exist this relation between any ordinate and its abscissa, viz. the former equal to a times the latter: thus CD=a AC, BE-a AE, &c. that is, agreeably to the notation already established, we shall have for every point in the line MN the relation y=ax.... (,) intimatin, that Nwhatev6r abscissa we take, a times that abscissa will be the value of the corresponding ordinate. Knowing, therefore, the constant a, we may, by giving arbitrary values to x, determine as many points in the line as we please, and, consequently, the line itself; equation 1 is hence called the equation of the straight line, MN. (5.) If the axes of coordinates. are rectangular, then the ratio CD CD or a expresses the tangent of the angle DAC, which the proposed AC line makes with the axis of abscissas. If this angle is obtuse, the tangent will be negative; hence, if AN (next fig.) be the position of the proposed line, in reference to the rectangular axes, AX, AY, the equation is y-ax.... (2.) where it must be observed that the sign - applies only to the number a, and not to x, the abscissa, for the sign of this depends upon its direction from A, the abscissas of every point in the proposed line, which is below AX, being positive, while for every point in the portion above the abscissa is negative.'Thus the abscissa AC, of the point P, is positive, for it is to the right of A; but N Y the abscissa, AC', of the point P" being on the op- P' posite direction, is negative., As regards the ordinates, it is plain, both from equation (2), and from C' C the diagram, that to positive abscissas belong nega- A tive ordinates, and to negative abscissas, positive ordinates. P (88 ANALYTICAL GEOMETRY. (6.) When the axes of reference are oblique, the coefficient a may still be represented by trigonometrical quantities, C) or aistesm sin. DAC for or aisthesameas; therefore, since the angle AC' sin. ADC ADC is equal to the angle YAD, if we represent the, angle NAX, which the proposed line makes with the axis of x by a, and the inclina.tion, YAX, of the axes themselves by 3, we shall have sin. DAC sin. o sin. ADC sin. (13-a) sinl. a and for the equation of the line AN, y x= __ll x sin. (13-a) In this equation the coefficient of x will obvious- NY X ly be negative, when a 7 1, that is, when AN takes the position in the annexed diagram to the left of the axis of y. We see, therefore, that, whether the axes be rectangular or oblique, the coefficient of x, in the equation of a straight line pass- A ing through their origin, will be positive, if the' portion of this line situated above the axis of x lie to the right of the axis of y, but the same coefficient will be negative, if it lie to the left. (7.) The equation to the straight line, which has just been exhibited, applies only when the line passes through their origin. Let us now suppose that this restriction is removed, and that the proposed line takes the position LM, cutting the axes in Cand B. N Let AN be parallel to LM, and from any point, P, in the latter, let the ordinate, PDE, be drawn.,ID Then, since AB=-DP, it follows that any ordin- c ate, PE, is equal to AB, plus the ordinate ED, /.7 /A E X of that point, D, in AN, which has the same ab- / scissa, AE, as the point P. Now this latter ordinate is always expressed by the equation y =ax, as we have already seen; consequently, if we put b for AB, the ordinate of the proposed line at the origin, we shall have for every point. in LM this relation between the coordinates, viz. y = ax + b, this, therefore, is the equation of the straight line in general. With regard to the sign of a, its changes have already been examined; and as to the sign of the ordinate b, we know that it will be positive so long as LM cuts AY above the origin, and negative awhen the intersection is below it. It may, however, be satisfactory to the student to have here exhibited the form of the equation for every possible position of the proposed line. NALYTICAL GEOMETRY. 89 1. Let the line, LM take the position shown in 7/M the annexed diagram, cutting the axis of x to left of the origin, and the axis of y above it, then a and h axe"both positive, and the equation is y=+ax+b. A 2. Next, let the proposed line cut the axes on the opposite sides of the origin, as here represented, then a will still be positive, but b will'be negative; the equation, therefore, in this position of the x/ line is A y= +.ax- b. LB 3. Thirdly, let the line cut the axis of y above, y and the axis of xato the right of the origin, then a becomes negative, and b positive, in this case, there- B fore the equation is y= -ax +- b. C L 4. Lastly, let the axis of y be cut below, and the Y axis of x to the left of the origin, then both a and b f\ will be negative, so that the equation becomes y --- ax - b. It thus appears that when both axes are intersected, the proposed line may take four different positions analytically represented by four distinct equations. There remain two other positions to be considered, viz. those in which the line is parallel to one of the axes. If it be parallel to the axis of abscissas, as it cannot then form an angle Y therewith c = 0 and therefore (6) a- 0, so that this position is characterized by the equation LI M y = Ox - b, or y =+b, intimating that, whatever abscissa be taken, the A X value of the ordinate remains the same..L When the line is parallel to the axis of ordinates, substitute for a, b in the general equation, its equal-, putting'c for X AC, -the distance of the intersection, with the axis of x from the origin, b representing, as usual, the -r distance of the intersection with the axis of y, in the I A C present case infinite, we thus have 12;H 90 ANALYTICAL GEOMETRY. b 1 b y =-Zxq -i b,' x=- t Y C C which, since b is infinite, is the same as x = 0 y F c, or x = T c, an equation which indicates that whatever be the ordinate, the abscissas are constantly equal to c. From, the preceding discussion it follows, that the general equation y = ax + b comprehends in it all those that can characterize straight lines, whatever be their position in reference to two assumed axes, any how inclined to each other, each particular position being denoted by the particular values given to a and b. As these quantities remain the same, while the coordinates x and y vary in value for every point in the same line, the former are called constants, and the latter variables. Since two constants enter into the general equation of a straight line, as many particular values as are given to them, so many particular lines will be represented. They may, therefore, take such values as will render the line whose equation is'expressed, subject to certain proposed conditions, provided such conditions are possible. Thus we may suppose such values given to the constants, that the line represented must of necessity pass through two given points, or only one of the constants may be fixed, and of such value, that all the lines represented by the equation shall'pass through a proposed point. We have already seen, that, if the origin of the axes were the proposed point, the constant, b, must take the particular value b =- 0 the equation y - ax representing all lines subject to this condition. Let us now proceed to a few determinations of this kind, and, as we are at liberty to assume any angle of inclination for the axes of reference, we shall, in general, for greater simplicity; assume them rectangular, excepting only in a few cases, to be hereafter pointed out, where oblique axes may be more advantageously employed. PROBLEM I. (8.) To find the equation of a straight line passing through a given point. Let us denote the coordinates of the given point by x' and y'; then, since the general equation for every point in the required line is y — ax b.... (1), it follows that for the particular point in question we must have the relation y' = ax' + b.~. b = y' - ct'... (2), hence, substituting this value of b in (1,) we have y- y'- a ((x-x- ), or y a= (x - +x) + f', which is the equation sought, and characterizes every straight line that can be drawn through the point (x', y'). By comparing this wittl ANALYTICAL GEOMETRY. 91 the general equation of the straight line, we find that the ordinate at the origin is y - ax'. If the given point were on the axis of x, then y' =0, and the equation would be y = a (x-x'). If it were on the axis of y, then x' = 0 arind. the equation would become y -y' - ax, or y = ax + y'. PROBLEM II. (9.) To find the equation to the straight line, which passes through two given points. Representing the given points by (x,' y',) and (x", y",) we have to subject the equation y -y = a (x - x') of a line passing through one of the points to the additional condition y' —y = a (x'- x"), which equation as x', x", y', y", are all given, determines for a the particular value a = Y —,,; substituting, therefore, this value of a in the former equation, the analytical representation of the required line is y y -" -,,(X- X/( ), in which equation all are constants, except x and y, the variable abscissa and ordinate of the line. y, _ yh,,/y// - y,/1 By writing the equation thus, y =-_ +- _, which however, is- less simple than the former, its identity with the equation y = ax +- b, for the particular case in question, is more distinctly seen, as it shows at once the value of the tangent a, and of the ordinate b at the origin. If, for instance, the coordinates of one of the given points (x', y',) be 4 and 6, and those of the other point (x"/, y//,) 3 and 5, then the equation of the line passing through them is y l- x + 2; therefore 1 being the trigonometrical tangent of the angle made by the line with the axis of x, this angle must be 45~, and the ordinate at the origin is 2. If (x', y',) is on the axis of x, y' = 0, and the equation is IfY ( X - X')f If it is on the axis of y, then =' 0, and the equation is y, - y/. Y -y = x" x. And lastly, if it be on both axes, that is, at the origin, then x' = 0, y' = 0, and the equation becomes y = Y/x. PROBLEM III. (10.) To find the equation of the straight line which passes through a given point, and is also parallel to a given straight line. aepresenting as before the given point by (x', y'), we have for every 92 ANALYTICAL GEOMETRY. line passing through it the general equation y - y' = a (x - x'), and among these lines we are required to distinguish that which is parallel to the given line, or, in other words, that which makes a given angle with the axis of abscissas; putting a' for the tangent of this angle, the equation of the line sought will be y - y' = a' (x - x'). PROBLEM -IV. To determine the point where two given straight lines intersect. Referring both lines to the same axis; let their equations be y = ax + b, y = a' +- b,' then, since at then, since atpoint of intersection the ordinate is the same for both lines, we must have for this particular point ax + b = a'x + b', whence we obtain for the coordinates of the interb'- b ab/ - a'b section __ a/ and y = ca a ca _a' If we suppose a = a', the expressions for z and y become infinite, as they evidently ought to do; since the lines, being in that case parallel, can meet only at an infinite distance. a --- a/ If b=b', then x=O, 0 and y = a =b b, showing that the ordia' nate at the origin belongs to the point of intersection. PROBLEM V. (11.) To find the expression for the angle of intersection of two given straight lines. Let the two lines be A'B and CD, P being their D 13 point of intersection; then it is required to find an ex- Y // pression for the angle A/PC. 3,/ Let the equations of A/B and CD be respectively, y= ax +b, and yz = za'x +; then a will be the tangent of the angle PA'X, and a' the tangent of the angle PCX, the lines being referred to the rectangular AkAC X axes, AX, AY. Now the angle A'PC is equal to the difference of the angles PCX, PA'X; hence, calling'the tangent of this difference a -a a/ -- a v, we have (Trig. p. v 1= - a, accordingly as the angle is to the' left or right of A'P. If the angle of intersection be a right angle, its tangent must be infinite, which the expression for v becomes, only when 1 -+ aa/is 0;I this condition, therefore, is necessary, in order that the proposed lines may be perpendicular to each other; so that, in this case, we must have aa' =- 1, or a -- a/' * Or, since the cotangent of a right angle is 0, and that the cotangent of an angle I 1 -f aa0' is the reciprocal of the tangent, we must have- or —, -- 0.'. 1 + aa' 0. ANALYTICAL GEOMETRY. 93 It follows from this, that, if y = a'x - b' be the equation of a given straight line, then will y = --- x-+ b be the equation of a line perpendicular to it. These perpendiculars may be innumerable. If we fix one of them by the condition that it may pass through a given point, (x', y',) then (Prob. 1) it will be characterized by the equation Y — — a (x —x'). When (x', y',) is on the axis of abscissas, y' 0, and the equation is y= —,-(x-X'). When it is on the axis of ordinates, then x = 0, and, therefore, the equation is y -y' = — -- x. And when the given point is at the origin x' and y','being both 0, the equation is, y = -- x (12.) If we, wish for the sine or cosine of the angle of inclination of two lines, instead of the tangent, they may be obtained thus. By trigonometry sin. (a — a') - sin. cos. a' - sin. a'/ cos. a, in which formula, if cc, a', represent the angles, whose tangents are a, a', we a 1 shallhave, sin. a -- -- --, cos. 1 a' 1 sin.' - 2 cos. a/- hence by substitution, a-a'lCs 1 +a/2 a -- a sin. (o — a') /(1 +a'2) (1 + d) In like manner, from the expression cos. (a -a') cos. a cos. ao' - sin. a' sin. a, we get, cos. (t a/) = aa PROBLEM VI. (13.) To find the equation of the straight line which passes through a given point, and which makes a given angle with a given straight line. Let the given line be represented by the equation, y = ax + b; then, because the required line passes through a given point, (x', y',) its equation will take the form y — y/ a' (x —x'); and we have to determine a', so that these lines may intersect in a given angle. Put v for the tangent, of this angle, then (art. 11) ~(a —d), a- v aor +a' 1+av- 1- av H2 94 ANALYTICAL GEOMETRY. hence the equation of the line sought is Y * Y =- a-V (x —')j or y-y- a+ v (x x'). 1 + av - av Two lines, therefore, may be drawn through the given point, fulfill, nfgl the required condition; the one forming the proposed angle, to the left, and the other forming it to the right, of the given line, conformably to the two expressions for its tangent, v. T'hus, in the annexed diagrams, the equa,.' ~' tion of PC forming the given angle to the left of the given line, AB, is a -v y- y'1a ( ); P / \. 1 + av but the equation of PC', forming an equal angle to the right, is y —y= 1 av (x-') A PROBLEM VII. (14.) To find the analytical expression for the distance of two given points. Let the given points be M, (x', y',) and N, (x", y/.) Y Draw Mp parallel to AX; then the axes being rec- N tangular, the distance, MN, will be 4 lMp2+Np2; but Mp = x" — x/, and Np = y" - y', therefore the M, expression for the distance, D, is D -- /(x"-/-)2 + (Y/ — y/)2 A X If one of the points, as (x', y), is at the origin, then x' 0,'and y' = 0; therefore, D = v xt//2 + y//2 The expression for D would have been much less simple, if we had chosen oblique axes of coordinates; for, if the angle MpN had been oblique, we should have had (Trig. p.,)* MN = V Mp2 + Np2 — 2Mp - rp cos. Mp N; or, putting A for M'pN we have cos. MpN = -cos. A, D = v I (x"- /x')2 - (y"- y)2+ 2(x"- X') (y/ - y') cos. A If (x', y',) is at the origin, D =, x"2 + y//2 + 2x" y" cos. A. PROBLEM VIII. (15.) To find the analytical expression for the distance of a point from a line. Let N be the point, and BC the line, then it is re- y quired to find the length of the perpendicular NM. l N If the coordinates of M be represented by x, y, and those of N by x/, y', we have by the preceding problem MN = V- (x'- x) + (y' - y)2 in which the - ~ values of x' - x and y/ - y, must be determined by - means of the equations to the lines BC, MN. * The Trigonometry usually referred to, is that of Dr. Gregory. ANALYTICAL GEOMETRY. 95 Taking for the equation of the former y = ax + b... (1,) that of the latter will be (art. 11,) y- y' = - _ (- x')... (2.) a In order to obtain from these equations the values sought, in the simplest manner, put equation (1) under this form: Y y- y'=a(x — X) —y')- + ax' + b.... (3.) by subtracting y' from one side, and its equal, y' + ax' — ax', from the other, in order that the equations (2) and (3) may both contain the unknowns x - x' and y - y', then the determination of the values of these becomes easy. Subtract (2) from (3) and the result is 1 o0 = (a +-) (x- x') - yl + ax' + b, (y azx/ - b). x --- XI a2+1 and by substitution in equation (3) ax' - b These expressions for x — x' and y y', by changing the signs prefixed to them, represent x'- x and y- y; but, as these latter enter into the expression for MN only in the second power, it is indifferent what signs are prefixed; we have, therefore, by substitution, V(a2 + 1) (y'-ax' -b)2 y- ax'- b (a2+ 1) V (a - 1) If the point N be at the origin, then x' = 0 and y' = 0; therefore b MN = / (a2 +1) expresses the distance of the proposed line from the origin. If the proposed line pass through the origin, then b = 0, and the -/ a/ value of the perpendicular upon it from (x', y',) is P = Y - a+ V(a2 ~1) (16.) By means of this last expression, we can immediately arrive independently of any trigonometrical property, at the values of the sine and cosine of the angle formed by the intersection of two given straight lines. For let AM, AM', through the origin, A, be pa- Y rallel to any two given intersecting straight lines, thus forming an angle at A, equal to the angle of intersection. Take any point, MR, on one of the lines, and, representing its coordinates by (x', y',)/ _ the value of the perpendicular from this point to theAg"other line will be / - ax other line will be M'M —, a being the tangenttof the angle MAX. Now, if we represent the tangent of the angle M'/AX by a, 96 ANALYTICAL GEOMETRY. we shall have at the point M/, y', = a'x/; hence, by substitution a/x'- ax/ V/(a2+1) Now M'M is the sine, and AM the cosine, of the. angle M/AM te radius AM/, whose length is expressed by the equation AM'/ _V A/2- + -y2 = v /2x/2 + ax2'2}=~' 1 + a/'2; hence, calling this radius R, we have x' ( +V (I +-a 2).~ M/M R (a/ — a) ~'.RM/MR(a' —a) = sine LZ A; and, subtracting the square of this from R2, and then taking the square root, we get AM R (aa/ -+- 1) = cosine /A;,/ I(a2 + l) ~(a/2 + 1)+ and these expressions are identical with those given at (12), where the radius, R, of the tables is unity. The preceding problems contain every useful particular relative to the straight line. They should be attentively studied by the student, who before closing this chapter, should be fully prepared to state the conditions under which any straight line is drawn, when its equation is given, or on the contrary, when the conditions are given to write the equation. We shall now devote a: short chapter to the solution of a few problemsj in which the principles here established will find their application. CHAPTER II. PROBLEMS IN WHICH THE EQUATION OF THE STRAIGHT LINE IS EMPLOYED. PROBLEM I. (17.) It is required to determine whether the perpendiculars drawn fiom the vertices to the opposite sides of a triangle meet in a point. Let the perpendiculars, AF, BE, CD, from the' vertices to the opposite sides of the triangle, be /k drawn and assume AB, AY, for rectangular axes / of reference. Let (x/, y') represent the given point, C, and for AB put c. Then the equation of AC, passing through the DA. — iB origin and the given point, C, is y = x; and the equation of BC ANALYTICAL GEOMETRY. 97 passing through the two given points, B,C, the former (c, 0) being on the axes of x, is y - (x _ c). X/ C Now BE, AF, beinlg respectively perpendicular to AC, BC, and passing each through a given point on the axes of x; their equations — l~ 3~ ~;/ — C are, of BE, y = —-7(-c); and of AF, y -,/ x. At the point where these intersect, the ordinates must be equal, so that at this point (x — c)= --- x, whence, x= x'; that is, x, the abscissa of the intersection of BE, AF, is equal to x/, the abscissa of the point -C; hence, the perpendicular, CD, passes through that intersection. PROBLEM II. (18.) It is required to determine whether perpendiculars from the middle of each side of a triangle meet in a point. Let M, M', M"', mark the middle points of the - sides of the triangle ABC. Let P be the point C where two of the perpendiculars,- MP, M'P, meet, and, as before, take the rectangular axes, AB, AY. M' -p Represent the point C by (x', y'), and the base, AB, by c; then the point M' will be (Q-c, 0), and the A points M", M//, having their ordinates parallel to the m ordinate of C, will obviously be ('- x', - y'), and ( - c -+ 2-x', l y) respectively, for the triangles AM" in//', M/M will be equal. Now the equation of AC, passing through the origin and the point, (x', y',) is y = x, and that of BC, through the points (c, 0) and ()','.,) is - Y (x —). Now PM", PM, being respectively perpendicular to these, a, -it the same time, passing the one through the given point M",'/ and theother through the given point M, we have for the equation of PM"/, X.. Y -Y -9 (x — x'/) and for the equation of PM, Y/ CY = X Vc +x )i. Now, if these tw6 lines meet in the perpendicular from M/, they must necessarily have a common ordinate for the abscissa, A M'/, or -c, otherwise the ordinates will be different. Substituting this value /2.M// X -cx of x, in the equation, for PM//, we find y'- and, 13 2y 9o- ANALYTICAL GEOMETRY. making the same substitution in the equation for PM, there results X/2 x/ Y Is +- 2y The two ordinates are, therefore, identical, and thus the three perpendiculars meet in a point. PROBLEM III. (19.) It is required to determine whether the straight-lines drawn from the vertices of a triangle to bisect the opposite sides, meet in a point. Let the lines, CM, BM', AM//, bisect the opposite sides of the triangle, CAB; and, let us, in this case, C employ the oblique axes, CM, MB. Since AM'/ is half AC, a parallel to CM, from the point M/, will Mgl /,' bisect AM, and be equal to half CM. In like manner, a parallel to CM, from the point M", will bisectA MB, and be also equal to half CM,-so that the coor- M dinates of the points M', M//, are numerically equal. Now, if (x/, 0) represent the point B, (-x/', 0) will represent the point A; and, if (z//, y"/) denote the point M"/, (-x'/, y/) will denote the point M'; hence the'equation of kAM", passing through the points (-x/, 0) and (x//, y//), is -- (x + /); and the equation of BM/, passing through the points (x', 0,) and (-x/, y//,) is, (X — X/) -x X Now, in order that these lines may intersect on the axis, MC, the ordinates of both at the origin must be the same, and this they evidently are, for the ordinate corresponding to x - 0 is, in both equations, y = x' = PM. Cor. Since x"/= -x, and y" -= CM, it follows that iCM PM =2"'= x CM. x 3 PROBLEM IV. (20.) To determine whether the lines bisecting the three angles of a triangle meet in a point. Let AF, CD, BE, bisect the three angles of the triangle CAB; and, as before, let the oblique lines, C CD, DB, be taken for axes of reference; then — DA will be the abscissa of the point A, and DB will be E the abscissa of the point B; consequently the equation of AF will be y- _ a (x - AD)...... (1.)-; and the equation of BE, y = a' (x- DB).. (2). ANALYTICAL GEOMETRY. 99 Now it appears from (art. 6,) that, as the axes ale oblique, the sin. FAB sin. 1A sin. -A value, of a will be 2n D value of a (CDB FAB) - sin.' (A + C) - cos. B sin. I B' In like manner, the valueof a' will be a' 2 hence, when cos. 2A x=0, the equations (1) and (2) become, by- these substitutions, sin. IA sin. B Y cos. B AD. (3), and y = A DB... (4); and it now Cos. I B cos. IA remains to inquire whether these two expressions for y are identical. sin, B For AD, in equation (3,) substitute its equal, viz. AD = A DB, sin. -A sin. A. sin. B and the equation becomes, = 2 DB, which is identical cos. 1B. sin. A with equation (4.) (Trig. p. )p Hence the three lines meet in a point. PROBLEM V. (21.) To express the area of a triangle in terms of the coordinates of two of its angular points. Let the triangle, BAC, be proposed; and let the rectangular axes originate at the point A; let | X i x', y', be the coordinates of B, and x"//, y" the co- ordinates of C; then we have for the equation of BC, passing through both these points, y -ye xy"-y//' " XX x/ Y-X X/ _ and for the length of the perpendicular AD, upon this line, from the -b origin, we have (art. 15), the expression P = - substituting here for a and b, their values as exhibited in the foregoing y' -" / — y/-x// equation, viz. a =-Y'bb -, and the expression for P becomes, after reduction, P -'(x-a - x/"/) - x")2+(y - Y Now the denominator of this fraction expresses the distance of the point (x", y"), from the point (x, y',) that is, it denotes the line BC; so that BC X P = - ( x'y' y'" ); consequently k PBCI X PI - - x'y" - y' x" = area of the triangle. The foregoing exercises on the equation of the straight line may suffice for the present; further applications will repeatedly occur in the succeeding chapters. sin. B sin. A'For, at art. 18, p. 43, we have sin. ~ and cos. AA s. 2 cos. JB 2 sin. 4A 100 ANALYTICAL GEOMETRY. (22.) At the commencement of the present section it was shown that a straight line, whatever its position, might be represented by a simple,indeterminate equation; and it will be proper, before we proceed to the circle, to show, conversely, that every simple indeterminate equation containing two variables is the analytical representation of some straight line. Let lMy - Nx + P be any simple indeterminate equation, contain: ing the two variable quantities, x and y, then we have N P N P y: x —+ -, or putting for si;mplicity A for- and. B fory=Ax+B.... (1) Now, draw any two straight lines, X'AX' XAY, intersecting at A, and make AB = B, AC-=-A" and through the points C, B, draw the straight line CBL, which will be the geometrical representation of the proposed equation. For the equation of this line, in reference to-the axes, AX, BA BA AY, is y = — C x AB. (2); but by the construction AC - B B 4- A- A, also AB = B, therefore the equations (1) and (2) are identical, each, therefore, is the analytical representation of the line CBL. (23.) The line which any equation represents, or in which the variable point (x, y) is always found, is called the locus of that equation, or of the point (x, y). Hence the locus of a simple equation containing two variables is a straight line. When the equation is given, and it is required to construct the locus, it will be sufficient to determine two points in it, since the locus will be the straight line passing through them. Now the two points most easily found are those where the locus intersects the axes. The abscissa of the one point will be furnished by the proposed equation, by making therein y = 0, and the ordinate of the other, by making x = 0. Let, for example, the locus of the equation 2y = 3x 5 be required. Making y = 0, there results for x the value x -; Y1 and, making x = 0, we have y = — ~; therefore hav- L ing assumed. the axes AX, AY, on the former take AC = -; and, on the latter, take AB = 5; then the straight line, BCL, drawn through the points B, C, willN be the locus sought.. This method of determining the locus can, however, / be applied only when the equation is of the form y = ax + b; for, if ANALYTICAL GEOMETRY. 101 it were of the form y = ax, then b being 0, the locus would pass through the origin, so that its intersection with the axes would fuirnish but one point: another, therefore, must be found, before we can determine the line; for this purpose, we may give to x any particular value, and this, with the resulting value of y, will be the coordinates of another p6int. It is-obvious that of every point in the locus of the indeterminate equation y' ax + b, the coordinates exhibit a geometrical solution; and. as an infinite number of these points may be taken, the locus supplies all the infinite solutions of the equation. If, therefore, any other indeterminate equation were susceptible of solutions that belong also to the former equation, and if the loci of both equations were to be constructed on the same axes, these common solutions would be geometrically represented by so many points being common to both, loci. CHAPTER III. ON THE CIRCLE. (24.) Let r represent the radius of a circle, the centre of which is 0. In the same plane as the cicrle, assume any rectangular axes, AX, AY; and let it be required to determine the equation of its circumference, or the analytical representation thereof, in reference to the assumed axes. Let the coordinates AB, OB, of the centre, be repre- Y { sented by a, /3; while the coordinates of any point, P, in the circumference, are denoted by the variables xz, 0 m y; then, drawing the radius, OP, and Omr, parallel to the axis of x, we shall have Om = x - a, and Pmn = y- /; consequently, since Om2 + Pm2 OP2, we A -B - have ( — a)2 + (y -- )2 = or2 - 2a- + y22 2- y +82 = r2... (1), which is the equation sought, and obviously subsists for every point, P or (x, y,) taken in the curve. For brevity, the equation is usually called the equation of the circle, the circumference, however, and not the enclosed surface, is to be understood. 102 ANALYTICAL GEOMETRY. If the origin of the axes be assumed on the cir-y; p cumference, as at A, in the) annexed diagram, the equation will be more simple in form; for then,. if AO be drawn, we shall have AB2 -+ B02 A= 2, that is' a2 +/C2 = r2; so that equation (1) reducesA — x to x2- 2ax +y2 21y= 0, or x2 + y2 2 (ax + -ly) = 0... (2), the equation of the circle, when the origin is on the circumference. If,-in this case, the axis of x pass through the centre, then a r, and j = 0, and equation (2)'be- o comes 2 + 2 2rx=0, or y2 ='(2r -x)x.. (3).A,X But, if the axis of y pass through the centre, then = 0, and f = r, and equation (2) takes the form x2 + y2 2ry = O *. (4)..A -x Y When the axes originate at the centre, the form P of the equation is still more simple; for, as in this case, both a, and / are 0, equation (1) becomes - + y2 = r2.,..(5), and this form, on account O u X of its simplicity, is most generally employed. Equation (3) may obviously be converted into this proportion, viz. x: y: y: 2r —x; that is, a perpendicular, PM, from any point in the circumference, to a diameter, AD, is a mean between the parts AM, MD, into which the diameter is divided by it. If in the foregoing cases the angle at A had been oblique, instead of right, the several equations would have been more complicated; equation (1) would then have taken the form (art. 14,) (x -)+ -(y - )2+2(x — a) (y - ) cos. A =2, and the simplest form of this equation, viz. that corresponding to equailon (5), above, the axes originating at'the centre, would be' x2 4-y2 + 2xy cos. A = r2. We shall now proceed to the solution of some problems relating to the circle, always referring the curve, for the sake of simplicity, to rectangulax axes. ANALYTICAL GEOMETRY. 103 PROBLEM I. (25.) To find the equation of the tangent at a' point in the circumference of a circle. Let the rectangular axes originate at the centre A,'A and let (x', y') represent the point, P; then we have to find the equation of the straight line, AB, p which touches the circumference in: this point; Let the radius, OP, be drawn,: then since it X passes through the point P, and is drawn from the origin, the equation of OP is y = x. Now a tangent,is perpendicular to the radius at the-point of contact, consequently, we have merely to express the equation to the line drawn through (x', y',) and perpendicular to that represented by the foregoing equation, the equation of the tangent is, therefore, y- y -' (x - x'), which, by reduction becomes yy -+ xx = y2 -- + x'2, or = H. The first form of the equation is that most frequently employed. The equation of the tangent may also readily be determined, independently of the geometrical'property referred to above, by a mode of inVestigation which is applicable to all curves whatever. *Thus, Let us first consider a secant to the curve, that is, a line cutting it in two points, (x', y/,) and' (x", y"). The equation of this secant is y -y' =Y Y-. (x-')... (1); and, as in the present instance,- both points are in the circumference of a circle, we must have x'/2- y"2 = r... (2), and x"2 + //2 = r2.. (3), equation (3), subtracted from (2), gives y/2 y//2 _ (x,2 -x,/2) that is, (y' - y") (y' -y") = - (x' -- x") (x' - "); y/ - u/1 I~/ + a/ wxhence ", _ =,,, consequently, by substitution, equawhenc X - y + &"' tion (1) becomes y-y/' - (x-x' ). (4) If now we suppose that the points through which this secant passes coincide, it will then become a tangent; we have only, therefore, to put in equation (4) x' x= a and y' = //, and there results for the tangent the equation y - y' =- (x - ), as before found. Since the equation of the line drawn from the origin through the point (x', y/) is y- x, it follows, from the foregoing equation, x 104 ANALYTICAL GEOMETRY. which is obviously that of a perpendicular to this, through the point (x/, y',) that the tanbgent through any point is perpendicular to the radius at that point, a property which was assumed in'the preceding investigation. PROBLEM II. (26.) To draw a tangent to a circle from a given point without it. Let (a, b,) characterize the given point, P, when referred to rectangular axes originating at the centre, then it is required to find through what point, (z', y/,) on the circumference the line must pass to be a tangent. The point (x', y',) being on the circumference, there must exist the relation'2 - y'2 r2... (1). Also, since the point (a, b,) is on the tangent, we must have, by substituting its coordinates for x and y, in the equation of the tangent, the relation ax/ + by' = r2.. (2). Now, from-these two equations, we may determine the unknown coordinates x/ and y'; and it is obvious that we shall arrive at two systems of values,'for the first of the equations above is of the second degree; we may infer, therefore, that there Will be two points in the curve, to each of which a tangent may be drawn from the given point. As the analytical representation of the coordinates of these points will be rather complicated, instead of obtaining them from the preceding equations, we shall determine the points geometrically; and to do this, we shall merely have to construct the locus of the simple equation ax + by -r'2, in reference to the axes of the circle, for this locals must of necessity intersect the curve in the two points sought, since, by virtue of equations (1) and (2); the coordinates x', y/, belong both to this locus and to the circle (see art. 23.) 22 For y = 0, the value of x is-, and for x - 0, the value of y is — a b therefore, on AX take AC =: and-on AY take AB then the straight line, BC, C/ drawn through the points B, C, will intersect / the circumference in the required points, M, M/. The value of x for y = 0, that is to say, A C N X 2/M AC, being, an expression independent of b, must be the same, whatever value b may take, that is, whatever be the length of the perpendicular, NP; we may infer, therefore, that in whatever point of this perpendicular P be situated, the chord joining the points of contact of tangents drawn from it, will always intersect the axis of x in the same point, C. This property may be thus expressed. ANALYTICAL GEOMETRY. 106 If fromn any number of points in a straight line tangents be drawn to a' cixcle, the chords joining each pair oj' tangents will all intersect the perpendicular,from the centre to the line, in the same point. If the proposed line cut the circle, the intersection of' the chords r2 will he without the circle; for as then a L r. x — must exceed r; but, if the line be wholly without the circles the intersection will obviousiy be within it. If instead of constructing the locus of the equation ax + by _ r2, we had actually solved the equations (1) and (2), we should have arrived at the following'expressions for the coordinates of the points of br2' ar contact; viz. y/= a 2+ b2-r2 (3) a2 + b2 a + b2 ar2 br and x' = + v + b2r2. (4). By means of these coordinates we can find the equation of the line passing through ththe two points to which they belong, and this equation we shall find to be ax + by = r2, as above. For the equation of a line passing through two points is of the form y -y'=a, -(x -x')....... (1), in which the coefficient a' is equal to the difference of the ordinates of the points divided by the difference of the abscissas; in the present 2ar case, the difference of the ordinates is +-2 - b2 a2 + b2 — r2, 2br and the difference of the abscissas, bea2 + b2 a b; so that, dividing the first of these expressions by the second, we have for the value of a', a' = — - and, consequently equation (1) is Y -y' = -b (x — x'), or by ax = by' + ax'; but, by the conditions of the problem, ax' + by' = r2, and.'. ax + by = r2, the equation required. PROBLEM III. (27.) A circle and a point being given, it is required to draw from the point a straight line through the circle, so that the part intercepted by the circumference may be of a given length. C Let ACBD be the given circle, and P the given point, the coordinates of which are x, y, / the axes being as before. Let the unknown" I distance of the point (x', y') from one of the points' of intersection be represented by z, we shall then have the following equations, viz. 14 106 ANALYTICAL GEOMETRY. X2+y2=r. (1) y -y =a(x — x.). (2) and 2 — (x -/)2 -+( y/)2 (3). the'first representing the given circle; the second, a straight line.through the given point; and the thild, the square of the unknown distance. Now at the points where the line and circle intersect; the same coordinates will belong to each, so at these'intersections the values of x and y will be the same, in each of the above equations. Hence, substituting for (y y/)2, in equation (3), its value a2(x x/)2, in equation (2), we have z2'=(x- x)2(1 + a2), Z az from which we get x -x =' - and y-' = A — z az consequently x' + - and y=y/+ +,}. Substituting'these values of x and y, in equation (1,) it becomes, after reduction, z2' 2( +ay ) Z +x2 y02 r. =... (4). The roots of this quadratic are. x --- - - 1 _ + a-} v (r2-x'2) a2 + 2ay/xt/+'r- Y2. Since the difference of these two values of z must express the given length, we have, by calling it 2m, 2m= (i- - V( /2)a2+2ax'y/+ r2-_y2t If now we square this expression, we shall obtain, after reduction, the quadratic (x/2 - n2_rA) a -) 2x/y'a +-'y/2 nz2 r2 = 0.(5 in which all the quantities are known, except a; this, therefore, may now be determined, and thence the required line y - y/ a (x -x/) drawn, and it is plain that, as the solution of the above quadratic will give two values for a, two lines may be drawn from P, fufilling the proposed condition. If the given point be upon the axis of x, then y' = 0, and equation (5), becomes (x/2-+m2 4Zr2) a2+ -mn- r2 = 0, so that the expres2 sion for a, which in this case is, a =... (6), -will be most simple, if we choose for the axis of abscissas a line from the centre through the given point. Let us now actually construct the'line from Its equation p = a (x - X/), which equation (2) becomes, when the axis of t passes through (x/, y/). The numerator x/1 _ —] 2.i of the coefficient a. in this equation, obviously expresses the side of a right-angled tri-an gle,, of which the hypothenuse is r, and the other side -M. Let, then, this line be constructed, and call it p., The denominator, al, o, of the fraction a being v/lx'2+m2 r2l, or x'",2- (r — i2) 5, expresses ANALYTICAL GEOMETRY. 107 the side of a right-angled triangle of which the hypothenuse is x', and the other side v/:r2 m-2, the line just constructed. If then this second line be also constructed, and represented by q, the equation to the required line will be y p (x x'). rake, therefore, on OP, the distance, PF = q, Y and on a perpendicular at the extremity, take \ FG =,p, then the straight line, GE/, through P, will be drawn as required; for the trigono-/ \ / metrical tangent of the angle FPG will be p'q. If in - 0, the proposed line will be a tangent to the circle, and, in, that case, equation (6) becomes a = r.+ - x/2-r2. This, therefore, is the value of the trigonometrical tangent, which the line from the given point to the centre must make with another line drawn from the same point, in order that this latter may touch the circle, hence we have an analytical solution to prob. 2. The same expression for the tangent of this angle might have been readily derived from equations (3) and (4), inproblem 2. For, since A tangent, PC, is perpendicular to the radius, OC, it follows that the angle, OCp, included by this radius and the ordinate of the point C, is equal to the angle OPC; so that the trigonometrical tangent of this angle will be expressed by dividing the abscissa, Op, of the point C by the ordinate, pC, the axis of x being supposed to pass through P. Now, on this supposition, b, in the equations referred to, is 0, so that the coordinates of C become r r2 / y= -Sa2 -;, X/=-' which expression a a - is the same as that above the abscissa of the point, P, being represented by a in this expression, and by x' in the former. From equation (4) two well-known theorems may be easily deduced, for representing the roots of that equation by z' z2/, we have, by the theory of equations, (.Ilg. p. 177), z/z//, or PE X PE-x/2+- y/2 -- r2; and as the values of x/, y/, and r, are quite independent of in, this equation subsists for every position of PE/; hence, when P is within the circle, we infer that chords intersecting each other in P are divided, so that the rectangle of the parts of each is the same; and, When P is without the circle, we conclude that. all ines drawn therefrom, and terminated by the concave part of the circumference, are so divided by the opposite part, that the rectangle of the whole line and the external part is the same in each. PROBLEM IV. (28.) To find the coordinates of the points of intersection of two circumferences. 108 ANALYTICAL GEOMETRY. Let the radii of the two circles be r, r', and the distance of- heir centres, d. Let the rectangular axes'originate at the centre of that circle, whose radius is r, and'let it pass through the centre of the other circle; then the equation of the former circle will be y2 + x2 = i;2 (1), and the equation of the latter, the coordinates of whose.centre is cc -d, 3=0, will be y2 + (x -- d)2 = r/2. (2). At the points of intersection, the values of x and y must be the same in both these equations. To determine them, subtract equation (2) from equation (1), and there reuts2x i r2,x= r/i+di results 2dx d2 _ -2_ r/- r.i x =- Substituting this 2d value of x, in equation (1), we have y - -r2 { +d whence y = - V / 4dir2I_- (2 r2 d c2)21. (3). Now we may observe of this equation that the expression within the brackets is the difference of two squares;it may, therefore, be replaced by two factors, the one denoting the sum, the other, the difference of the roots of these squares, that is by the factors, (2dr + 2 +4 d2 r/'2) (2dr - 2 -d2+ - r'2). Here again it occurs that each factor is the difference of two squares, the first being (r + d)2 - r/", and the second, r/2 - (r' d)2; hence, by decomposing each of these into factors, equation (3) is finally reduced to Since y has here two values, numerically the same but with contrary signs, it follows that the line joining the centres of two intersecting circles bisects at right angles the line joining the intersections. The form under which we have just exhibited the expression for-y is very convenient for the examination of the circumstances of the problem, which examination will lead us to the theorems relative to intersecting circles already established in Elements of Geometry.'As the first factor under the radical is necessarily positive, the whole expression must also be positive, provided that all or only one of the remaining factors' are likewise positive. Now two of the remaining factors, at least must be positive; for if one, (r + d - r/) for instance, be negative, then (r + d) L_ r'; consequently r Z r', and also d v r', which proves that the other two factors must be positive. There can, therefore, be but two cases to examine, viz. that in which all the factors are positive, and that'in which one is negative. In the first case the values of y will be real, in the second, they will be imaginary. In the first case, there must obviously subsist the conditions r + d-7 r', r + r' 7 d, r' + d 7 r, which prove that if two ANALYTICAL GEOMETRY. 1.09 circu.iferences cut, the distance oj' their centres must be less than the sum, and greater than the diference of the radii. In the second case where y becomes imaginary because- of a negative factor, we must have one of the conditions d/' r'-r, d 7 r+r/, d' r - re; so that two circumferences can have no point in conmmon, if the distances of the centres be less than the difference or greater than the sum of the radii. Lastly, let one of the three last factors be 0, which can happen only when d is equal either to the sum or difference of the radii. In this case, y O, showing the circumferences have but one point in cornmon, and that this is on the axis of x; so that- two cirumlfren.ces touch, vwhen the distance of the centres is equal to the sun or difference of the radii. (29.) In addition to the problems here given, a variety of others relating to the circle might be proposed, which would conduct to other properties of this curve. But, as the circle occupies so large a portion of elementary geometry, where its most important properties are unfolded with the utmost simplicity and elegance, it would be superfluous to dwell upon it at any great length here. Indeed, the theorems established in elementary geometry are, for the most part, obtained with less ease and simplicity by analytical processes than by pure geometrical reasoning. This fact the student has, no doubt, had occasion to remark, in some of the foregoing investigations; these however, it would not have been proper wholly to have omitted, on this account; their introduction has not only furnished the student with the means of applying the fundamental principles of analysis, but has, at the same time, given him confidence in those principles, by conducting him to results previously known to be true. In the remaining sections of this work, which will treat of curves not within the limits of elementary geometry, the great advantage of analysis will be more distinctly seen. Many important properties of these curves will be obtained with the utmost ease and facility, which could not be established by common geometry but by very lengthy and.elaborate reasonings. (30.) We shall terminate this section with one or two problems on loci; first, however, showing that, as the general equation of a circle, when referred to rectangular coordinates, is (x - a)2 + (y P)2:r. or, x2 - 2c0x + a2 + y2 2/3y + /2 - r2 = 0, so conversely, every equation of the second degree of the form x2 + y2 + Ax + By+ C=0 (1)- will be the equation of a circle:'this may be proved as follows. Take any rectangular axes, AX, AY, and find the point 0, whose abscissa is - iN and ordinate — AB; then, from this point as a centre, A2 q- B2 with a radius equal to V A + - C I, describe a circumference; his circumference will be the locus of the proposed equation. K 1 10 ANALYTICAL GEOMETRY, For the equation of this circumference beingy z — )2 - (y -_ )2-r2, where, by construction, - — A 25 —' —B, andr2' (A2+B2) - C, it is the same as (x.+ IA)2 +(y + 1 -B)2= (A2+B2)- C, which reduces to x2 + y2 + Ax + By + C = O; hence the circumference just described is the locus of this equation. - A. Suppose, for. example, it were required to construct. the locus of the equation 2x2 + 2y2 - 3x 4y- 1 =0, which reduces to x2 + y2 - zx + 2y - ~-0. First, then, find a point 0, having 3 for its abscissa, and 1 for its ordinate; then from this point, as a centre, with a 322 radius equal to 1(2 +22) + - I = 33, describe a circumference, which will be the locus sought.'It must be observed that the coefficients of the proposed equation may be related so as to render (A'2 + B2) = C, in which case the equation will represent a circle whose radius is 0, that is, merely a point; such is the equation x + y2 _ 3x _ 2y + 1+ 3 = 0, which represents a point whose coordinates are x -- and y - 1. The coefficients may also be so related that the equation may have no geometrical representation, as in the equation x2 - y -2 + 4x 2y + 7 = 0, which, for every possible value of x, gives an imaginary value for y'; so that no real line or point can be represented by the equation. In such equations, the expression for the square of the radius will always be negative; when, therefore, we say that equation (1) represents a circle, we must be understood. as meaning, that no other line can be represented by it; so that, when the locus is not a circle, the geometrical representation is impossible. We shall now add a few questions, leading to indeterminate equations of the second degree, the loci of which will furnish every geometrical solution. PROBLEM V. (31.) Given the base and the sum of the squares of the sides to determine the triangle. Let AB be the base, and put m for the sum of the Y squares of the sides, AC, BC. Let the perpendicular, OY, from the middle of the base, form, with the base, the rectangular axes; then putting a for AO, or OB, and (Y, y), for the point C, we shall have the equations y2+(x + a)2=AC2. (1) o B and y +- (x - a)2 - BC2. (2). Adding these'together, 2y2 + 2x2 + 2a2 = AC2 + BC2. (3); therefore, putting m for AC2 + BC2, y2 + x2 = 2(in - 2a2) which equation represents a circle, of which the centre is the origin, 0, and the ANALYTICAL GEOMETRY. 1ll radius V/(Im( a2); so that, if this circle be described, and lines be drawn from A, B, to any point in its circumference, a triangle having the proposed conditions will always be formed, and thus, when the base and sum of the squares of the sides are constant, the locus of the vertex is a circle. Since y2 + x2 OC2, it follows, from equation (3), that we have 2A02 + 20C2= AC2 +-'BC2; that is, in aay triangle, the sumi of the squares of the sides is'equivalent to twice the squares of half the base, and of the line from the vertex to the middle of the base. (Geom. p. 38.) PROBLEM VII. (32.) Given the base and the vertical angle to determine the triangle. Let b represent the given base, AB, and put v for y the tangent of the given angle; then, taking AB, AY, for rectangular axes, we shall have, for the equation'of any-line, AC, drawn from the origin y — ax.... (1), and the equation of another line, BC, drawn from the point (b, 0), and making theA proposed angle with the former line will be y = + (x -b). (2), (x, y) being the point of, intersection, C. I -av Now, for a, in equation (2), substitute'Y its value in equation (1). y+ v and we thus obtain the equation y —-- (x- b), which reduces to b2 b- _2 _ ~ y _ bx ~ O, the equ ol. y2 + x2 - y- bx -O, the equation of a circle of which the coordib b b2 b2 natts of the centre are / g and a and radius/(-2 2/(3 +U2) From the foregoing expression for the radius, it is plain that the circumference passes through the origin; and, since for y 0 the above equation gives x = b, it follows that it also passes through B; hence the base of the triangle subtends the are of which the vertex is the locus. If the given angle is right, then its tangent, v, is infinite, and therefore i -- 0; so that, in this case- the centre of the circle is at the middle of AB. If the angle be acute, e must be positive, for v will be so in this case; therefore, the centre is situated above the base but, if the angle be obtuse, then jS will be negative, and the centre 112 &NALYTICAL GEOMETRY. will be below the base. Hence the segment containing a right angle must be a semicircle, and the segment will be greater or less than a semicircle, according as the angle contained in it is acute or obtuse. It must be remarked, that no part of the arc c below the base belongs to the locus, which we ha've determined, because equation (2) requires that the angle be formed to the right of AC, (see art. 13,) fixing the intersection above AB. Asi B however, there is io-restriction of this kind in the problem, we may admit, the proposed angle to be formed to the left of AC, in which case the equation of BC will be y + (ix-b), and, the 1 +av locus,9y2 + 2 + -y - bx = 0, representing a circle of the same ra. b dius as before, the coordinates of the centre being /3' and ox= b;.so that, in strictness, the locus consists of two equal arcs, situated the one below the other, as in the annexed diagram. The three problems next following resolve themselves into the preceding. PROBLEM VII. (33.) Given the base and vertical angle of a triangle to6determine the locus of the intersection of perpendiculars from the angles to the opposite sides. Let AB be the base, P the intersection of the per- c pendiculars, AF, BE, on the sides BC, AC; then, since the sum of the angles of a quadrilateral amount E to four right angles, the angle P must be the supplement of the angle C, and therefore constant, because -\ C is. Hence we have the base, AB, and vertical angle P of the triangle PAB, to, find the locus of P. By the preceding problem, this locus is the arc APB, whose chord, AB, subtends an angle equal to the supplement of C. An equal arc below AB also belongs to the locus,'PROBLEM VIII. (34.) To find the locus of the centre of the, inscribed circle when the base and vertical angle of the triangle are given. The centre of the inscribed circle is at the inter- e section of the lines bisecting the angles at the base, and, as the sum of these angles is constant, because A the vertical angle is, the'half sum must be con- / r stant, so that the triangle APB, formed by the given base, AB, and the lines AP, BP, to the centre. ANALYTICAL GEOMETRY. 1 13 of the inscribed circle, has the vertical angle P, constant, and equal to two right angles, minus half the sum of the angles at the base of the proposed triangle, that is, to one right angle plus half the given angle; hence the required locus is an arc described on AB, containing this angle. A similar arc below AB, belongs also to the locus. PROBLEM IX. (35., The base and vertical angle of a triangle being given to find the ocus of the intersection of- the straight lines, drawn from the angles to the middle of the opposite sides. By art. (19.) if P be a point of intersection, its distance from M, the middle of the base, AB, will beequal to half its distance from the vertex C. If, therefore, Pm, Pm', be drawn respectively pa- c rallel to AC, BC, we shall have Min, Mm', each equal to one third of AM or BM, and the angle P equal to the- angle C; that is to say, the base, nm, / \ and vertical angle, P, of the triangle, mPm/, are constant, the locus of P is, therefore, the are mPm'/, A m,7 To construct it, we shall have to trisect the base, and to describe upon the middle portion an arc to contain the given angle, or rather, two such arcs, one on each side of mm'. We shall give another solution to this problem. Take the rectangular axes, AB, AY, then, since PM = kCM, the perpendicular from P to Y C the base will be one third of that from C; hence, representing the base by b, the vertex by (x, y), and the point Pby (X, Y), we shallhave y=3Y; -: = 3X- b; and, substituting these values of x and y in the locus of (a, y), as represented by'its GHE1l A B equation in Prob. 6, it becomes y2 + X2 F bX + 2 b2 = the equation of a circle, of which the' coordinates of the centre are a = b b "2=~ 6'6v Let Y = O, then the corresponding values of X, given by the solution of the quadratic, are X = -b; X- = b b, showing that the locus intersects the base in two points, at these distances from A, thus intercepting the middle of three equal portions. f615 hnK2 114 SECTION III. ON, LINES OF THE SECOND ORDER. (36.) The only curve which we have as yet considered is the circle, whose equationwe have found to be of the second degree. Be sides this there are three other curves, which, like the circle, are each represented by an equation of the second degree. These three curves we propose iin this section to examine, first determining the equation and form of each, and then proceeding to investigate its properties. We shall afterwards show, that every equation of the second degree, with two variables, whatever be its form, can represent no curve, but the circle, or one of these three; and, from this circumstance of-their equations being all of the second degree, these four curves, are called lines of the second order. Before proceeding to the three new curves, of which we have just spoken, the student should attentively read the following preliminary chapter. CHAPTER I. ON THE TRANSFORMATION OF COORDINATES.,(37.) Every equation which characterizes a line, whether straight or curve, will depend for its simplicity upon two circumferences; the relative position of the axes to which the line is referred, and the absolute position of the origin. This fact has already been observed, as regards the two classes of lines which have hitherto been examined, the straight line and. the circle. It was seen in the case of the straight line, that when the axes were oblique, and their origin not upon the line, the equation which characterized it was far less simple than when rectangular axes were employed;' the relative position, therefore, of the axes affected the equation. It was moreover observed, that if, in addition to the axes being rectangular, their origin were upon the proposed line, the form of its equation became still more simple. Similar remarks apply to the'circle, the equation of which is -much more complicated when the axesof reference are oblique, and originate without the curve, than when they are rectangular, and originate at the centre. It may, therefore, readily be conceived, that, with regard to other curves, there may also exist certain positions for the axes, and certain points for their origin, lby assuming which, the curve may be susceptible of a more comm6dious analytical representation, than when the axes and origin are chosen at random. Now the object of this chapter is to show that when a curve, is represented by an equa ANALYTICAL GEOMETRY. 115 tion, in reference tO any system of axes, we can always transform that equation into another, which shall equally' represent the, curve, but in reference to a new system of axes chosen at pleasure. This is called the tralsformation of coordinates: it may consist either in'altering the relative position of the axes, without displacing the origin; in removing the origin, without disturbing the relative position of thie axes; or, lastly, it may be found necessary to alter both the direction of the axes and the situation of their origin. By means of these transformations, we may often simplify the equation of a curve, and many of its properties, not readily derivable firom its equation in one form, may frequently be obtained with great facility by a transformation of it into another, as will be repeatedly seen in the course of the subsequent chapters. (38.) Let the axes, AX', AY, be those to which any liie, MM'M", is related by its equation, and let A'X/, A/'Y' be the new axes, to which it is proposed to refer the same line. Let the coordinates of any point, M, in.. the line, relative to the primitive axes, be y M x and y, and the coordinates of the sam/ne point, referred to the new axes, A/P'', / and P'M -y'. Draw A/X" and P/'I r / each parallel to AX, and P'K parallel to3 K/ AY, then we shall have x = AP = BA/' 2 x + A'/K + P'H, and y = PM - AB + KIP/' + M. In these equations BA/, AB, are known, being the coordinates of the origin, A', of the new axes, when referred to the primitive. It remains, therefore, to determine the other terms, and for this purpose let us represent the known coordinates of the new origin, viz. BA', AB by a, b; the angle X'A'X/, which the new- axis of abscissas makes with the old, by a, and the angle Y'A'/X/" which the new axis of ordinates makes with the old axis of abscissas, by c'; then the inclination of the new axes will be Ge - a. Let also jd represent the angle Y/A'X", the inclination of the primitive axes; then will,' — a be the angle formed by the new axis of Z and old axis of y, and M - a' will' be the angle formed by the new and old axes of y. Now, by trigonometry, the value of A'K is __ A'P"'(sin. A'P'K) x/ sin. () —,) A'K -P sin. API-K) - -xI sin(. (, observing that A'P/K = Y"A'X'/, or C - a, and that Y/"A/X", or 2, is the supplement of A'KP'. S In like manner, for KP/ we have A/P' (sin. P/A'K) x' sin. oc sin. A'KP' = sin.' Also, in the triangle MP'H, P'M (sin. P'MfI) _ y' sin. (8 - a') we have, for P/H, -PH P' and sin. MHP/- sin. 2 116 ANALYTICAL GEOMETRY. HP'M (sin. MP'H) y' sin. a' for HM, LHM sin —MHP' - sinf Hence, for the vaz x' sin. ( -- a) +y' sin. (- a/) luIes of x and y, we have x a + - sin. ( - x sin. a - y' sin. a' and y =b +. These then are the values sin. -3 which must be substituted in the equation of the curve, when related to the piimitive axes, AX, AY, in order to transform it into the equa, tion which the same line must have when referred to the new axes /X/,- A'Y'. The above general expressions become modified in,particular cases, the -principal of- which we shall here exhibit. 1. When the new axes are parallel to the old. In this case, the inclination of the axes remaining unaltered, while the origin is removed, we have a - O, - a' O; hence the above expressions become x = a + x' and y = b + Y/.... (1). 2. WFhen the primitive axes are rectangular, and the new ones oblique. -Here sin. ( - a) = cos. a, and sin. (3 - a') = cos., therefore...s z=a +' cos. a~+ os. a'} y= b zx sin. a - y sinm. 0+ 8. When both systems are rectangular. Here a' 90~ + —a,.'. sin. a'= cos. a; also sin. (/3- a') = sin. ( -) - sin.a;;; and sin. (/3-a) = cos. a; \ i hence the expressions become, in this case, x = a - X/ cos. a-y' sin. a }. y = b +x'sin. a+y'cos. a 4. When the primitive axes are oblique, and the new ones rectanI gular. In this case, a/'=90~ +a.. sin. a'=cos. a also, since the complement of /3 - a'is 90~"- (/3 — a') \ / = 180~- (/3 ),W e have sin. ( --') - = \ a' cos. (3 - a), so that the formulas for this case axee ~ a ANALYTICAL GEOMETRY. 117 x/ sin (/3-)- cos. (/3- a) y' sin./ b x/ in. a + y/cos..a At In the first of the marginal figures, the two systems of axes are so placed that, / - is less than 90~; its sine and cosine are, therefore, both positive; and the expression for x shows that y' times the latter is to be subtracted from x' times the former. In the second figure the two -systems are placed so that /3 a- is greater than 90~; its' sine, therefore, is positive, as before, but its cosine is negative; and the' formula shows that in this case y' times the cosine, is to be taken positively, that is, it must be added to x' times'the sine. It must be remembered that, in each of the preceding cases, when the transformation is confined merely to the direction of the axes, the origin remaining' fixed. the' terms a, b, becomre 0. It must further be remarked, that, when the axes of x in the two systems coincide, then - = 0, and when the axis of y in both systems are identical, then 3 = ao'. Suppose, for example, we wish to pass from an oblique to a rectangular system, the origin and axis of x remaining undisturbed, the formulas (4) will give y/cos./3 _ x y z- =x i / y' cot. /3and y - y- cosec. a8. sin. - sin./~ (39.) Throughout the whole of the preceding investigation the angle or, or that which the new axis of x makes with the old, is supposed to be positive, that is, we have uniformly conceived this angle to be situated above the primitive axis of x. If, on the contrary, it be supposed negative, or to fall below the same axis, then its sine will be negative, but its cosine will continue positive. Hence, in this position of the new axis of x relatively to the old, the preceding formulas willrequire some modification. We shall, therefore, in order to complete this theory, here present them with the necessary changes. For the first of the preceding cases the formulas are the same. For the second they are (2/).. a + x cos. a y cos.'.' = b - x/ sin. a + y' sin. a'o For the third,. (3')... A; x = a + x cos. a + y' sin. a. y = b - x sin. a + y cos. a. x' sin. (/3-+) -y' cos. (8-a) For the fourth, (4/).. sin. Cos. cc - x sin. o~ Ycsaymb + asin./3 or, since the angle (/ + — a) is here the same as a/ in figs. I and 2, we may write these last expressions thus: a' expressing the inclination of the old axis of y and new axis of x, and a/ - a the inclination 118 ANALYTICAL GEOMETRY. -/ sin. o/ - y/ cos. a. b+ of the.old axes of x and y, (4').. i sin.-(_na. sin. (a"- a). It may, perhaps, be satisfactory to the student to verify this last formula, which is the most complicated, by determining the values of x' and y! from formulas (2) we shall thus have' the values of the oblique coordinates in terms of the rectangular, as above. Omitting the constants a and b, in (2), and multiplying the first equation by sin. a' the second by cos. a', and subtracting the latter result from the former, we have, x sin. a! —/y cos. a' - x/ (cos. a sin. a' — Cos. a' sin. a) x sin. a/ — y cos. a' = x' sin. (a/ -a).X'. sin.In like manner, sin.,(a/ a) by multiplying the first equation by sin. a, the second by cos. a, and proceeding as before, we get y/ y covalues which sin. (W - a) verify the preceding formulas. (40.) Some authors employ a different notation in these formulas; thus, instead of using the letters a, a/, &c, to denote the angles about A/', they employ the sides which contain them,' the, primitive axes being denoted by X, Y, and the new ones by X/, Y', so that [X, X/] is put for a, [Y/, X] for or, &c. By adopting this notation, the first class of formulas, where [X, X'] is positive, will be 1. When the primitive axes are rectangular, and the new ones oblique. x a + x cos. [&, X/] + y' cos. [X, Y/] y = b x' sin. [X, X1'] A- sin. [X, X'] in. X, Y']. 2. When both systems are rectangular. x = a +-t- x cos. [X,X/'] - y' sin. [X, X'] y= b -+ x' sin. [X, X/] + y' cos. X, X/1]. 3. When the primitive axes are oblique, and the new ones rectangular. x' sin. I[X, Y] - y/ cos. [X', Y] sin. [X, Y], sin. [x, ]x, + y' cos. [X, x'] " + sin. [X,-Y]' The second class of formulas, where [X, X'] is negative, will be 1. When the primitive axes are rectangular, and the new ones oblique. x = a + X, cos. [X, X/] + y' cos. [X, Y'] y = b- x sin. [X, X/] + y/ sin. [X,'Y/] 2. When both systems are rectangular. x = a + x' cos. [X, X'] + y' sin. [X, X'] y = b —x' sin. [X, X/] +' cos. [X, X/']. 3. When the primitive axes are oblique, and the new ones rectangulatr. X/= a sin. [X'/,] - y's. [X', Y] sin. [X, Y] ANALYTICAL GEOMETRY. 119 yb + cos. [X, X/] - x/ sin. [X, X/] sin. [X, Y] Under this form of notation, the angles introduced are more distinctly marked, and therefore more readily recognized, than when they are each represented by a single letter. Still, however, as these formulas are less brief, and, consequently, less commodious in calculation, the form of notation first given is generally adopted in preference. CHAPTER II. ON THE ELLIPSE. Its equation and Properties. (41'.) An ellipse is a curve from any point, P, in which, if straight lines be drawn to two fixed points, F', F, their sum will always be the same. The points F/, F, are called the foci of the ellipse r and the distance F'P -or FP of either from a point / in tle curve, is called the fJbcal distance or radius vector of that point. X Froin the definition of an ellipse, the curve may be readily- described mechanically; thus, to the two fixed points, F', FY, let. the extremities of a cord be fastened, let this cord be stretched into a loop, F'PF, by means of a pencil, P, then the motion of this pencil, still keeping the cord stretched, will evidently describe an ellipse. The cord must obviously exceed in length the distance F'F. Let us now seek the equation of this Y curve, and, for this purpose, let us take, OX, OY, for rectangular axes, the origin O being-placed at the middle of F/F, and let the sum of the distances of any point, P,A IA - -X in the curve from the foci, be represented by X' 2A'. Put also c for OF or OF/; then, if - PM be drawn perpendicular to OX, we shall have OM x, PM-y and, consequently, y2(x —c)2P —F2, (1), y2+ (x + c)2=PF'2, (2), also P?- -PPF/ = 2A,. (3). Hence, by addition and subtraction, we get 2y2 + 2x2 + 2c2 — PF'2 -T PF/2', (4), and 4cx = PF'2 -P F2 = (PF/ + PF) (PF/ - PF) =2A (PF/ - PF), ~. PF' - PF - 2cx - A; hence, combining this last equation with equation (3) there results PF/ A + - and PF c 7A A; and, if these values be substituted in equation (4), we A' 120 ANALYTICAL GEOMETRY ey -- xI -- c2 A c x have y2 + x + C2 = A2 +-A-, which finally reduces to A2y2 + (A2 C2) 2 A2(A2_-c2) (5). In this equation, x andl y are the coordinates of any point in the curve, and the other terms are all constant; this, -therefore, is the equation of the curve. (42.) Let us now inquire at what points the curve cuts the axes. For this purpose, put y = 0, in equation (5), and there results for the abscissa of the point where the curve cuts.qOz, z = +_ A. We hence learn that-x: has two values, viz. x -- A, and x - A, so that the curve cuts the axis of x in two points, B, A, each at the distance, A, from the brigin, the one being to the right, the other to the left, Suppose, now, x 0, in the same equation, and there results the ordinate of- the poinlt'when the curve cuts the axis of y, y-= t vA2 -- c2. Since this value admits of being taken positively or negatively, we infer that the curve cuts the axis of y also in two points, C, D, equi-distant from 0, the one above and the other below it. Hence the two chords, AB, CD, are mutually bisected at the point 0. As the former chord is represented by 2A, let us denote the latter by 2B, that- is, put 2V' A2- c2i = 2B, and then the equation of the ellipse (5), in terms of the two chords AB, CD, when these are taken for axes, assumes the more simple form, A2 y2 + B2 2-= A2 B2 ory2I- (A2 X2), (6), and.y= y A — A2- x2; A 2 A' A x = + 5B - y2 2,. From these expressions for y and x, it appears that for the same value of x, there are two values of y numerically equal, but having contrary signs; hence the chord AB bisects all the chords drawn parallel to CD. In like manner with regard to x; this also has two values numerically equal, but differing in sign for one value of y; therefore, the chord CD bisects all the chords drawn parallel to AB. Moreover, since,, for x = A, or x - A, the corresponding value if y is 0, it follows- that parallels to the axis of y drawn through the points B, A, of the curve, meet it in no other point, that is, they are tangents to it at that point. In like manner, for y = B or y = - B, the corresponding value of x is 0; hence it may in the same way be inferred that parallels to the axis of x through the points C, D, of the curve, are also tangents to it. It appears, likewise, that at no point of the curve can the abscissa exceed A, for when x 7 A, y is imaginary. (43.) Let us now seek the expression for the distance of any point (x, /) in the ellipse, from the origin, 0. We shall merely have to B2 substitute for y2 in the expression, D = lx2 + - Y2 its value A(A2' x2), as given in equation (6), and the expression sought will ANALYTICAL GEOMETRY. 121 M2 B2 be D- = IB2- + A2 - x2, (7). This expression for the distance will'obviously remain the same, whether x, y, be positive or negative. or the one positive and the other-negative; -hence, those points in the curve, whose coordinates have respectively the same numerical values, however the signs thereof may differ, are equally P' distant from 0, so that if OM — = x, PM = y, and OM/ — x P''= — y be the coordinates ofA two points, P, P', then will OP be equal to OP,' \ /M and, consequently, the angles M/OP,'MOP, are equal; therefore, M, M/, being a straight line, PP' must be also a straight'line, so that OP' is only the continuation of OI' Since, to every abscissa there belongs two ordinates, equal in length, but of different signs, it follows that, if the ordinates of P', P. be produced' to p', p these points will'be expressed by (- -, y,) (x, - y); they are, therefore, at the same distance from O as the former, and p, p', will, in like manner,' be a straight line, and equal to P'P'. Because every chord passing through O is thus bisected in that point, O is called the centre of the curve, and the chords passing through it, diameters. It appears, from the above, that diameters PP', pp/, which make equal; angles, POM, p/OM/, with the axis of x, are equal. Since A is necessarily greater than B (art. 42), the preceding expression for D must increase or diminish, accordingly as x increases or diminishes: D will, therefore, be greatest when x is greatest, or equal to A, and it will be least when x is least, or equal to 0; so that OB or OA is the greatest distance of the centre, 0, from the curve, and OC or OD is the least distance. Hence, of all the diameters, AB is the greatest and CD the least, and, for this reason, AB is often called the major diameter, and. CD the minor diameter; when spoken of together, they are called the principal diameters, or the prib cipal axes of the ellipse; consequently, the equation A2y2 +- B2x2 - A2 B2, (8) is the equation of the ellipsr related to its principal diameters, the origin being at the centre. When A = B, this equation characterizes a circle, for it then becomes y2 + x2. A2. If we wish for the equation of the ellipse, when the origin of the axis is removed to A, the vertex of the principal diameter, A, it will be obtained by a very simple transformation. We shall' merely have to substitute in the primitive equation, x -- A, for x, (p. 116), because the coordinates of the new origin are A, 0, hence the transformed equation is A2y2 + B2x2 2AB2'- 0 or BQ y2 2-B(2 Ax —x2), (9), the equation of the ellipse, when the origin of A2 the rectangular axes is at the vertex of the major axis. 16 L 122 ANALYTICAL GEOMETRY. It is sometimes convenient to introduce the quantity, c, denoting the distance of either focus from the centre, into the equation of the ellipse, that is, to substitute for B2 its equal, A2 - c2. The quantity, c, is called' the eccentricity; hence, from equation (6), C2 y = (! - (Al x2), or putting e for y2 (1-e2) (A2 2), (10), tle equation of the ellipseas a iJnction of the eccentricity. We shall now proceed to examine more attentively the foregoing equations, for the purpose of deducing from them the principal properties. of the curve. Properties of the Ellipse as related to its principal Diameters. (44.) Referring to equation (6), we find that the second'form of that equation reduces to (A- A which furnishes the (A +x.) (A- x) A' proportion y: (A A ) (A x):: B2: A2. Now A --, A - x, are the two portions of the major diameter, into which. the ordinate, y: of any point, (x, y,) divides it; hence the above -proportion shows that'the square of any'ordinate is to the product of the parts into which it divides the major diameter, as the square of the minor.diameter, is to the square of the major. Consequently, the squares of the ordinates are as the products of the parts into which they divide'the major diameter. If we suppose, in equation (6), the axes of reference to be transposed, that is, x to' become y and y to become x, we must then, in the foregoing theorems, substitute major. diameter for minor, and minor for mcjor; so that the theorems hold good, whichever diameter be taken for the axis of x. They are true, also, of the circle, which the ellipse becomes, when B - A; hence, if a circle be described on a-principal diameter of the ellipse, any ordinate in the ellipse will be to the co, responding ordinate in the circle in a constant ratio, viz. as B to A, or.as A to B,. accordingly as the major or minor diameter is employed. Thus, in the annexed diagrams, CI ED:::C: C'D': AB. From.I equation (10) we have A2y2(, _ ca) (A- i)); hence, Ahen A- 1B x c, Ay 2- c2 = B2, E. 2y: B:: 2B: 2A,"where 2y /v is the double ordinate through the o focus,-and is called the parameter of the major diameter; it is alsc sometimes called the latus rectum. Fromn this proportion it appears that the parameter is a third proportional to the major and minor diameters Calling the parainete.r p, we have, therefore, 2Ap' = 4B2, or dividing by 4A', A = 2P' hence,..~~~~~~~~~~~A ANALYTICAL GEOMETRY. 123' by substitution, in equation (6), y,2 = (A2 — X). In like manner, from equation (9), y2-_pz-A X2. The following problem will conduct us to some other properties. PROBLEM I. (45.) To find the expression for an angle inscribed in a semi-ellipse. Let APB be a semi-ellipse, and P an angle Y inscribed in it. Then the equation of AP P passing through the point A, whose coordinates are x -A —, y = 0, is y - a (x + A)..a_-Y (1) and the equation of BP B x+A passing,through the point B, whose coordinates are x=A, y = isy a'(x-A).'. a'=- - X(2), in which equations a, a', represent the trigonometrical tangents of the angles PAX, PBX, respectively. Now the expression for the angle P, which theseA B a'-a lines form when they meet, is tan. P 1 + - which, by substitut2Ay ing fora, a', their values in(1)'and(2),becomes, tan. P= Ay (3). y'+-A2 But the lines in question not only meet, but they meet in the curve, the equation of which gives x2 - A - A2 (4) hence, by subB2 BAy;e2AB2 [stitution, (3) becomes, tan. P = 2 - (5). A 2 (A2- B2) () As this expression will remain the same, though x be negative, it follows that there are two points in the (semi-ellipse, at which the diameter subtends the Same angle, viz. (x, y,) and (- x, y). If A is greater than B in this expression, that is, if the angle be subtended by the major diameter, the tangent of P is negative, the angle is, consequently, obtuse. Now an obtuse angle increases as its tangent diminishes; when, therefore, the tangent is least, the angle is greatest; and the above expression will evidently be least when its denominator is greatest, that is, when y = B. We inferi, therefore, that all the angles subtended by the major diameter are obtuse, and the greatest is that whose vertex coincides Fwith the extremity of the minor diameter. The expression for this maximum angle is tan. P= (Asd) 124 ANALYTICAL GEOMETRY. If B is greater -than A in equation (5), that is, if the angle be subtended by the minor diameter, then the expression for the tangent is positive; the angle is, therefore, acute, and diminishes as its tangent diminishes; it is least, therefore, when y is greatest, that is, when y = B. Hence all the angles subtended by the minor diameter, are acute, and the least is that whose vertex coincides with the extremity of the major diameter. The expression for this minimum angle is 2BA tan. P -B- As this expression differs from the former only in its sign, we, conclhide that these two angles are supplements of each other. From the foregoing theorems it appears, that, if an are of a circle be made to pass through the extremities of one principal diameter, and a vertex of the other, it will be wholly within the ellipse, if its chord be the major diameter; and wholly without it, if it be the minor diameter. Also,. if upon the major diameter there be described a semicircle, or any. greater arc, such are will be entirely without the ellipse; but if, on the contrary, it be described on the minor diameter, it will be entirely within the ellipse. (46.) Returning to equations (1) and (2) in the preceding problem, we find for their product aa' = t2YA2, or, substituting for B2 x2 -A2, its value in (4,) ad = -A2 This equation shows that, the product of the trigonometrical tangents of the two angles, formed by lines drawn from the extremities of a principal diameter to meet in B2 the curve, is constant and equal to -A A representing the half of that diameter which subtends the inscribed angle. PROBLEM II. (47.) To find'the expression for the distance of any point in the ellipse from the focus. (See Diagram to Art. 41, p. 119.) Let (x, y) be any given point; P, in the ellipse. Then FP2 = (x- c)2 + y+2 = xm 2cx + c2 + y2; but, from equation (10), p. 122, we have y2 = (1 -e2) (A2 x2); hence FP2 x= - 2cx + c2 (1 -e2) (A2- x2) = A2 2Aex + e2 x2. FP A -ex (1); and,- since FP + -FP must be equal to 2A, we therefore have F/P = A + ex (2); hence, (1)'and (2) are the expressions sought; and we may conclude that the radius vector of any point in the ellipse is always a rational function of the abscissa of that point. For the difference of the focal distances we have F/P - FP= 2ex; and for their product, FP ~ F'P -A2 - e2 x2. ANALYTICAL GEOMETRY. 125 Properties of the Ellipse when related to its conjugate Diameters, (48.) The foregoing are the principal properties of the ellipse, which are directly deducible from its equation; when the principal diameters of the curve are taken for axes of coordinates. By referring the curve to other ssystems of coordinates, we shall obtain equations leading to other properties; we shall, therefore, now inquire what form the equation assumes when the axes of coordinates do not coincide with the principal diameters, but make any angles whatever with them, the origin remaining the same. It appears from equation (2), p. 116, that in order to transform the coordinates from rectangular to oblique without displacing the origin, we must substitute for x and y in'the equation of the curve, the values x = x cos. a + y cos. a', and y =.x sin. a + y sin. a/; making, therefore, this substitution in the equation A2 y2 + B2 x2 A2 B2, it becomes A2sin.2' I Y2 +2A sin. a sin, fX a xy + A2 sin.2 a x AB B2 cos.2 a'' 2B2 cos. a cos. a./'B2 cos.2 o Such is the general form of the equation when the oblique coordi^ nates x, -y, make any proposed angles, a, a', with the major diameter of the curve. It is obvious that, if the term containing zy "were abb sent from this equation, it would then correspond in form to the primitive equation; and, in order, therefore,' that this correspondence may take place, the angles, a, a' must be' so related that we may have! A2 sin. a sin. a/' + B2 cos. G cos. a' = 0O; or dividing by cos. a cos. a', A2 tan. a tan. a' + B2 = 0; the relation, must be such that B2 tan.-a'= -a''2_ tan. (1) It hence appears that the transformed equation will have the same form as the primitive, whatever be the angle, a, provided that the other angle, a', be subject, to the condition (1). The angle a, therefore, being arbitrary, it follows that the systems of axes that may be chosen so as to render the transformed equation of the form (A2 sin. B. (A2 in B2 COs.2 ()x2=-` A2B2 (2) are limited in number. This equation will be more concisely expressed, and, at the same time, its analogy to the primitive equation more distinctly shown, if -we put A'2 for aA2 snd A2 sin'2 a + B2 cos.2 a A9B2 Bt' for 2 sin -B2 cos a' because then equation (2) may be put under the form 422 xy2+ As XI = 1, as will appear by making the foregoing substitutions for A/2, B'2, in this equation: hence, finally, A/'2 y2 -- B'2 = A/'2 B/'2 (3). Such, then, is the equation'of the 126 ANALYTICAL GEOMETRY. ellipse related to oblique axes originating at the centre of the curve, the angles a, a/, at which these oblique axes are inclined to the primitive axis of abscissas being related to each other, as in equation (1)..(49.) If, in this last equation, we put x = 0, the resulting value ot y will be the ordinate of the point where the curve meets the axis of y; and, if we put y =-0 the resulting value of x will be the abscissa of the point where the curve meets,the axis of x; in other words, these particular values of y and x will be the value of those sermi-diameters of the ellipse, which have been taken for axes. Now for x 0, y =~ —+ B', and for y = 0, x = + A'; hence the semi-diameters in question' are A"'and B'/. We may therefore conclude that the equation (3) is not only similar to the equation A2 + B2 x2=A2B2 in form, but also that in the same manner as the semi-diamieters A, B, enter this equation, so do the semi-diameters A', B'; enter the former. Equation (3) furnishes the following properties, viz. 1. Each diameter, 2A', 2B', bisects all the chords drawn parallel to the other, as was shown of the principal diameters, 2A, 2B2 (42.) Diameters possessing this property are called conjugate diameters, hence the equation A'2 y- + B'2 x2. A'2 B'2 is the equation of the ellipse reJfrred to conjugate diameters. To distinguish the principal diameters from the other systems of conjugate diameters, they are generally called the conjugate axes of the ellipse; the longer is also sometimes called the transverse axis, and the shorter its conjugate. 1. Straight lines drown at the extremities of a diameter, 2A/, parallel to its conjugate, 2B', are tangents to the curve (42.) 3. The squares of chords drawn parallel to one of two conjugate diameters are as the rectangles of the parts into which they divide the other (44.) These properties are established as in the articles referred to. 4. If any number of parallel chords be drawn, the line which bisects them all will be a straight line passing through the centre. For if a diameter be drawn parallel to the chords, that which is conjugate to it must bisect the chords; and, therefore, must coincide with the former line. Hence the solution of the two following problems, viz. 1. A diameter being given, to find its conjugate. Draw a chord parallel to' the given diameter, and the line bisecting both will be the diameter sought. 2. To find the centre of an ellipse. Draw a line to bisect any two parallel chords, and we shall thus have a diameter; bisect this diameter, and the centre will be determined. (50.) From equation (1) we may infer that no system of conjugtate diameters can bed perpendicu lar to each other, except the principal dia ANALYTICAL GEOM ETRY. 127 meters. For, that the diameters may be perpendicular to each other, we must have the condition, (11) tan.'- t, but, by equation (1), this cannot be, unless A = B, that is, unless the curve ceases to be all ellipse, and becomes a circle; this teaches us, however, that each system of conjugate diameters in a circle includes a right angle. As to the principal diameters of the ellipse, the equation - = B2 — in which the above condition is implied, tan. a' - A2 tan. a does subsist, for then, a being 0, this equation is the same as ccn = c. Equation'. (1) moreover shows that if one of the tangents, tan. a, tan. a', be positive, the other must be negative; consequently, accordingly as the axis of x is above or below the major diameter of the curve, so will the axis of y be to the left or to the right of the minor diameter of the curve; for it can make an obtuse angle with the major diameter only in the former position, and an acute angle only in the latter. Again, from the same equation, it follows, that the product of the tangents of the two angles which a system of conjugate diameters make with a' principal diameter is constant, for that equation gives tan. a B2 "an. a'/ -- But it has been shown (46) that if a, a', be the tangents of the angles which two straight lines drawn from the extremities of a principal diameter to a point in the curve make with that diameter, we B2, B2 must have a a' - or a' =- It follows, therefore, that A" r~2a' if from the vertices of a principal diameter two lines be drawn to meet in the curve, the diameters parallel to these will be conjugate, and conversely, chords drawn from the vertices of a principal diameter parallel to a system of conjugates meet in the curve. Art. (50) might obviously nave been inferred from this property. Chords drawn from a point in the curve to the extremities of a diameter are called supplemental chords Hence (45) of all systems of conjugate diameters, those contain the greatest angle which are- drawn parallel to the equal supplem?.ental chords from the extremities of the miajor diameter. These last diameters are also equal, for they form equal angles with the major diameter (43). When the conjugate diameters, 2At, 2B', are equal, the equation of the ellipse related to them is y2+-i x = A/2, which corresponds in form to the equation of the circle, and which curve it would characterize, were it not that here the axes of reference are oblique. (51.) It is obvious, from the condition (1); that if any diameter of 1 28 ANALYTICAL GEOMETRY. an ellipse be represented by the equation y = ax, the conjugate B2 thereto- will be represented by the equation y = x, the principal diameters being taken for axes. If from the extremities of any dia. meter, 2A', supplement chords be drawn, and they be referred to the semi-conjugates, A'9 B', as axes, their equation will be y = nm (x+ A') and y =m' (x- A'), since the' one chord passes through the point (- A', 0), and the other through the point (A'/, 0). Consequently, by imitating the steps in art. (46), where the principal axes were employed, we arrive at the analogous property when any system of B'2 conjugates are taken for axes, viz.'nM' - in which equation it is to be observed, that the symbols m, m', do not denote the tangents of the angles which the chords form with the axis A', but they denote the respective ratios of the sines of the two angles which each chord makes with the axes A', B', these being oblique. (52.) The several properties of the ellipse; which we have just noticed all' immediately, flow from the equation of the curve, when referred to the oblique, axes, A', B'. If now we return from these to the original rectangular axes, A, B, by a transformation of'the equation, other properties will unfold themselves: let us, therefore, effect this transformation. As we here propose to.pass from oblique axes to rectangular, we must substitute for x and y, in,the primitive equation, the values in equation (4/), (art. 39).x sin. a' — y cos. a ycos.as.- x sin. a sin. [A', B]' and sin. [A', B'] the new axis of x bleing situated below the primitive axis. Making, therefore, these substitutions, in equation (3), p. 125, the transformed will be A'2cos. ay2-2A'2 sin. a co. a x+A'2 sin.2 a x2='2 B sin.2.[A', B'] B'2cos.2a' -2B'2 sin. a' cos. a' B'2 sin.2 a' Now this equation is to be identical with A2 y2 + B2x2 2 A B2B; hence there must exist the following relations, viz. A/2 cos.2 a + B'2 cos.2a' A2 (1). A/2 sin.2 a + B'2 sin.2 a' B2 (2). A/2 sin. a cos. a + B'2 sin. a' cos. a/'- O.=. (3) A' B'2 sin2 [A'; B'] A2B2.... (4). By adding together equations (1) and (2), we obtain the property A'2 +.B/2 = A2 + B2 (5), that is, the sum of: the squares of any systen -of conjugate diameters is equal to the sum of the squares of' the principal diameters. (53.) From equation (4) there results 4A' B' sin. [A/ B/] +4AB. (6). Now the first member of this equation expresses the surface of a parallelogram, of which the adjacent sides are equal to the conjugate diameters, 2A/, 2B', and included angle equal to the angle [A', B'i; ANALYTICAL GEOMETRY. 129 and the second member of the equation represents the rectangle of the principal diameters, 2A, 2B. Now, if through the extremities of each of the two conjugates, A'B', C/D/', parallels be -Irawn to the other, they will be tangents to the c' ellipse, and the. angle 0' will be equal to the B angle 0, that is, to the angle, [A' B']; hence the parallelogram formed by these tangents is equal to the rectangle of the principal diameters. Equation (6) therefore expresses this theorem, viz. Any parallelogram circumscribing an ellipse, and having its sides parallel to a system of conjugate diameters, is equivalent to the rectan gZe of the two axes. Since systems of conjugate diameters are unlimited in number (48), it follows- that an infinite number of circumscribing parallelograms mnay be found all equal in surface. Of these but one will be a rhombus, viz. that of which the sides are parallel to the equal conjugate diameters; and but one will be a rectangle, viz., that of which the sides are parallel to the principal diameters. It can be shown, conversely, that if a parallelogram circumscribint an ellipse-be equivalent to the rectangle of the principal diameters, its sides must be parallel to a system of conjugate diameters. For let PR be a circumscribing parallel- - ogram, the sides of which are not parallel (' to a system of cdnjugate diameters. Draw. a diameter,'/B/, parallel to one of the sides 0 so PQ,, and let C'D' be the conjugate to this diameter, and complete the parallelogram P'W', having its sides parallel to the system p of conjugates just drawn. Then, since A/B' is parallel to PQ, but does not meet the parallels SP, ROL, it is less than PQ,, but it is equal to.P'o/; therefore P'Q' is less than PQ;o and as both parallelograms are between the same parallels, PQJ, S'/; P'R' must be less. than PUR, but P'R' is equiivalent to the rectangle of the principal diameters; hence PR is greater than that rectangle, It follows, therefore, that of all parallelograms -ci7-cumscribing an ellipse,. those about co y'tgate diameters are least in surface - AB:From equation (6), A/B' = and adding twice this sin. [A, B/] to equation (5), and extracting the square root we have 2AB v(',,2 +- 2A'B/ - B/2) or A." -- B/ = (A2 + B2 sin., consequently A'+ B' is greatest when sin. [A', B'] is least, that is, when the obtuse angle [A/, B/'] is greatest, and A' -B' is least, when sin. [A/, B'] is greatest, that is when [A/, B'] is a right angle; and hence of all systems of conjugate diameters, the sum of th which are 17 130 ANALYTICAL GEOMETRY. equal is the greatest, and the.sum of those which are rectangular is the least. We shall terminate this division of the present chapter with the following additional problems. PROBLEM III. (54.) The axes of an ellipse and the vertex of any diameter being given to find the length of that diameter, and of its conjugate. Let (x', y/) represent the given vertex of the diameter 2A', the length of which is required. Then the distance of (x', y') from the centre of the ellipse is A/2 x/'2 + y'/2; but by the equation of the B2 curve, y/2 = B2 x/2 hence, by substitution,'2=B 2A2 X/2=B2+ C /2= B2+e2'2..A= V(B2 +- e2 ); also since A-2+ B'2 = A2 + B2.. B/2 = A2 e2 x/2.. B/ =/ (A2 - e2'2), ad: these values of A' and B' are the expressions sought. If the radii vectores of the point (x', y') be F'P, FP, then (47) F/P' FP = A2 - e2X/2. Hence the product of the radii vectores of any point is equal to the square of the semi-diameter conjugate to thai passing through the point,that is F/P FP = B/2. PROBLEM IV. (55). The axes and the inclination of a system of conjugate diameters being given to determine them in length and direction. Let A', B' represent the semi-conjugates; then, from equations (5) 2AB' and (6), p. 128, we have (A' + B')2 = A2 + B2 + 2AB 2AB and (A'- - B')2 = A2 + B2 -]; Consequently, by addition and subtraction, 2AB ~ 2AB' A-' t It S aA + -t- B2 + 1,I A2 B2 sin. A', B/] 2 sin. [A/, B-] 2AB VA+B_ 2AB B' ~'/A-+B2+ [A -'B' 2 a sin. [A/, B/] These are the expressions for the lengths of the semi-conjugates. It remains, to determine at what angles these are inclined to the major diameter, 2A. We have seen (51) that if the tangent of the angle at which A' is B2 inclined to the major diameter be a, then will- -a be the tangent of the angle at which B/ is inclined to it; hence for the angle [A/, B'], at which the.conjugates themselves are inclined, we have, if we denote its tangent by v, the expression (1 1), ANALYTICAL GEOMETRY. 131 a_ -a_ (A2 e2~ B2) 1 —+ aa_ (Aa q- B2) - Hence, by reduction, we obtain 1- aat - (A' — B') a' B2 B2 the quadratic a2 + (1 -A) va =- A which, solved, gives a — A2 B(i2 ) B)v:i v; (A2- B2)2%2 _4AIB2I l 2AB If the given tangent, v, be less than _2B2, abstracting from the sign of v, the question will be impossible, that is, no system.of conjugate diameters can have the, proposed inclination, a fact which the above expression for a plainly intimates, and which we previously 2AB knew from (45). If v be equal to A2,B2,then will A', B', be the equal semi-conjugates (45, 50), in which case a = - B - A, consequently, a' — B + A, the inclinations being supplements of each other. For the lengths of the equal semi-conjugates we have, from the property (5), p. 128, 2A'2A2 + B2. A' -- / }(A2 + B2). The geometrical construction of the. preceding problem is very simple: On the major diameter, AB, of the ellipse describe a circular arc, capable of containing the proposed angle, if it be obtuse, or its supplement if acute. Then from' ca either of the' points, P, P', in which it intersects the,~ ellipse (45), let chords be drawn to A, B; then dia- < meters drawn parallel to either of these systems of supplemental chords will be conjugate, and will include the given angle. The two values of a in the preceding analytical expression agree with the two systems of conjugates here constructed. If the arc pass through C, it will touch the ellipse at that'point, so that P, P/, will coincide, and the conjugates sought will be equal. Properties of the Tangent to the Ellipse. (56.) In order to obtain the equation of the tangent to the ellipse, let us first, as in the circle, consider a secant to the curve, or a straight line cutting it in two points, (x/, y') and (x", y"). The equation of this, secant is y y- =,,, - ), (1)y; and as both points' are on the curve, there. subsists the equations A/2yn2 -+ -B'2x'22 = A/2 B2, (2), A/2yf2 //2 + -B/2 /2 - A2.. (3). Hence, subtracting (3) from (2), A'2 (y + j') (yi) = — B',.+2 (X,', (' -X) and from this we get therefore equation( X KA2' f + y therefore equation (1) 132 ANALYTICAL GEOMETIRY. B/2 x/q- XI/ becomes, by substitution, y y - A/ 2. (-x') ~ (4) A2,y y/1 This, then, is the equation of the secant passing through the two points (x/, y'),(x", y") of the ellipse; whose equation is A'/2y2 + B/'22 =A/2 B/2 If now we suppose these two points to coincide, the secant will become a tangent; making, therefore, x/ = x/, and y' = y/=; 8 /2 x' we have y - - y'=, (x-x'); or more simply, by reduction, A'/2y' y+ B'/2 x'X= A'2B'2, (5), the equation of the tangent related to any system of conjugate diameters. The second form of this equation is very easily retained in. the memory, from its resemblance to-the equation of the curve' the only difference is that, in the equation of the tangent, a' x occurs instead of a2, and y'y instead of y2. Connected with the tangent are several other lines, which it is requisite to consi-, Y der. Thus, if AB, CD are the principal diameters of an ellipse, and the tangent N PR be referred to them as axes, then the A f of -JpBj j X distance, MR, of the ordinate of the point of contact from the intersection of the tan- gent with the axis of a, is called the subtangentI the perpendicular, PN, to the tangent from the point of contact, is called the normal, and NM, the distance of its intersection with the axis of x from the ordinate, is called the subnormal; moreover, in. estimating the, length of the tangent, we consider only the portion PR, included between the point of contact and the axis of x. Hence on one side of the ordinate of the point of contact are situated the tangent and subtangent, PR and MR; and on- the other side the normal and subnormal, PN' and MN. We shall now proceed to deduce the equations of these lines, and to determine analytical expressions for their several lengths. (57.) For the tangent the equation has already been exhibited when any system of conjugates are employed for axes. For the.rectangular system, 2A, 2B, the equation is therefore B2 x/ Y y — A. y/ (a -'/); or A2 yy/ + B2 x;2x A2B2. B2 x' The coefficient. in the first of these forms, which,is that most frequently referred to, is the trigonometrical tangent of the angle PRX; it is obviously negative for positive values of x, and positive for negative values of x, since the angle is- obtuse in the former case, and acute in the latter. For the normal the equation is immediately deducible from that of the tangent. We shall have merely to characteiize a perpendicular to the latter, drawn from the point of contact ANALYTICAL GEOMETRY. 133 (x', y'). The equation of this line is therefore y y;=27, (x'-x').' The length of the subtangent is easily derived from the equation of the tangent, for supposing in that equation y = 0, which is the case at the point R, the resulting value of x will express the length, OR; that is, xZ =OR = A2 X/ and, if from this we deduct x', or OM, we shall have, for the length of the subtangent, A2 2 MR -. In like manner, if, in the equation of the normal, we suppose y; 0, which is the case at the point N, the resulting value of x will express the lengrth, ON; and this subtracted from OM, or x', will give the length of the subnormal, MN. We have, therefore, only to express the value of x/ - x in the equation of the norB2 mal when y = 0, which value is MN = x. From these expressions for the subtangent and subnormal are obtained those for the tangent and normal. Thus, since PR = v IMR2 + PM2I we have by substitution, PR = V/(A2 2)2 Y2}, or because )2 = BA2z') this becomes PR =' (A2- X) B) X/2 A2(A2Likewise, since PN = v /MN2 + PM2}, we have, by substitution, B4 B2 B A2 PN = 4 x A —2 (A2_ —xX2) E I( E 1) x'/2 + A2} = - (A2- e2 x'/2). (58.) For more convenient reference, let us now collect the pre-, ceding formulas together. The equation of the tangent is y -y/= / (x-x /). A2y''rhe equation of the normal is y- y' = BY2' (x -x). The length of the tangent is T / (A2- + B2 (A A2 X/2 The length of the subtangent is T, - B2 The length of the subnormal is N. X. (59.) We may now by aid of these formulas deduce some other properties of the ellipse. And, first, we may remark that, as the 134 ANALYTICA'L GEOMIETRY. expression for the subtangent is independent of B, it remains the same for all the values of B, so that in every ellipse described upon the axis 2A, the subtangent is the same for the same abscissa. This is true also when B = A, that is, when the ellipse becomes a circle; and, hence is suggested a method of drawing a tangent to an ellipse from any given point in the curve. Thus, let P be the given point; then, having described a semi-circle on AB, draw P'PM perpendicular to AB, and at P/ draw a tangent, P/R, to the circle; then - ~ draw the line PR, and it will be the tangent re-,A quired; for the points P, P' have the same subtangent, MR. It is further obvious, from the manner in which the expression for the subtangent has been obtained, that if the tangent had been referred to oblique conjugates, as in (56), instead of rectangular, the expression would have preserved the same form, that is, it would have been A'2 -/2 T. = and this independently of the sign of y/; so that a tangent to the curve through the point (x', y/), and another through the point (x', - y'), both meet the axis of x in the same point; and thus from any point without an ellipse two tangents may be drawn to the curve. If the point, of contact of one of the tangents drawn from a given point be known, and it is required to find the point of contact of the other, it may be done as follows: Draw a diameter to pass through the given point, and parallel to its conjugate draw a chord from the known point of contact, then the other extremity of this chord will be the other point of contact. It ought to be noticed here that we must not, as in the preceding case, conclude that because the value of the subtangent is independent of B/, and that this may therefore be equal to A', that the ellipse may become a circle, and yet the subtangent for the same abscissa remain the same. For the conjugate diameters of the ellipse, which are here taken for axes, are not conjugate diameters of the circle described upon either of them; because, in the circle, the systems of conjugates are all rectangular (50). (60.) By multiplying the foregoing general expression for the subtangent by x' we have T, X' = (A' +x') (A' - x'), which shows that the rectangle of the subtangent and abscissa of the point of contact is equal to the rectangle of the parts into which the diameter is divided by the ordinate. Thus, in the annexed diagram, OM ~ MR -=A'M MB'. ci This- property furnishes a commodious and expeditious method'of drawing a tangent to an ellipse Dock M j fro-'a.. point without the curve, when the centre of A. the ellipse only is given. ANALYTICAL GEOMETRY. 135 Let R be the given point, and draw RB'A' through the centre, and let CD be the diameter conjugate to -"nA' B'. Upon-A'B' and OR, as diameters, c/' \ describe arcs intersecting in m, m,' then - the line joining min, m/, will cut from OB' A m the abscissa, OM, of the point of contact sought, so that the parallel, P, P' to OC, through this point'will intersect the curve at the points where the tangents from R must touch it. For, as the circles upon A'B', OR intersect in the- line mMm', it follows that nmM Mmn' = A/M MB/ = OM MR; hence, fiorn' the above property, the tangent from P must -have the subtangent MR. (61.) By referring to (47) we find that the rectangle of the radii vectores of any point, P, in an ellipse, is F'P~ FP = - A2 e2.', and comparing this with the expression for the normal, we find, therefore, B2- F'P FP N A that is, the rectangle of the radii vectores of any -42 point in an ellipse is to the square of the normal as the square of the major axis is to the square of the minor. Also, since (54) A2- e2x2' B2, we have A ~ N- = B ~ B', that is, the rectangle of the major axis and the normal is equivalent to the rectangle of the minor axis and the semi-diameter parallel to the tangent. Thus in the annexed diagram, - F'P FP: PN2': AB2: CD2, and AB PN = CD ~OC'. A. D (62.) Let us now examine the equation of the tangent as exhibited in (56). If in this equation we make x = 0, the resulting value of 9 will be the ordinate, OT, at the origin; hence we obtain the property OT ~ y' or OT PM OC=. In like manner, by making y-=:O, int' T the same equation, we shall have for X,_ _t the resulting value of x the abscissa OR; hence OR" x' or OR' OM = OB2; hence, we infer, first, That the rectangle of the ordinate of the ellipse at the point of contact, and the ordinate of the tangent at the'centre, is equal to the square 6f that semi-diamzeter which is taken for axis of ordinates; and second, that the rectangle of the abscissa of the point of contact, and of the point where the tangent mneets the axis of abscissas, is equal to the square of that semi-diameter which is taken for axis of abscissas. Both these properties are also derivable from the general expression 136 ANALYTICAL GEOMETRY. for the subtangent, as will appear by adding x' thereto for the second property, and changing x' into y' for the first property. (63.) Let us now suppose, in the same general equation of the tangent, that x _ A' instead of 0, then the resulting value of y will be the ordinate B' t, which is also a tangerint to the curve at B', since it is parallel to the diameter conjugate to A' B' (see preceding diagram); In like manner, if instead of A''we put A' for x,'the corresponding.rditiate will be the tangent A' T'. Making then these successive B'/2 sulbstitutions for, we obtain the values y = B't- AY (A' ),.B/2 B/4 But, from the equation of the curve, y - (A'-xf); hence, by.. 2 substitution, A'T' B't = B', that is, if at the extremities of any diameter, lines parallel to its conjugate be drawn terminating in any., tangent to" the curve, their rectangle- will be equal to the square of'the semiconjugate to which they are parallel.' Still confining ourselves to the same general cr M, equation of the tangent, the conjugates A'B/,' C'D', being the axes of reference, let the supple- z mental chords, B' M, A/ M, be drawn, the former B parallel to the tangent at the point P, or (x', y'); D then we already know (51) that in the equations of these chords, viz. I m (x — A'), and Y = m/ (x + A'), the coefficients m, I', must B"2 be so related that m m'- =W —- (1). But, since B'M is parallel to the, tangent, the coefficients of x must be the same in the equations of these lines, that is, we must have mn- A2. It follows, therlefore, from the relation (1), that m' = y' + x'. Let now OP be drawn to the point of contact, and let ON be parallel to B'M, then the equation of OP is y = ax, and for the point (si','y"), y' ax';, therefore a = y' ml- x --', consequently OP is parallel to A'M, and-OP, ON are semi-conjugates, ON being parallel to the tangent at the' vertex. of OP. Hence we may conclude that diameters cdraiwn parallel to any systemn of supplemental chords are conjugate, which is an extension of the theorem at (50). As the principal diameters are the only conjugates, which contain a right angle, it follows that any systemof supplemental chords which include a right angle must be parallel to the principal diameters.' These latter. therefore, can always be found when we know'the centre of the ellipse: it will be necessary merely to describe a semi-circle on any diameter, ANALYTICAL GEOMETRY. 137 and to draw supplemental chords from the point where it cuts the curve, the diameters parallel to these will be those sought. It appears, moreover, that a semi-circle described on any diameter of an ellipse can cut it in but one point. (64.) From what has just been shown, it follows that, when any system of Conjugates are taken for axes, if the equation of any diamneter referred to them be y?nx,' the equation of its conjugate, will B/2 be y -- 2 x. Bearing this in mind, suppose that at the vertex of the diameter taken for the axis of x, a tangent is drawn, it must be parallel to the axis -of y; hence the part thereof intercepted by the line y = mx will be the value of y, which its equation gives for x = A', viz. y = mA'. In like manner, the part intercepted by the conjugate to this will be the value of y, which the equation of this conjugate gives for x= A', viz. yA5m Now the product of these two values is - B', it follows, therefore, that'if a tangent be'drawn at anypoint of an ellipse, the square of half the diameter conjugate to that from the point of contact will be equal to the rectangle of the two portions of the tangent intercepted between the point of contact, and any system of conjugates whatever. Thus, if TPt be a tangent, and OC" a semi- X diarneter, parallel to it, then OB', OC', being any system of semi-conjugates, we shall always have, OC"2 = PT Pt. 0 If the'tangent be drawn through the vertex of one of the least conjugate diameters, and be terminated by the'principal conjugates pro. duced. it will be bisected at the point of contact. For the least conjugates bisect the principal supplemental chords, and, as one of these is parallel to the tangent, both this and the tangent must be bisected by the same conjugate; it follows, therefore, that this tangent is equal to the least conjugate diameter. It is, moreover, evident that this tangent is the shortest that can be included between the prolonged principal diameters, for every other tangent included between them must exceed its parallel diameter, since the rectangle of two unequal parts of it is equal to the square of half that diameter.. It follows also that, if a tangent be drawn through the vertex of one of the least conjugatles, the portion intercepted by the prolonged principal diameters will be less than the portion intercepted by any other system of conjugates. (65.) If a tangent pass' thro'ugh the extremity of the latus rectum, and be referred to the principal diameters, the ordinate at the point of contact will be y' p2- B'' A, so that the equation of this tanB2 X'B2 ~- x'2 -' gent will be, y- = -;(X-X/), and ~ B+ — x A A A 18 M2 ];38 ANALYTICAL GEOMETRY. Or, since x' = C, y A A — ex. But (47,) A ez expresses the distance of the focus'from that point in the ellipse whose abscissa is x; hence the length. of any orcdinlte to a tangent through the extremity ofthe latus rectm is equal' to the distance of the fccus fromn the point wliere this ordinate- intersects the c, urve, therefore, the tanogent the ough the vertex ofl-he latus, rectum cuts from the tangent through thPe vertex of the major dialneter, a part equal to the distance of thefocusJ Jnom the vertex. Thus, in the annexed diagram, MN FP, also AT AF and BT' =- BF. (66.) If in the equation of the normal we put y = 0, the resulting value of x will express the distance, ON, (see the diagram at p. 132,) that is, we shall haveON- -T x' e2 x'; ihence, adding c or Ae to this, there results F'N -e(A exn'). Now, in the triangle PF'F, the sumn of the sides PF' + PF is to the base F' F, so is one of the sides, PF', to the distance intercepted between F' and the line bisecting the vertical angle, P,. (Geom. p. 90;) but 2'A; 2Ae:- A + ex': e(A + ex/) = F' N. It follows, therefore,'that the normal at'any point of the ellipse bisects the angle formed by the radii vectores qf that point,. and consequently the radii vectores are equally inclined to the tangent, also the- angle included by one radius vector and the prolongation of the other is bisected by the tangent. We can arrive, at the same conclusion without the aid ofthe-geometrical property here employed;. thus: Since the' equation of FP, passing through the points (c, 0) and (x/, y'), isy =' — (x-c), XI~- c y the coefficient - must express the tangent of the angle PFR, and - x- c we already know, from the equation of PR, that - is the tanA2 y' gent of the angle PRX. Hence, in order to obtain the tangent of the angle FPR, we shall merely have to substitute the preceding expressions, for a and a', in the formula (1I),v = which then be1+aa" come B2 cx'-(A2'2 x) In this expression if we sub(A2 - B2) xy, y' — A2 cy stitute for A2 y'2 + B2 Z,2 its equal A2 B", and for A2 - B2 its equal ci, B2 we shall finally obtain v = -. If the other radius vector, iA'P, C' ANALYTICAL GEOMETRY. 139 had been'employed, c would have been negative, and therefore the tangent of its inclination to PR would differ from that here deduced only in its sign; thus showing that the angles F'PR, FPR are sup. plements of each other, and therefore, that the angles FPR, F'PR', are equal. The same property admits of a simple geometrical proof; thus: Let one radius vector, F'P, be produced, till PG is equal to the other, FP, then the line PR, bisecting the angle GPF will be a tangent to the curve at the point P. For join NX' FG, then, since PR bisects the vertical angle of the isosceles triangle, PFG, it also bisects the base, FG, at -right an-. OR gles; therefore from whatever point, N, in PR, lines be drawn to F and G, we shall always have NF - NG ( Geom. p. 20); consequently, if there existed any point, N, in PR, besides P which was common to the curve, we should have F'N + FN = 2A and therefore F'N + — NG =- F'G, which is impossible,'so that PR, which bisects the angle FPG, is a tangent to the curve, and therefore, conversely, the tangent at P must' bisect the angle FPG. IfOMbe drawn from the centre to the middleof /'. G FG, it will be parallel to F'G, because O is the " I middle of F'F, therefore FO FF':: OM: F'G; ( / hence OM = -F'G, but F'G = 2A by construc-' A B tion, consequently OM = A, therefore, if from either focus a perpendicular to any tangent be E' - drawn, its intersection therewith will -be always at the same distance from the centre, viz. at the distance, OB, in other words, the locus of these intersections is the circumference of the circle described on the major axis as a diameter. If MO be produced to meet this circumference in E, and F'E be drawn, the triangles F'OE, FOM, having two sides, and the included angle in each equal are themselves equal, therefore F'E is both equal and parallel to FM, and is therefore the continuation of the perpendicular, F'M', from the us, F, to the tangent; hence, from the property of the circle ~ F'E -- F' NMF = AF' F'B, but AF' F'B = ( -c) (A + c) =A2 -c2 = B2; hence M'F' ~ MF = B2, trout is the rectangle of the perpendiculars from the foci upon any tangent to the curve is equal to the square of half the miner axis. Since the triangles F'PM', FPM, are similar, we have the equation FP which, multipled by that just deduced, gives MF- FP' BFTP A + ex' M/F _ B2 __ _ and, multiplying its reciprocal FP A- ex' 140 ANALYTICAL GEOMETRY. FP A ex/ by the same, we have MF2 = B2 B2 * A- X being the F'P A-[- exb abscissa of the point P. (67.) The property in (66) furnishes a simple method of drawing a tangent to the ellipse from a point either in the ( curve or without it, when the foci are known. -P-4 Thus, if the point be in the curve, as at P, then. / it will be necessary merely to draw the lines —- FP, F'PG, and to bisect the angle FPG. If the point be without the curve, as at N, then,, from the focus, F', as a centre, with a radius equal to the principal diameter, describe an arc and from the given point, IN, as a centre, with its distance from the other focus, describe another arc, intersecting the former in the point G, then the line F'G will cut the curve in'the point through which the required tangent must pass. For, let NPR be drawn through this point and draw PF, NG, NF; then, since, by construction, F'G - FP + PF, PF PG, also NF - NG, consequently the triangles NFP; NGP, are equal, therefore the angles NPF,1 NPG are equal, as also their supplements FPR, GPR, hence NPR is a tangent to'the ellipse. As the distance, F'N, of the centres from which'the arcs intersecting in G are described is less than the sum, and greater than the difference, of the radii, (Geom. p. 19,) it follows (28) that these arcs intersect also in another point, and thus two tangents may. be drawn from N. We shall terminate the' present chapter with the following problem. PROBLEM V. (68.) Pairs of tangents to an ellipse being always supposed to intersect at right angles, to find the locus of the points of intersection. Let MT, NT, be any pair oftangents intersecting at right angles in and, parallel thereto, draw FP, F'P', from the pT pl foci, then the points P,'P''will be in the circum-. ference of a circle described upon the major /, diameter, AB. Produce PF to meet this circum- /'/\, ference again in M', draw M'T', FM, each per- A' F F0 i pendicular to the tangent TT', and lastly, draw. the chords PP', M, M', the points M, M' being G' —. obviously on the same circumference as the points P, P'. Then, since TM' is a rectangle, PT- M'T', and on account of the parallels PM/, P'M, the arcs PP/, MM/ are equal, their chords are therefore equal; hence the rectangles PP', MMA/' are equal, and TP' =-TM - FM', c-n.sequently TM TP/ = PF' FPM', but, by the property of the circle, PF P FM/ = AF' FB and' TM TP/'.= Tt2; Tt being a tangent to the circle, therefore Tt2- = AF FB. a constant quantity ANALYTICAL GEOMETRY. 141 and, as the radius, Ot, is also constant, the distance, OT, must be constant, therefore the locus of T is a circle of which the radius OT is Vi(A2 + A2- c2) _ (A92 + B=2). CHAPTER III. ON THE HYPERBOLA. Its equation and Properties. (69.) An hyperbola is a curve from any point, P. in which, if two straight lines be drawn to two fixed points, F, F', P their difference shall always be the same. The given points, F, F', are- called the foci of' the hyperbola, and the lines, FP, F'P, drawn there- PA B from to any point, P, in the curve, are called the -' r radii vectores or focal distances of that point. This curve may be described, by means of points, thus: From one of the foci, F, as a centre, with any assumed radius, describe an arc, and from the other focus, F', with any other radius exceeding the former, describe a second arc, intersecting the first in two points, P, p. Let this operation be repeated with two new radii, taking care that the second of these shall exceed the first by.the same difference as before, and two new points will be.determined; and this determination. of points in the curve may thus be continued till its tract becomes obvious. That the locus oF-these points will be an hyperbola is plain from the definition, since the distance of any one- of them from F' always exceeds its distance from F by the same constant difference. *If of each pair of intersecting arcs employed- to determine the several points, we had supposed the greater to have been described fror F, and the less from F'/, the same constant -difference being preserved, we should obviously have obtained a series of points, P', p', &c. equally belonging to the hyperbola, although none would be situated in the locus of the former series. It appears, therefore, that the hyperbola consists of two separate branches, PBp, P'Ap'. Any portion of this curve may be described by continuous motion,.by employing a ruler and a cord. Thus, let a ruler, FR, be fixed -to F, so that it may be turned round this point, in the plane-whereon the curve is to be described; then having assumed any other point, F'/, in this plane, connect, it by means of a cord shorter than the ruler to the extremity, R; then a pencil, P, keeping this cord always stretched, and at the same time pressing against the edge of the ruler, will, as the'ruler * The analytical investigation of this problem will be given hereafter. 142, ANALYTICAL GEOMETRY. revolves round F, describe an arc of an hyperbola, of which F, F', are the foci, for the difference of the distances of the describing point, P, from the fixed points F, F/, will be always the same. (70.) Let us now seek the equation of this curve, by means of its characteristic property. C, Draw F'F, and let 0, the middle point of / this line, be taken for the origin of the rectangular a.xes, and let the constant difference of F 1 the radii vectores of any point in the curve be F represented, as in the ellipse, by 2A. Put c for OF, or OF', and x, y, for the coordinates of any point, P, in the curve. Then we shall have these equations, viz. y2 + (x _c)2 2_ pF2, (1), y + ( + C)2 = PF/2 (2) and FPt - PF - 2A, (3). Hence, by first adding and then subtracting equations (1) and (2) we have 2y-+2X2+22c2-=PF'2_+-pF2 (4), 4cx(PF/' + PF) (PF/ - PF),.. PF + PF 2cx + A,+ Combining this with equation (3), we have, PF' A and PF=CA- A, and these values substituted in equation (4) give: y2 -4- 3 c2 = -D- + A2 whence A2 y2- +(A2 -c2) x2=-A2(A2-C2), (5). This A2 equation would be identical with that at p. 120, and would thus characterize an ellipse, were it not that here A represents half the difference, instead of half the sum of the radii. vectores. With this condition, therefore, equation (5) is the equation of the hyperbola. (71.) Iny order to determine the points of intersection with the axes, suppose y =0, in equation (5), and there results for x the value - =: A, which intimates that the hyperbola, like the ellipse, intersects the; axis of x in two points, B, A, equidistant from 0, the one to the right, and the- other to the left, and that A expresses this distance. If, in the same equation, we suppose x 0,- we have for the corresponding value of y the expression y _.v A2- c2I. Now since 2c is the base, and 2A the difference, of the sides of a triangle, PF'F, it follows (Geomn. p. 19) that c must exceed A, -and consequently the value of y, in the foregoing expression, is impossible; the curve, therefore, can never meet the axis.of ordinates. In this respect, therefore the hyperbola differs entirely froml the ellipse. Let us, however, mark on- the axis of ordinates two points, C, D, each at the distance of v/ C2- A2I from 0, and, calling this distance B, we shall then have the equation of the hyperbola in a form analogous to B2 that of the ellipse, viz. A2 y2- B2 x2 - A2B2 or y2 X2(x-2 A2) 6. The only difference between this equation and that of the ellipse ie that here the sign of B2 is negative. The general expressions, there. fore, for the coordinates of any point in the curve is ANALYTICAL GEOMETRY. 143 x =-~ A/iy2_[_ B2I;y A z/ - X2_ A"3. From the first of these expressions it appears, that there are two values of X numerically equal, but differing in sign for every value of y. ~ We conclude, therefore, that chords drawn parallel to AB are bisected by CD, or its production. In like manner, in the second expression, we are furnished with two values of y numerically equal, but differing in sign for every value of x. If, however, x be assumed numerically less than A, the resulting value of y will'be imaginary; now x is less than A for every ordinate drawn -between the points A and B, hence none of these ordinates can meet th6 curve; but, if.x = A, then, since y = 0, we infer that parallels to CD, drawn through the points A and B, are tangents to the curve. It further appears, from'this expression, that so long as x is assumed numerically greater than A, there will always correspond a possible value of y, which'will increase as x increases; hence the'two branches of the'curve are unlimited, proceeding onwards, in opposite directions, to infinity. Let us actually suppose, that x takes a succession of values from t A to infinite, then, putting the.expression for y under the form y = -- 1 -.A we see that as x increases, the fraction A2 + xa diminishes, so that the values of y go on approaching to'the value - Bx A, which value, however, is never reached till x becomes infinite, rendering the fraction A2 + x nothing. It is obvious from this, that if through M' K the origin, 0, two, straight lines, KL, MN, be drawn, making angles, KOX, MOX; with AB whose tangents are respectively + B + A and -B - A, these two lines will continually approach the curve, and yet can never meet it. For'B B the equation of these lines are y= -x'and y= — x, and it has just been seen that in the curve y can never be so-great as ~ Bx+-A, till x becomes infinite, although throughout the course of the curve, y continually approaches to this value, that is, the differences between the ordinates of the curve- and those of the lines just drawn, for the same abscissas continually diminish as the abscissas increase. The two lines KL, MN, are called asynmptotes to the hyperbola. (72.) Let us now, as in the'ellipse, seek the' expression for the distance of any point (x, y) in the curve from the origin, 0. For this purpose, we must substitute for y2 in the expression B.2 D = lx-+y2l', its value A (x' —A2), which reduces it to D = +/ A' x2'-Ba'. This expression is obviously independ 144 ANALYTICAL GEOMETRY. ent of the signs of x and y, and, therefore, it may be shown here in precisely the same way as the. fact was established with regard to the ellipse, that every chord passing through 0 is bisected at that point, and hence 0 is called the centre of the curve, and chords drawn through it, diameters.' Indeed, all lines drawn through the cenltre of the hyperbola are, for the sake of uniformity, called diameters, although an infinite number may be drawn, so as not to meet the curve, viz. all those comprehended between the asymptotes, or that inclined to the axis of x at an angle of which the tangent is not numerically less than B A. The asymptotes may, therefore, be regarded as separating those diameters which are chords, called transverse diameters, from those which are not chords, called second diameters. Of the former, least is AB, since the above expression for D is least when x is least, that is, when x = ~ A; but, as there is no limit to the value of x, the other transverse diameters increase from this least value to infinity. (73.) As in the ellipse, so here, 2A., 2B, that is, AB, CD, are called the principal diameters, or the principal axes of the curve. The distinction of these diameters into major and minor, as in the ellipse, cannot, however, be here used with propriety, for in the expression v c2 — A2f, which B represents, it is only necessary that c exceed A (p. 142,) so that B may either be greater or less than A, or indeed equal to it. From these remarks then it appears that the equation B2 A2 y2 B2x2 = -A2 B2 or y2 _ (X2- A2), (6), is the equation of the hyperbola related to its principal diameters. If we suppose B = A, the equation is y2- x2 - A2. In this case, the hyperbola is called equilateral, on account of -the equality of its principal diameters. Thus the same modification transforms the equation of the common hyperbola into that of the equilateral hyperbola, that changes the equation of the ellipse into that of the circle. We may here remark that, in the equilateral hyperbola, since B + A = 1 = tangent of 450, the angles which the asymptotes make with the axis of x are 45~, and 900~ 45~; hence, in this case, the asymptotes are perpendicular to each -other. (74.) By removing the origin of the axes of coordinates from the centre, 0, to the vertex, A, of the transverse axis, by a transformation similar to that employed in the ellipse, the equation becomes B2 -2y2 B2x2 + 2AB2 x = 0, or y2- (x2 — 2Ax) (7), the equation of the hyperbola when the origin is at the vertex of the transverse axis (75.) If we wish for the equation in terms of the eccentricity, c, we may obtain it from equation (6), by substituting for B2 its equal c2A, which gives y = A (a - A ) or putting e for A ANALYTICAL GEOMETRY. 145 y2= (e- 1) (x2 - A2) the equation of the hyperbola as a fitnction of the eccentricity. From the intimate, analogy which subsists between the equations of the ellipse and those of the hyperbola, it may easily be conceived that the principal properties of the former curve belong also to the latter. This is in fact the case; and hence most authors exhibit the theory of these two curves in conjunction.'With a view to simplicity, it has, however, been here thought preferable to consider these curves separately; but, as we propose to develope the properties of the hyperbola by imitating the steps which led us to those of the ellipse, we shall frequently have occasion to refer to the preceding chapter for details, which need not be repeated in this. Properties of the Hyperbola related to its principal diameters. y2 B' (76.) From equation (6), x - 2 hence: (A ) (x -A):: B2A2. Now x+-A.and x —A are the distances of the ordinate, y, from the vertices of the transverse axis, hence the square of any ordinate is to the product of:its distances from the vertices of the transverse axis as the square of the conjugate axis is to the square of the transverse;, consequently the squares of the ordinates are as the products of the parts into which they divide the transverse axis. If the hyperbola is equilateral, that is, if B A, then y2 _ (x + A) (x- A); so that, in the equilateral hyperbola, the square of any ordinate is equal to the products of the parts into which it divides the transverse axis, a property analogous to that of the circle. (77.); From equation'(8) the parameter or double ordinate through the focus is easily determined, for, putting c for x, in that equation, and extractin'g-the square root of each side, there results Ay c2 A2 = B2, in which equation y is the semi-parameter; therefore, calling the parameter p, we have p = 4B2 + 2A that is, the parameter is' a third proportional to the transverse and second axes. Hence the equation of the hyperbola, as a function of the parameter, is'obtained by substituting, in equation (6), p + 2A' for its equal, A2 so that (x _ A2), is the equation of the curve, in terms of the parameter. In the equilateral hyperbola, since 4B2 2A = 2A, we have p = 2A, that is, the parameter is equal to the transverse axis. PROBLEM I, (78.) To find the expression for the angle contained by supplement. al chords drawn from the extremities of the transverse axis. XN 19 146 ANALYTICAL GEOMETRY. Referring to the corresponding problem on the j ellipse, we find that the tangents of -the angles which the supplemental chords, meeting'in the point (x, y), make with the axis of x are A a —+-Y- and a -' (1), and that, consex + A x A quently, the general expression for the angle P, formed at the point (i, yU,:is tal. P 2A = (2).' Now the point (i, y) being in'(x,i2 - A2 the hyperbola, we must have, from the equation of the curve, X2 - A2 = A2y2 + B hence, by substitution, equation (2) becomes tan. P= ( +- BI).'. (3), the expression sought. As this result is independent of the sign of x, it-follows that the angle is the same, whether the chords be drawn to (x, y) or to (- x, y). Since the preceding expression for the tangent is always positive for any point in the curve above the axis of X, it follows that all the angles subtended by the transverse axis are acute. These angles diminish as y increases, till they become 0, when y is infinite. Multiplying the expressions (1) together, we have ya 2 A B2 aa/ =c,,Y A2 — A2 (4), therefore the product of the trigonometrical tangents of the two angles formed by the transverse axis produced to the right, and its supplemental chords, is constant and equal to B2 + A2. As this product is positive, we conclude that the two angles must be either both acute or both obtuse. A A If the hyperbola be equilateral, then tan.- P = (2 A) and aa' 1 1 or a - but, since tan it follows that a,-a' must represent the'tangent and cotangent of the same angle, consequently, in thte equilateral hyperbola, the angles which the supplemental chords make with the axis of x are together equal to a right angle. PROBLEM II. (79.) To find the expression for the radius vector of any point in the curve. C Let x, y be the coordinates of any point, P, in the curve, then FP2 (x —c)2 y2 = x2 — 2c a_ /// +- c2 + y2. Now equation (8) p. 145 gives 0' 0 x 2= (e- 1) (x2-A.2}) hence by substitution, FP'2 = x2 - 2ex q+ c'2 (e2 1) (x2 -- A2)- D'x2 - 2Aex + A2..P P ex- A, and since FTP - FP = 2A, it ANALYTICAL GEOMETRY. 147 follows that F'P ex -;- hence the radius vector of any point is always a rational function of the abscissa. For the sunrl of the radii vectores we have FP + FTP = 2ex, and for their product, FP F'P = ex2 - A2. Properties of the Hyperbola, when related to its conjugate diameters. (80.) By transforming the equation of: the hyperbola from rectangular axes to oblique, the origin still remaining at the centre of the curve, we shall have the equation A2 sin.a' y2 + 2A2sin a sin.. a' xy + A2 sin.2 t X= - A2 B2 -r2 cos.2 a' - 2B2 cos. a cos. o' - 12 cs.2 a where a, a' denote the angles which the oblique coordinates x, y make with the primitive axis of x. Let us now, as in the ellipse, determine the relation which must subsist between the angles a, cc', in order that the term containing xy.may'disappear from this equation. That such may be the case, the coefficient of xy must obviously be assumed equal to 0, or dividing this coefficient by the expression 2 cos. a cos. a', we must have the equation A2 tan. a tan. a' B2 = 0. Hence the relation between the B2 angles a, a' will be thus expressed, viz. tan. a/ A (1), so A2 tan. a that one of the angles being chosen at pleasure, the other will be determined by this equation, and thus an infinite number of oblique axes exist that will render the transformed equation of the form (A2 sin.2 a' -B2 CoS.2 C') i2 ~ (A2 sin.2 -32os.s.2 a) x2 - A2 B2.... (2), This equation will be simplified- by putting A 2for, A sin., andB'2 for a' sin.2 a- "cos.Aa A2 sin.2 a' B2 cos.2 a for then equation (2) may be put under'the form B x2 =- 1A as will readily appear, by making the proposed substitutions. Hence finally, A/2y2_ Bx 2 — 2 - An Ba (3). This, therefore, is the equation of the hyperbola, when the oblique axes of reference originate at the centre and form angles a, a' with the primitive axes of x, related as in equation (1). If in this equation we put x =0; y= -/(-B') = B'/ v ( —1), and y = 0; x = + A'. Since the value of y for x = 0 is impossible,. it follows that the curve does not meet the axis of y, but + A' being the value of x for y = 0, it follows that the curve meets the axis of x at the extremities:of the diameter, 2A'. We must here remark that the imaginary expression + B V (- 1), denoting the ordinate of the origin, merely indicates that such ordinate aoes not belong to the curve, or in other words, that the point determined' by its extremity has no existence therein; we are not, therefore, to infer that the ordinate itself has no existence, for its absolute 148 ANALYTICAL GEOMETRY. value considered independently of the curve, is = B', since the abso-. lute value of its square is B'2. Now, as lines through the centre are diameters, whether they meet the curve or not, it is plain that, if we assume 2B/' for the length of the diameter, which coincides with the axis of y just in the same way as we before assumed, 2B for the length of the principal-second diameter, equation (3) will be analogous to equation (6), p. 144, for the -semi-diameters, A/, B'/, enter into the former equation in the same manner that the semi-diameters, A, B, enter into the latter. (81.)'By reason of this analogy in the forms of these two equations, analogous properties of the curve are deducible from each. The following are obvious. 1. Each diameter, 2A', 2B', bisects the chords drawn parallel to the other, as was shown of the principal diameters, 2A, 2B, (art. 71.) Such are called conjugate'diameters, and the equation A'2y2- B2n = — _A2 Bt2, is the equation of the hyperbola related to conjugate diameters. Hence (72) of every system of c6njugate diameters, one is a transverse and the other a second diameter. 2. Straight lines drawn at. the extre2mities of a transverse diameter parallel to its conjugate are tangents to the curve, (71). 3. 1 transverse diameter is divided by an ordinate parallel to its conjugate into two parts, such that' their rectangle is to the square of the ordinate as the square of the transverse is to the square of the conjugate, (76)., 4. The line which bisects parallel chords is a straight line. For the conjugate to that diameter which is parallel to the chords must bisect them.' Hence the method of finding the centre of an hyperbola, and of determining the conjugate to any given diameter are analogous to those already given for the ellipse (art. 49). From equation (1) it is obvious that the principal diameters are the only system of conjugates which are rectangular, for in that equation there can never be tan. a' =-, except when a = 0 in which tan. a case we must have a' = 90~. It is further evident, from the same equation, that both the tangents tan. a, tan. a/, must have the same sign, so that accordingly as any semi-transverse is above or below the principal semi-transverse, so will the semi-conjugate to the former be to the -right or the left of the principal semi-conjugate, the one pair being always included between the other. (82.) Since, from equation (1), tan. a tan. a'= -B2 A2, and from (4), p.. 146, aa- = B2 A-, a, Ic' being the tangents' of the two angles:formed with'the diameter, 2A, by supplemental chords from its extremities, it follows that diameters tdrawn parallel to a system of ANALYTICAL GEOMETRY. 149 supplemental chords from the principal transverse diameter are conjugate and conversely. From what has been shown of the supplemental chords (78), it is obvious that the angle included by a system of, conjugate diameters may be any magnitude fromu 0 to, 180~. (83.) Referring again -to equation (1), we find that, if the equation y = ax represent any diameter of an hyperbola, when referred to its principal axes,, then will y = B2 2 a be the equation of its conjugate. If the problem at (78)- be solved with regard to the transverse diameter, 2A/, we shall have fnm/ B/2. A/2 where, as in the ellipse, i, m/, are the coefficients of x, in the:equations of two supplemental chords drawn from the transverse, 2A/. (84.) Having established these properties, let us now, as in the ellipse, return from oblique to the original rectangular conjugates, by substituting, in the equation A'2y2 - B'2 x2 -= A/2 B/2 the values for x and y already employed at (52),' and for the transformed equation we shall have A/2cos.2y&y2 — 2A/2sin.acos. alxy+A-2sin.2ajlx=- A'2B'2sin2[A',B'] — B/'2COS,2cl ~-2B/2sin.e'cos.a' _-B/2sin.2'o/l Therefore, this being identical with A2y2 — B2x2'= - A2 B2, we have the following equations, viz. A/2 cos.2 u - B'2 cos.2c/ = A2 (1), A'2 sin.2 a-B/2 sin.2 a'- B2(2),-A/2B'2 sin.2 [A/, B/] =-A2 B2(3). By adding (1) and (2), A'/2 -B'2 = A2- B2 (4), that is the difference of the squares of any system of conjucgate diameters is equal to the diQference of the squares of the principal diameters. From equation (3) there results 4A/ B/ sin. [A' B'] = 4AB. Hence, as in the ellipse, the parallelogram constructed on any system of conjugate diameters is equivalent to the rectangle on the axes of the curve. These parallelograms are said to be inscribed in the hyperbola, as they are included between the two branches of the curve, the sides-parallel to that diameter which does not meet the curve being tan- -' gents at the extremities, of the conjugate thereto A / (81).. The property (4) shows that if A'= B'/ / then A = B, and conversely, so that there can-/ \ not be a system. of equal conjugates, except in the equilateral hyperbola, and in this, each system consists of equal conjugates. Hence, in'the equilateral hyperbola, all the inscribed parallelograms are rhombusses, but in the common hyperbola none are. (85.) It has been shown'(71) that the equation of the asymptotes, when referred to the principal diameters, is y = t Bx. A; and in precisely the same manner may it now be shown that when any sys. tem of conjugates, 2A', 2B', are substituted for the principal diame& ters, the equation of the asymptotes will be y f B/x A/. N2 150 ANALYTICAL, GEOMETRY. If, in this equation, we give to x the value A', the resulting expres, sion for y will be the length of the tangent drawn at the extremity of the diameter 2A', and terminated by the asymptote. This length is therefore ~t B', and thus we obtain, a correct notion of the absolute iength of any imaginary diameter of the hyperbola. Thus, if A'B' be any transverse diameter, and 0, and a point, if B2 — 4AC < 0. In the case B2-4AC L 0, the locus will be an imaginary curve, provided F/ be positive. ElXAMPLES. (128.) 1. Construct the locus of the equation y - 2xy+- 3x2+2y-4x-3 0. Comparing this with the general equation (1.), we find A = 1, B = - 2, C = 3, D 2, E - 4, F =- 3; hence, substituting these values in the first class of formulas, above, we have a = -, b — - F' = - -; therefore the equation of the curve, when the origin is removed to the centre, is y- 2 xy+3x2- _ O. By the second class of formulas we have tan. 2a=- 1, M= 2 + V/2, N _ 2 -,v, 2, P -= 9; hence the equation of the curve, when related to its principal diameters, is (2 +- / 2) y2 - (2- V2) x2 = -. To determine the values of the diameters 2A, and 2 B, we have, by supposing y = 0, in this equation, x2= = 4-(2 +V/2)~ A- 3/ (2+ V2) 2(2-V 2) = 2' 7, also for x=0 we have Y2= (2 9 - (2v 2) B The(2 construo2) 2(2 1f s 1.w The construction- of the curve, is therefore, as fol(,ws: Let AX, AY be the original axes, to which YY' X" the curve is referred.. Make AC =-, CO= —}, then 0 will be the centre of the curve, and OX/, OY', parallel to the primnitive axes, will be the axes to which the first transformed equation re- / fers the curve. From O draw the straight line" C OB, making with OX' an angle of which the D tangent is- 1, that is, an angle of 135~. Bisect this angle by- the line OX//, then OX" and the perpendicular to it, OY"/, will be the axes Vo which the second transformed equation refers F ANALYTICAL GEOMETRY, 179 tne curve; therefore, taking on- these axes OE, OF, each equal to 2 7, and OD, OG, each equal to 1 ~ 1; the principal diameters of the ellipse will be determined, and thence the curve easily traced.@ 2. Construct the locus of the equation 2y2 — 2xy - x2 + y + 4x -10-0. HereA- 2, B= -2, C —, D-= 1,E=4, F = -10, and, since B2 — 4AC 7 0, the locus is an hyperbola; also a-3, b= —, F' -_ 247; hence, when the origin is removed to the centre the equation is-2y —2xy -x2 - 7-'0. Again, tan. 2c-, M= -1 13, N = 1- - V 13, P = 4, therefore the equation of the curve, when referred to its principal diameters, is ( +- v/ 13 ) x-+ (2 2 VMY - ) I + AV,/ 1or ( 1 + 13 ) x2 +V 2 74 27 When y 0, x2 = 213 ~ 1 2'93= A2; 2(1+ -V 13)27 when x = 0, y2- = 3 ~ 5 18= h2 2( 1- 1 13)5 18 B2. Hence, as in the preceding example, there are given the axes of the curve to construct it. 3. Determine the axes of the curve of which the equation is 2y2 4xy + — 5x- 3x = O.T -J.ns. The curve is an ellipse, whose axes are 3 and 2/ 2. 4. Required the axes of the curve which is the locus of the equation 5y2 + 2xy + 5x2'+ 2y-2x —- 0..is. The curve is an ellipse, whose axes are 2 v ~ and 2 V ~5. Required the axes of the curve which is the locus of equation y -6xy + x + 2y y- 6 + 5 = 0. J-tns. The curve is an hyperbola, whose axes are 2 a/ 2 and 2. 6. What is the locus of the equation y2-6xy+x2-2y-6x+-1' 0?.ins. Two straight lines, characterized by the equation y= =v ~. 7. What is the geometrical representation of the equation y2_ 4xy -+ 5x2+ 2x + 10. ns. A point (- 1, -2). S. What is the locus of the equation 2x2-+2y2-3x+4y-1 =0 -.ns. A circle, whose radius is 4 V/ 33. 9. What is the locus of the equation y2 - xy-2x2 - 2x+4=0.? Ans. An imaginary curve. 10. What is the locus of the equation y2 -_ 2x - 2x'2 - 4y - x + 10 = o?. ns. An hyperbola, in'which the, second axis is takent for the axis of abscissa. The axes are 1' 7 and 2.2. 11. -What is'the geometrical representation of the equation 2y2-+ 3x2 -3x -2y +- 2 = 0?.ins. The equation has no geometrical representation. 1.2. What is the locus of the equation 3y' - 6x -- 24x -4- 6 = 0, ~ The expression for the distance between the centre and focus is given at (p. 170.). t The solution isgiven in a Key just published. 180 ANALYTICAL GEGMETRY. tie axes of reference being oblique?. ns. An ellipse in which the semi-conjugates parallel to the axes of reference are V 3 and ~ 6. (129.) Before we proceed to the construction of parabolas, we shall remark, that if, in the equations of condition (2), (3), p. 177, we substitute for the constants a, b, the variables x, y, the equations 2Ay -- Bx + )D = 0, (1), 2.x + By + E = 0, (2), will characterize two straight lines, passing through the centre of the locus, as is evident, since a, b, the coordinates of this'centre, satisfy both equations. Were we to suppose these lines to be parallel, or the centre of the locus to be infinitely distant, as in the parabola, we could infer, forn the equations (1), (2), that in the equation of the locus there mustbe B2- 4AC =. For these equations give.~Bx~+ D 2Cx + E) Y-AY' — andy (3); and since, when the lines are parallel, the difference of the ordinates corresponding to every abscissa must be constant, we have, by reducing these expressions to the same denominator, B2x -+ BD - 4ACx- 2AE = constant or (B2 - 4AC) x + BD 2AE = constant, which can only happen when B2 — 4AC = 0. If the lines (1) and (2) coincide, we must conclude that the locus has an infinite number of centres, and all situated in the line (1) or (2). The locus therefore can be no other than. a system of parallel straight lines, as GH, KL, equally distant from the line through the centres, for then every chord of the locus must be bisected by this line. G We already know that'this locus is a variety ofX the parabola. Equations (1) and (2) will- also show this to be the case, and will'moreover furnish an additional criterion, whereby we may readily ascertain, by inspecting the coeffi cients of the proposed equation, when that equation characterizes a system of parallels, and when it does not. For, since,' in this case, equations (1) and (2) represent the same line, we have, equation (3), Bx+D -- 2Cx+ - E 2A +B% -= E..'4- B2+ -- BD -4ACx - +2AE, whatever be the value of x; consequently (d/g. p. 156.) B2 = 4AC, and RD - 2AE; so that, when the indeterminate' equation of the second degree represents a system of' parallels, there must'exist among the coeffi cients the conditions B2 —4AC = 0, and BD - 2AE = 0. These lines may be at once determined from the given equation, for, being parallel, the' coefficient of x must necessarily be the same in: the equation of each, that is, these equations will be of the form y + p + q = 0 Oand y - px + r= 0; so that the proposed equation, after having freed?/2 from its coefficient, mav always, in the case,se are considering, be decomposed into two factors of this form ANALYTICAL GEOMETRY. 181 where p, the coefficient of x, in each must be equal to half the coefficients of xy, in the proposed equation, after this has been divided by A the coefficient of the first term; hence p =. With regard to q and r, it is plain that their sum must be equal to D - A, the coefficient of y, in the proposed; and their product must make the last term, F +- A. Having thus the sum and the product of q and r we shall get their difference, by subtracting four times the product from the square of the sum, and extracting the square root, that is, D2' 4F 1 q-r= V(A vD2 — 4AF}; therefore, adding the half sum to the half difference, we have,.for the greater, q=AA (D +- v/ D2 _-4AF); arid, by subtracting the same, we get the less. Now it is plain, from this expression, that, if D2 = 4AF, then q-'=r; hence the two equations (129) become, in this case, identical, and the parallels therefore coincide, and become a single, straight line. If D2 / 4AF, then the values of q and r become imaginary, so that, in this case, the locus of (1, 2,) is two imaginary lines. From what has now been said, we may conclude that the equation Ay2 + Bxy + Cx-+ Dy + Ex + F = 0, represents a system of parallel lines when B2 — 4AC = 0 and BD - 2AE - 0, these coincide, and form but one straight line; when also D2 - 4AF = 0, and they become imaginary when, instead of this, D2 - 4AF L 0. When the lines are real, their equations are B 1 B 1 y + z 2A (D +- V D2 - 4AF ) = 0, and y + 2A - x D -- v'D2 4AF ) 0. When they coincide the equation is DB' (130.) As, in the case we are discussing, the factors of the original equation consist of the sum and difference of the same two quantities, their product must be the difference of the squares of these quantities; hence, when the proposed equation represents two straight lines, it must consist of the difference of two squares, or at least of one square, minus a number. On the contrary, when the lines represented are imaginary,' the equation must consist of the sum of two squares, or at least of one square and a number. -And the equation will be a perfect square when only one straight line is represented. Hence we may frequently discover at a glance when the equation denotes a variety of the parabola, without even trying whether BD - 2AE = 0. Thus we see at once that the equation y2 - 2xy + x2 - 1 - 0, is the difference of two squares, viz. (y - x)2, and 1, the locus of it, is therefore two parallel straight lines, the equations of which are y- x +- 1= 0, Q. 182 ANALYTICAL GEOMETRY. and y — - I 0. Also the equation y2- 4xy + 4x2 + 9 = 0, is immediately seen to consist of the two squares (y- 2x)2, and 9, therefore, the locus, is imaginary. In like manner, since the -equation y2 + 4xy + 4x2 + 2y + 4x + 1 = 0, is obviously a perfect square, viz. (y + 2x + 1)2, its locus is a straight, line, the equation of which is y + 2x + 1 = 0. Let now the equation y2 + 6xy + 9x2 - 2y - 6x - 15 = 0, be proposed, which is a variety of the parabola, because B2 -'4AC = 0; and since, moreover, BD- 2AE = 0, this variety is a system of parallels, of which the equations are y + 3x -- 3 =-0 and y + 3x - 5 - 0. Lastly, let the equation be y2 - 4y + 42 + 2y - 4x + 4 = 0, the coefficients of which furnish beside the conditions above the relation D2 -4AF L 0, therefore the locus is imaginary. (131.) We shall now proceed to furnish formulas for the construction of parabolas, as we have already done for the central curves. In the present case, our object will be first to remove xy from the equation, and afterwards to remove the term containing the first power of one of the variables, and the absolute number. The first transformation, as we have already seen (122), brings the equation to the form My2 + Ry'+ Sx +- F 0,. or Nx2 + Ry + Sx+ F = 0, where, R = Dcos. a — E sin., S-D sin. a + E cos. a. Now, since, 1.+ cos. 2a 1- cos. 2a cos., 2 - /, and sin. a / 2-. And since also 2 the expression given at (124) for the cos. 2a0-becomes, when B2 =!A C A C 4AC, cos. 2a + + or ( + (A+C)7 -(A + C) accordingly/,as B is negative or positive, we have, by substitution, the following expressions for R and S, viz. when B is negative, DVA-EVC DV/C+EfAR A S C - A + C ), and whenB is positive, x/V(A + C) s/ (A + C) (132.) The values of M and N have already been determined (124), as also the values of a and b, employed in the second transformation (122); hence, collecting these formulas together, we have 1. PROPOSED EQUATION OF THE CURVE Ay2 +'Bxy + Cx2 + Dy + Ex + F = 0. Formulas to be employed for removing the Terms containing the Product of the Variables and the Square of one of them. B 1. When B is negative, tan. 2 — A C M= A + C, N-0, D VA-_E V/C D v/C + E VA v (A~+ C)' v(A + C) ANALYTICAL GEOMETRY. ] q3 II. RESULTING EQUATION, The rectangular axes being parallel to the axes of the curve, My2 + Ry + Sx - F = 0. Formulas for the removal of the terms Ry and F. R Mb2 + Rb + F R2 — 4J1fi' b- a4MS III, RESULTING EQUATION, The axes being those of the curve, My2 + Sx 0. 2. When B is positive, B The first class of formulas is, tan. 2a = A-C N = A + C, 0; R -DC —E/VA DDVA —EAvC _ i/ EAC ) S /(A C) and the result. ing equation, Nx2 + Ry + S~ + F = 0. The second class of forS' S2-4NF mulas is, a - 2Nb= 4NR; and the final equation is Nx2 + Ry = 0. EXAMPLES. 1. Construct the curve, which is the locus of the equation y2 -4xy + 42 + 2y -7x — 1 = O. Here A =, B = -4, C = 4, D = 2, E =-7, F = -1, and, as B is negative, we must employ the first collection of formulas, which givetan. 2 - - 4, M 5,N=0,R 1-6V5, S = —3 5. Hence, when the axes are parallel to those of the curve, the equation becomes, 5y2 — 5 _/ y --- / 5x — 1 =. Again, formulas I1 give, b = —28 / 5, a = -7 2/ 5; these, therefore, are the ordinate and abscissa of the principal vertex of the curve, the equation to which, in reference to the axes of the curve is, 5y2 — 3 v 5 x = 0, or y2' 35y v 5x. The construction of this curve is therefore as follows: Let X, AY.be the primitive rectangu-, lar axes. On the former take AC = 1, and Y x X make the perpendicular, CD -4. Draw 3. Y DAB, and bisect the angle BAX = 2a by. the line AX';;then the rectangular axes, X AX', AY', are those to which the first transformed equation refers the curve. Again take AE -7 / 5, and the perpendicular EA = —8 5, then the g axes A/X", A'Y", parallel'to the former, will be those of the curve. Having thus the axes and the parameter 25- V/5, the focus and directrix are readily determined,' and thence the -curve constructed. 184 ANALYTICAL GEOMETRY. 2. Construct the locus of the equation y2+2xy + x2- 6y~- 9=-0. Here A —1, B=2, C=-1, D=-6, E=0, F=9, and, as B is positive, the second collection of formulas must be used, which give -2a tan. 2a= - M =O, N=2, R — 3/.2, S= 31/2. Hence the equation of the curve, when the axes are parallel to those of the curve, is 2x2-3/2y-3 312 x+9 =O; and when they coincide with the axes of the curve, the coordinates of whose origin are, a=_- 4 / 2, b=- 2, the equation is 2x2- 3 / 2y=0, or x2=5 V 2y. Hence, as in the preceding example, the curve may be constructed. In the first transformation of axes, since tan. 2c is infinite, 2c is a right angle, so that in this transformation the new axis of x will be 450 below the primitive. 3. The equation of a parabola being y2- 4xy + 4x2 8y + 3x - 2 = 0, what will it become when the curve is referred to its axes? sns. y2=131/5x - 25. 4. Required the principal parameter of the parabola whose equation is 4y2 - 4xy + x2- 2y - 4x + 10= 0. ins. p= / 4. 5. What is the principal parameter of the. parabola represented by the equation y2 - 2xy + x-2 3y 0= ins. p = 2 2 (133.) The student must bear in mind, that the various formulas given in this chapter for the construction of lines of the second order, apply only when the different equations refer the curves to rectangular axes, which, indeed, are in most cases employed. With regard, however, to the varieties of the three curves, the tests by which they maybe discovered, and the formulas for their construction, apply generally, because in discussing these varieties we have considered the axes to have any inclination whatever, and because moreover the criteria (pp. 175, 176), by which the three classes of curves are distinguished, apply for every inclination of axes, as we are about to show in the following chapter, Which has for its object the determination of the locus of the general equation when the axes are oblique. CHAPTER II. DISCUSSION- OF THE GENERAL EQUATION. Ay2+Bxy+Cx2 +Dy+Ex+F= 0, By the separation of the variables. (134.) This equation may be put under the form x +D which, E F A" $- y t A X XX+ 0, which, solved as a quad.~ — X- y %X~+7 ANALYTICAL GEOMETRY. 185 Bx I- D! ratic, gives for y the expression y 2- B (B - 4AC) x2 - 2 (BD - 2AE)x D2 -4AFt (1). In like manner, we have for x, in terms of y, the expression By~ E 1 By + E V(B2 —4AC ) y. + 2 ( BE-2CD )y + 2C -2C E2- CFcl... (2). Either of these expressions will furnish an indefinite number of points in the locus, when this is not imaginary, since the first will give the ordinates corresponding to any assumed abscissa, and the second will give the abscissa corresponding to any assumed ordinate. If we wish to determine points in the curve from the equation (2)1, we must first, for any assumed abscissa, x, draw an ordinate Bx lD D P equal to - -. determining some point, P; then, Y if this ordinate be prolonged, and the distances PM, PM' be taken thereon, each equal to the line represented by the remaining part of the expression for y, two points of the locus will thus be determined. P therefore is the middle of the chord M, M/. The same construction for another value of x will determine another point, P/, and two new points, m,', of the curve, which will, as before, be equally distant from P/. Hence, calling the variable ordinates of these points, P, P', &c. Y, since we must always have Y - 2 it follows that the locus of these points is a straight 2A line, which, because it bisects all the chords in the curve drawn parallel to the axis of y, is called a diameter of the curve. Similar reasoning applied to the expression (2) will show that the straight line By + Ei e represented by the equation X a diameter, bisecting -the chords drawn parallel to the axis of x.' These diameters are obviously the same as those represented at (129). IHaving thus the equations of two diameters, we can always readily find the centre of any locus of the second order, to whatever axes it be referred; for, representing the coordinates of the centre by a, b, we shall have, by substituting these for the coordinates in each of the preceding equations, and solving them, as at (126), the values 2AE -- BD 2CD -BE B2 4AC - B2 -4AC' (135.) From these remarks it appears that the nature of the curve depends upon the irrational part of the expression (1), or (2), and that it cannot exist when this irrational part becomes 0, or imaginary for every value of the variable it contains. Let us examine the circumstances under which these irrational expressions can become real, 24 Q02 186 ANALYTICAL GEOMETRY. imaginary, or nothing. We shall first take the expression (1), and shall suppose that the quantity under the radical is decomposed into two factors, each containing x; in other words, we shall suppose the 2(BD —2AE) D2 4AF solution of the equation x2 - lB -) + _4A -'2 - 4AC) (B2-4AC) (3) to be effected, and that the resulting values of x are x = f and x -'/, then ve know, (A/g. p. 175-6) that the multiplication of. the factors (. - /3), (x -/3') willproduce this equation, and consequently the quantity under the radical (1.) will be (B2 - 4AC) ( -(/3) (x —'/3) (4). If, however, B2 - 4AC = O0, then' the expression under the radical will have only one factor containing x, discoverable DV - 4AF by solving the simple equation x 2 (D 4AE) 0; so that, 2 (BD - 2AE) - putting 8 for the value of x, in this equation, the expressions under the radical will be 2(BD - 2AE) (x - 6). The form (4) therefore only exists when B2 - 4AC Z 0, or B2 - 4AC 7 0; let us examine the'expression in the first case, viz. (136.) When B2 - 4AC L 0. There are three circumstances to consider in this case: 1. When th6 roots _i, f/, are real and unequal. 2. When the roots are real and equal. 3, and lastly, When they are imaginary. Suppose, first, that the roots are real and unequal, /3 being greater than /', then (i3g. p. 180) if in the expression (B 2- 4AC) (x- 1)' (x - /3) any quantity greater than /, or less than /3', be substituted for x, the product (x- /3) (x- /') will be positive, and since, by hypothesis, B2 -- 4AC is negative, the whole expression will be negative, and therefore, for all such values of x, the expression for y will be imaginary. But, if we substitute for x any value between,/3 and /3', then the product (x - /3) - /3) will be negative, and consequently the expression (4) will be positive; for all such values of x, therefore, there correspond real values of y. From this discussion it follows, that, under the conditions we have supposed, the curve always exists, and' that it is comprised between, or limiled, by two parallels to the axis of ordinates drawn at the respective distances of /3' and /3 from the, origin, for between these parallels all the values' of x which give possible values for y are comprehended. By applying precisely similar reasoning to the expression (2), it would result that the curve is also limited by two parallels to axis of x; as, therefore, these, parallels meet the former, and form a parallelogram, circumscribing the curve, it followvs that the curve must be limited in all directions, as in the annexed diagram. The curve, therefore, must necessarily be an ellipse. / ANALYTICAL GEOMETd lI. 187 Suppose, secondly, that the roots /3,3 are real and equal, then the expression (4) is (B2 -4AC) (x-3) (x -//); where, since it is tmoossible to substitute any valuefor x between the roots /3 and /', it is, by the preceding reasoning, also impossible to render the expressioni for y real by any substitution for x,/except in the single case x- 3, which renders the irrational part of the expression 0, and reduces the value of y to y hence, when the roots, /3' are equal, the curve is reduced to a point, of which the coordinates B/3,'+ D are /3 and 2AIf, lastly, the roots be imaginary, thef whatever value we substitute for x, in tile equation containing. them, the result will be positive (bl/r. p. 173); hence every value -of y will be imaginary, so that, in this case, the curve cannot exist. We may now therefore infer, that when, in the general equation, B2- 4AC Z 0 whatever be the inclination of the axes, the locus is an ellipse, if the roots of the irrational part of the expression for y be real and unequal; but it is merely a point, if these roots be equal, and it is imaginary, if the roots be so. (137.) Let us now discuss the equation upon the second hypothesis, WVhen B --- 4AC 7 0. Resuiming the expression (B2 — 4AC) (x /) (x- /3/), and reasoning as before; in reference to the roots / and /', we find that here, when these roots are real, and 23 greater, than /3', every value of x greater than,(3, or less than 3', will, because B2 - 4AC is positive, render the expression for y real; while, on the contrary, every value comprised between the limits 3/ and /3' will render the expression for. y imaginary. As, therefore, without these limits x may increase indefinitely, both positively and negatively, it follows that the curve must consist of two infinite detached branches, proceeding in opposite directions, and separated fromn each other by the distance between two parallels to the axis of y, of which the abscissas are respectively / -and /3', for within these limits there exist no possible value of y. This curve therefore is the hyperbola. If the roots /3, /' are equal, the expression above is (B2 — 4AC) (x -3)2; and hence the value of y Bx+ -I- D (4- -/A): becomes y- 2 —— A- (x 2A 2A -B _.:,/B2D-4AC D h e VB-4AC or y 2A 2A, hence the locus is a system of twoo straight lines, which intersect, since the coefficient of x is not the same in both. When the roots /, /' are imaginary, then, since every value given to x, in the equation containing 188 ANALYTICAL GEOMETRY. them, gives a positive result, the whole expression under the radical will be positive, and, therefore, the value of y will be always real. As, therefore, x may take any value from 0 to infinity, in both directions, it follows that the curve is unlimited in both directions. It moreover consists of two distinct branches: for as each double ordinate, or chord drawn parallel to the axis.of y, is bisected by the diameter whose equation is, y- 2 one half of the curve must be 2A situated entirely below this line, and the other half above it; neither can have a point in common with this diameter, because the irrational'part of the value of y can never vanish; hence the curve must be an hyperbola. (137}.) There is a particular form of the general equation which ought here to be noticed, it is that where the squares of the variables are absent, when the equation becomes Bxy+Dy+Ex+F=O, (5), Ex -- F which gives for y the expression, y = and, as this vaBx +D lue, of y will always be real, whatever be the value of x, it follows that the curve extends indefinitely in opposite directions. As each value of x furnishes but one value of y each ordinate meets the curve in but one point. If the value - D.- B be given to x, the corresponding value of y will be infinite, that is, if a parallel to the axes of y be drawn at the distance of - D _ B from the origin, it will never meet the curve; but, as every parallel drawn on either side of this must necessarily meet the curve, because no abscissa but x - D + B can render the ordinate infinite, it follows that the curve consists of two distinct branches, separated by the parallel whose abscissa is- D + B. The curve, therefore, is an hyperbola; and the parallel, whose abscissa is - D - B, is obviously one.of the asymptotes, as this parallel has been seen to be the only one which does not meet the curve. By solving the equation (5), with regard to x we have x =Dy+F E By+-E' in which expression- B is the only value that can be given to y, that will render x infinite; hence we infer here, that a parallel to the axis of x, of which the ordinate is - E - B, isthe other asymptote of the curve. Hence equation (5) represents an hyperbola whose asymptotes are parallel to the axes of coordinates, the coordinates (x,'y') of the point of intersection of the asymptotes being x = - y'-B. The asymptotes are therefore easily determined when the equation of the hyperbola takes the form (5). ANALYTICAL GEOMETRY. 189 If the term Ax2 had appeared in the equation (5) the sanme reasoning with regard to the expression for y would apply; so that then also. the parallel to the axis of y, of which the abscissa is - D + B, Is an asymptote. If Cy2 appear in the equation, instead of Ax2, then, reasoning as above on the expression for x; we find that a parallel to the axis of x, of which the ordinate is - E _ B, is also an asymptote. If both C = O and D = 0 the axis of y coincides with an asymptote. If both A =O and E = 0, the axis of x coincides with an asymptote. If both D = 0 and E = O; the origin coincides with the intersection of the asymptotes; and when, in addition to this, the squares of the variables are absent, both axes coincide with the asymptotes, and the equation takes the form Bxy + F 0 O. To determine the asymptotes from the general equation, let us actually extract the root of the expression under the radical, in the general value of the ordinate (1), we shall find this root to be of the BD —2AE K K/ formx I B2 —4AC +( C + - B - &c. therefore Bx+D~Vl2- %-4A BD-2AE K X+2ACB2 1 -2A 2A 2A- B2 -- 4AC - 2As K/ + 2A-2 - &c. Now it is here obvious that as x increases the term 2A and all that follow will diminish, while those that precede 2Ax will increase, and to these first terms the expression is finally reduced, when x becomes infinite. Hence the curve continually approaches the two straight lines denoted by, -Bx +B D V(B2-4AC) 1BD -- 2AE 1yA_ ABx — v(B —4 AC) (x ~ BD24ACE ) these lines -A -A B2 2AC are therefore the asymptotes to the curve. Comparing this equation with the equation at p. 188, which represents the locus when it'becomes a system of two straight lines, we shall find them to be identical. For, as that equation takes place only when the roots /, /3' are equal, it follows that then, must be equal to minus half the coefficient of x, in the equation (3), which contains BD-2AE them, that is, we must haveC - Cwhich value of a renders the equation (p. 188) identical with that above for the asymptotes. We may therefore say that, when the equation represents a system of straight lines, the hyperbola degenerates into its asymptotes. (138.) We already know that the asymptotes intersect at the centre, this is also readily ascertained from their equation above: for since at their intersection the two values of Y coincide, we must have 1 90 ANALYTICAL GEOMETRY. 2AE —BD for x, at that point, the value, x — which (134) is the abscissa of the centre, and the corresponding value of Y is Y - Bzx-+D 2CD - BE 2A+ — B- 4AC = -, which (134) is the ordinate of the centre. Hence, when we wish to construct the asymptotes, when the equation of the hyperbola appears under the general form, we shall have first to determine the centre from these formulas, and'then to draw through this point two straight lines inclined to the axis of x, at angles a, -a,' -B +V 1~B13 4ACf whose tangen.ts* are respectively tan. a = - B +- IB22A and tan. a' =-B-V B2-4AC. The product of these two tan2A - gents is tan. a' tan. a — 4AC = which, when C = - A, be4A2 -A w comes tan. a tan. ac = — 1, an equation which indicates (when the axes of reference are rectangular) that the asymptotes are perpendicular to. each other (11). Hence, if, in the general equation, B24AC 7 0, and C -- A, when the locus is referred to rectangular axes, we may conclude tthat the equation represents an equilateral hyperbola.. It must be here remarked, that, when A = 0, the preceding expression for tan. a becomes o, which is not a definite result; but, by multiplying numerator and denominator by B -- / {B2 - 4AC I, it reduces to tan. a B B 4AC - when A = 0. B +V ~B2 - 4AC:, w (139.) We shall now examine the general equation upon the third hypothesis, viz. When B2 - 4AC - 0. Under this condition, the general expression for any ordinate of the xz -- D I locus isy 4= i -2 V 2(BD-2AE) - D2 —4AF 2A 2A D2 4AF If we put /3 for - AE the quantity under the radical will be 2(BD - 2AE) (z -/), in which the factor 2(BD - 2AE) -may be either positive, negative, or nothing. If this factor be positive, the whole expression will be positive for every value of x greater than /3, but negative for every value less than 3; hence, in this case, the locus extends indefinitely to the right of a parallel to the axis of y drawn- through the abscissa x /; therefore this parallel is a tangent to the curve, to the left of which no point in the locus can be situated. * We are here supposing the axes to be rectangular; if they are oblique, then for tan. substitute ratio of the sines. ANALYTICAL GEOMETRY. 191 If the factor 2(BD - 2AE) be negative, then, on the contrary, the locus would extend indefinitely to the left of the parallel, whose abscissa, is 2, and no point:in the curve could be situated to the right of it. In each of these cases, therefore, the curve will be limited to one direction, but unlimited in the opposite direction; it must therefore be a parabola. If, lastly, 2(BD — 2AE)- 0, then the, expresBx- D 1 sion for y becomesy=- 2 2-A / D2 4AFI, denoting a system of parallel straight lines, which, however, coincide, when D2 - 4AF = 0, and which become imaginary, when 1D2 - 4AF L 0. Because the condition B2 = 4AC or B= 2 / ACI characterizes the parabola and its varieties, the three first terms in the general equation of this curve will always form a perfect square, viz. (y VA+x 2/C)2 Ay2 +2V ACI ~ xy + CX2. (140.) We might now proceed to inquire into the form of the general equation when it represents one of the varietiesof the three curves, and thence derive, as in the preceding chapter, criteria by means of which these varieties may be distinguished. For the varieties of the parabola the tests of their existence which have been given in the preceding chapter are the simplest that can be employed, and may be readily applied in any case of doubt. But for the other curves,. the shortest and most direct way of proceeding will generally be to solve the equation, with regard to one of the variables, and then to find the roots of that part of the resulting expression which is under the radical, the nature of these roots will make known the nature of the locus conformably to the preceding discussion. The examples we shall here give will further illustrate this. Construction of Curves of the second order. EXAMPLE. I. To determine the position of the curve of which the equation is y2 2xy + 3+ 2y-4x- 3 = 0. As, in this example, B2- 4AC /_ 0, the curve must be an ellipse; let us therefore first proceed to determine its limits. For this purpose let us put the, equation under the following form, viz. y2 - 2 (x -1) y = — 3x2 ~ 4x + 3 (1), which, solved first for y and then for x, gives y-x-1 ~ = - 2x2 +2z+4 (2). Z= + ~,7 V - 2y _ 2y +- 13 (3). Equating the irrational part of (2) to 0, we have x-x — 2 =0. x= 192 ANALYTICAL GEOMETRY. consequently, the roots of this equation being real and unequal, viz J = 2, and //- -1, we know (136) that _ K F M the curve exists, and that it is included between two parallels, LL', MM', to the axis of y, of D N which the abscissa, AG,, of the one is equal to E - 1, and the abscissa, AIH, of the other equal / A a to 2. Solving, in like manner the equation 2y2 + -2y 13- 0, we obtain for the roots the values i -- -+ /27 -1-n 27 1 1,, -and, therefore the F'KI k I' 2 2 curve is also comprehended between two parallels to the axis of x, of which the ordinate, AK, of the one is -1+ 27I -— V27 2-'4 -, 2 and the ordinate, AK/, of the other 2-V' The curve is therefore circumscribed by the parallelogram LM'. To find the points of contact of the parallels LL/, MM', we must construct the diameter, Y = -- 1, (134); for, as the abscissas / and'/ of these points render the irrational part of the equation (2) nothing, the corresponding ordinates must belong as well, to the diameter as to the curve. This diameter cuts the axes in the points x I1 and y =- 1; if, therefore, through these points the line II', be drawn, the two points of contact will be determined. Constructing also the second diameter X-_ 3(y + 2), which cuts the axes in the points x = 2, and y- 2, we'obtain the other two points of contact, F, F'. To find the points where the curve intersects the axis of y, suppose x = 0, the equation (2), and we have for the ordinates of those points y = 1 and y -3; hence these points, D, D', are readily determined. In like manner, supposing y 0, in equation (3), we have for the abscissas of the points E, E', where the curve cuts the axis of xa, the values of x = - + Y/ 13 and x =- - 13. The eight points thus determined are amply sufficient to make known the position of the curve. But there is another mode of proceeding by which an indefinite number of points in the curve may be determined. Thus: Draw, as before, the parallels LL',- MM/, and then construct the diameter, IF', from its equation, Y _ x- 1 The middle point, 0, of this diameter is the centre of the curve, therefore the abscissa, Am, on +'/~ 2 1 of the centre_ is = - because 3 and /f are the 2 2 2 a abscissas of the extremities of the same diameter. Hence, drawing the ordinate nrN, we shall have the direction of the diameter conjugate in II', since this ordinate will be parallel to the tangent at the vertex of that diameter; therefore, putting for x the value z = -, in the expression (2), the irrational part gives for the semi-diameter, ON, ANALYTICAL GEOMETRY. 191 -2x+ 2x~__ 2 - i4- -ON; hence we have a system of conjugate diameters given in length and direction to construct the ellipse. This construction is as follows: On the given diameters, AB, CD, taken as principal axes, construct an ellipse; then, if the d double ordinates, ab, CD, cd, &c. of this ellipse 4.be inclined to AB, in the given angle, while A- * their length remains unchanged, their extremi- mities, a- b', C', D', c', d', &c. will be all upon l' the required curve, which may therefore be drawn through. them. The truth of this is obvious, for the curve thus traced will, by construction, be such, that the squares of the chords-parallel to one diameter, C' D', are as the rectangles of the parts into which they divide the other, AB, and AB, C/D/,' are the given conjugates, both as to the length and direction, EXAMPLE II. To construct the curve of which the equation is 2 -2xy — 3x — 2y 7x -1 = 0. Since here B2 - 4AC 7 0, thecurve is an hyperbola. We shall proceed first to determine the asymptotes, because, when these are known, and a single point in the curve found, we can easily obtain as many more points in, the curve as we please (97). The equation of the asymptotes is given at (137); but as it is adviseable to proceed independently of the general formulas, we shall here deduce the equation of the asymptotes from the given equation of the curve, which furnishes for y the value K K' y=c+ I +i 2}4z2_5x F +~ 2}x+l 2 (2x-a+ + -2 4- &c.) Hence, for the two asymptotes we have the equation Y = x + ~ (2x -). For r = O we have Y -., and Y2+; \ therefore, making AS= -i-, and AS' -- 2~, the s points S, S' will be those in which the asymptotes cut the axis of y. In like manner, for Y - O we havexr =l -j- and x -- 2~; therefore, making As: v, and As' = 21, the points s, s' will be those in which the asymptotes cut the ) +y' + z x'-+ + z a'2x + -- x + + y-2+z:: m: n, or n (x:: y+z')_ 2Z n(a ( — x x +y? +' z ), or supposing ns> m we have (n-m) X (xa y +2) — 22naxma m,or x y 2 + 2ma ma 2max on, i mz + -.x:= -x,or X m- + y + - + n-?n n-n —m n- (n-m) n-m nma- ninma ma nma+ _ or (x - y- z- which is the (n-qm)5 (n-sm) n-n (n-M) equation of the surface sought, it being a spherical surface whose radius as/(nm) -a(nm), and the distance of its centre from the point A on the line AB n-m mae ma bm- n-rn,U-~ which shows that the centre is on the negative side of the n-m m-n origin. If m=n we have a' —2ax 0O, or x-= -.'. the locus is a plane at right angles to the middle of AB. Note.-This question might have been made much more difficult by considering the obliquity at which the light strikes the elements of the surface sought; but as this does not appear to have been the author's intention, we shall here leave the subject at present for those who may feel disposed to prosecute the inquiry. (See WILLIAMS' Key.) Problems 18, 20, 22, and 23, in art. 159, are not solved, for want of room; and also in art. 267, Prop. 16 and 17, p. 283.-Ed.