PLANE AND SOLID Analytic Geometry FREDERICK H. BAILEY, A.M. (Harvard) AND FREDERICK S. WOODS, Ph.D. (Gottingen) Assistant Professors of Mathematics in thb Massachusetts Institute of Technology Boston, U.S.A., and London GINN & COMPANY, PUBLISHERS 1897 Copyright, 1897, by F. H. BAILEY AND F. S. "WOODS ALL RIGHTS RESERVED PREFACE. EMS Lib. QA This book has been prepared primarily for the use of the students in the Massachusetts Institute of Technology, but it is hoped that it will be found adapted to the needs of other technical schools and colleges. While the authors have restricted themselves to subject- matter properly belonging to a first course, they have, never- theless, endeavored to give a complete and rigorous treatment of all questions discussed. The memorizing of a mass of formulas has been discouraged, the attention of the student being directed rather to the methods employed. At the end of the book a collection of formulas has been made, which, it is believed, is sufficient for all needs of the student. In the Plane Geometry considerable space has been devoted to the general forms of the equations. In particular, the equation of the circle is treated in its most general form. The conies have been approached from tlieir general defini- tion, and the student is led to discuss the general equation of the second degree in which the xy term is missing, and to find the equations of the tangent, the normal, and the polar for the equation of this form. By this method of procedure much time is gained when the special equations of the parab- ola, the ellipse, and the hyperbola are discussed, and it is believed that the student's grasp of the subject is thereby strengthened. Finally, the general equation of the second IV PREFACE. degree has been studied with the view of enabling the stu- dent to determine most rapidly and easily the nature and the position of the locus of any equation of that degree. In the Solid Geometry, besides the plane and the straight line, the cylinders and the surfaces of revolution have been noticed, and all the quadric surfaces have been studied from the simplest forms of their equations. This study includes the treatment of the tangent, the polar, and the diametral planes, conjugate diameters, circular sections, and rectilinear generators. The examples in the earlier parts of both the Plane and the Solid Geometry are mainly numerical, but later they are largely purely geometrical. They have been arranged at the end of each chapter, that the teacher may take them up in any order preferred. Many of them have been prepared especially for this book, and for the others the authors wish to acknowledge their indebtedness to other writers, especially to Professors Eunkle, C. Smith, and Todhunter. In conclusion, the authors wish to thank Mr. H. K. Burrison, who drew the diagrams for Chapters II and IV of Part II, and other friends who have kindly aided them by suggestions. Massachusetts Institute of Technology, February, 1897. OOIl^TENTS. PART I. PLANE ANALYTIC GEOMETRY. CHAPTER I. THE POINT. ARTICLE PAGE 1. Direction of a Line ......... 1 2. Cartesian Coordinates 2 3. Distance between Two Points and Slope of Joining Line . 4 4. Point of Division ......... 7 5. Area of Triangle 9 0. Area of Polygon 11 7. Polar Coordinates 13 8. Distance between Two Points and Slope of Joining Line, in Polar Coordinates . 14 9. Area of Triangle in Polar Coordinates 15 10. Relation between Cartesian and Polar Coordinates . . 16 Examples 17 CHAPTER IL IiOCI. 11. Relation between Equation and Locus 12. Equation of the First Degree . 13. Equations of Degree Higher tlian the First 14. Transcendental Equations 15. Equations in Polar Coordinates 16. Locus Defined by Geometric Property 17. Intersection of Loci (Case I) . 18. Intersection of Loci (Case II) . 19. Limiting Cases of Intersection Examples 22 23 24 28 30 32 32 35 38 40 vi CONTENTS. CHAPTER III. THE STRAIGHT LINE!. ABTICLK PAGE 20 45 21. Equation of the Straight Line in terms of its Slope and Liter- cept on the Axis of y 45 22. Equation of the Straight Line in terms of its Intercepts on the Axes 47 23. Equation of the Straight Line in terms of its Normal Distance from the Origin and the Angle the Normal makes with the Axis of X . . . . . . . . . .49 24. Any Equation of the First Degi'ee represents a Straight Line . 50 25. Lemma ........... 52 26. Determination of the Parameters of any Straight Line . . 52 27. Equations connecting the Parameters of the Straight Line . 55 28. Equations of the Straight Lme in Oblique Coordinates . . 56 29. Lemma ........... 58 30. Equations connecting the Parameters of the Straight Line in Oblique Coordinates 59 31. Equation of the Straight Line in Polar Coordinates . . 60 32. Distance of a Point from a Straight Line . . . .60 33. Equation of a Line through any Given Point with a Given Slope 63 34. Equation of a Line through Two Given Points . . .64 35. Angle between Two Lines 65 36. Equation of a Line through a Given Point Perpendicular to a Given Line 68 37. Problem 69 38. Equation of a Line through the Intersection of Two Given Lines 70 39. Equations of the Bisectors of the Angles between Two Lines . 72 40. Equations of Higher Degree than the First representing Straight Lines 74 41. Condition that the Equation of the Second Degree represents Two Straight Lines 75 Examples 77 CHAPTER IV. TRANSFORMATION OP COORDINATES. 42. Meaning of Transformation of Coordinates . . . .86 43. Translation of Origin 87 CONTENTS. Vll ABTICLE PAGE 44. notation of Rectangular Axes 88 45. Transforuiatiou from a Set of Rectangular Axes to a Set of Oblique Axes having the Same Origin .... 89 40. Transformation from one Set of Oblique Axes to a new Set of Oblique Axes with the same Origin .... 90 47. Double Transformation of Coordinates 92 48 93 Examples .......... 93 CHAPTER V. THE CIRCIjE. 49. Definition and Equation of the Circle 97 5(4. General Form of the Equation of the Circle .... 98 51. Determination of the Centre and the Radius of a Circle from its Equation 99 52. Determination of the Ecjuation of a Circle Passing through Three Given Points 100 53. Determination of the Tangent to a Circle when its Slope is Given 101 54. Equation of a Tangent to a Circle when the Point of Contact is Given 108 55. Normal 105 50. Subtangent and Subnormal 1()(; 57. Length of a Tangent Drawn from a Point to a Circle . . 107 58. Chord of Contact 1(»H 59. Definition and Equation of the Polar 110 00. Other Properties of Poles and Polars 113 01. Diameter 115 62. Circle through the Points of Intersection of 'JVo Circles . 118 03. Radical Axis 119 04. Radical Centre 121 65. Polar E(iuation of the Circle 122 Examples .......... 123 CHAPTER VI. THE CONIC SECTIONS. 60. Definition and Equation 131 67. The Parabola [e = 1] ........ 133 Vlll CONTENTS. ARTICLE PAGE 08. The Ellipse [e < 1] 135 69. The Hyperbola [e > 1] 140 70. Generalization 144 71. Limiting Cases of Conic Sections 149 72. Recapitulation 150 73. Sections of a Cone 151 74. Polar Equation of the Conic Section, the Focus being the Pole 157 Examples 157 CHAPTER VII. TANGENT, NORMAL, AND POIjAR. 75 161 76. Tangent 161 77. Normal 165 78. Pole and Polar 106 79. To find the Tangents from a Point not on the Conic Section 169 80. Properties of Poles and Polars 171 Examples 174 CHAPTER VIII. THE3 PARABOLA: y^ = 4px Constructions Angular Properties of Tangent and Normal Perpendicular Tangents meet on the Directrix Perpendicular to Focal Chord Perpendicular from Focus to Tangent . Diameter ...... Properties of the Diameter Parabola with Focus at the Origin Parabola referred to a Diameter and the Tangent as Axes .... Parabola in Polar Coordinates Examples Corresponding 177 177 179 181 182 183 186 188 189 191 193 193 CONTENTS. IX CHAPTER IX. THE EHiIilPSE!: X.- y2 AKTICLE 92. . 93. 94. 95. Parallel to To find the Foci when the Axes are givei Focal Distances Sum of Focal Distances 96. Constructions 97. Angle between Tangent and Focal Radii 98. Point of Intersection of a Pair of Perpendicular Tangents 99. Perpendicular from the J"ocus upon any Tangent 100. Equation of a Diameter .... 101. Conjugate Diameters 102. The Tangent at the Extremity of any Diameter is the Conjugate Diameter .... 103. Supplemental Chords 104. Given the Extremity of any Diameter, to find the Extremities of the Conjugate Diameter 105. Lengths of Conjugate Diameters . 106. Angle between Conjugate Diameters 107. Parallelogram on Conjugate Diameters 108. Auxiliary Circle .... 109. Eccentric Angle .... 110. Eccentric Angles and Conjugate Diameters . 111. Ellipse referred to Conjugate Diameters as Axes 112. Polar Equation of the Ellipse Examples PAGE 199 199 200 201 202 202 203 205 206 208 209 210 212 213 214 215 216 217 218 220 222 223 CHAPTER X. THE HYPERBOIiA: 113. Conjugate and Equilateral Hyperbolas . 114. To find the Foci when the Axes are given 115. Focal Distances 116. Difference of Focal Distances 117. Constructions. ..... 118. Angle between Tangent and Focal Radii 119. Equation of a Diameter .... 229 231 232 233 233 234 2.30 X CONTENTS. ARTICLE PAGE 120. Conjugate Diameters 237 121. Propositions on Conjugate Diameters 238 122-126. Asymptotes 240 127. Equation of an Hyperbola referred to its Asymptotes as Axes of Coordinates 244 128. Polar Equation of the Hyperbola 246 Examples 246 CHAPTER XL THE GENERAIj E5QUATION OF THE SECOND DEGREE 129. Statement of Problem 251 130. To Remove Terms of the First Degree 252 131. Case I. AB-H2>0 253 132. Case II. AB-H2<0 258 133. Casein. AB - H-^ = 260 134. Summary 262 135. Rule for Handling Numerical Equations .... 263 136. Oblique Coordinates 267 137. Conic through Given Points 268 Examples .......... 271 PAKT II. SOLID ANALYTIC GEOMETRY. CHAPTER I. THE POINT. 1. 2. 3. 4. 5. 6. Rectangular Coordinates in Space . Polar Coordinates in Space .... Distance between Two Points Point of Division Direction Cosines Projection . 273 . 275 . 277 . 278 . 279 . 281 CONTENTS. Xi ARTICLE PAGE 7. Angle between Two Lines 282 8. Translation of Origin 283 9. Transformation of Coordinates from One Set of Rectangular Axes to a New Set of Rectangular Axes having the Same Origin 284 10 286 11. Relation between Rectangular and Polar Coordinates . . 286 Examples 288 CHAPTER II. INTERPRETATION OP EQUATIONS. 12. Single Equation 13. Two or More Equations . 14. Cylinders 15. Surfaces of Revolution . Examples 290 292 293 294 298 CHAPTER III. THE PLANE AND THE STRAIGHT IiINB. 16. Any Equation of the First Degree Represents a Plane . . 301 17. Normal Equation of the Plane . . . . . . 302 18. Equation of a Plane in Terms of its Intercepts . . . 303 19. To Find the Equation of a Plane which Passes through Three Given Points 304 20. Angle between Two Planes 304 21. Perpendicular Distance from a Given Point to a Given Plane 305 22. Any Two Simultaneous Equations of the First Degree Repre- sent a Straight Line 306 23. Equations of a Straight Line in Terms of its Direction Cosines and Any Known Point upon it 306 24. To Place the General Equations of the Straight Line in the Form [14] . ' 308 25. Equations of a Straight Line Passing through Two Given Points 309 26. Plane through a Given Line and Subject to One Other Condition 310 Examples 311 XU CONTENTS. CHAPTER IV. EQUATIONS OF THE SECOND DEGREE. AKTICLE PAGE 27. The Cones : -, ± f^ ± ^ = 316 a^ b^ c^ 28. The Central Quadrics : -, ± ^, ± ^ = 1 . . . .320 a- b- c2 X^ y'2 -^1 29. Relation between the Cone -: 4- 7-, ; = and the Hyper- a? b- c^ boloid ^ + ^ - ^ = 1 328 a^ b- c^ 30. The Unparted Hyperboloid is a Ruled Surface . . . 329 31. The Paraboloids :-, ± ^" = 4pz 330 a- b^ 32 330 33. Tangent Plane 332 34. Normal Line 333 35. Polar Plane 334 36. Diametral Plane 335 37. Diameters 337 38. Auxiliary Line of the Ellipsoid 339 39. Parallel Sections 340 40. Circular Sections 341 41. Umbilic 343 Examples 345 FORMULAS 351 ANSWERS 361 PAET I. PLANE ANALYTIC GEOMETRY. CHAPTER I. THE POINT. 1. Direction of a Line. The line AB, A B, joining the points A and B, may be regarded as having two directions, according as it is drawn from A to B or from B to A. To distingnisli between these two directions we will adopt the following notation : If the line is drawn from A to B, if shall he called AB ; if it is drawn from B to A, it shall be called BA. The first letter, then, denotes the beginning, or the origin, of the line, and the last letter denotes the end, or the term, of the line. If, now, A, B, and C are any three points on the same straight line, A B C A C B B A C C A B Fig. 1. it follows that AB + BC = AC (1) For the resnlt, in both distance and direction, of going from A to B and then from B to C is the same as the resnlt of going from A to C, as shown in Fig. 1, which illustrates some of the cases which may occur. 2 PLANE ANALYTIC GEOMETRY. If, however, the third point C coincides with A, (1) becomes AB+ BA=^ AA=0, whence BA = — AB. That is, if one direction on a straight line is jiositive, the oppo- site direction is negative, so that AB and BA are each the negative of the other. From (1) we may derive the formula, BC=AC-AB; (2) i.e., if three j^oints lie iij/on the same straight line, the line joining any one of them to a second is equal to the line join- ing the third to the second decreased by the line joining the third to the first. 2. Cartesian Coordinates. Since the first element of geometrical discussion is the point, let us now find a method by which we can determine the relative position of a point in a plane with respect to some fixed positions in the plane. We will first take the method invented by Descartes. Y Fig. 2. THE POINT. d Let XX' and YY' be two straight lines intersecting at 0. Distances from YY', measured parallel to XX', shall be de- noted by X, X being positive, if measured in the direction OX, and negative, if measured in the direction OX'. Dis- tances from XX', measured parallel to YY', shall be denoted by y, y being positive, if measured in the direction OY, and negative, if measured in the direction OY'. Then there can be but one value of x and one value of y corresponding to any point in the plane. For tlirougli the point we can draAV only one line parallel to YY', and only one line parallel to XX', and the respective distances of these lines from YY' and XX', measured according to the rule laid down above, will be the respective values of x and y for the point. On the other hand, if any simultaneous values are given to X and y, these values determine the position of a single point in the plane. For example, let x ^= a and y ^ b, wliere a and b are positive quantities. Then the point is in a line parallel to YY', and a units to the right of it, and also in a line parallel to XX', and b units above it. These two lines meet in a single point, Pi, as shown in Fig. 2. Or, again, let x^ — a, y = — b; then the point is in a line parallel to YY', a units to the left of YY', and in a line parallel to XX', b units below XX', and the point P2 is determined, as sliown in Fig. 2. Thus we see that, given simultaneous values of X and y, we can find one and only one point corresponding to them. Practically, the position of this point may be found, or, as we usually say, the point may be j^lotted, by first laying off X along XX', and then measuring y on a line parallel to YY'; for example, in Fig. 2, by laying off OMi, and then Ml Pi. The distances x and y are called respectively the abscissa and the ordinate of the point ; or, taken together, they are called the coordinates of the point. The lines XX' and YY' are the axes of abscissas and ordinates respectively, or the 4 PLANE ANALYTIC GEOMETRY. coordinate axes, and their point of meeting is called the origin. Tlie axes may meet at any angle, and we have rectangular coordinates, or oblique coordinates, according as the axes are or are not perpendicular to each other. Both classes of coor- dinates, however, fall under the class of Cartesian coordinates, named from the inventor Descartes, the Latin form of whose name is Cartesius. If no angle between the axes is specified, we shall assume it to be a right angle. Instead of writing x = — a, y = — b, we shall always write P ( — a, — b) or ( — a, — b), for the sake of brevity, the abscissa always being written first and separated from the ordinate by a comma. 3. Distance between Two Points and Slope of Joining Line. Now that we can determine the position of a point in the plane, let us see how we can use the coordinates of points in a geometrical problem. The first problem is that of find- ing the distance between two points, which shall be Pi (xi, yi) and P2 (x2, y2). Fig. 3. THE POINT. O From P2 and Pi draw lines respectively parallel to XX' and YY', meeting at R. Then, OMi=:x„ MiPi = yi, OM,= Xo, M,P. = yo. By (2) § 1, P,R=M,Mi = OMi-OMo ^Xj-x., RP^= MiPi- MiR= MiPi- MoP, = yi-y,. If we denote the distance PiPo by d, we have, from the right triangle PiPoR, whence, by substituting the values of PoR and RPi, d-V(Xi-x.)^' + (y:-y,)^. [1] If 6 denotes the angle made with the axis XX' by the line RPi . Yi — ya Pi P2, tan = H-B» or tan = -• r 2r» Xi — Xo Tan $ is called the slope of the line, and is usually denoted by m. Since 6 and m are quantities constantly occurring in the following work, it is important to notice carefully how 6 is to be measured. It is always the angle above the axis of x and to the right of the line P1P2, the line being produced, if necessary, so as to cross XX'. Its angular magnitude is, there- fore, always between 0° and 180°. Consequently, m may have any value whatever, and is positive when 9 is an acute angle, and negative when is obtuse. Conversely, when m is known, 6 is determined without ambiguity. In Fig. 3, from which we have derived formulas [1] and [2], both Pi and P2 are in the first quadrant, but the reasoning will apply to any position of the two points. For example, it applies to the following diagrams, Figs. 4 and 5, without a single change of wording. PLANE ANALYTIC GEOMETRY. P, IR Y Fig. 4. O M, Y Fig. 5. Ex. Find m and d for line joining tlie points (2, 3) and {— 1, 2). d = V[2 - (- 1)]2 + [3 — 2]2 = VlO, 3 — 2 2-(-l) If the axes make an angle w with each other, we shall have to find new formulas for d and m. Accordingly, let XX' and YY', in Fig. 6, meet so that Z XOY = (o. Fig. 6. THE POINT. 7 Pi (xi, yi) and P.2 (xo, 72) are the two points. Draw PiR parallel to YY', PoR parallel to XX', PjS perpendicular to PoR, and P.M., parallel to YY'. Then, 0M., = X2, M,P., = y.3. .-. P.,R= M,Mi = 0Mi-0Mo = xr-X2, RP^=M,Pi-MiR = yi-y.,. In the triangle PiRP,>, P^P,=zAyp,R-+RP,'-2 P.,R.RPi cos Z PaRPi- As cos ZP2RPi^cos (180° — co) = — cos w, we have by substitution d = V(xi — X.,)" + (yi — yo)- + 2 (xi — Xo) (yi — y^) cos u>. (1) Now, RS=RPi cos w=(yi — yo) cos w, so that PoS = PoR + RS = (Xi — x,) + (yi — y2) cos w ; and SPi ^ RPi sin w = (y, — y.,) sin w. SP But m = tan $ = p— ' ; r 2^ . (y, — ya) sin o) ^ (xi — x2) + (yi — yo) cos It is evident that, if w = 90°, (1) and (2) reduce respec- tively to [1] and [2], as we should expect. 4. Point of Division. The next problem that naturally occurs is that of finding a point which divides a line in a given ratio. We will find, then, the coordinates of the point P, on the line joining Pi (xi, yi) and P, (x., y.,), such that P,P : PP^^ I, : L PLANE ANALYTIC GEOMETRY. Y HR Fig. 7. Draw PiMi, PM, cand P2M2 all parallel to YY', and PiR aud PS parallel to XX'. Then, P^R= MiM=: OM — OMi = x— x^, PS =MM, = OM,-OM =Xo-x, RP=MP -MR =MP -MiPi=y-yi, SP,= M2Po-MoS=M,Po- MP = y,-y. The triangles P^RPand PSP2 are similar, since their sides are respectively parallel; therefore their homologous sides are proportional, and PiP_ PiR^RP PP2~ PS~SPo" Substituting the respective values we have found, we get i, whence x== X — xi ^ y Xo — X y llXo + loXl I1+I2 y and y Iiy2 + l2yi I1 + I2 [3] Ex. Find the coordinates of a point dividing tlie line joining (1, 4) to (3, — 2) in the ratio 2 : 3. THE POINT. y If we let (1, 4) be (xi, yi), aud (3, — 2) be (xo, yc), then li = 2, and I2 = 3, and the coordinates of the required points are ^ 2(3) + 3(1) ^ 9 2 (- 2) + 3(4) ^ 8 2 + 3 5' ^ 2 + 3 5' If the point bisects the line P1P2, Ii = l2j and the formulas become If the point is one of external division, the lines PiP and PP2 will have opposite directions, and hence their ratio must be negative. We can make it negative by putting a minus sign before I2, in which case our formulas become ^ _ liX2-i2Xi _ l iy^ - l2yi r-n ii 12 11 "2 As our proof has depended entirely upon the properties of similar triangles, it is evident that it would be exactly the same for oblique coordinates, so that our resulting formulas, [3], [4], [5], are true for oblique coordinates. And it may be noted here, that, if no use is made of the right angle between the coordinate axes, the derived formula will be a general formula, holding for all cases of Cartesian coordinates, even though in the figure used XX' is perpendic- ular to YY'. 5. Area of Triangle in terms of the Coordinates of the Vertices. Let the vertices of the triangle be Pj (xj, yi), P2 (xo, y2)> and P3 (X3, ys)- Draw PiN parallel to XX', and P3M and P.N parallel to YY', thus forming the right triangles P1P3M and P1P2N, and the trapezoid MP3P2N. Then the area of PiP2P3 = area of P1P3M -|- area of M PgPaN — area of PiPoN. 10 PLANE ANALYTIC GEOMETRY. Y x' ^' / \ 1M2 \y^-^ M3 1 1 p, 1^ /I yi' Y Fig. s. Area of P1P3M = iPiM.MP3, '' " MPsPoN^iMN (MP3+NP2), u u PiP,N = iPiN.NP2. Now, PiM = MiM3 = OM3 — 0M,. = X3 — xi, MP3=M3P3- M3M = M3P3-MiPi = y MN =M3Mo = OM2-OM3 = Xo-x3, N P, = M,P, - MoN = M,Po — MiPi = yo - yi, P,N= MiM2 = 0Mo — OMi = Xo— xi; .•.by substitution, the area of PiP2P3 = h (^3 — xi) (ys — yi) + h (^2 — X3) (ys — yi + y2 — yO — h (x2 — xi) (y2 — yi), whence we get, in reduced form, area of PiP2P3 = ¥{(xiy2 — X2yi) + (xoy3 — X3y2) + (xsYi — Xiy3)}. [6] The following is an easy method of solving numerical ex- amples in accordance with the above formula : Write xj yi down the coordinates of the successive vertices in a Xo y2 column, repeating at the bottom of the column the X3 ya coordinates of the first vertex ; multiply each x by Xi yi the y in the line below it, prefixing the positive sign. 2, 3 1, 5 -1, — 2 THE POINT. 11 and each y by the x in the line below it, prefixing the negative sign, and take one half the algebraic sum of all the terms thus formed ; the result is I {xiy2 — Xoyi + xoys — xsya + xsyi — xiya} , which we have proved to be the expression for the area. It is to be noted that this last is only a working rule, giving a true result, and is in no sense a proof of the formula. Ex. Find the area of the triangle of which the vertices are the points (2, 3), (1, 5), and {— 1, — 2). Area = i \ (2)(5) - (1)(3) + (1)(- 2) - (- 1)(5) 2, 3 +(-l)(3)-(2)(-2)|=-V-. If the expression for the area of any triangle comes out negative, the absolute value of the result is the area, and the negative sign shows that we have lettered the vertices so that in going from Pj to Pa and then from Pg to Pg, we have the triangle on our right, while, as we have lettered Fig. 8, we have it on our left. That such is the meaning of the negative sign may be shown by interchanging P2(x2, y^) and P3(x3, yg) in Fig. 8, and finding the formula for the area of the triangle, the result differing from that already found in the negative sign only. Note. Fig. 8 is necessarily drawn for some chosen position of the triangle, but the proof as given will apply to all cases, if the vertices be appropriately lettered. It is suggested that the student make other figures from which to derive the formula, as in § 3. 6. Area of a Polygon in terms of the Coordinates of its Vertices. We may, of course, find the area of any polygon of four or more sides, by dividing it into triangles by diagonals, and taking the sum of the areas of the triangles thus formed. 12 PLANE ANALYTIC GEOMETRY. But we can deduce a formula for the area of any polygon, exactly similar to that for the area of a triangle, and giving the same working rule. This causes a great saving of time over the method first suggested. We will illustrate the method by finding the formula for the area of the quadrilateral P1P2P3P4, the coordinates of the vertices being respectively (xj, yj), (x^, y^), (xs, ys), and (X4, y^). Let P (x, y) be any point within the quadrilateral, and draw the lines PPi, PPg, PP3, and PP4, forming the triangles PP1P2, PP2P3, PP3P4, and PP^Pi. By [6] §5, Area of PPiPo = i{xyi — Xjy + Xiy, — xsyi + x.y — xyz}, • " " PP2P3 = |{xy2 — x.2y + x2y3 — X3y2+X3y — xys}, " " PP3P4 = i{xy3 — xsy + xgyi— X4y3+X4y — xy4}, • " " PP4Pi = i{xy4 — X4y + X4yi — Xiy^ + xiy — xyj. Adding and reducing, we get Area of Pi P2 P3 P4 = i {xiy2 — X2y 1 + Xays — Xsy- + xsy* — X4y3+ X4yi — Xiy4}. (1) THE POINT. 13 Trying a working rule like that for the triangle, we get xi yi Area of PiP2P3P4= i{xiy2 — Xoyi + x^ys — X3y2 + X3y4 X2 y2 — X4y3 + x^yi — x,y4}, X3 y,3 a true result agreeing with (1). X4 y4 This may be extended to a polygon of any number Xi yi of sides, the resulting fornuda being, if the vertices are (xi, yi), (X2, yo) . . . . (x„, y„), h {(xiy2 — xoyi) + (xoya — xsy-) + • • • + (x„yi — xiy„)} . (2) 7. Polar Coordinates. So far, we have determined the position of the point in the plane by two distances, i.e., by x and y. We may, however, determine the position of the point by a distance and a direc- tion, as follows : Fig. 10. Let 0, called tlie origin or pole, be a fixed point in the plane, and OM, called the initial line, be a line of fixed direction in the plane. Take P any point in the plane, and draw OP. Denote OP by r, and the angle MOP by ^ ; then r and 6 will be called the polar coordinates of the point, and, when given, will com- pletely determine the point. 14 PLANE ANALYTIC GEOMETRY. For example, let r = 2 and $ = 15° ; then the point is on the circumference of a circle of radius 2, the centre of which is at 0, and on a straight line drawn from 0, making ^ MOP = 15°. These two lines meet in one, and only one, point, i.e., P, the position of which is completely determined. For brevity, we may write P (2, 15°) to designate this point, for which r = 2 and 6 = 15°. The point P is j^ioUed by laying off the angle 15°, OM being taken as the initial line, and measuring from a distance 2 along the terminal line. The angle 6 may have any magnitude, and may be either positive or negative, OM being regarded always as the initial line of the angle, as implied by its name. Generally, r will be positive, but we may give r a negative value, if we make the following convention : After the angle $ is constructed, r shall be positive if the point is on the terminal line of 6, and negative if the point is on the extension of the terminal line of 6, the terminal line being extended backward through the origin. Thus the coordinates of Pi, Fig. 10, may be (2, 195°), or (2,-165°), or (-2, 15°), or (-2, —345°). 8. Distance between Two Points and Slope of Joining Line, in Polar Coordinates. /. ,-^ Fig. 11. Let Pi (ri, ^i) and P2 (r2, O^) be the two points, and draw OPi and OP2. THE POINT. 15 Then in the triangle PoOPi, Z PoOPi = ^i — ^2, so that, by Trigonometry, p;p;'=op;'+op"/— 20P1.OP2COS (^1—^2); .". if the length of P1P2 is denoted by d, d = Vri^ + r2- - 2rir2 cos {6^ — 6^). (1) Draw PjRjand P2R2 perpendicular to OM, and P2S perpen- dicular to PiRi- Then, if we detine the slope as the tangent of the angle the line makes with OM, it is evident that SPi ^=p7s- But SP,= RiPi— R2P2 = ''i sin B^—r^ sin 6^, PoS^RsRi^ORi— 0R2=ri cos ^1 — r2 cos ^2; ri sin c^i — r.i sm 6^2 .-. m = —• ri cos Bi — To cos d-i 9. Area of Triangle in Polar Coordinates. (2) Fig. 12. Let the vertices of the triangle be Pi (ri, ^1), P2 (ro, B^, and P3 (""3, ^3). Connecting the origin with the vertices of the triangle, we shall form three triangles, P1OP2, PaOPj, PsOPi, 16 PLANE ANALYTIC GEOMETRY. such that area of PiPoPs^area of PiOPa + areaof P2OP; area of P3OP1. By Trigonometry, Area of PiO Po = ^ nr. sin (0, — O^), " " P,OP3 = ir,r3 sin (^2-^3), " " P30Pi = ir3ri sin (^1-^3); .-.Area of P1P2P3 = H^i^^ ^in {O^ — e^) + x^x^ sin {62 — O3) + XsXi sin (^3 — Oi)} . Note. It is better to regard formulas (1) and (2) of § 8 and (1) of § 9 as the solutions of their I'espective kinds of problems, and to follow the same methods in solving similar numerical problems, rather than to memorize them and substitute numerical values for ri, x^, etc. (1) 10. Relation between Cartesian and Polar Coordinates. As the position of the same point may be determined either by x and y or by r and 6, there must be some equations, connecting x and y with r and 6. Let the origin of polar coordinates be the origin of rect- angular coordinates, and the initial line of polar coordinates coincide with OX. Y Fig. 13. THE POINT. 17 If the rectangular coordinates of P (r, 0) are x and y, by Trigonometry, X = r cos 0, y = r sin 9- From [7], we may derive the formulas [7J r = Vx- + y-, e-tan-'^- ^^^ X Ex. 1. Find the rectangular coordinates of P(o, 30°) ; by [7], 3V3 X = 3 cos 30° = — — , y = 3 sin 30° = - ■ Ex. 2. Find the polar coordinates of P(— 2, + 3) ; by [8], r =1 V4 + 9 = Vl3, e = tan-i (— 3) = 180° - tan-i (l) = 180° -50° 19'"= 123° 41'. If the angle XOY between the axes of the Cartesian coordi- nates is w, it is not difficult to prove the following formulas : X = r cos 6 — r sin ctn w, ^ r sin ^ y (1) y = sm w r = Vx" + y' + ^ xy cos to, ^ |_x H-y cos wj J (2) EXAMPLES. 1. Find the length and the slope of the line joining (3, 4) and ( — 5, — 6). 2. Find the length and the slope of the line joining ( — 5, 2) and (3, — 1). 3. Find the lengths and the slopes of the sides of the triangle of which the vertices are (1, 2), (—3, 4), and (2,-3). 18 PLANE ANALYTIC GEOMETRY. 4. Find tlie lengths and the slopes of the sides of the tri- angle of which the vertices are ( — 3, — 4), (0, — 2), and (4, 3), 5. Prove that the point (1, 5) is on the line joining the points (0, 2) and (2, 8) and is equally distant from them. 6. Prove the triangle having its vertices at the points (2, 2), (— 2, — 2), and (2 \/3, — 2 V3) an equilateral triangle. 7. Find the perimeter of the triangle having as its vertices the points (a, b), ( — a, b), and ( — a, — b). 8. Prove the quadrilateral having its vertices at the points (2, 3), (—1, 2), (—2,-3), and (1,-2) a parallelogram. 9. Show that the points (2, 3), (—2, 3), (—2, —3), and (2, — 3) lie upon a circle of which the centre is the origin. 10. Find the coordinates of P such that PiP : PP2 = 3 : 5, where Pj is (2, 2) and P^ is (4, 3). 11. Find the coordinates of the point which divides the dis- tance from ( — 3, 4) to (— 5, — 6) in the ratio 5 : 7. 12. Find the coordinates of the point f of the distance from (3, -4) to (-2, 5). 13. Find the coordinates of the points trisecting the line joining (2, 3) and (4, 5). 14. Find a point which divides the distance from the origin to the point (4, 3) externally in the ratio 6 : 5. 15. On the straight line passing through (1,3) and (4, — 6) find a point such that its distances from the given points respectively shall be in the ratio — |. 16. To what point must the line drawn from (— 2, — 3) to (3, 7) be extended in the same direction that its length may be trebled ? 17. Given the three points A (4,-1), B (1,-4), and C (— 1, —6) upon a straight line. Find a fourth point D such fi . AD AB '^^"* DC =^ - BC- 18. One extremity of a line is at tlie point (3, —5) and a point I of the distance to the other extremity is (4, 3). Find the other extremity of the line. THE POINT. 19 19. Find the lengths of the medial lines of the triangle (-5,3), (-3, 7), (1,-1). 20. How far is the point bisecting the line joining the points (5, 5) and (— 3, 7) from the origin ? What is the slope of this last line ? 21. Find the length and the slope of the line joining the points ( — 1, 3) and (2, 5), tlie axes being oblique and to = 60°. 22. Find the area of the triangle (0, 0), (3, 1), (—2, 0). 23. Find the area of the quadrilateral (—2, 3), (—3, —4), (5,-1), (2,2). 24. Find the area of the pentagon (1, 2), (3,-1), (6,-2), (2, 5), (4, 4). 25. What is the ratio of the areas of the triangles (2, 3), (1, 2), (3,-1) and (5,-4), (2,-1), (-1, 3)? 26. The vertices of a triangle are the points (1, 2), (3, — 5) and ( — 2, 1). What is its area? What is the area of the triangle formed by joining the middle points of its sides in succession ? 27. If the angle between the axes is 30°, plot the points (1, 2), ( — 2, — 4), and (3, — 5), and find the perimeter of the triangle formed. 28. Find the rectangular coordinates of the following points: (2, 30°); (-3,45°); (l-"!)- 29. Find the polar coordinates of the following points : (2,5); (3,4); (-5,12). 30. Find the lengths of the sides and the area of the triangle (5, 15°), (3, 75°), (6, 135°). 31. Find the perimeter and the area of the triangle (1, 30°), (3, 60°), and (5, 90°). 32. Show that the distance between the two points (r,, d{) and (ro, O.2) is a maximum when 6.. — ^i = 180°, and a mini- mum when Oi=^0.2, ri and r., remaining unchanged. 33. In the triangle ABC, A (1, 2), B (3, -2), C (2, 5), a line is drawn bisecting the adjacent sides AB and BC. Prove this new line to be parallel to AC and half as long. 20 PLANE ANALYTIC GEOMETRY. 34. Prove that the area of a triangle, in oblique coordinates, is K^iYa — ><2yi+x2y3 — X3y2 + X3yi — xiYa) sin w, 0) being the angle between the axes. 35. Pind the coordinates of a point equally distant from the points (2, 3), (—2, —3), and (1, 4). 36. A point is distant 7 units from the origin and equally distant from the two points (1, 2) and ( — 2, — 1). What are its coordinates? 37. A point is distant 5 units from the origin, and the slope of the line joining it to the origin is f . What are its coordinates ? 38. Assuming that the medial lines of a triangle meet in a point I of the distance from each vertex to the opposite side, find this medial point of the triangle (3, 8), ( — 2, 7), (1, — 4). 39. Show that the medial point of the triangle (x„ y,). (X. y.), (X., ya) is ( ''■+;'+^ y+i'+y- y 40. Show that the lines from the vertices to the medial point of the triangle (3, — 8), (—4, 6), (7, 0) divide it into three triangles of equal area. 41. Show that the line joining any two vertices of the tri- angle (xx, yi), (x2, y2), (xs, ya) to the medial point form with the adjacent side of the triangle another triangle of one third the area of the given triangle. 42. Given four points Pi (xi, yi), P^ix^, ya), Psi^s, ya), P4(x4)y4); fiiid the point half way between Pi and Pj, then the point one third of the distance from this point to Pg, and finally the point one fourth of the distance from this point to P4. Show that the order in which the points are taken does not affect the result. 43. Prove analytically that the line joining the middle points of two sides of a triangle is parallel to the third side and equal to one half of it. 44. Prove analytically that a line which divides two sides of a triangle proportionally is parallel to the third side. THE POINT. 21 45. Prove analytically that, if in any triangle a median be drawn from the vertex to the base, the sum of the squares of the other two sides is equal to twice the square of half the base, plus twice the square of the median. 46. Prove analytically that the diagonals of a parallelogram bisect each other. 47. Prove analytically that the lines joining the middle points of the adjacent sides of any quadrilateral form a paral- lelogram. 48. Prove analytically that the above parallelogram (Ex. 47) is equivalent to one half the quadrilateral. 49. Prove analytically that in any right triangle the straight line drawn from tlie vertex to the middle point of the hypot- enuse is equal to one half the hypotenuse. 50. Show that the sum of the squares on the four sides of any quadrilateral is equivalent to the sum of the squares on the diagonals, together with four times the square of the line joining the middle points of the diagonals. CHAPTER n. LOCI. 11. Relation between Equation and Locus. We have seen, in the previous chapter, that two conditions are necessary to fix the position of a point in a plane ; for example, x ^ a, y = b. If only one of these conditions is given, the point is not fully determined, but may lie any- where upon a certain straight line or curve. For example, X = a gives a condition which is true for all points upon a straight line parallel to the axis of y and at a distance of a units from it. We say that the equation x = a represents this straight line. In the same way, the equation y = b represents a straight line parallel to the axis of x, at a distance of b units. As particular cases, x == represents the axis of y and y = represents the axis of x. The above are simple examples of the interpretation of an algebraic equation in Analytic Geometry. In the same manner, any equation between x and y will be true for the coordinates of an infinite number of points, which will be found to lie upon a certain curve or locus. We say that the equation represents the locus, or that it is the equation of the locus. Thus, the equation x = y is true for the coordinates of all points which lie upon the straight line bisecting the angles of the first and the third quadrants between the coordinate axes. It is, therefore, the equation of this straight line. Thus, also, x^ + y- ^ 25 is the equation of a circle, the centre of which is at the origin and the radius of which is 5 ; for the equation is evidently true for the coordinates of all points upon this circle, and for those of no other points. This rela- tion between the equation and the locvis is fundamental for LOCI. 23 the work of Analytic Geometry, and because of its great im- portance we restate it for easy reference as follows : All 2^oints the coordinates of ivhich satisfy a given equation lie upon a certain curve, called the locus of the equation ; and, conversely, if a point lies upon the locus, its coordinates satisfy the equation of the locus. 12. Equation of the First Degree. It folloAvs from the the above that the locus of an equation may be empirically constructed. We need only, by trial, to find a sufficient number of points the coordinates of which satisfy the equation, in order to obtain a dotted outline of the curve. The rest of the curve may then be tilled in free-hand, provided the points are near enough together. Ex. Let us take, for example, the equation 24 PLANE ANALYTIC GEOMETRY. We may find any number of points on the locus by giving to x succes- sively any number of arbitrary values and computing the corresponding values of y from the equation. For example, giving x the value — 3, we have — 6 + oy = 0, whence y = -1. The values, x = — 3, y = 4, satisfy the equation, and hence the point (—3, 4) lies upon the locus. In this way we find the following points: (-3, 4), (-2, -\% (- 1, »), (0, 2), (1, f), (2, |), (3, 0), (4, - |). If these points are plotted, it will be found that a straight line can be drawn through them, as in Fig. 14. We assume, in the example, that the locus is a straight line. The student should motice, however, that this is really an assumption. Our work gives us no information concerning any points of the locus except those actually found. We will draw a smooth curve through these points, however, thus obtaining an approximation to the shape of the curve, the accuracy of the approximation depending, in general, upon the nearness together of the points actually found. In the next chapter, we shall prove that any equation of the first degree represents a straight liyie. Hence, if we wish to find the locus of an equation of the first degree, it is necessary to find only two points and to draw a straight line through them. The two points most readily found are those in which the straight line cuts the two axes. The point B in which the line cuts the y-axis is the point for which x = ; i.e., in the example, the point (0, 2). Similarly, the point A in which the line cuts the x-axis is the point for which y ^ ; i.e., the point (3, 0). The distances OA and OB are called the intercepts on the axes of x and y respectively. 13. Equations of Degree Higher than the First. If the equation to be considered is of a degree higher than the first, the work becomes more difficult. It is not easy to give specific directions which will apply to all cases, but the following plan of work may be followed in the simpler cases at least. LOCI. 25 (1) Find the points in which the curve cuts the coordinate axes. (2) Solve the equation for one of the coordinates in terms of the otlier, let us say for y in terms of x. (3) Find what values of x, if substituted in the equation, make y imaginary. These values do not correspond to any real points of the curve. (4) Assume values of x, not excluded by (3), and compute corresponding values of y. (5) Plot the points thus found and draw a smootli curve through them. The above directions are given as hints to the student and not as rigid rules to be invariably followed. Special prob- lems often require modification of the method. It should be remembered that the whole object is to find, by trial, enough points of the curve to outline it, and the method whicli will do this most quickly is the best. In particular, we should notice tliat it will sometimes be found more convenient to solve the equation for x instead of y. In that case, tlie letters X and y are to be interchanged in reading directions (2), (3), and (4). No definite instruction can be given as to how near together the points should be taken. When the curve changes its direction slowly, they may be far apart ; but if the shape of the curve is doubtful or peculiar at any place, the points must be taken close enough to make the proper shape per- fectly clear. Ex. 1. Take, as the first example, 4x-+ 9y-=36. (1) riacing X = 0, we have 0y2 = m, y =±2. Hence the curve cuts the axis of y in the two points (0, 2) and (0, — 2). Placing y = 0, we find x = ± .']. Hence the curve cuts the axis of x in the two points (.'1, 0) and (— 3, 0). (2) The equation is readily put into the form y = ± ^ V9 - x2 = ± 2 V(3 - X) (3 + X). 26 PLANE ANALYTIC GEOMETRY. (3) It is apparent that, it x < — o or x > ;J, tlie quantity under the radical sign is negative, and hence y is imaginary. Therefore there can be no point of the curve above or below that portion of the x-axis which lies to the left of — 3 or to the right of + 3. (4) Using values of x which lie between — 3 aud_+ 3, we find the following points of the curvej (-3, 0), (-2, ±|V5), (-1, ±|V8), (0, ±2), (1, ±|V8), (2, ± 3V5), (3, 0). (5) The plotting of the points leaves the shape of the curve somewhat doubtful in the extreme right and left hand portions. Therefore we com- pute also the points (—2^, ± Wll), (2.1, ±iVll), and then draw the curve through all the points found. The curve is an ellipse. (Fig. 15.) We have seen that the curve cannot lie to the left of x = — 3 or to the right of x = + 3. Similarly, we may show that it cannot lie above y = +2 or below y = — 2. This we do by solving the equation for x We have, then, X = ± I V4 - y2 = ± 3 V(2 - y) (2 + y). From this it appears that x is imaginary, if y is >»2 or < — 2. This finding of the limits of the curve in the vertical as well as the horizontal direction is not necessary to the plotting, but is often useful. LOCI. 27 It will be noticed that the curve is symmetrical with respect to the axis of X. This follows from the fact that to each value of x correspond two values of y, equal in absolute magnitude but opposite in sign. Similarly, since to any value of y correspond two values of x, equal in magnitude but opposite in sign, the curve is symmetrical with respect to the axis of y. It is useful in plotting to notice such symmetry when possible. Ex. 2. Take, as a second example, the equation 4x2 _ oy'2 _ i6x + 18y - 29 = 0. (1) When x = 0, 9y2 — 18y = — 29, 3 ± V-'20 whence y — ^ . This is imaginary. Hence the curve does not cut the axis of y. When y = 0, 4x2 _ lox = 29, 4±3V5 . ,, , „, or X = = 5.35 or — 1.35. Hence the curve cuts the axis of x in the points (5.35, 0) and {- 1.35, 0). (2) By solving the equation for y, we obtain 3±2Vx2-4x-5 3±2V(x-5) (x + 1) . y= 3 = 3 • (3) The value of y will be imaginary when the product (x — 5) (x + 1) is negative; i.e., when x has values which cause the factors to have oppo- site signs. Such values of x are seen by inspection to be those which are > — 1 and < + 5. Hence there is no point of the curve above or below the portion of the x-axis between — 1 and + 5. (4) Giving x in succession the values —4, —3, —2, — 1, 5, 0, 7, 8, we find the following points of the curve : (-4,:±W.). (-,s,lf^), (-.■^'), (-,.). (5) The plot results as in Fig. 10. There are no limits to the value of y. For, if we solve for x, we have 4±3Vy2-2y + 5 x = and no value of y will make the quantity under the radical sign negative. 28 PLANE ANALYTIC GEOMETRY. As regards symmetry, the curve is evidently not symmetrical with respect to the coordinate axes. But from the equation y = l±|V(x-5) (x+l), it appears that the curve is symmetrical with respect to the line y = 1. Similarly, from X = 2 ± I Vy2 — 2y + 5, it appears that the curve is symmetrical with respect to the line x = 2. ^ / 5 N s 4 / /' s \ 3 / / ^ ^ 2 / ^ \ 1 / y' -5 -4 -3 -2/ ■10 1 2 3 4 5 \e 7 8 / / -1 \ ^ y / -2 \ s ? / -3 s ^ -4 -5 Fig. 16. 14. Transcendental Equations. The equation of a locus is not necessarily algebraic, but may be transcendental ; that is, it may involve the non- algebraic functions. A few examples will make clear the methods of handling such equations. Ex. 1. Take the equation y = sni X. We consider the angle x expressed in circular measure, and lay off first the distance, 7r = 3.1416, and its multiples. Each division of the LOCI. 29 axis we then subdivide into convenient intervals, say sixtlis, and obtain thus upon tlie x-axis the points It Tt 7t 2Tt bit ItC ^TC ?>Tt ^ —, —1 — ' — ^' — ' 7t, — ' -^' — ^' etc. G' 3 2 3 G ' G 3 2 By means of a table of natural sines, the value of y corresi>onding to each X is readily found. The plot results as in Fig. 17 : It is necessary to plot only that portion of the curve between x = and X = TT. For sin (x ± tt) = — sin x, and sin (x ± 2;r) = sin x. Hence the curve y = sin x is composed of an indefinite number of arches extend- ing both to the left and to the right, the arches being alternately above and below the axis of x, and each arch being of the same shape as that between x = and x= tt. Ex. 2. Consider, as another example, y = 5 log X. By use of a table of logarithms, any number of points are readily found with the following result, as shown in Fig. 18. 30 PLANE ANALYTIC GEOMETRY. Fig. 18. It is to be noticed that the curve approaches the axis of y in the nega- tive direction, but never reaches it. Tliis arises from the fact that the logarithm of a quantity increases negatively without limit as the quantity approaches zero. It is also to be noticed that we have no portion of the curve corre- sponding to the negative values of x, owing to the fact that the logarithm of a negative quantity is not a real quantity. 15. Equations in Polar Coordinates. The equation of a locus may also be given in polar coordi- nates (r, 6). To plot sucli a curve, it will be found convenient to lay off first a number of radii at convenient angles. Then the corresponding values of r may be computed from the equation, perhaps with aid of a table of trigonometric func- tions. These values of r are then laid off upon the corre- sponding radii, with careful attention to the meaning of the negative sign, as explained in § 7. For example, r — 2a sin 6. We compute the following table of corresponding values : LOCI. 31 61 = 0°, r = 0; 6= 15°, r= .52a; 61 = 30°, r = a; ^ = 45°, r = 1.41a 61 = 60°, r= 1.73a; 61 =75°, r= 1.93a; 61 = 90°, r = 2a; 6i = 105°, r = 1.93a 61 = 120°, r = 1.73a; 6*= 135°, r= 1.41a; tf = 150°, r=a; 6i = 165° r = .52a; ^ = 180°, r = 0; 6* = 195°, r=— .52a; ^=210°, r=— a 6* = 225°, r= — 1.41a; 6i = 240°, r=- 1.73a; 61 = 2.55°, r= — 1.93a (?=270°, r = -2a; 6* = 285°, r= — 1.93a; 61=300°, r= — 1.73a ^ = 315°, r=- 1.41a; 61=330°, r= — a; d* = 345°, r=-.52a e - 300°, r = 0. The plot results as in Fig. 19. For angles in the third and the fourth quadrants r is negative, and is, therefore, measured along the backward extension of the terminal line. For example, if ^=195°, r= —.52a. We measure .52a units away from the 195° point and toward the 15° point. We obtain, therefore, the same point of the curve for 9 = 195° as we obtained for 6= 15°. In the same way, all angles of the third and the fourth quadrants give the same points as angles of the first and the second quadrants. Hence we obtain the curve above the initial line twice. The curve is a circle. 32 PLANE ANALYTIC GEOMETRY. 16. Locus Defined by Geometric Property. A locus is often defined by means of a property which is common to all points upon the locus, but to no other point. The object of Analytic Geometry is then to express this prop- erty by means of an algebraic equation between the coordi- nates of any point upon the locus, and then to find the shape and the properties of tlie locus from the equation. For example, let it be required to find the locus of a point at a dis- tance of 5 units from the origin. By [1], § 3, the distance of a point (x, y) from the origin is Vx^ + y^. For the locus, therefore, we have Vx- + y- = 5, or x2 -f- y2 = 25. This is the equation of a circle of which the centre is at the origin and the radius is 5. Again, let it be required to find the locus of a point, the sum of the distances of which from the points (— 1, 0) and (1, 0) is equal to 6. By [1], § 3, we have V(x-h 1)2 -hy^ + V(x — l)2-}-y2 = 6, which reduces readily to 8x2 ^ f,y2 =: 72. This locus is readily plotted and found to be similar to that of Ex. 1, § 13. 17. Intersection of Loci. It is often necessary, when the equations of the two loci are given, to find the coordinates of the point or points in which the loci intersect. Now, a point of intersection of two loci is a point which lies upon both, and hence by § 11 its coordinates satisfy the equation of each. The problem is, therefore, to find those values of x and y which satisfy both equations, or in other words : To find the coordinates of the jjoints of intersection of two loci, solve their equations for simultaneous values of x and y. Two cases are of special importance in elementary work. Case I. When both equations are of the first degree. LOCI. 33 There is, in general, one and only one pair of values which satisfies the equations. The two loci intersect, therefore, in one point, as is evident if we accept the statement in § 12 that each locus represents a straight line. For example, consider the equations 2x + .3y = 5, 4x — 5y = 8, the solution of which is x = f|, y = ^\. The loci intersect in the point (II) T'r)' ^s shown in Fig. 20. Fig. 20. An exceptional case arises when the two equations are contradictory ; for example, 2x + % = 4, 2x + ;5y = 7. Since we can find no values of x and y common to these two equations, it follows that the two loci do not intersect. In fact, a plot shows that the equations represent parallel straight lines. 34 PLANE ANALYTIC GEOMETRY. A rigorous discussion of the principle here involved may be given as follows : Let ax + by = c, (1) and a'x + b'y = c', (2) be two equations, each representing a straight line. By elimination, we find (ab' — a'b)x = b'c — be', and (ab' — a'b)y = ac' — a'c. Hence, unless ab' — a'b = 0, we find as the solution of the two equa- tions, _ b'c — be' ab' — a'b' /^\ _ ac' — a'c ^ ~ ab'-a'b ' and these are the coordinates of the point of intersection of the two lines. If ab' — a'b = 0, the above solution is impossible. A meaning may be given to equations (3), however, by use of the method of limits. To do this, let us suppose, first, that the coefficients of (1) and (2) are such that ab'— a'b is a very small quantity, although not zero. The values of X and y in (3) are then possible, but are, in general, very large. If, now, the expression ab' — a'b can be made to approach zero as a limit, the values of x and y will increase without limit, except in the case men- tioned below. Now ab' — a'b can be made to approach zero in various ways; for example, we may keep (1) unchanged, and alter step by step the coefficients of (2). If each step is accompanied by a plot, we see that the line (1) is kept fixed, but the position of (2) is altered. The point of intersection of the two lines is at each step more remote. Hence the case ab' — a'b = may be regarded as the limiting case of two lines, the point of intersection of which has receded indefinitely from the origin. But the limiting position of two such lines is evidently one of parallelism, and hence it follows that if ab' — a'b = 0, the two lines are parallel. (See also § 35.) The above reasoning fails, if at the same time that ab' — a'b = 0, the numerators of the fractions in (3) are also zero. But, in that case, we have evidently a _ b _ c a' b' c' ' and the equations (1) and (2) are identical. We have, therefore, only one line in the problem. LOCI. 35 18. Intersection of Loci. Case II. When one equation is of the first degree and the other of the second. We shall find it convenient to eliminate by substituting in the quadratic equation the value of one of the unknown quan- tities as given in terms of the other by the simple equation. Let us suppose we eliminate y in this way. There results, in general, a quadratic equation in x, let us say ax'-+bx + c = 0, (1) the roots of which are the abscissas of the required points of intersection. When the values of x have been found, the corresponding values of y are found hy siihsthuting in the simple equation. In the solution of the above quadratic equation, one of three cases will arise, as illustrated in the following examples : Ex. 1. (1) To find the intersections of y -t- 2x = 3, and X- + y2 = 5. (2) 36 PLANE ANALYTIC GEOMETRY. Eliminating y by substitution from (1) into (2), there results 5x2-12x + 4 = 0, which gives x = 2 or |. The corresponding y's found from (1) are —1 or y-. Therefore the points of intersection are (2, — 1) and (|, -U-). (Fig. 21.) Ex. 2. To find the intersections of y + 2x = 5 and x''^ + y- = 5. Y Je Eliminate y, and there results 5x2 _ 20x + 20 = 0. This equation has the equal roots 2 and 2, to which correspond the equal y's, 1 and 1. The loci are tangent at the point (2, 1). (Fig. 22.) Ex. 3. Find the intersections of y + 2x = 7 and x^ + y2 = 5. Eliminating y, we have 5x2 _ 28x + 44-0, 14±2V-6 whence x = • 5 Since this is imaginary, the two loci do not intersect. (Fig. 23.) LOCI. 37 We may now state the three cases which may occur in the solution of the general equation (1) on p. 35, as follows : (1) If b^ — 4ac > 0, the two roots are real and uneqnal. The two loci intersect in tivo distinct points. (2) If b"^ — 4ac:=0, the ttvo roots ore real and equal. The tivo points of intersection become coincident ; the two loci are therefore tangent. 38 PLANE ANALYTIC GEOMETRY. (3) If h"^ — 4ac ■< 0, the roots are imaginary. The two loci fail to intersect. It is sometimes said in such a case as the last, that tlie loci intersect in "imaginary points." The beginner will under- stand this as equivalent to saying that the loci do not inter- sect at all. The meaning of the phrase is that imaginary values of x and y may be found which satisfy both equations, and the advanced student finds advantage in recognizing this fact. 19. Limiting Cases of Intersection. In the process of eliminating y as outlined in the last article, it may happen, in exceptional cases, that the term containing x^ cancels out. The loci then intersect in only one point. Ex. 1. Consider By elimination, whence 24x - 72 = 0, X = 3, y = 0. The loci intersect in one point (3, 0). (Fig. 24.) LOCI. 39 Again, it may happen that the elimination causes both x*^ and X to disappear, thus showing the equations to be contra- clictory. We have then no point of intersection. Ex. 2. Consider 4x-'_()y2_8x — ,32 = 0, 2x- :ly — 2 = 0. Substituting from the second equation in the first, we obtain - 36 = 0. This shows the equations to be contradictory, and the loci do not inter- sect. (Fig. 25.) We may examine these cases by the method of limits as we did the exceptional case of § 17. In general, if we eliminate y between a given equation of the second degree and a given equation of the first degree, we obtain a quadratic equation of the form ax- + bx + c = 0. The roots of tliis equation are X2' - b -h Vb'^ -4ac 2a - b - Vb-'^ — 4ac 2c 2a b — Vb^ — 4ac' 2c b-f- Vb2 - 4ac" 40 PLANE ANALYTIC GEOMETRY. If, now, a approaches zero as a limit, X2 increases without limit and xi approaches the value — , . If, then, a is made to approach zero by altering the position of one of the given loci (see § 17, fine type), we may say that the case in which we find only one solution of a quadratic and a simple equation, repre- sents the limiting case of the intersection of two loci, when one point of intersection is indefinitely remote. (See Fig. 24.) If both a and b approach zero as a limit, both xi and X2 increase with- out limit. Hence the case in which the quadratic and the simple equa- tion are contradictory, represents the limiting case of the intersection of two loci, when both points of intersection have receded indefinitely from the origin. (See Fig. 25.) EXAMPLES. Plot the following straight lines : 1. 5x + 7y + 15 = (). 3. 3x 4" 5y = 0. Plot the following curves of the second order : 4. x2 + y2 = 49. 6. y2 = 8x. 6. 25x2 + 4/-' = 100. 7. x^ — y^ = 4. 8. xy = 12. 9. x2 + 4x— 4y + 4 = 0. 10. x-'+y2 — 4x + Gy — 3 = 0. 11. 9x2 ^ igy2 _|_ i^^ _ (54y _ 71 ^ 0. 12. 9x2 _ 4y2 _ 3Q^ _^ 24y - 3(3 = 0. Plot the following curves of the third order : 13. y = x". 14. f = xl 15. y2 = X (x — 3)2. LOCI. 41 16. y2=x (x + 2) (x-3). ^ x- + 4a 18. y^ = ^^^^. Plot the curve of the sixth order : 19. 16y- = 4x*-x''. Plot the f olio win ar transcendental curves 20. y = cos x. 21. y z=: tan X. 22. y ^ sin~'x. 23. y = 10\ ot the following curv 24. r = ae. 25. r = a sin 20. 26. r = a cos 30. 27. r = a (1 — cos 0). 28. r = a tan 0. 29. ■ 3^ r = a snr -• 30. r = a- cos 2^. 31. Find the equation of the locus of a point of which the distance from the axis of x equals 4 times the distance from the axis of y. Plot the curve. 32. Find the locus of a point wluch, if connected with the points (— 1, 3) and (— 3, — 3), forms a triangle of the same area as if connected with the points (1, 2), (2, 2). 33. A point moves so that its distances from the two fixed points (2, — 3) and (— 1, 4) are equal. Find the equation of the locus and plot. 34. A triangle of area 25 has two vertices at the points 42 PLANE ANALYTIC GEOMETRY. (5, — 6) and ( — 3, 4). Find the locus of the third vertex and plot. 35. Find the equation of the locus of a point 5 units from the point ( — 1, — 2) and plot the locus. 36. A point moves so that its distance from the axis of y equals its distance from the point (4, 0). Find the equation of its locus and plot. 37. A point moves so that its distance from the origin is equal to the slope of the straight line joining it to the origin. Find the equation of the locus in polar coordinates and plot. 38. A point moves so that its distance from the origin equals a constant times the angle which the line joining it to the origin makes with the initial line. Find the equation of the locus in polar coordinates and plot. 39. A point moves so that its distance from the axis of x is 1 its distance from the point (0, 4). Find the equation of the locus and plot. 40. A point so moves that its distance from the point (1, 3) is to its distance from the point ( — 4, 1) in the ratio of 2 : 3. Find the equation of its locus and plot. 41. A point so moves that its distance from the axis of y is to its distance from the point (3, 2) in the constant ratio 2 : 3. Find the equation of its locus and plot. Find the points of intersection of the following loci, noting points of tangency if such exist : 42. 5x — 7y + 13 = 0,, 2x + fiy + 3==0. 43. x + y — 7 = 0, 2x + 3y + 2 = 0. 44. y-3x-3 = 0, 5y-2x-2 = 0. 45. 3x + 2y = 0, x — 4y = 0. 46. x2 + y2-4x + 6y-12 = 0, 2y = 3x + 3. 47. x2 + y2 — 4x + 6y — 12 = 0, x — 3 = 0. 48. y2-10x — 6y — 31 = 0, 2y-10x — 47 = 0. 49. 3x'^ + 7x + 2y + 2 = 0, x + y -3 = 0. 50. 2x-4-3y- = l, 2x-3y + 3 = 0. LOCI. 43 51. 4x2 + 9y2=r36, x + 3y = 5. 53. 9x" — 4y^ + 54x-lGy + 29 = 0, 2y — 3x + 5 = 0. 54. 4x'-' — 25y-=100, 2x-5y = 0. 55. xy + 3x-2y-8 = 0, y + 3 = 0. 56. x2 + y2=r41, xy = 20. 57. X' + y- — Gx - 2y - 15 = 0, 9x- + 9y'^ + 6x - Gy - 27 = 0. ^^' 25 + 9 ~^' 25 9-^- For what values of b will the following curves be tangent? 59. x- + y2 = 49, y = 3x + b. 60. 5 + 4=1, y = ;;x + b. 61. y2 = 8x, y==3x + b. What must be the values of m in order that the following loci may be tangent ? 62. x2+y2 = 4, y = mx + 3. 63. 9x2-16y2 = 144, y = mx + 3. 64. x2 + y2 = 25, y = mx + 3. 4 65. Show that y ^= mx -| is tangent to y-^16x for all values of m. (The following examples are to be solved by finding the intersection of two loci each of which contains the required point.) 66. Find a point equidistant from the points (— 3, 4), (5, 3), and (2, 0). 67. A triangle of area 10 is so constructed that two vertices / are at the points ( — 2, 2) and (0, 0), while the third vertex is on the line x + 3y = 0. Find the third vertex. 44 PLANE ANALYTIC GEOMETRY. 68. Find a point on the axis of x which is equidistant from (5, 7) and (3, - 4). 69. Find a point on the line x — 4y + 7 =; 0, the distance of which from the axis of x equals its distance from the axis of y. 70. Find the points on the curve x^ + y^ = 25, such that the slope of the line joining each of them to the origin is 2. 71. Find the points which are 5 units distant from (2, 3) and 4 units distant from the axis of y. CHAPTER III. THE STRAIGHT LINE. 20. In the last chapter, we have seen that, corresponding to every single equation containing x and y, there is a locus such that the coordinates of every point on the locus satisfy the equation, and the coordinates of any point not on the locus do not satisfy the equation. As the straight line is the simplest locus, we will find its equation first. Now a straight line is determined, if it is made to pass through two points, or through one point in a given direction. According as we determine the line in one way or another we shall expect to get different forms of the equation. In the first three forms of the equation Avliich we shall find, and which may be regarded as the fundamental forms of the equation, i.e., [9], [10], [11], the line [9] and [11] is determined by being passed through a given point in a given direction, and the line [10], by being passed through two given points. 21. Equation of the Straight Line in terms of its Slope and Intercept on the Axis of y. Let AB (Fig. 26) be the given line, having the slope m, and cutting off on YY' the intercept OB, denoted by b. Let P (x, y) be any point on the line. From P draw PR parallel to YY', and from B draw BR parallel to XX'. Then BR == x, and RP = y - b. RP Since, by Trigonometry, ^-^= m, we have, by substitution, y-b = m, X or y = mx + b. [9] 46 PLANE ANALYTIC GEOMETRY. Y As P is any point of tlie line, [9] is an equation which is satisfied by the coordinates of every point of the line, and which may be shown to be not satisfied by the coordinates of any point not on the line. For if any point not on the line be joined to B, the slope of this joining line will not be m, as it must be if the coordinates of the point are to satisfy the equation. Therefore, [9] is the required equation of the line. As the line has been drawn in Fig. 26, m and b are both negative ; but if the line is drawn in a different direction, or cuts off a different intercept on the axis of y, m and b will have different values. And since m is the tangent of an angle between 0° and 180° and b is a distance, m and b can have all possible values from minus infinity to plus infinity. Ex. Find the equation of the straight Ime making an angle of 30° with the axis of x, and cutting off an intercept 5 on the axis of y. Here m = tan 30° = — ^' and b = 5 ; V3 •. the required equation is y = — = + 5. THE STRAIGHT LINE. 47 In equation [9], m and b are constant, and x and y vary as different points on the line are taken. If we give new values to m and b, as rrii and bj, y ^ rriix + bi will represent a new straight line, in the equation of which x and y alone vary. Now quantities such as m and b, which are constant for any given curve, but may be changed so as to obtain the equation of a second curve of the same kind as the first, are called parameters. It follows, then, that the parameters do not determine the kind of curve, but do determine which one of a particular kind of curves has been selected. What will be true of all the lines for which b = 0? What will be true of all the lines for which m = ? If the equations of a series of lines have the same b, but different m's, what is true of all the lines? If the equations of a series of lines have the same m, but different b's, what is true of all the lines ? What is true of the line for which m is infinite ? What will be the form of equation of this last line ? 22. Equation of the Straight Line in terms of its Inter- cepts on the Axes. y Y Fig. 27. Let AB be the given line, cutting the axes of x and y at points A and B respectively. Denote the intercepts OA and 48 PLANE ANALYTIC GEOMETRY. OB by a and b respectively. Let P (x, y) be any point on the line and draw PM parallel to YY'. Then AM = x — a. M P = y, AO = — a, and OB = b. The triangles AOB and AMP are similar, since their sides are respectively parallel, and hence their homologous sides are proportional ; . 0B_ MP •"■ A0~ AM whence, by substitution, b ^ y — a X — a' By reduction, we get the required equation, 1+1=' M Ex. 1. Find the equation of the straight line making intercepts 3 and 5 on the axes of x and y respectively. X V Here, a = 3 and b = 5, so that the equation is ^^ + r = L o o Ex. 2. Find the equation of the straight line making intercepts 3 and — 5 on the axes of x and y respectively. Here, a = 3 and b = — 5, so that tlie equation is Q + r = 1 or - - i = L 6 — do In equation [10], it will be observed that a and b are the parameters, since they are constant in the equation of any given line, and, when varied, will give a new line determined in the same way as the previous lines of the series were determined. If a remains constant and b varies, what is true of the resulting sys- tem of lines ? If a and b vary in tlie same ratio, what is true of the resulting system of lines ? How is the line situated for which a is finite and b infinite ? Througli what quadrants in the plane will the line pass, for which a and b are botli positive ; for which a is negative and b is positive ? "What will be the values of a and b for the line which passes through the origin ? THE STRAIGHT LINE. 49 23. Equation of the Straight Line in terms of its Normal Distance from the Origin and the Angle the Normal makes with the Axis of x. /t)u ■ C.«~ Vb'"' + a^ - 2ba cos w 60 PLANE ANALYTIC GEOMETRY. 31. Equation of the Straight Line in Polar Coordinates. Fig. 32. Let AB be the given line, OD, equal to p, be the normal distance of the line from the origin, and let /_ XOD = a. Let P(r, G) be any point on AB. Then Z DOP = ^— a, and our required equation is rcos(e — a) = p. [12] It is to be noted that in this equation is variable and has no relationship to the angle the line AB makes with the initial line OX. What is true of the line under the following conditions : a = 0° ; a = 0° and p = ; a = 90° ; a = 90° and p = ? What angle will the line r cos (B — 45°) = 5 make with the initial line ? What will be its intercept on the initial line ? 32. Distance of a Point from a Straight Line. AB shall be the given line, its equation being written in the normal form, x cos a + y sin a = p. Through the origin draw the line CD parallel to AB. Then there will be three cases to be considered, according as the point, the distance of which from AB is required, is above AB, between AB and CD, or below CD. THE STKAIGHT LINE. Y 61 N N V ^xP, \^^^ ^V '^N^ ^\ "-X ^ \ ">-. v' '^v^^) ** N N S ^ / N, "^ X- P2^\ ^ ^^V /O -k. \a X / -^ N \ V ^^^x ''"^^^^ ^ "^s ^. '-N "-.(3) ^^ / Fig. 33. Case I. Let Pi(xi, yi) be the given point, its distance from AB being pi. Tlie normal distance from the origin of a line through Pi parallel to AB will be p + pi, and the angle the normal makes with the axis of x will be a. Therefore its equation Avill be X cos a 4" y sin a = p + pi. (1) But Pi is a point of this line, so that Xi cos a + yi sin a ^ p + pi, whence pi ^= x, cos a + yi sin u — p. Case II. Let P2(x2, yz) be the given point, its distance from AB being p.,. The normal distance of a line through P., parallel to AB will be p — p^, and the angle the normal makes with the axis of x will be a. Therefore its equation will be X cos a + y sin a = p — po. (2) But P2 is a point of this line, so that X2 cos a + yo sin a ^ p — p.i, whence — po =■ y... cos a + y^ sin a — p. 62 PLANE ANALYTIC GEOMETRY. Case III. Let P3(x3, y^) he the given point, its distance from AB being pg. The normal distance of a line tlirough Pg parallel to AB will be pg — p, and the angle the normal makes with the axis of x will be 180° + «• Therefore its equation will be X cos (180° + a) + y sin (180° + a) ^ pg - p, or X cos a + y sin a = p — Pg. (3) But Pg is a point of this line, so that Xg cos a -}- yg sin a = p — pg, whence — Ps = Xg cos a + yg sin a — p. From the results of these three cases we may formulate the following working rule : Put the equation of the line i?i the form X cos a + y sin a — p =^ and substitute for x and y the coordinates of the 2^oint. The resulting value of the left-hand side will he the distance of the point from the line, and will be positive if the point and the origin are on opposite sides of the line, and ivill be negative if the p)oi7it and the origin are on the same side of the line. Ex. Find the distance of the point (2, 3) from the line 2x - y + 2 = 0. Writing the equation in the form x cos a + y sin a — p = 0, we have V5 V5 VS ,y. • 1 r . 2(2) ,3 2 _ 3 . '. the requu-ed distance — ^ H ^ -p. = ; ■\lb ■\ro yJl V5 3 i.e., the point is distant — = from the line, and on the same side of the V5 line as the origin. THE STRAIGHT LINE. 63 33, Equation of a Line through any Given Point with a Given Slope. Fig. 34. Let AB be the required line, having the given slope m, and passing through the given point Pi (xj, yi). The equation of AB call, therefore, be written in the form y = mx + b, (1) where at present b is unknown. But since the line passes through Pj, Xi and yi satisfy equation (1) ; •■•yi = mxi+ b, whence b = yi — mxi. Substituting in (1) the value of b thus found, and reducing, we get the required equation y — yi = m (X — Xi). [13] Ex. 1. Find the equation of tlie line through tlie point (1, —3) making /_ 45° with OX. Here m = tan 45° =1, xj = 1, yi = — ^5. ... y + .3= l(x-l) or y- x + 4 = is tlie eqviation of the line. 64 PLANE ANALYTIC GEOMETKY. Ex. 2. Find the equation of the line through (— 1, — 2) parallel to the line x + 2y — 1 = 0. The required line will obviously have the same slope as the given line. X + 2y — 1=0 may be transformed to the equivalent form, -1,1 ■ - y--^x + -- .-. m-- and the equation of the required line is 1 ^, XI =-1, yi 2, y + 2= — -(x + 1) or 2y + x + 5 = 0. 34. Equation of a Line through Two Given Points. Y Let Pifxj, yi) and P2(x2, yg) be the two given points, and let P(x, y) 1)6 any point on the line through tliem. DraAv PiMi, PM.^, and PM all parallel to YY' and PjS par- allel to XX'. Then the triangles PjSP and P1RP2 are similar, their sides being resjiectively parallel, so that their homol- ogous sides are proportional. RP. PiS PiR' THE STRAIGHT LINE. 65 whence after the signs of the denominators are changed, y — yi^ x — x, Yi-yj xi — X2' which is the required equation of tlie line. [14] Ex. Find the equation of the line through the points (1, — 2) and (-2,4). Letting (1, — 2) be (xi, yi) and (—2, 4) be (xo, ya), we get V 4- 2 X — I We must now consider some special cases to which [14] does not apply, since the construction fails. (1) If y2 =^ yi, the denominator on the left hand side be- comes zero, and hence the left hand member seems to become infinite. But if ya ^ yi, two points of the line are at the same distance, i.e., yi, from the axis of x and hence the line is parallel to that axis and its equation is y = yi. (2) If X2 = Xi, we can prove in the same way that the equa- tion of the line is x = Xx. (3) If yo == yi and Xj = Xj, both denominators become zero. But in this case there will be no line determined, as the two points which were to be used in its determination are coinci- dent, and through any one point an infinite number of lines can be drawn. Therefore we should not expect to find any equation. 35, Angle between Two Lines. I and II (Fig. 36) shall be two lines, making angles 0^ and O2 respectively with XX', and f3 shall be one of the angles made by them at their point of intersection. As we liave taken /?, it will be the internal angle of the triangle formed with the axis of X by the two linos. .■./3 = o,-e„ J ^ ^ tan 61 — tan Oo and tan B = j—, — ^ . 1 + tau di tan 62 66 PLANE ANALYTIC GEOMETRY. But tan ^1 = nrii and tan d^ = ma, where rrii and mj are the respective slopes of I and II. tan p mi — ma l + mims [15] In applying this formula, the value of tan ^ may be found to be negative. There are two cases in which this may occur : (1) As chosen, /8 may be an obtuse angle, when its tangent will necessarily be negative. In this case /3' is the acute angle between the two lines, and )8' := 180° — /3, so that tan /?' = tan (180° — ^8) = — tan /5. We can now say that the acute angle between the two lines is the angle of which the tangent is numerically equal to the value found by formula [15]. (2) If we do not plot the lines, ^ may be acute, but we may have written ^ = O2 — Oi when we should have written I3=0i — 62, thus finding tan (— /S) instead of tan /?. But tan ( — )8) = — tan yS, so that in this case, also, we may take for the acute angle between the two lines the angle of THE STRAIGHT LINE. 67 which the tangent is numerically equal to the value found by formula [15]. To sum up : ive may alwai/s take rrii and m^ fvotn the equa- tions, substitute in formula [15], and take the numerical value of the result as the tangent of the acute angle between the lines; but ive shall have to plot the lines if ire wish to know icliich is the acute and whicli is the obtuse angle. Ex. Find the acute angle between the two lines y = 2x + 1 and 2y + 3x — 1 = 0. Putting these equations in the form y = mx + b, we have y = 2x + 1 o 1 o and y = — ;^x + -, whence mi = 2 and mo = — ^* -(-1) 7 . •. tan /3 = 1 + and the acute angle between the two lines is tan K-l) ' . If the lines are parallel to each other, tan fi^=0, and mi = m2. [16] If the lines are perpendicular to each other, tan (3 = tan 90° = co ; .•. 1 -|- miiTia = 0, and mi = -— • [17] rn2 As the cases of parallelism and perpendicvdarity are the most important, we will now show how to determine by inspection of their equations when two given lines are par- allel or perpendicular to each other. Let the lines be Ax+ By + C = 0, A'x + B'y + C' = 0. 68 PLANE ANALYTIC GEOMETRY. These equations may be written A C y = -R'^- B B B'"" B' .-. mi = — - and m. = — — • A A' .'.if the lines are parallel, — = —, or the ratio of the coefficients of X and y in the two equatio72s is the same ; and if the lines A B' are perpcndicidar to each other, — ^^ — -r-,, or the ratio of the coefficients of x and y in one equation is the negative of the ratio of the coefficients of ^ and x in the other. 36. Equation of a Line through a Given Point Perpendicu- lar to a Given Line. Let Pi(xi, yi) be the given point, and let the equation of the given line be written in the form y = mx + b, (1) The equation of a line through Pj with slope mi is by [13] § 33, y — yi = noi (x — xi). (2) But if lines (1) and (2) are perpendicular to each other, 1 mi=: m .•.our required equation is y-yi = -^(x-xi). [18] Ex. Find the equation of a line through the point (— 1, — 3), and perpendicular to the line x + 2y — 1 = 0. X + 2y — 1 = is equivalent to y = — - x + -• . •. y + 3 = (x + 1) or y — 2x + 1 = is the required equation. THE STRAIGHT LINE. 69 37. Problem. To find the equations of the two lines which pass tlirongh the point (2, 3) and make an angle tan""^ (^) Avith the line 3x + y - 3 = 0. Y Fig. 37. Let AB be the given line, and let lines (1) and (2) be the lines making the required angle, tan~^ (■^), with AB. Then, if we denote tan~^ (^) by /3, and the angles made by lines (1) and (2) with XX' by 6i and 6^ respectively, we have from triangle ADF, 6i = $-\- ^, the exterior angle of a tri- angle being the sum of the two opposite interior angles, and from triangle C^E,6., = 6—ji since 6 = 6.,-\- (3. (Z ACE = /3.) tan -\- tan /? ■.tan 6i = -1 and 1 — tan 6 tan (3 tan 6 — tan /3 ^ ^'''' ^' ~ l + tan^tan;8~ '' the equations of the lines are : (1) y_3^_i (x-2) or y + x-5 = 0, (2) y — 3 = 7 (x — 2) or y — 7x + ll =0. 70 PLANE ANALYTIC GEOMETRY. On reviewing the work of this article, we see that any problem of this kind can be solved as follows : Draw the given line and the required lines of such lengths as to form two triangles, in each of which the sides are respectively the axis of X, the given line, and one of tlie required lines. Then, since vertical angles are equal and the exterior angle of a triangle is the sum of the two opposite interior angles, we can always express the new angles, 6i and 62, in terms of and (3, and hence determine the slopes of the new lines. It may happen, however, that we can construct only one, or, perhaps, neither of these triangles. These special cases will occur when one or both of the required lines pass through the point of intersection of the given line and the axis of x, in which cases one or both of the triangles will be reduced to a point. An examination of the corresponding figures below shows that no new difficulty arises. X X Fig. 38. h = 0-/3. + (3, e-(3. 38. Equation of a Line through the Intersection of Two Given Lines, Ax + By + C = (1) and A'x+B'y + C' = (2). If these two lines meet at point (xi, yi), we find by solving (1) and (2) simultaneously that THE STRAIGHT LINE. 71 _ BC'-B'C _ CA'- C'A * '^ ~ AB' - A'B ""''"^ y^ - AB' - A'B- Having found this point of intersection, we can write down the equation of tlie line passing through it by one of the for- mulas, [13], [14], [18], according to the other condition im- posed upon tlie line. Ex. Find the equation of the line through the point of intersection of the lines y — 2x = 5 and y = ox -f 6, and the point (1, 1). The given lines are found, on solving the equations, to intersect at (- 1, 3). .-. the equation of the new line will be, by [14] § 34, There is another method of dealing with this same class of problems which is, mathematically speaking, much more ele- gant. It is as follows : If I and k are any two arbitrary midtipliers, independent of X and y, we may form the equation l(Ax + By + C) + k(A'x + B'y + C) = 0, (3) which is the equation of a straight line, since it is an equa- tion of the first degree. If (xj, yi) is the point of intersection of lines (1) and (2), Axi + Byi + C = and A'x, + B'yi + C = 0, whence l( Ax^ + Byi + C) + k(A'x, + B'y, + C; = 0. Hence, for any and all values that nwiy be assigned to I and k, (3) is the equation of a straight line passing through the point of intersection of lines (1) and (2). We may accord- ingly impose any condition we wish upon I and k. wlicn the line will be completely determined. * It is to be noted that, if the two lines are parallel, A B' — A'B = 0, and no values of xi and yi can be foimd. For a further discussion of this case see § 17. 72 PLANE ANALYTIC GEOMETRY. Applying this method to the problem just solved by the other method, we have l(y - 2x - 5) + k(y - 3x - 6) = 0, a line passing through the point of intersection of the lines y — 2x — 5 = and y — 3x — 6 = 0. As the line is to pass through (1, 1), 1(1 - 2 - 5) + k(l - 3 - 6) = or I = - - k. .-. - ^k(y-2x -5) + k(y- 3x - 6) = 0, or y + X = 2 is the required line. It is to be noted that by this method there is no need of finding the point of intersection of the given lines, and that it holds even if the two lines are parallel, in which case, of course, the required line is also parallel to the two. 39. Equations- of the Bisectors of the Angles between Two Lines. Y Fig. 39. Let the equations of the given lines be written in the forms X cos ai + y sin ai — Pi = 0, I. X cos a2 + y sin a, — p.^ ^ 0. II. THE STRAIGHT LINE. 73 Let (1) and (2) be the two bisectors, (1) being called the internal bisector since it passes through that one of the angles made by I and II which includes the origin, and (2) being called the external bisector since it does not pass through the angle including the origin. We know by Plane Geometry that every point in the bisector of an angle is equally distant from the sides of the angle. Therefore, if (x, y) is any point of bisector (1), we have, by §32, X cos ai + y sin aj — pi = x cos ao + y sin an, — p.i, (1) since any point of this line is either on the same side of both lines as the origin or on the opposite side of both lines from the origin, so that the values of its distances from the two lines would both be negative or both be positive. If (x, y) is any point of line (2), by § 32, X cos ai -]- y sin ui — Pi = — (x cos uo + y sin a^ — p-^, (2) since any point on this line is on the same side of one line as the origin and on the opposite side of the other line from tlie origin, so that the values of its distances from the two lines will be each the negative of the other. It is evident that the bisectors pass through the point of intersection of the given lines ; for, if we take I = 1 and k = — 1 in the work of § 38, we get (1), and if we take I = 1 and k = 1, we get (2). Ex. Find the equations of the bisectors of the angles made by the lines X + oy — 7 = and ox + y + 2 = 0. We must rewrite these equations in the form x cos a + y sin a — p = 0, i.e., X , ^y^ 7_ _ 3x y 2_ _ ^ VlO VlO VlO ' VlO VlO VlO . •. the internal bisector is ^ J. 3y 7 3x y 2 . I < r _ n -= + —^ — = -;_ — or 4x + 4y - 5 - 0; VlO VlO VlO VlO VlO VlO 74 PLANE ANALYTIC GEOMETRY. and the external bisector is vio Vio Vio V Vio Vio Vio/ 40. Equations of Higher Degree than the First Repre- senting Straight Lines. We have seen, iu our work up to this point, that every form of the equation of the straight line is of the first degree, and conversely, that every equation of the first degree represents a straight line. It may happen, however, that an equation of higher degree than the first will have as its locus two or more straight lines. For example, the equation Ax2 + 2 Hxy + By- + 2 Gx + 2 Fy + C = may have been formed by multiplying out the left-hand side of the equation (rx + sy + t) (r'x + s'y + t') = 0. Now tliis last equation is satisfied if either factor is zero, the other factor remaining finite. But if x and y have such values as to make rx + sy + t ^ 0, x and y must be the coordinates of a point on the line rx + sy + t = 0. Similarly, if x and y have such values as to make r'x + s'y + t' = 0, they must be the coordinates of a point of the line r'x + s'y + t' = 0. Therefore all points of the straight lines rx + sy + t = and r'x + s'y H~ t' ^= ^ ^^^^ points of the locus of the equation (rx + sy + t) (r'x + s'y + t') == 0. Moreover, any point not on either of these two straight lines will have coordinates such that neither factor will be made zero by their substitution in the equation, hence the product will not be zero, and the point is not a point of the locus. Therefore the locus of the equation (rx + sy + t) (r'x + s'y + t')=0 is the pair of straight lines rx + sy + t == 0, r'x + s'y + t' =^ 0. THE STRAIGHT LINE. 75 It follows that Ax^ + 2 Hxy + By=^ + 2 Gx + 2 Fy + C = will represent a pair of straight lines, if the left hand side can be written as the product of two linear factors ; that an equa- tion of the third degree will represent three straight lines, if the left-hand side can be written as the pi'oduct of three linear factors, etc., the right-hand side being zero in each case, of course. For example, x2 — xy -f- 5x — 2y -F 6 = is the equation of the pair of straight lines x -f 2 = and x — y -f- 3 = 0, since it can be written in the form, (x -|- 2) (x — y -f 3) = 0. 41. Condition that the Equation of the Second Degree represents Two Straight Lines. Following the usual method of factoring the expression Ax2+2 Hxy+ By- + 2 Gx + 2 Fy + C, by eqiiating it to zero and finding the value of either x or y, in terms of the other and the coefficients of the expression, we get, if we solve for x. _-(Hy + G)±V(H^-AB)y^4-2y(HG-AF) + (G--CA) ^_ _ That we may be able to factor into rational factors, the above values of x must not be surd, at least with y under the radical sign ; hence the expression under the radical sign must be a perfect square. The condition that (H^- AB) y- + 2y (HG - AF) + (G" - CA) shall be a perfect square is 4(HG- AF)2-4 (H^- AB) (G--CA) = 0, or ABC + 2 FGH- AF^- BG^- CH-=:0. 76 PLANE ANALYTIC GEOMETRY. This is, therefore, the necessary condition that Ax2 + 2 Hxy + By2 + 2 Gx + 2 Fy + C = shall represent two straight lines. If the above condition is fulfilled for any quadratic equation, it may be factored into two linear factors. The three cases which may occur are as follows : (1) If the two factors are distinct and the coefficients are real, the equation represents two real straight lines. (2) If any of the coefficients of the factors are imaginary, i.e., contain the square root of a negative expression, there is no real locus corre- sponding to the equation. We may say, from analogy, if we choose, that the equation represents tivo imaginary straight lines. It is a remarkable fact that one and the same real point may be found on each of the two lines. (.3) If the two factors are equal, each factor represents the same straight line. We sometimes say, in this case, that the equation repre- sents two coincident straight lines. Similar considerations hold in the case of an equation of degree higher than the second, so that, when an equation can be factored into linear factors, the line corresponding to any factor may be real or imaginary. Note. If A = 0, the above method of factoring is impossible, but if we had solved for y, instead of x, in the course of the above proof, we should have found the same necessary condition. If, however, both A and B are zero, the above type of proof will be unavailable, though the condition is equally true. In this case the general equation reduces to 2 Hxy4-2Gx-|-2 Fy+ C = 0, and it is evident that, if the left-hand member can be separated into linear factors, these factors must be of the type rx -j- t and s'y + t'. Multiplying rx + t by s'y + t', we have rs'xy ■+■ rt'x + s'ty -f- tt'. Therefore, if 2 Hxy-l-2Gx + 2 Fy-F C =0 represents the two straight lines rx -I- t = and s'y -|- t' = 0, rs' = 2 H, rt' = 2 G, s't = 2 F, and tt' = C. THE STRAIGHT LINE. 77 But (ilL(lt:)_(ttX(rs:)^^ or 2FG-CH = 0. Therefore, 2 FG — C H = is the necessary condition that 2 Hxy + 2Gx + 2 Fy + C = shall represent two straight lines. But if A and B are both zero, the original condition reduces to 2 FGH-CH-i = 0, or 2 FG-CH =0, since H cannot be zero at the same time with A and B if we are to have an equation of the second degree. Hence we see that the necessary condition first found is the necessary condition in all the cases which can occur. EXAMPLES. 1. Find the equation of the line of which the slope is 5 and the intercept on OY is 3. 2. Find the equation of the line passing through the point (0, — 2) and making an angle of 135° with OX. 3. Find the equation of a line making an angle of 60° with OX and cutting off an intercept — 1 on OY. 4. A line making a zero intercept on the axis of y makes an angle of 150° with the axis of x. What is its equation ? 5. A line making a zero angle with OX cuts OY at a point — 2 units from the origin. What is its equation ? 6. What is the equation of the line of which the intercepts on the axes of x and y are respectively 2 and — 4 ? 7. What is the equation of the line of which the intercepts on the axes of x and y are respectively — 3 and — 5 ? 8. A line is distant 5 units from the origin, and its normal makes an angle of 60° with OX. What is its equation? 9. The normal to a line, which is distant 3 units from the origin, makes the acute angle tan-i|- ^[^i^ ^^q axis of x. What is the equation of the line ? 78 PLANE ANALYTIC GEOMETRY. 10. A line distant 2 units from the origin makes an angle of 45° with the axis of x. What is its equation? 11. The normal to a line, which passes through the origin, makes an angle tan~|- with OX. What is the equation of the line ? 12. The normal to a line, distant 7 units from the origin, makes an angle of 180° with OX. What is the equation of the line ? Determine the parameters a, b, p, m, and a for the fol- lowing lines : 13. y = 3x-l. 14. ^-y-1. 2 5 15. ^X+^y-2:::.0. 16. 2x + y = 0. 17. x + 3 = 0. 18. 2y — 1 = 0. 19. Find the equation of a line making an intercept 3 on OY and an angle of 30° with OX, when the angle XOY is 60°. 20. Find the equation of a line making an intercept — 2 on OY and an angle of 15° with OX, when the angle XOY is 45°. 21. Find the equation of a line making a zero intercept on OY and bisecting the angles of the second and the fourth quadrants, when the angle XOY equals 45°. 22. The normal to a line distant 2 units from the origin makes an angle of 45° with OX, the angle XOY being 60°. What is the equation of the line ? 23. The normal to a line distant 5 units from the origin makes an angle of 30° with OX, the angle XOY being 75°. What is the equation of the line ? 24. Show that the equation of the straight line may be written THE STRAIGHT LINE. 79 X COS a + y COS (3=p, where p is the length of the normal from the origin and a and ^ are the angles which this normal makes with the axes of X and y respectively, the angle between the axes being of any magnitude whatever. 25. Determine tlie parameters of the line 3x — 2y + 1 = 0, the angle between the axes being 30°. 26. The normal to a line distant 3 units from the origin makes an angle of 75° with the initial line. What is the polar equation of the line ? 27. Find the polar equation of a line distant 4 units from the origin and making an angle of 37° with the initial line. 28. Find the distance of (2, 3) from the line y = 6x — 1. On which side of the line is the point ? 29. How far distant from the line x + 3y + 7 = is the point (1, — 4), and on which side of the line is it ? 30. How far distant from the line 2x — 3y = is the point (1,5)? 31. Find the distance of the point of intersection of the lines y — 2x — 5 = and 2y — x — 7 — from the line y + 3x + 1=0. 32. Find the locus of the points at the constant distance 3 from the line 3x + 4y — 5 ^ 0. 33. Find the locus of the points distant 2 units from the line 2x + y + 2 = 0. 34. Find the locus of the points equally distant from the lines x + 3y — 7 = and 3x — y + 1 = 0. 35. Find the equation of tlie line through the point (—3, — 5) parallel to the line 4x — 3y + 1 = 0. 36. Find the equation of a line through (2,-1) parallel to the line x — y = 6. Where will the new line cut the axes of x and y ? 37. Find the equation of a line through (— 1, — 3) parallel to the line 2x + 5y — 3 = 0. How far is each line from the origin ? How far apart are they ? 80 PLANE ANALYTIC GEOMETRY. 38. How far apart are the parallel lines 3x + 4y — 10 = and 3x + 4y - 7 == ? 39. Find the equation of a line parallel to the line 3x — 2y + 3 = and bisecting the line joining (1, — 2) and (3, 4). 40. Find the equation of the line passing through the points (— 3, — 5) and (0, 6). 41. Find the equation of the line through the points (2, — 3) and (4, 1). 42. Find the equation of the line through the points (—1,6) and ( — 1, — 3). 43. Find the equation of a line through (— 5, — 3) parallel to the line joining (5, 2) and (— 2, — 5). 44. Find the acute angle between the lines 4x — oy+'^=0 and X + 3y + 2 =: 0. 45. Find the acute angle between the lines x + 3y — 1=0 and 3x + y - 3 = 0. 46. Find the equation of a straight line passing through the point (— 5, 7) perpendicular to 3x — 5y + 6 = 0. 47. Find the equation of the line through the origin perpen- dicular to the line 6x + 3y — 7 = 0. 48. Find the equation of a line perpendicular to 2x — 3y = 7 and bisecting that portion of it which is included between the axes. 49. Find the equation of the perpendicular bisector of the line joining (— 2, — 4) and (— 4, 3). 50. What are the equations of the lines which pass through the point (— 1, 0) making an angle tan-i i with the line 2x-5y + 2==0. 51. What are the equations of the lines which pass through the point (— 5, 4) making an angle tan" ^ i with the line 15x _ 3y + 10 = 0. 52. Find the equations of the lines which pass through (3, — 1) making an angle tan" ^ f with 5x — 6y + 3 = 0. 53. Find the equations of the lines passing through (1, 1) and making an angle tan- ' f with 4x — 3y + 4 = 0. THE STRAIGHT LINE. 81 54. Find the equations of the lines passing through the point (3, — 5) and making an angle tan" ^ f with the line x = 4. 55. Find the equation of a line through the origin and the point of intersection of the lines y - 2x - 5 = and x - 2y + 5 = 0. 56. Find the equation of a line through (5, — 5) and the point of intersection of the lines X — 2y + 3 = and 3x + 4y + 1 = 0. 57. Find the equation of the line passing through the point of intersection of 2x + 5y + 6^0 and 3y — 2x + 6 = 0, and parallel to x — y + 1 ;= 0. 58. Find the equation of the line through the point of intersection of 3y — 2x + 5 =^ and 4y + 5x + 6 = 0, and per- pendicular to 2x + 7y + 8 ^ 0. 59. Find the equation of a line through the point of inter- section of y-7x + 36 = 0, (1) X - 3y - 8 = 0, (2) and parallel to the line 4x -}- 3y = 0, without finding the point of intersection of (1) and (2). 60. Find the equation of a line through the point of inter- section of 2x - 3y - 5 = 0, (1) 3x - 4y - 6 = 0, (2) and perpendicular to x + y -f 1 = 0, without finding the point of intersection of (1) and (2). 61. Find the equation of a line through the point of inter- section of 2x - 5y - 41 = 0, (1) 7y-3x + 58 = 0, (2) and the point (— 2, — 9), Avithout finding the point of inter- section of (1) and (2). 82 PLANE ANALYTIC GEOMETRY. 62. Find the equation of a line thvougli the point of inter- section of X + y + 5 = and 2x + L'y + ''^ = and tlie point 63. Find the equ.ation of tlie line joining the points of intersection of the lines 2x — y - 5 = O; ) , ( 2x + y — 13 = 0, ^ ' and ' ^ 3x + y-5 = 0, j (3x-y-2 =0. 64. Given the four straight lines, x-2y + 2 = 0, x + 2y-2 = 0, 3x-y-3 = 0, x + y+() = 0, find the equations of the three other lines, each of which con- tains two of the six points of intersection of the given lines. 65. Determine the value of m so that the line y := mx + 3 shall pass through the point of intersection of the lines y = X + 1 and y = 2x + 2. 66. What are the equations of the bisectors of the angles between the lines 3x — 4y + T = and 12x + 5y — 6 = 0? 67. What are the equations of the bisectors of the angles between the lines V3x -1- V2y = and x + 2y + 5=^0? 68. What are the equations of the bisectors of the angles between the lines 3x + y = and 4x — 3y ^ ? 69. Find the bisectors of the angles between the lines x == and y = 0. 70. Show that llx + 3y + 1 = bisects one of the angles between the lines 12x — 5y + 7 = and 3x + 4y — 2 = 0. 71. The vertex of a triangle is the point ( — 3, — 4), and the base is the line joining (2, 4) and ( — 1, — 3). Find the lengths of the base and the altitude. 72. Find the vertices and the angles of the triangle, tlie equations of the sides of which are respectively y = 0, 7x — 3y — 14 = 0, and x + y — 12 = 0. The vertices of a triangle ABC are the points ( — 3, — 5), (2, -4), and (7, 6): THE STRAIGHT LINE. 83 73. Find the lengths of the medial lines of ABC. 74. Find the equations of the sides of ABC. 75. Find the lengths of the perpendiculars drawn from the vertices to the opposite sides of ABC. 76. Find the equations of the perpendicular bisectors of the sides. 77. Where do these bisectors intersect ? The equations of the sides of a triangle ABC are respec- tively y + 4x — 34 = 0, 3x + 4y — 19 = 0, 2y-5x + 23 = 0: 78. Find the perimeter and the area of ABC. 79. Find the equations of the medians. 80. Find the point of intersection of the medians. 81. Find the equations of the lines drawn from tlie ver- tices perpendicular to the opposite sides. 82. Find the equations of the lines joining the middle points of the adjacent sides. 83. Find the simplest form of the polar equation of a straight line through the origin. 84. What locus is represented by the equation sin 3^ =: ? 85. Prove that the equation of the line through the two points (ri, Oi) and (r^, 62) may be written r [ri sin (O^ — 6) — r^ sin {9. — 6)~\ = r^r. sin (^1 — ^.,) or more symmetrically rri sin (6 — 0^) + x,r. sin {6^ — 6.^ + r,r sin (Oo — 6)=0. 86. Find the point of intersection of the loci r cos ( ^ — - j = a, r cos ( ^ — - 1 = a. 87. Find the points of intersection of the loci r cos I ^ ~ 9 ) = ~r ''^"^ r = a sin d. 84 PLANE ANALYTIC GEOMETRY. 88. One diagonal of a parallelogram joins the points (3, — 1) and ( — 2, 5). One extremity of tlie other diagonal is ( — 3, — 4). Find its equation and length. 89. Prove by calculation that the perpendicular bisectors of the sides of the triangle (2, 3), (— 1, — 2), and (— 3, — 4) meet in a point, and find the point. 90. Given the triangle 4x — 3y + 12 = 0, 3x — 4y — 6 = 0, 12x + 5y + 60 = ; prove by calculation that the bisector of any angle divides the opposite side into segments proportional to the adjacent sides. 91. Given the triangle 4x — 3y + 12 = 0, 3x + 4y — 6 = 0, 12x + 5y + 30 := ; prove by calculation that the bisectors of the interior angles of the triangle meet in a point, and find the point. 92. Prove that the bisectors of the angles formed by two intersecting straight lines are perpendicular to each other. 93. Prove analytically that the locus of points equally dis- tant from two points is the perpendicular bisector of the line joining them. 94. Prove analytically that the medial lines of a triangle meet in a point. 95. Prove analytically that the perpendicular bisectors of the sides of a triangle meet in a point. (Take one vertex of the triangle as the origin, and one side as the axis of x. Then the vertices are (0, 0), (xi, 0), and (xg, y2).) 96. Prove analytically that the perpendiculars from the vertices of a triangle to the opposite sides meet in a point. (Take axes as in Ex. 95.) 97. Prove analytically that the perpendiculars drawn from any two vertices to the median from the third vertex are equal. (Take axes as in Ex. 95.) 98. Prove analytically that the bisectors of the interior and the exterior angles of any triangle meet by threes in four points, of which one lies inside the triangle. (Take the THE STRAIGHT LINE. 85 origin inside the triangle and write the equations of the sides in the normal form.) 99. Prove analytically that the lines joining the middle points of the opposite sides of a quadrilateral bisect each other. 100. Prove analytically that, if two medians of a triangle are equal, the triangle is isosceles. 101. Show that the locus of a point which moves so that the sum of its distances from two given straight lines is con- stant is a straight line. 102. What two straight lines are represented by the equa- tion 2x2 + 3xy + y2 — X — y = ? 103. What two straight lines are represented by the equa- tion 3x2 _^ 8xy — 3y2 + 6x + 8y + 3 = ? 104. Plot the locus of the equation 4x2 — gy2 __ q 105. Plot the locus of the eqviation 2x2 _^ 7xy ^ 3y2 _ X _ 3y = 0. 106. Plot the locus of the equation 9x2 _^ 9y2 _ 6x + 24y - 19 == 0. 107. Show that 56x2 — 441 xy — 56y2 — 79x — 47y + 9 = is the equation of the bisectors of the angles between the straight lines represented by 15x2 - 16xy — 48y2 - 2x + 16y - 1 = 0. 108. Find the angle between the two straight lines repre- sented by the equation Ax2 + 2 Hxy+ By2==0. 109. Prove that any homogeneous equation in x and y repre- sents a system of straight lines passing througli the origin. CHAPTER IV. TRANSFORMATION OF COORDINATES. 42. Meaning of Transformation of Coordinates. So far in our work we have dealt with the Cartesian coor- dinates of any point in the plane on the supposition that to a given point corresponds one, and only one, pair of coordinates ; and, conversely, that to any pair of coordinates corresponds one, and only one, point. This supposition is true when the axes of coordinates are once fixed, hut a moment's thought will make it evident that the coordinates of the point are relative quantities, dependent on the selection of the axes, and will be different for different axes chosen. Moreover, it is obvious that the axes may be chosen at pleasure, but that there must be some relations between the coordinates of a point with respect to one set of axes and the coordinates of the same point with respect to a second set of axes. If these relations are known, we may use them to find the coordinates of any point with respect to a new set of axes, its coordinates being given with respect to the original axes. We are then said to make a trmisformation of coordinates, and the equations expressing the relations are called formulas of transformatioii. It must be borne in mind that a transformation of coordi- nates never alters the position of the point in the plane, the coordinates alone being changed because of the new standards of reference adopted. Note. In all the work that follows, OX' and OY' should be carefully distinguished from the OX' and OY' used up to this point. From now on, any set of axes will be lettered only at the origin and at the respec- tive positive extremities of the axes. TRANSFORMATION OF COORDINATES. 87 43. Transformation to a New Origin, the New Axes being respectively Parallel to the Old Axes. Y' ,Y Let OX and OY be the original axes, and O'X' and O'Y' the new axes meeting at 0', the coordinates of 0' with respect to the original axes being Xq and yo- Let P be any point in the })lane, its coordinates being x and y with respect to OX and OY, and x' and y' with respect to O'X' and O'Y'. Draw PM'N parallel to OY. Then OM == x,,, MO' = yo, ON = x, NP = y, 0'M' = x', M'P = y'. But ON=OM + MN, NP= NM'+ M'P; .•.x = Xo + x', y = yu+y'- [19] These formulas of transformation are obviously true whether the axes are oblicpie or rectangular. Ex. 1. The coordinates of a certain point are (2, — 1). What will be the coordinates of this same point with respect to a new set of axes parallel to the old, and meeting at the point (3, — 2) with respect to OX and OY ? 88 PLANE ANALYTIC GEOMETRY. Here xo = 3, yo = — 2, x = 2, and y = — 1. .-. 2 = 3 + x' and — 1 = — 2 + y', so that x' = — 1, y' = 1, Ex. 2. Transform the equation f + 4y - 8x + 28 = to a new set of axes parallel to the original axes and meeting at the point (3,-2). The formulas of transformation are x = 3 -f- x', y = — 2 + y', by [19]. Therefore our equation becomes (— 2 + y')2 + 4(— 2 + y') - 8 (3 + x') + 28 = or y'^ - 8x' - 0. As no point of the curve has been moved, the curve is changed in no way whatever. Its equation is different because referred to neio axes. After the work of a transformation has been completed, the primes may be dropped. Therefore the equation of this example may be written y2 — 8x = 0, the new axes being now the only ones considered. 44. Transformation from one Set of Rectangular Axes to another Set of Rectangular Axes having the same Origin, and making an Angle with the Original Axes. Y Fig. 41. OX and OY are the original axes, and OX' and OY' are the new axes making Z. with OX and OY respectively. Let P TRANSFORMATION OF COORDINATES. 89 be any point in the plane, x and y being its original coordi- nates, x' and y' its new coordinates. Draw PM parallel to OY, PM' parallel to OY', M'N' parallel to OY, and M'N parallel to OX. Then Z NPM' = 9, and OM ^x, MP = y, OM'=x', M'P = y'. But OM = ON'- MN'=ON'- NM' = OM'cos^— M'Psin^, MP=MN+ NP= N'M'+ NP = 0M' sin 6+ M'P cos 6. .'. X = x' COS 6 — y' sin 6, y = x' sin 9 + y' cos 9. Ex. Transform the equation x + y — 1 = from one set of rect- angular axes to a new set having tlie same origin and bisecting the angles made by the original set. y' . [20] Here V2 V2' V2 V2 ('^)^(^')— »»■■ — ^- 0. 45. Transformation from a Set of Rectangular Axes to a Set of Oblique Axes having the Same Origin. Y 90 PLANE ANALYTIC GEOMETKY. Let OX and OY be the original axes at right angles to each other, and OX' and OY' be the new axes, ^^ XOX' being denoted by 6 and /_ XOY' by 6'. The oblique angle included by the new axes will accordingly be 6' — 6. Let P be any point in the plane, its original coordinates being x and y, its new, x' and y'. Draw PM parallel to OY, PM' parallel to OY', M'N' parallel to OY, and RM'N parallel to OX. Then ZRM'P = ^'. .•.OM = ON'+ N'M = ON'+ M'N = OM' cos 0+ M'Pcos 6', MP=MN + N P = N'M'+ NP = OM' sin 6^ M'P sin 9'. .'. X = x' cos 9 + y' cos 9', y = x' sin 9 + y' sin 9'. [21] 46. Transformation from one Set of Oblique Axes to a new Set of Oblique Axes with the Same Origin. y Fig. 43. OX and OY shall be the original axes, Z XOY being w. The new axes shall be OX' and OY', making angles 6 and & respectively with OX. TRANSFORMATION OF COORDINATES. 91 Let P be auy point in the plane, its original coordinates being x and y, its new, x' and y'. Draw PM parallel to OY, PM' parallel to OY', M'N' parallel to OY, and RM'N parallel to OX. Then ZN'M'0 = w-^, ZON'M' = 180°-co, ZM'NP = ) sin sm CO and M'P sin ^' y'sin^' N P = -. = -—. . sm w sm w But OM = ON'- MN'=ON'- NM', and MP=MN + NP= N'M'+ NP. _ x^ sin (ci) — 9) + y^ sin (o) — 9^ sm 0) x' sin 9 + y' sin 9' y = ^ ' sm 0) [22] It is obvions that this last set of formulas includes all the transformations that can be made, when the origin is not changed, different values being assigned to w. Q, and B\ 92 PLANE ANALYTIC GEOMETRY. 47. Double Transformation of Coordinates. We now have, on tlie one hand, a formula by which we may transform from one set of axes to a new set of axes having a new origin but retaining the respective directions of the old axes, and, on the other hand, formulas by which we may trans- form from one set of axes to a new set of axes having the same origin. But it often happens that we wish to transform to a new set of axes having a new origin and not parallel to the original axes. Formulas for these classes of transformations may be worked out, but it will be preferable to make such transfor- mations by a combination of the formulas already found, as shown in the following example. Transform the equation 7x2+ i3y2_6V3xy- (6V3-f-28)x+ (26 + 12 V3)y -|- 12 Vs -|- 25 = to a new set of rectangular axes, making an angle of 30° with the original axes and meeting at the point (2, — 1) with respect to the original axes. First transform to the new axes O'X" and O'Y" meeting at 0' (2, — 1) and respectively parallel to OX and OY. TRANSFOKMATION OF COORDINATES. 93 By [19], § 43, X = 2 + x", y = - 1 + y" will be the formulas of transformation, and the equation becomes 7(2+ x")2+13(-l+ y")2 — 6V3(2+ x") (- 1 + y") - (6V3 + 28) (2+ x")+ (2G+ 12 VS) (- 1 + y") + 12V3 + 25 = 0, or 7 x"2 + 13y"2 - 6 ^/3x"y" - 16 = 0. Now by formulas [20], § 44, we can transform to the required axes O'X' and O'Y', the formulas of transformation being „ x'Vs y' „ x' , y'VS '^ =-^-2'y =2+ ^• — 16 = or x'- + 4y'2 -4 = 0. 48. In reviewing this chapter, we see that the expressions for the original coordinates in terms of the new are all of the first degree, so that the result of any transformation cannot be of higher degree than that of the original equation. On the other hand, the result cannot be of lower degree than that of the original equation ; for it is evident that, if we transform any equation to a new set of axes and then back to the original set of axes, the equation must resume its origi- nal form exactly. Hence if the degree had been lowered by the first transformation, it must be increased to its original value by the second transformation ; but this is impossible as we have just noted. It follows that the result of any transformation, hence of several successive transformations, is of the same degree as that of the original equation. Finally, in connection with this chapter the student should read § 10, on transformation from rectangular coordinates to polar coordinates, and conversely. EXAMPLES. 1. Transform the equation 3x + 4y + 14 = to a new set of axes parallel to the original axes and meeting at the point (2,-5). 94 PLANE ANALYTIC GEOMETRY. 2. Transform the equation x" + y- + 4x — 6y + 3 = to a new set of axes parallel to the original axes and meeting at (-2,3). 3. Transform the equation x" + 12x^ — y- + 47x — 2y -j- 59 = to a new set of axes parallel to the original axes and meeting at ( — 4, — 1). 4. Transform the equation x^ — 2xy + y^ — 8 V2x — 8 V2y = to a new set of axes making an angle of 45° with the original axes without changing the origin. 5. Transform the equation x^ — y^ = 16 to new axes bisect- ing the angles between the original axes. 6. Show that the equation x"^ + y^ = a^ will be unchanged by transformation to any pair of rectangular axes, if the origin is unchanged. 7. Transform the equation x'' — 3y^ = 1 to a new set of axes having the same origin as the original axes, the new axes of X and y making the angles — 30° and 30° respectively with the original axis of x. 8. What will be the equation of the ciirve represented by X V - — [- — ^ 1 in rectangular coordinates, if new axes of x and y 3 / 3 are taken making the positive angles tan~^ - and tan^^ | — - respectively with the original axis of x, the origin not being changed ? 9. After transforming the axes to a new origin (2, — 1) and then rotating them through an angle of 45°, find the resulting form of the equation x^ + 2xy +y2 + (4 V2 - 2)x - (2 + 4V2)y + (1 - 12 V2) = C- 10. Transform the equation 3x2 _^ 10 ^3xy _ 7y2 _ ^i^ _ 30 V3)x - (42 + 30 V3)y - (42 + 90 V3) = to a new set of axes meeting at the point (3, — 3) and making an angle of 30° with the original axes. TRANSFORMATION OF COORDINATES. 95 11. Wliat will be the equation of the curve represented by 3x- + 4y'- — 3x + 4y — 11 = 0, if new axes, parallel to the original axes, are taken so that there shall be no terms of the first degree in the transformed equation ? 12. Transform the equation 3x-' + 6x-l-4y — 5 = to a new set of axes parallel to the original axes, so choosing the origin that there shall be no constant term and no term of the first degree in x. 13. Transform the equation 3x- + Sxy — 3y-^0 to the straight lines it represents, as axes. 14. The equation of a given locus is 7x^ — 2xy + 4y'^^5, the angle between the axes being 60°. What will be the equation of tliis locus if the axes are revolved through an angle of 30°? 15. Prove that the formulas for the length and the slope of a straight line and for the area of a triangle are not changed by a transformation to a new set of axes parallel to the old. Transform the following equations to polar coordinates : 16. X cos a + y sin a = p. 17. xy = 7. 18. i+t=^- a- b- 19. X- + y- — Say — Sax = 0. 20. y- (2a — x) = xl 21. x^+ x-y- — ay-^O. Transform the following equations to rectangular coordi- nates : 22. r cos fo-'^\=- 10. 23. r cos fe-'^\-\-r cos f6-\-^,] = 12. 24. r = a sin 6. 25. r = a (1 - cos 6). 96 PLANE ANALYTIC GEOMETRY. 26. r = a sin 20. 27. r^ = a' cos 26. 28. r = a (cos 2^ + sin 20). 29. Prove the formula for distance in polar coordinates by transforming to polar coordinates the same formula in rect- angular coordinates. 30. Transform to polar coordinates the formula for the area of a triangle in rectangular coordinates. CHAPTER V. THE CIRCLE. 49. Definition and Equation of the Circle. Although we shall see later on that the circle is only a special case of a more general class of curves called the conic sections, so that naturally it would not be studied by itself in detail until after the general consideration of those curves, nevertheless, we will take up the circle here for the following reason. We are acquainted with the circle and its properties from our study of Plane Geometry ; and hence for us it is the simplest curve in the plane, and the easiest on which to practice our new methods, lohicli can all be learned in this chapter. For the sake of learning how to treat a curve analytically and of acquiring power to deal with new curves by practice with a familiar curve, the student is urged not to solve prob- lems by his knowledge derived from Plane Geometry, if he can in any way solve them analytically, even though this requires more time. We shall adopt the usual definition of a circle, i.e.. The circle is the locus of a point which vioves so that its flis- tance from a fixed point, called the centre, shall he constant, this constant distance being called the radius. Let P(x, y), Fig. 45, be any point of the circle of radius r, of which the centre is the point C(d, e). Then by [1], § 3, the equation of the circle will be , (x-di-'+(y->j^'^=r2. [23] There is one particular f(jrm of this equation, which is of almost as much importance as [23] by reason of the uses we 98 PLANE ANALYTIC GEOMETRY. can make of it, i.e., the equation of the circle with its centre at the origin. Here d = e = 0, and the equation is X' + y- = r\ [24] Fig. 45. I. Find the equation of a circle tangent to OX. 2. Find the equa- tion of a circle tangent to both OX and OY. 3. Find the equation of a circle tangent to OX and having its centre on OY. 4. What are the parameters of a circle ? 5. What will be true of all the circles having the same r, but different d's and e's ? 6. What will be true of all the circles having the same d and e, but different r's ? 7. All circles of a system have the same d and r, but different e's. What is true of them ? 50. General Form of the Equation of the Circle. The most general equation of the second degree is Ax2 + 2 Hxy+ By2 + 2 $x -\-2^y -\- (^j=0. a) The most general equation of the circle, when expanded, is x2 + y2-2dx-2ey + (d? + e'-r2) = 0. (2) By comparing these two equations, we see tliat the equation of the second degree represents a circle, if H =: and B = A, THE CIRCLE. 99 i.e., if no xy term is present and the coefficients of x^ and y^ are equal, for then we can divide through by A and get a form exactly like (2). Therefore Ax- + Ay" -\-2 Gx-)-2 Fy + C = always repre- sents a circle ; or, as we shall prefer to write, after dividing through by A, and replacing^, — , -~ by a new G, F, and C respectively, i^ x^ + y^ + 2 Gx + 2 Fy + ^- [25] always represents a circle. " '^ This equation can be written (x2 + 2 Gx + G^) + (f + 2 Fy +F'^)--^G-'+ F'^ - C, or (x + G)'^ + (y + F)'^ = ( VG^ + F^ - C)'^ which is in the form (x — d)" + (y — e)" ^ r^. There are three cases which can occur : (1) If G^ + F^ — O 0, r is real, and we have a real circle. (2) If G^ + F^ — C = 0, r = 0, and the circle reduces to a point, the coordinates of which are — G and — F. (3) If G^ + F- — C ^i + Gf + (y,-^ Ff - {G' + F' - C), or PiP-Vxr + yi^ + 2 Gxi + 2 Fyi + C. Therefore, to find the length of the tangent draivn from a given 2>oint to a circle, we have only to tvrite the equation of tJie circle in the form x- + y^ + 2 Gx + 2 Fy + C = 0, and take the square root of the result of substituting the coordinates of the 23oi7it in the left-hand meynher of the equation. Ex. Find the length of the tangent drawn from the point (7, 2) to the circle x2 + y2 — 4x + 6y + 8 = 0. Length = Vt^ + 2- - 4 (7) + G (2) + 8 = V45 = 3 V5. Note. If the length of the tangent as found by the above rule is zero, the point is on the circle ; and if the result is imaginary, it is evident that the point is within the circle, and that, hence, no tangent can be drawn from it to the circle. 58. Chord of Contact. Frotn any 'point outside a circle two real tangents rnay he draivn to the circle, and the straight line joining their i'esjjec- tive poiiits of contact is called the chord of contact. Let PiPo and P^Pg be the tangents to the circle x2 + y2 + 2Gx + 2 Fy + C = from the point Pi(xi, yi). Then P2P3 is the chord of contact. Assume the coordinates of P2 and Pg to be respectively (x2, ya) and (x^, ys). THE CIRCLE. 109 Fig. 50. By [26], §54, tlie equations of PjPa and P1P3 are resi)ec- tively x.x + y.y + G (x + X,) + F (y + y,) + C = 0, (1) X3X + y^y + G (x + X3) + F (y + y3) + C - 0. (2) Since Pi is a point of both these lines, x.xj + y,yi + G (xi + X,) + F (y^ + y.) + C =- 0, (3) X3X1 + ysyi + G (x, + xa) + F (yi + y,) + C =- 0. (4) From inspection of (3) and (4) let us form the analogous equation xXi + yy, + G (xi + x) + F(yi + y) + C = 0. (5) Of this equation Ave may assert two things : (1) It is of the first degree, and therefore represents some straight line. (2) It is satisfied by the coordinates of both P., and Pg by virtue of equations (3) and (4). Therefore, since two points are sufticient to determine a straight line, (5), or after re-writing. 110 PLANE ANALYTIC GEOMETRY. XiX + y.y + G (X + Xi) + F (y + yO + C = [27] is the equation of tlie chord of contact. As tlie equation of the tangent determined by the point of contact and the equation of the chord of contact are identical ill form, one can distinguish between them only by noting whether or not Pi(xi, yi) is on the curve. For example : Fiud the equation of the tangent to the circle x2 + y2 — 4x + 6y + 8 = 0, passing through the point (1, — 1). Since (1, —1) satisfies the equation of the circle, it is the point of con- tact, and by [26], § 54, the equation of the tangent is Ix - ly - 2 (X + 1) + 3 (y - 1) + 8 = 0, or X — 2y — 3 = 0. As a second example, we will find the tangent to the same circle, passing through the point (7, 2). Since (7, 2) does not satisfy the equation of the circle, 7x + 2y - 2 (x + 7) + 3 (y + 2) + 8 = 0, or X + y = is the chord of contact of tangents drawn to the circle from (7, 2). Solving the equations x + y=0, x2 + y2 _ 4x + Gy + 8 = 0, we find, as the points of contact of the required tangents, (1,-1) and (4, - 4). .-. by [26], § 54, the tangents are Ix- ly-2(x + 1) + 3(y - 1) +8 = 0, . 4x - 4y - 2 (X -I- 4) + 3 (y - 4) + 8 = 0, or X — 2y — 3 = 0, 2x' — y — 12 = 0. 59. Definition and Equation of the Polar. It may seem strange that the equation of a tangent and the equation of a chord of contact are of exactly the same form, and such coincidence of form would be strange if the two classes of lines were not more closely related than has been indicated in the last articles. As a matter of fact, they are both but special cases of a type of locus, called a THE CIRCLE. Ill polar, wliicli may be defined with respect to a curve for a point which is called the pole. The definition is as follows : If through a fixed j^oint P^ any number of secants be draxmi to a circle, and tangents to the latter be draivn at the 2^0 mts of intersection of each secant with the circle, then each pair of tan- gents ivlll intersect upon a straight line called the polar of P^. Reciprocally, Pi is called the pole of the straight line. Fic. 51. Let the circle be x- + y- + 2 Gx + 2 Fy 4- C = and Pi(xi, yO be the pole. Let (1) be any secant througli P^ the tangents at its ex- tremities meeting at P2(xo, y.,). Then, by [27], § 58, the equation of (1) is X2X + W + G (x + X,) + F (y 4- y.) + C = 0. (1) But Pi is a point of line (1) ; therefore X2X1 + y^yi + G (xi + X,) + F (yi + y,,) + C = 0. (T) Similarly, secant (2) through Pi determines a new point PsCxs, ys) of the locus, for which the following equation is true: X3X, + y,y, + G (x, + X3^ + F (y, + y,.,) + C = 0. (II) 112 PLANE ANALYTIC GEOMETRY. Comparing equations (I) and (II), we see that xix + y,y + G (xi + x) + F (yi + y) + C = 0, is a type of equation satisfied by the coordinates of every single point of tlie locus. Hence x.x + yiy + G (X + xO + F (y + yi) + C = [27] is the equation of the polar of Pi(xi, yi) with respect to the CI rclc xM-y' + 2 Gx + 2 Fy + C = 0, (1) If Pi is on the circle, it is evident that the polar becomes the tangent to the circle at Pj, for one of the tan- gents which meet on the polar is always the tangent at Pj, so that all points of the polar are on this tangent (Fig. 52). Hence the polar and the tangent coincide. Fig. 52. (2) If Pi is without the circle, its polar becomes the chord of contact of the tangents to the circle from Pi, for the limit- ing positions of the secants from Pi are the two tangents P1P2 and P1P3 (Fig. 53), and as the secant approaches these tangents as limits, the corresponding point of the polar will THE CIRCLE. 113 approach P2 and P3 respectively as limits. Therefore the chord of contact P2P3 is the polar of Pi, since these two straight lines coincide at two distinct points. Fig. 53. (3) If Pi is within the circle, its polar will nut cut the circle at all (Fig. 51), as is readily shown by showing that its distance from the centre of the circle is greater than the radius of the circle. 60. Other Properties of Poles and Polars. (1) The polar is perpendicnla?' to the line jobilng the pole to X + G the centre. For the sloi)e of the polar is ^—. — - and the yi + r • Vi -f- F slope of the line joining the pole to the centre is • (2) The polar of the centre of the circle is at an infnite dis- tance from the centre. For the distance of any point (x, y) from the polar is, by § 32, 114 PLANE ANALYTIC GEOMETIIY. xix + yiy + G (X + xQ -f F (y + yO + C ^ V(x,+ G)^+(yi+F/ ' and this distance becomes infinite if Xi = — G and yi = — F, as tliey must if the centre is the pole. (3) The polar of any pohit of a straight line j^asses through the pole of that line. Fig. 54. Let Pi(xi, yi) be the pole of AB. Therefore the equation of AB is xix + yiy + G (X + X,) + F (y + yO + C = 0. (l) If P2(x2, y2) is any point of AB, then X1X2 + y^y^ + G (x.2 + xO + F (y^ + y,) + C = 0. (2) The polar of P2 will be X2X + y.y + G (x + X2) + F (y + y2) + C = 0. (3) By virtue of (2), the coordinates of Pi satisfy (3) and hence this line passes through P^. (4) The polar of the intersection of two straight lines is the straight line which joins their poles. {Proof by (3).} (5) A triangle ABC is given, and a second triangle A'B'C is draivn, B'C beiiig the p)olar of A, C'A' the polar of B, and A'B' THE CHICLE. 115 the polar of C. Then A', B', C are the respective p)oles of BC, CA,a7idAB. {Proof by (4).} Two such triangles are said to be conjugate, and it may be proved that A A', BB'^ and CC meet at a common point. (6) If in any triangle each vertex is the pole of the opposite side, the triangle and its conjugate coincide, and it is said to be self-conjugate. {Special case of (5).} 61. Diameter. The diameter of a circle is the locus of the middle 2)oints of a system of parallel chords of the circle. Accordingly let us find the equation of a diameter of the circle x^ + y^=r^ (1) bisecting a system of parallel chords of slope m. Fig. 55. Let y = mx + b (2) be any one of the chords, intersecting tlie circle at the points Pi(xij yi) and P2(x2, y2)- Then if K(x8j ya) is the middle point 116 PLANE ANALYTIC GEOMETRY. of the chord, i.e., a point of the diameter, xi + X2 yi + y2 ,Qx ^3 = — ^ , y3 = ^-y— • (3) Substituting for y from (2) into (1), we get the equation x'H" (mx + b)"= r, or (1 + m^) x'-^ + 2m bx + (b- — r) = 0, of which the roots are x^ and Xo. Therefore Xi + X2 = — ^Zl. — 2' ^^ that, by substituting in (3), we have Proceeding in similar manner, substituting for x from (2) into (1), we find Equations (4) and (5) give us the coordinates of the middle point of any one of the system of chords, according as differ- ent values are given to b. Hence, dividing (5) by (4), we get y^ _ _ 1 X3 m' or ya = X3, (6) an equation independent of b, which must, therefore, be satis- fied by the middle point of any one of the chords, i.e., by any point of the diameter. Therefore y = x (7) is the required equation of the diameter. But (7) is not a general equation, since it is true only for the circle having its centre at the origin. We will now find the general equation of the diameter of the circle (x-d)^ + (y-e)^ = r2, ^g^ bisecting chords of slope m. THE CIRCLE. 117 Fig. 56. Transform (8) to a new set of axes parallel to the old and meeting at the centre of the circle, Fig. 56, by tlie formulas X =: d + x', y = e + y'. The equation of the circle becomes x'- + y'- = r, so that the equation of the diameter will be, by (7), y' = x', for the slope of the chords will be the same for both sets of axes, since OX' is parallel to OX. Transforming back to the original set of axes by formulas x' = X — d and y' = y — e, we have the equation of the circle in its original form, (x-d)^+(y-e)^=r^ while the corresponding equation of the diameter becomes y — e = (x — d). [28] Examining this equation we find that : (1) The diameter is a straight line. (2) The diameter passes through the centre of the circle. 118 PLANE ANALYTIC GEOMETRY. (3) Wie diameter is perpendicular to the chords which it bisects. Any of these properties may be used in determining a diameter. Ex. 1. Find the diameter of the circle x2+ y2 + 4x — Gy + 2 = 0, bisecting chords parallel to the line 2x + y — 1 = 0. The centre of the circle is (—2, 3) and m = — 2. . '. the diameter is y — 3 = — ^7 (x -|- 2) or x — 2y + 8 = 0. Ex. 2. Find the diameter of the same circle, which passes through the point (1, — 2). As before, the centre of the circle is (— 2, 3). .*. by properties (1) and (2), the diameter is ^^i^lOrSy + .« + >=0. By property (3), it bisects chords of slope f . 62. Circle through the Points of Intersection of Two Circles. Let x2 + y24-2 Gx + 2 Fy + C = 0, . (1) x^ + y^ + 2 G'x4-2 F'y + C' = 0, (2) be the two circles. If I and k are two arbitrary multipliers, l(x^ + y^ + 2Gx + 2Fy + C) + k(x'^ + y2 + 2G'x+2 F'y + C') = ^^ will be the equation of a circle, since it is of the second degree, contains no xy term, and has the coefficients of x^ and y^ the same. Moreover, this circle will pass through the points of inter- section of circles (1) and (2), if they intersect. For if Pi(xi> yO is a point of both circles, Xi^ + yi^+2Gxi + 2 Fyi + C = and Xi^ + yi2 + 2 G'xi + 2 F'yi + C' = 0, whence equation (3) will necessarily be satisfied by (xj, yj). and therefore circle (3) passes througli the point Pj. THE CIRCLE. 119 As I and k are given different values, we shall get different circles through the points of intersection of (1) and (2). As a special case, let I = 1 and k = — 1, when (3) becomes 2 (G-G') X + 2 (F-F')y + (C-C') = 0. (4) But this equation is of the first degree and hence is the equation of a straight line.* In fact, it is the equation of the common chord of the two circles (1) and (2). It can readily be proved that (4) is perpendicular to the line of centres of the two circles (1) and (2). Note. In § 38 we have seen that I (Ax + By + C) + k (A'x + B'y + C) = is a straight line passing through the point of intersection of the two straight lines Ax + By + C = and A'x + B'y + C = 0. Just above we have seen that I (x2 + y2 + 2 Gx + 2 Fy + C) + k (x^ + y2 + 2 G'x + 2 F'y + C) - is a circle passing through tlie points of intersection of the two circles x2 + y2 + 2 Gx + 2 Fy+ C = 0, x2 + y2 + 2 G'x + 2 F'y + C = 0. These are only two special cases of a more general theorem, which may be stated as follows : If U and V are any expressions involving x and y, U = and V = will he the equations of two curves ; and, if I and k are any two arbitrary constant multipliers, I U + kV = will he a curve passing through all the points common to the two curves and intersecting the curves at no other points. The proof, which is exactly like that used in the two special cases, is left to the student to formulate. 63. Eadical Axis. If the circles do not intersect, we can, nevertheless, form the equation of the straiglit line, * (4) is the limiting case of a circle, for the arc of a cii-cle approaches the straight line as a limit when the radius of the circle is indefinitely increased. 120 PLANE ANALYTIC GEOMETRY. 2 (G - G') X + 2 (F - F') y + (C - C) = 0, which will be perpendicular to the line of centres ; but it will not be the equation of the common chord, for the circles have no common chord. We shall call this line the radical axis of the two circles. A.S obviously this equation can always be found, every pair of circles have a radical axis, which is their common chord if they intersect. We will now prove a property of the radical axis, which might be used in defining it. We will find the locus of points such that the tangents drawn from them to the two circles shall be of equal length. FlO. 57. Let Pi(xi, yi) be any point of the locus, a and b being the tangents to the circles (1) and (2) respectively. If Lj and Lz are the respective lengths of a and b, by the hypothesis, U = U, or Li^ - U^ = 0. By §57, Lr = Xi^ + yi^ + 2Gxi + 2 Fy, + C, L/ = Xi2 + yi2 + 2 G'x, + 2 F'yi + C, whence 2 (G - G') Xi + ^ ( F - F') y^ + (C - C) = 0. THE CIRCLE. 121 Therefore Pi is a point of the radical axis, since its coordi- nates satisfy tlie equation of the radical axis. Conversely, the tangents drawn to the two circles from any point of the radical axis may be proved of equal length, so that we may now define the radical axis of two circles as the locus of 2)oints such that the tangents from them to the two circles are of equal length. 64. Radical Centre. If we have given the equations of three circles, x^+y2 + 2 Gx + 2 Fy + C = 0, I. x" + y- + 2 G'x + 2 F'y + C' = 0, XL x^ + y2 + 2 G"x + 2 F"y + C" = 0, III. these three circles taken in pairs determine three radical axes, 2 (G - G') X + 2 (F - F') y + (C - C) = 0, (1) 2 (G' - G") X + 2 (F' - F") y + (C - C") = 0, (2) 2 (G" - G) X + 2 (F" - F) y + (C" - C) = 0. (3) 1(2 (G - G') X + 2 (F - F') y + (C - C) } + k{2 (G' - G") X + 2 (F' - F") y + (C - C")} = (4) represents a straight line passing through the point of inter- section of radical axes (1) and (2). In (4) let I = — 1 and k ^ — 1, and it becomes 2 (G" - G) X + 2 (F" - F) y -f (C" - C) = 0. But this is (3); therefore radical axis (3) passes tlirough the point of intersection of radical axes (1) and (2). This common point of the three radical axes is called the radical centre of the three circles. It is evident that, in general, from the radical centre six equal tangents can be drawn, two to each circle. Ex. Find the radical centre of the circles x2 + y2 + 2x — 2y + 1 = 0, x2 + y2 - 8x + 7y — 4 = 0, and X- + y- + 5x — .'!y + 2 = 0. 122 PLANE ANALYTIC GEOMETRY. The radical axes are lOx — 9y + 5 = 0, 13x — lOy + = 0, and 3x — y + 1 = 0. The radical centre is ( — -, — j and the common length of the tangents is 65. Polar Equation of the Circle. P Fig. 58. Let C(ri, ^i) be the centre of the circle of radius a and P(r, 0) be any point of the circle. Then Z COP = ^ — ^i. By Trigonometry, OP' + OC' — 2 OP.OC cos ^ = CP' or r- + r/^ — 2rri cos (9 — Gi) = a". [29] For any particular value of 9, the resulting equation in r is of the second degree, so that a line drawn from the origin will meet the circle in two distinct points, two coincident points, or not at all, according as 4 ri^ cos^ {6 — e,)—'i (r-" — a^) is >, = , or < 0. If 4 r,^ cos^ {9 -6,)-^ (r,^ - a") = 0, ri2[l-cos2 (e-e,)^ = a% sin (0 — e{) = ± - and $ = $^±1 sin~^ -. THE CIRCLE. 123 Therefore 6 must be as great as Oi — sin ^ - , and as small as ^1 + sin~^ -, if the corresponding line from the origin is to meet the circle at all. (1) Find the equation of tlie circle under the following conditions : centre on the initial line ; centre on the initial line and the circle passmg through the origin ; centre on a luie at right angles to the initial line, and the circle passing through the origin. (2) What will be true of all the circles having the same parameter ri, but different parameters a and Ox ? (3) What will be true of all the circles the equations of which agree in ^1, but differ in ri and a ? EXAMPLES. Find the equations of the following circles : 1. Centre at (3, — 4) and radius 5. 2. Centre at (1, 3) and radius 2. 3. Centre at (— 2, 7) and radius 4. / 2 4\ 4. Centre at ( — - , — 77 j and radius 6. 5. What is the equation of the circle constructed on the line joining (1, 5) and (—3, 7) as diameter? 6. Find the equations of the circles of which the line join- ing the points (2, — 3) and (3, — 1) is a radius. 7. Find the equation of the circle having the line joining (— 4, 3) and (1, — 2) as a diameter. 8. Find the equation of the circle having the line joining (a, — b) and (— a, b) as a diameter. 9. Find the equation of the circle having as a diameter that part of the line x — 2y + 4==0 which is included be- tween the coordinate axes. 10. Find the equations of the circles of radius a which are tangent to the axis of y at the origin. 124 PLANE ANALYTIC GEOMETRY. 11. Find the equations of the circles of radius a which are tangent to both coordinate axes. 12. Find the equation of the circle having its centre at the origin and tangent to the line 3x + 4y + 7 = 0. 13. Find the equation of the circle having its centre at the point ( — 1, 2) and tangent to the line 2x — y + 3 = 0. 14. Find the equation of the circle having its centre at (3, — 2) and tangent to the line x + 5y + 1 = 0. 15. The centre of a circle which is tangent to the axes of x and y is on the line 2x — y + 2 = 0. What is its equation ? Find the centre and the radius of each of the following circles : 16. x2 + y^-6x-2y-l = 0. 17. x2 + y- + 4x — 10y + 25 = 0. 18. x^ + y2-4x + 6y + l=0. 19. 3x2 + 3y2 — 6x + 9y-2 = 0. 20. 36x2 ^ 3gy2 _ losx + 48y — 155 = 0. 21. 5x2 _|_ r,y2 _ 2x + 4y + 1 =: 0. 22. Ax^ 4- Ay- + 2 Gx + 2 Fy + C = 0. 23. Find the equation of the straight line joining the centres of the circles x^ + y^ — 4x — 6y — 12 = and x' + y- + X = 0. 24. Do the circles 4x2 + \f + 4x — 12y + 1=0, 2x2 + 2y2 + y = intersect ? Prove your answer. Find the equations of the circles determined by the follow- ing sets of points : 25. (0,0), (1,-1), (-2, 3). 26. (2,-2), (1,2), (-1,1) 27. (0,2), (2,0), (-2,-2) 28. (a, b), (b, a), (- b, - a) 29. (-1,2), (3,-1), (4, .3) THE CIRCLE. 125 30. Find the equation of the circle circumscribing the triangle (0, 3), (— 1, 0), (0, -3). 31. Find tlie equation of a circle passing through the point ( — 2, 3) and concentric with the circle x2+y2-3x + 4y-2=i0. 32. The centre of a circle which passes through the points (2, — 1) and (— 1, 3) is on the line 8x — 4y + 5 = 0. What is the equation of the circle ? 33. The centre of a circle which passes through the points (2, — 3) and (— 4, — 1) is on the line 3y + x — 18 = 0. What is the equation of the circle ? 34. A circle of radius V85 passes through the points (2, 1) and (— 3, 4). What is its equation ? 35. A circle of radius Vl3 passes through the points (2, — 3) and (1, 4). What is its equation ? 36. Find the equation of the locus of the centres of the circles of radius 3, all of which pass through the point (—1, 7). 37. Find the equation of the locus of the centres of the circles which pass through (a, b) and ( — b, — a). 38. Find the equation of the locus of the centres of all circles which pass through (a, b) and (a', b'). 39. Find the equation of a circle in oblique coordinates. 40. The centre of a circle which is tangent to the lines X — 5 = and y — 10 = is on the line 2x + 3y — 5 = 0. What is its equation ? 41. The centre of a circle which is tangent to the lines 17x + y — 35 = and 13x + lly + 50 = is on the line 88 x + 'J'Oy + 15 = 0. What is its equation ? 42. Find the equation of the circle circumscribing the tri- angle of which the sides are the lines-> y + 2 = 0, 3x - 4y + 1 = 0, 3x + 2y - 5 = 0. 43. Find the equation of the circle inscribed in the triangle the sides of which are the lines 8x-f y-17 = 0, 7x-4y + 29 = 0, x + 8y + 17=0. 126 PLANE ANALYTIC GEOMETKY. 44. Find the equation of the circle inscribed in the triangle (16, 3), (-14, 3), (l,-^^^ 4 45. Find a tangent to the circle x^ + y^ = 12, making an angle of 30° with the axis of x. 46. Find a tangent to the circle x^ + y'-^ -)- 4x ~ 6y + 9 = 0, making an angle of 60° with the axis of x. 47. Find a tangent to the circle 4x2 + 4y2 _ 4x + 8y — 139 = 0, making an angle tan~^ | with the axis of x. 48. Find a tangent to the circle 4x2 + 4y2 _ 24x - 12y + 25 = 0, parallel to the line x — 2y + 8 = 0. 49. Find a tangent to the circle x^ -}- y^ — 2x + 4y — 4 =: 0, perpendicular to the line 2x — 3y + 6 = 0. 50. Find a tangent to the circle 9x2 ^ 9y2 _|. 24x + I8y = 119, perpendicular to the line 4x + 3y — 17 ^= 0. 51. What will be the slope of the tangent to the circle x2 _}_ y2 _ 4x -|_ 2y + 1 = 0, which passes through the point (-2,1)? 52. Find a tangent to the circle (x — d)^ + (y — e)^ = r^, making an angle tan~^ m with the axis of x. 53. Find a tangent to the circle x'-\-f + 2 Gx+2 Fy + C = 0, making angle tan~^ m with the axis of x. 54. A tangent and a normal to the circle x^ -|- y^ = 13 pass through the point (2, — 3). What are their equations ? 55. A tangent and a normal to the circle x2 + y2 — 4x + 2y + l=0 pass through (2, — 3). What are their equations ? 56. Find a tangent to the circle x^ -f- y^ — 8x — 6y — 75 = 0, which passes though (— 10, 5). 57. Find a tangent to the circle Sx^ + 3y2 — 2x + 4y = 0, which passes through (2, 1). THE CIRCLE. 127 58. Find a tangent to the circle 2x' + 2y2 -f Gx — 6y — 9 = 0, 1-1 fi 1 /^l 3 + 2 V 5^ which passes through I ~i — 59. Find the chord of contact of the tangents drawn from (— 1, 2) to the circle x- + y^ — 3y + 2 = 0. 60. What is the length of the tangent drawn from (5, 4) to the circle x^ + y^ — 2x — 4y + 2 = ? 61. What are the lengths of the tangents drawn from (^'2)'''''^ (2'"^) to the circle 4x- + 4y- — 20x + 12y + 18 = 0? Explain the second answer. 62. Find the polar of the origin with respect to the circle 5x2 _^ 5y2 _ lox + 7y - 2 =r 0. 63. Can real tangents be drawn from (2, 3) to the circle 9x2 + 9y' + 6x - 12y + 4 = ? Show why your answer is correct. 64. Prove that, if the equation of the circle be written in the form (x — d)^+ (y — e)^= r^, the tangent at Pi(xi, yi) is (xi-d)(x-d) + (yi-e)(y-e) = r2. 65. Does the polar of (1, — 2) with respect to the circle 3x2 _^ 3y2 _|_ 4x _ 3y _ -^r _ Q iiitersect the circle ? Q>^. The polar of a certain point with respect to the circle 2x2 + 2y- + 4y + 7 = is 3x + y + 3 = 0. What are the co- ordinates of the point ? 67. The length of the tangent from the point (x, y) to the circle x-' -f y^ -[" y — 1 = is twice the length of the tangent from the same point to the circle x^ + y^ + 3x — 2y + 2 = 0. Show that the point lies on the circle x- + y- + 4x — ;;y-l-3 = 0. 68. Find the angle at which the circles x' + y' + 6x — 2y + 5 = and x^ + y^ + 4x + 2y - 5 == intersect. (The angle between two curves is the angle be- tween their respective tangents at the point of intersection.) 128 PLANE ANALYTIC GEOMETRY. 69. Show that the circles x^ + y^ + 4x — 6y -|- 8 = and x^ + y^ — 2x — 14y -|- 30 = are orthogonal, i.e., cut each other at right angles. 70. Find the length of the chord cut from the line 2x-y + 10 = by the circle x^ + y^ + 4x — 6y + 11 = 0. 71. The middle point of a chord of the circle x^ + y^ =^ 49 is (1, 3). Find the equation and the length of the chord. 72. Find the diameter of the circle x2 + y2 — 2x + 4y + 1 = 0, bisecting chords parallel to y — 2x + 3 = 0. 73. Find the diameter of the circle 6x-^ + 6y2 + 2x — 3y — 5 = 0, bisecting chords which make an angle of 60° with the axis of x. 74. Find the diameter of the circle x2 + y2 4- 2x — lOy + 10 = 0, making an angle of 45° with the axis of x. 75. Find the diameter of the circle x^ + y" — x -|-y = 0, bisecting chords perpendicular to 3x 4" 2y + 1 = 0. 76. A diameter of the circle 4x'-' + 4y- + 8x - 12y + 1 = passes through the point (1, — 1). What is its equation and the slope of the chords which it bisects ? 77. Find the radical axis of the two circles 2x2 _|_ 2y2 + 3x — 4y + 1 = and 3x=^ + 3y2 - 6x + 1 = 0. 78. Find the radical axis and the line of centres of the two circles x2 + y2 — 4x + 6y-2 = and 2x^ + 2y2+6x-8y + l = 0, and show that they are perpendicular to each other. 79. Find the point of intersection of the radical axis and the line of centres of the two circles x^ -|- y^ = 25 and x- + y- — 6x — 8y = 24. THE CIRCLE. 129 80. Find the radical centre of the circles x' + y' + 6x + y — 5 = 0, 2x2 + 2y2-x-3y + l = 0, x" + y2+7x + 4y-2 = 0. 81. Find the circle which passes through (0, 4) and the points of intersection of the two circles x^ + y' = 10 and x2 + y2-8x + 2y + 8 = 0. 82. Show that the circles x2 + y2 + 2 Gx + 2 Fy + C = and x:' + y2 + 2 G'x + 2 F'y + C' = are tangent to each other, if V(G - G')^ + (F - F')- = VG- + F^ - C ± VG'^ + F'- - C. 83. Prove that the radical axis of any two circles is per- pendicular to the line joining their centres. 84. Prove that two circles will be concentric if their equa- tions differ only in the absolute term. 85. If A, B, C, and D are four points on a circle, and the tangents at A and B meet at P, and the tangents at C and D meet at Q, and the chords A B and C D meet at R, prove that R is the pole of PQ. 86. The polars of A(xi, yi) and B(x2, y2) are given with respect to the circle x^ + y^ = rl A P is the perpendicular from A to the polar of B, and BQ is the perpendicular from B to the polar of A. If C is the centre of the circle, prove fi f CA CB ^^^'"' AP=BQ- 87. Lines are drawn from ( — a, 0) and (a, 0) at right angles to each other. Find the equation of the locus of their point of intersection. 88. Prove that the square of the tangent drawn from any point of one circle to another circle is proportional to the perpendicular from that point to their radical axis. 89. Find the locus of a point such that tangents from it to two concentric circles are inversely as their radii. 130 PLANE ANALYTIC GEOMETRY. 90. A point moves so that the sum of the squares of its distances from any number of fixed points is constant. Show that the locus is a circle. 91. A point moves so that its distances from two fixed points are in a constant ratio k. Show that the locus is a circle, except when k = 1. 92. A line of length a moves with its extremities upon the coordinate axes. Find the locus described by its middle point. 93. is a fixed point and AB is a fixed straight line. A straight line is drawn from meeting AB at P, and in OP a point Q is taken so that OP. OQ =^ k'. Find the locus of Q. 94. A point moves so that the length of the tangent from it to a fixed circle is always equal to its distance from a fixed point. Find the locus. 95. A point moves so that the sum of the squares of its distances from the four sides of a square is constant. Show that the locus is a circle. 96. A point moves so that the square of its distance from the base of an isosceles triangle is equal to the product of its distances from the other two sides. Show that the locus is a circle. 97. A point moves so that the sum of the squares of its distances from the sides of an equilateral triangle is constant. Show that the locus is a circle. 98. Find the locus of a point, the square of the distance of which from a fixed point is proportional to its distance from a fixed straight line. CHAPTER VI. THE CONIC SECTIONS. 66. Definition and Equation. A conic section is the locus of a point ivhich moves so that its distance from a fixed point, called the focus, is in a constant ratio to its distance from a fixed straight line, called the directrix. The constant ratio is called the eccentricity of the conic section, and will be denoted thronghout this book by the letter e. Its value is of great importance, not only in lixing the shape of any one conic section, but also in determining to which one of three kinds the curve belongs. Jfe 1, it is called an hyperbola. As the name literally indicates, the conic sections may be obtained by cutting a right circular cone by a plane, and it was from this standpoint that the curves were first studied and named. For the purposes of Analytic Geometry, how- ever, it is more convenient to use the definition given above. We Avill later prove that the curves thus defined are identical with those obtained from the cone. In anticipation of this proof, we will continue the use of the name " conic section," or more shortly, " conic." We can readily deduce the equation of the conic section from the definition. Let RS (Fig. 59) be tlie directrix and F the focus. Take RS as the axis of y, and a line through F perpendicular to RS as the axis of x. Then D, the intersection of these two lines, is the origin of coordinates. Call the distance DF, 2p. Then the coordinates of F are (2p, 0). 132 PLANE ANALYTIC GEOMETRY. c 5 M P / • D f/ N Fig. 59. Let P be any point on the conic section, DN its abscissa, x, and NP its ordinate, y. Draw FP, and PM perpendicular to RS. Then, from the definition, PF_ PM~^' or PF' = e^ PM'. But PM = X, the ordinate of P, and by [1], § 3, PF'=(x-2p)2+y2. Therefore (x-2p)^ + y' = eV [30] is the required equation of the conic section. If we place x = in equation [30], we find the corre- sponding value of y to be imaginary. Hence, a conic section never i7iter sects the directrix. The farther discussion of the conic sections is best made by considering each case separately. THE CONIC SECTIONS. 133 67. The Parabola, [e = 1] S Fig. 60. If we place e = 1 in equation [30], it becomes (x-2p)2 + y■^=x^ or y" — 4px + 4p-^0. By placing y = in this equation, we find that the curve intersects the axis of x in one point, namely x = p. This point is called the vertex (A) of the parabola, and the line (AF) drawn through the focus and the vertex is called the axis. Remembering that the distance from the directrix to the focus is by definition 2p, we see that the jxirahola /las one vertex half way between the directrix and the focus. The equation of the parabola will take a simpler form, if the origin of coordinates be transferred to the vertex, the axis of X being unchanged, but a new axis of y being taken through A. This transformation is effected by the substitu- tion [19], § 43, 134 PLANE ANALYTIC GEOMETRY. x = x' + p, y = y'. There results y'2 = 4px', or, dropping the primes, y' = 4px. [31] This is the simplest equation of the parabola. From this equation, the parabola may be readily plotted by computing ordinates corresponding to values of x. Among these ordinates, the one passing through the focus is to be especially noticed. The abscissa of the focus is p, tlie corresponding ordinate is ±2p. Hence the line LL' = 4p. This line, the chord through the focus perpendicular to the axis, is called the latus rectum of the curve. If p is positive, the equation y^ = 4px represents a curve as in Fig. 61 (a). If p is negative, the same equation represents a curve as in Fig. 61 (b). If we had called the directrix the axis of x at the outset, the equation of the parabola would have reduced to the form x2 = 4py. In this case, the curve would be in the position represented in Figs. 61 (c) and 61 (d), respectively, according as p is positive or negative. In connection with any parabola, we should notice its axis, directrix, and focus. If the equation is in the form y^ = 4px, the equation of the axis is y — ; the equation of the direc- trix is X = — p ; and the focus as at the point (p, 0). If the equation is in the form x^ = 4py, the equation of the axis is x = ; the equation of the directrix is y ^ — p ; and the focus is at the point (0, p). Ex. Find the equations of axis and directrix, and the focus of the parabola y^ = — 7x. THE CONIC SECTIONS. 135 Here .: Focus is at (— |, 0) ; Equation of axis, y = ; Fig. 61. (b) (d) 68. The Ellipse, [e < 1] In interpreting equation [30], we have to remember tliat e is less than unity. Placing y i= 0, we have two values of X, corresponding to two vertices A and A', Fig. 62, namely, 2p 2p 1-e' -^^ 1 + e' Since e < 1, Xi < 2p and Xo > 2p. Hence, the ellipse has two vertices lyiiKj on the same side of the directrix but on oiqyo- site sides of the foctis. The line A A' is called the major or transverse axis of the ellipse, and its length is denoted by 2a. 136 PLANE ANALYTIC GEOMETRY. Fig. 62. Then whence 2a = DA' -DA, _ 2p 2p 4pe 1 — e l + e~l-e2' _ 2pe_ (1) The point C, half-way between A and A', is called the centre of the conic. Its distances, both from the directrix and from the focus, are readily found. For DC DA + AC, l + e _ 2p , 2pe ^ 2p l + e^l-e^ l-e^' Hence Also Hence DC = FC=DC 9 8 DF, (2) -^^_2p-lP^ l-e^ FC= ae. 1-6^ (3) THE CONIC SECTIONS. 137 The equation of the ellipse will take its simplest form if the origin of coordinates is transformed to C, the direction of the axis being unchanged. For that purpose, we substi- tute in equation [30], the values e .«. and also, from (1), There results y = y Zp^= ae. e (x' + ae)'^ + y"^ = eM x' + e Whence, by expanding, uniting like terms, and dropping the primes, (l-e^)x^' + y^-a'^(l-e^), a"^+aMl-e^)^^- If, now, we introduce a new quantity b, such that b2 = a2 (1-e'^), (4) we have, finally, $ + $=h [32] the simplest equation of the ellipse. From this equation the ellipse may be plotted, and the shape given in tlie figure may be verified. The meaning of the quantity b is found by placing x =: in [32]. There results y = ± b, and hence the line BB' is equal to 2b. It is evident from (4) that b < a, since e < 1 ; hence BB' is called the minor axis, or sometimes the conjugate axis. When the semi-axes a and b are given, the eccentricity is readily computed. For, from (4), we have Va^ - b'^ ... 6= . (O) 138 PLANE ANALYTIC GEOMETRY. From [32] it is evident that the curve is symmetrical with respect to the axis of y. Hence, if the left-hand portion of Fig. 62 is folded over upon tlie axis of y, it will come into exact coincidence with the right-hand portion of the figure. It follows that the ellipse has another focus F' and another directrix R'S' symmetrical to F and RS. The latus rectum of an ellipse is defined as the chord through the focus perpendicular to the major axis. To find its length, we place x == ae = Va^ — b^ in [32], and find b' XT , , , 2b-' y = ± — . Hence LL — — . a a We have discussed the equation of the ellipse up to this point on the assumption that the directrix RS was taken as the axis of y in the deduction of the original equation [30]. It would have been equally allowable to take RS as the axis of X, however, and to call the major axis 2b, the minor axis 2a. We should be led to the same final equation a-' ^ b^ as before, but in all the intermediate steps the quantities x and y, and a and b would interchange their roles. This result is at once distinguished from [32] from the fact that here b > a, whereas in [32], a > b. For convenience, we collect our results as follows : I. Corresponding to the ellipse where a !> b, are the following formulas : Major axis, y = ; Centre, (0, 0) ; Va^* - b^ . a Foci, (± ae, 0) ; THE CONIC SECTIONS. 139 Vertices, (± a, 0) ; ^. , a Directrices, x = ± - : e 2b2 Latus rectum, — . a II. Corresponding to the ellipse a^ ' b _2 I^U2 -^' ivhere a < b, are the following formulas : Major axis, x = ; Centre, (0, 0) ; Vb^ - a^ b ' Foci, (0, ± be) ; Vertices, (0, ± b) ; b Directrices, y :=±-: ^ e 2a2 Latus rectum, 7—. b If a = b, the equation of the ellipse reduces to x^ + y^ = a^ the equation of a circle. By substituting b ^ a in the above formulas, we find e = ; foci, (0, 0) ; directrices, y = ± - = 00. The last result must be interpreted by the method of limits, and we say : the circle is the limiting form uihich the ellipse approaches as the eccentricity approaches zero and the directrix recedes indefinitely from the origin. Consider, as example, i)x- + 4y'- = 36. This is equivalent to — + — = 1 whence a = 2, b = 3, and a < b. 140 PLANE ANALYTIC GEOMETRY. We take the second set of formulas with the results : Major axis, x = 0; Centre, (0, 0) ; e=iV5; Foci, (0, ± V5) ; Vertices, (0, ± 3) ; Directrices, y = ± | Vs ; Latus rectum, f . 69. The Hyperbola, [e > 1] D' Fig. 63. The equation [30] is to be interpreted on the hypothesis that e is greater than unity. Placing y r= 0, we have for the vertices 2p -P ' 1 + e' ^ 1-e Since e > 1, Xi < 2p and Xg is negative. Hence, the hyperbola has tivo vertices lying on the same side of the focus and on opposite sides of the directrix. The line A'A is called the transverse axis of the hyperbola and its length is denoted by 2a. The point C, half-way between A' and A, is the centre of the curve. Then 2a = -DA'+DA, 2p 2p _ 4pe ■~ l-e^l + e~e^^=^' THE CONIC SECTIONS. 141 whence a = .-, '^ - • (1) e- — 1 DC = DA + AC, _ 2p 2pe _ 2p ~l + e e^-l" e--l' a e FC=FD+DC, whence DC =: (2) e o, 2p _ 2pe^ -1~ e'-r whence FC ^ — ae. (3) The origin may be transferred to C by placing - a e and 2p = ae e There results, as in the case of the ellipse, a2^a2(l-e'-^) The denominator nnder y- is now a negative quantity, since e > 1. We will, therefore, place — h^=a'(l-e''), or b''=a-(e^-l). (4) The equation then becomes r!-^=l, [33] a b the simplest equation of the hyperbola. From this equation the hyperbola is easily plotted. It will be found to consist of two infinite branches, cutting the axis of X at the points A and A'. When X = 0, y = ± b V— 1. This is imaginary, but it is convenient to lay off the ordinates CB and CB', each equal to 142 PLANE ANALYTIC GEOMETRY. b, and, from analogy to the ellipse, to call the line BB' the conjugate axis. This axis appears at first sight to have no connection with the hyperbola, but the following discussion shows that it has an important part in determining the shape of the curve. Consider a line, y = mx, passing through the centre. It cuts the hyperbola in two points, the abscissas of which are ab x = + Vb^- a-m'' These values of x are real if m-<— , and imaginary if 3." a m'^> -5. The dividing case between these is when m : a'' corresponding to the two straight lines b , b y ^ -X and y = x. a ■'a If, now, we construct a rectangle of which the centre is at C and the sides are equal to 2a and 2b, respectively, these two lines will be its diagonals. Any line the slope of which is less than that of these diagonals will cut the hyperbola in two points ; any line the slope of which is greater than that of the diagonals will not intersect the hyperbola. Fig. 64. THE CONIC SECTIONS. 143 Hence these diagonals 'tnay he used as guide lines in draiving the branches of the hyperbola. They are called the asymptotes of the hyperbola. If the semi-axes a and b of the hyperbola are given, the eccentricity is given by the formula ^_ Va^+b^^ (5) a as follows at once from (4). The symmetry of the hyperbola shows the existence of a second focus and directrix at the left of the centre. The formulas thus far obtained depend upon the original assumption of the directrix as the axis of y in the derivation of formula [30]. If the directrix had then been taken as the axis of X, the simplest equation would be a^^b^ ' and in all the discussion x and y, and a and b would in- terchange their roles. We have, therefore, the following summary of our work. I. Corres2)onding to the hyperbola a^ b'^ ' are the folloiving formulas : Transverse axis, y ^ ; Centre, (0, 0) ; Va'^ + b^ ^ = 1~' Foci, {±L ae, 0) ; Vertices, (db a, 0) ; ' Directrices, x = ± - ; Lotus rectum, — . a 144 PLANE ANALYTIC GEOMETRY. II. Corresponding to the hyperbola are the following formulas : Transverse axis, x = ; Centre, (0, 0) ; Vb^ + a^ e = b Foci, (0, ± be) ; Vertices, (0, ± b) ; b Directrices, y = ± — ; e Asymptotes, y =: ± -x ; a 2a^ Latus red ion, -r- • b 70. Generalization. We have seen that, by proper choice of coordinate axes, the equations of the three conic sections may be reduced to forms of the types y2 = 4px, ^ 1 y!_i a==^b^~-^' r_y_- If these three equations are transferred to new axes parallel to the old, the new origin being at the point (xq, yo), by means of [19], §43, they become (y' + yo)^ = 4p(x' + xo), (x'+xo)^ , (y' + yo)^ ^-. a^ ' b'^ ' (x'+x„y^ (y'+yoy^ ^., a- b^ THE CONIC SECTIONS. 145 If the above equations are simplified, they will all be found to be of the general form Ax^ + By2 + 2 Gx + 2 Fy + C = 0. Conversely, an equation of the form just written can be put in the form of the equation of one of the conic sections, except in certain exceptional cases. The theoretical discus- sion will be found in chapter XI ; it is the purpose of the present chapter to give methods by which the student may accomplish the required results with the least possible de- pendence upon formulas. This will best be done by consider- ing a few numerical examples. It will be seen that the process is a modification of that called "completing the square " in the solution of quadratic equations. Ex. 1. Take the equation 2x2 + 3y^ -h 8x - 6y + 5 = 0. It may be written 2(x2 -I- 4x) -F 3 (y2 - 2y) = - 5. We may now complete the square of the quantity in eacli parenthesis, adding to the right-hand side of the equation the same quantities that are added to the left-hand side. There results 2 (x'-i -t- 4x -f- 4) -f 3 (y2 - 2y 4- 1) = - 5 -h 8 + 3, or 2 (X -I- 2)2 + 3 (y - 1)2 = 6. If we now place X -f- 2 = x', y - 1 = y'> thus transforming the equation to new axes parallel to the old ([19], § 43), the new origin being at the point (— 2, 1), we have 2x'2 + 3y'2 = 6, x'2 , y'2 the equation of an ellipse, in which a = Vs, b = \/2. Since a >> b, we apply the first set of formulas of 8 68, with the following results.- Centre, (x' = 0, y' = 0) ; Major axis, y' = ; 146 PLANE ANALYTIC GEOMETRY. Vertices, (x' = ± V3, y' = 0) ; Foci, (x' = ± 1 , y' = 0) ; Directrices, x' = ± 3. If we now transfer these results back to the old axes x and y, we have, finally. Centre, (- 2, 1) ; Major axis, y = 1 ; e = V|; Vertices, {— 2 ± Vs, 1) ; Foci, (- 1, 1) and {- 3, 1) ; Directrices, x = + 1 and x = — 5. Ex. 2. Let us consider 2x2 _ 3y2 -|- gx - 6y + 11 = 0. As before, we have the following work : Now place we have 2 (x2 + 4x) - 3 (y2 + 2y) = - 11, 2 (x2 + 4x + 4) - 3 (y2 + 2y + 1) 2(x + 2)2-3(y + 1)2 = -6. X + 2 = x', y + 1 = /; 2x'2 — 3y'2 = — 6, 11 + 8-3, The curve is therefore an hyperbola, with the tranverse axis parallel to OY. We find by § 69 the following values : Centre, (x' = 0, y' = 0) ; THE CONIC SECTIONS. 147 Transverse axis^ x' = Oj_ e = Vf = WlO; Foci, (x' = 0, y' = ± V5) ; Vertices, (x' = 0, y' = ± V2) ; Directrices, y' = ± ? Vo ; Asymptotes, x' = ± Vfy'. These results, transferred back to tlie old axes OX and OY, are Centre, (- 2, — 1) ; Transverse axis, x =^ 2 ; e rr 1 VlO ; Foci, (— 2, — 1 ± V5) ; Vertices, (- 2, — 1 ± V2) ; Directrices, y = — 1 ± | V5 ; Asymptotes, x + 2 = ± Vi(y + 1). Fig. 66. Ex. 3. Finally, we will consider y2 — 4y + 8x + 28 = 0. There is here only one quadratic term. By completing the square, we have 148 PLANE ANALYTIC GEOMETRY. y2 — 4y + 4 = — 8x — 24, or (y - 2)-2 = - 8 (x + 3). Placing • y - 2 = y', X + 3 = x', we have y'2 - - 8x'. This is the equation of a parabola, in which 4p = — By § 67, w^e have the results : Vertex, (x' = 0, y' = 0) ; Axis, y' = ; Pocus, (x' 2, y'-O); Directrix, x' = 2. Transforming hack to the original axes, we have Vertex, (- 3, + 2) ; Axis, y = 2 ; Focus, (— 5, 2) ; Directrix, x = — 1. Fig. 67. THE CONIC SECTIONS. 149 71. Limiting^ Cases of Conic Sections. There are certain cases where the process of completing the squares, as used in the previous article, leads to a result which is not one of the equations of the ellipse, hyper- bola, or parabola, although bearing a strong similarity to one of them. These cases are as follows : 1. Similar to the Ellipse. On completing the squares, we may be led to the form a-^ ^ b^ . This equation is satisfied by no real values of x' and y', since the sum of two squares cannot be negative. There is, there- fore, no real curve corresponding to the equation ; but, because of the form of the equation, it is sometimes said to represent an ^' imafjinary ellipse.'''' In the second place, on completing the squares, we may find the right-hand side equal to zero. We have, then, — 4-^ = 0. a'' b- Tliis is satisfied, as far as real quantities are concerned, only by x' = 0, y' = 0, for the sum of two squares can be zero only when each of the squares is zero. Hence the equation represents a sinyle point.* We may say, therefore : The iynaginary ellipse and the point are exceptional cases of the ellipse. 2. Similar to the Hyperbola. On completing the squares, the right-hand side of the equation may be zero. * This equation may also be said to represent two imaginary straight lines intersecting in a point. See § 41 and § 133. 150 PLANE ANALYTIC GEOMETRY. The equation is tlien of the form 0. X y_: _ This is factored into n')(^-0=«. and each factor represents a straight line passing through the point (0, 0). Hence : Two intersecting strut (jltt lines are an exceptional case of an hyperhola. 3. Similar to the Parabola. An exceptional case arises here when the equation con- tains only one of the coordinates, let us say y. In this case the left-hand side may be factored into two linear factors, (y--)(y-/3) = o. Each factor represents a line parallel to the axis of x, and the two are distinct, coincident, or imaginary, according as a and /8 are real and distinct, equal, or imaginary. Hence : Two parallel straight lines, which may he real, coin- cident, or imaginary, are excejitional cases of the parabola. 72. Recapitulation. The results of the foregoing articles lead to the following statement : A quadratic equation of the form Ax- + By- + 2 Gx + 2 Fy + C = always represents a conic section. (1) If f^ and B have the same sign, the conic is an ellipse, a circle, or, in exceptional cases, an imaginary ellipse or a point. (2) If A and B have opposite signs, the conic is an hyperbola, or, in exceptional cases, tivo intersecting straight lines. THE CONIC SECTIONS. 151 (3) If either A or B is zero, the conic is a parabola, or, in excejitional cases, two ^ja>'a^^eZ straight lines which may be real, coincidejit, or imaginari/. The student should notice that the above equation is not the most general equation of the second degree, since it lacks the term 2 Hxy. The discussion of the general equation is found in chapter XI. 73. Sections of a Cone. In this article we mean to prove the statement made in § 66 of this chapter that every plane section of a right circular cone is one of the curves we have called the conic sections. In so doing we need to distinguish the case in which the plane that cuts out the section passes through the vertex from that in which it does not. In the latter case we shall prove the theorem : Any section of a rigid circular cone made by a jjlaiie not pass- ing through the vertex is a circle, an ellij^se, an hyperbola, or a parabola. Let a right circular cone with vertex and base AB be cut by a plane, not passing through the vertex ; to prove the section a circle, ellipse, hyperbola, or parabola. If the cutting plane is parallel to the base, the section is a circle by Solid Geometry. If the plane is not parallel to the base, let the plane OAB be taken perpendicular both to the base of the cone and to the cutting plane. This plane necessarily contains the axis of the cone, and hence passes through the centre of each circular section of the cone. Let MN be the intersection of OAB and the cutting plane. We have then to distinguish three cases. Case I. ll'/u'/i MN 7)ieets both of the elements OA a?id OB. Take C, Fig. 68, the middle point of M N and Q any other point, and pass planes parallel to the base through C and Q. 152 PLANE ANALYTIC GEOMETRY. Fig. 68. The new sections are circles of which the lines FG and H K are diameters. QP, the intersection of the planes FG and MN, is perpendicular to the plane OAB and hence perpen- dicular to the lines FG and MN by Solid Geometry. For similar reasons, CR is perpendicular to HK and MN. Now, since the sections FG and H K are circles of which the lines FG and HK are diameters, QP'=FQ.QG, and CR'^HC.CK. But, from similar triangles, FQ^MQ HC~ MC QG^QN CK CN" and QP- IQ.QN CR' MC.CN (1) THE CONIC SECTIONS. 153 Let us place CR = b, and MC = CN-=a. We have, then, QP' MQ.QN b- a- (2) Finally, in the nates, and M N as plane MN, take C as an the axis of x. origin of coordi- Then, for the point P, MQ: QH-- QP=y, CQ = x, = MC + CQ = a + x, = CN-CQ = a-x. Hence equation (2) becomes y^ a- — X- b^ ~ a' ' or 2 2 - + ^ =1 Therefore, the section MN is an ellipse. Case II. IV/ien MN meets one of the elements, Ok, produced. Let the construction be made exactly as in Case I. Then, from the section FG, Fig. 69, QP'=FQ.QG. But from similar triangles, FQ^MQ KG MC' and QG_NQ. ^"""^ HC~CN Hence Q^'= MCXN ■^^*^- ^^^- ^^^ Now place MC = CN=a, and KC.HC=b2. 164 PLANE ANALYTIC GEOMETRY. / n^ \ Ae— T Pig. 69. Equation (1) becomes QP'^%^ MQ.NQ. di (2) If, finally, in the plane MN we take C as the origin of coor- dinates and M N as the axis of x, then MQ=MC + CQ = a+ X, NQ= CQ-CN = x-a, and equation (2) becomes y2 = ^(x^'-a^, or — — — = 1. a"' b Hence, the section is an hyiierhola. THE CONIC SECTIONS. 155 Case III. When M N is parallel to one of the elements OA. Fig. 70. Take Q, Fig. 70, any point on the line MN, and N, the ex- tremity of M N, and pass planes parallel to the base of the cone, as before. Then QP'=FQ.QG. But FQ=HN, hn~oh' and Q^'-^-NQ. (1) Now take N as the origin of coordinates, and MN as the axis of X. Then QP = y, __ NQ=x, HN' and TTTT) since it is constant, may be represented by 4p. OH *156 PLANE ANALYTIC GEOMETRY. Equation (1) is then and the section is a parahola. Limiting Cases. If we consider now the case in which the cutting plane passes through the vertex, we have from Solid Geometry, the theorem : Any section of a right circular cone, made hy a plane j^assinff through the vertex, is a point, two intersecting straight lines, or two coincident straight lines. In particular, if the plane which cuts out a circle or an ellipse be made to approach the vertex, the circle or ellipse grows constantly smaller, until finally it becomes a point. Hence, the point is a limiting case of the ellip)se or circle. If the plane which cuts out an hyperbola approaches the vertex, the hyperbola approaches two straight lines. Hence, two intersecting straight lines form a limiting case of the hyperbola. If the plane which cuts out a parabola is made to pass through the vertex, the plane becomes a tangent plane, inter- secting the cone in two coincident straight lines. Hence, two coinciderit straight lines form a limiting case of the parabola. Finally, if the vertex of the cone is made to recede indefi- nitely from the base, the cone approaches a cylinder as a limit, and the parabolic section approaches two parallel straight lines. Hence, tioo parallel straight lines form a limiting case of the parabola. We have obtained thus all the loci discussed in the previ- ous articles, except the " imaginary ellipse " and " imaginary straight lines." Of course, these loci cannot be obtained from a real cone, since they have no real existence. We agree to class them with the conic sections, because of the similarity of their equations. THE CONIC SECTIONS. 167 74. Polar Equation of the Conic Section, the Focus being the Pole. The equation of the conic section, y^+(x — 2p)^ = eV, may be transformed to the focus as origin, by placing y = y^ x = x' + 2p. It then becomes y'^' + x'^^ = e^'(x' + 2p)l By [7], § 10, to transform to polar coordinates, we have to place x' = r cos 6, y' =: r sin 6. There results r^ = e- (r cos 6 -\- 2py, or r = ± e (r cos 6 + 2p). Whence, by separating the two values of r, 2p ^_ 2p 1 — 6 cos 1 + e cos EXAMPLES. Determine the nature and equations of the following conies : 1. Focus, (4, 0) 2. - (-5,0) 3. " (0, 2) 4. - (1, 2) 5. - (3,-4) 6. - (-2,3) directrix, x = '' x = '' y=0 '^ X = u y = 1 e — J. e = 2. 6 = 1. P = 1 e — 3. e = «. X = — 4; 6 = 1. Find the equations of the following parabolas from the given data : 7. Focus, (0, 1) ; vertex (0, 0). 8. - (0,-1); - (0,0). 158 PLANE ANALYTIC GEOMETRY. 9. Focus, (-3,0) vertex, (0, 0). 10. a (4,0) a (0, 0). 11. ii (0,0) li (0, 2). 12. u (-2,3) ii (3, 3). 13. a (1,2) (I (1, 1)- Find the equations of the following ellipses, the axes of coordinates being parallel to the axes of the curves : 14. a = 3 ; b = 1 ; centre, (0, 0). (0, 0). (1, 2). 15. a — 2 j b — ^ ; 16. a = l; b = i; 17. a =2; b = 4; " (-3, 0). -18. a-f; b = i; "(-2,-1). 19. Vertices, (± 5, 0) ; one focus, (3, 0). 20. Foci, (± 3, 0) ; directrices, x = ± 4. 21. Vertices, (— 1, 0), (5, 0) ; one focus, (0, 0). 22. Minor axis, 6 ; one focus, (— 4, 0) ; centre, (0, 0). 23. Minor axis, 12 ; e = y% ; centre, (0, 0) ; minor axis along OX. 24. Major axis, 6 ; e = ^ ; centre, (0, 0) ; major axis along OX. 25. Centre, (0, 0) ; passing through (3, 4) and (1, 5) ; major axis along OX. Find the equations of the following hyperbolas, the axes of coordinates being parallel to the axes of the curves : 26. a = 1 ; b = 6 ; centre, (0, 0); transverse axis along OX. 27. a = 3 ; b = 4-; centre, (0, 0) ; transverse axis along OY. 28. a = § ; b = i ; centre, (0, 0); transverse axis along OX. 29. a = 1 ; b = 2 ; centre, (1, — 2) ; transverse axis parallel to OX. 30. a = Vj ; b = V|^ ; centre, (2, 3) ; transverse axis par- allel to OY. 31. Vertices, (± 4, 0) ; e = f . 32. Foci, (± 7, 0) ; directrices, x = ± 5. THE CONIC SECTIONS. 159 33. a = 5; centre, (0,0); e = 1^ ; transverse axis along OX. 34. Centre, (0, 0) ; passing through (1, 3) and (— 2, — 4) ; conjugate axis along OX. 35. Centre, (0, 0) ; passing through (1, 1) ; transverse axis along OX ; asymptotes having slopes ± |. 36. Passing through (1, 4) ; having same asymptotes as 9y2 ^ x^ == 8. Determine the nature of the following conies. If the conic is a parabola, find (1) vertex, (2) focus, (3) axis, (4) directrix. If the conic is an ellipse or an hyperbola, find (1) centre, (2) equation of major axis, (3) lengths of semi-axes, (4) ver- tices, (5) eccentricity, (6) foci, (7) directrices. If the conic is an hyperbola, find also (8) the asymptotes. 37. 4x- + 49y2 = 196. 38. 3x- + 2y-^ = l. 39. x2==-7y. 40. ^-t=:-l. 16 9 41. y'^ — 16x = 0. 42. 2x2-3y^==6. 43. x^^lSy. 44. 36x2 _ isy2 _|_ 36x + 24y - 35 = 0. 45. x2 — 4x — lly — 7==0. 46. 3x2 + y2 — 6x-6y + 9 = 0. 47. 4y-2-4y + 6x + 3 = 0. 48. x2 + 2y2 + 6x + 8y + 4 = 0. 49. 12x2 _ igy2 _ 3gx _ i2y -f 31 = 0. 50. 3x2 _ 7y2 _ 6x — 28y — 46 = 0. 51. 9x- + 25y2 — 36x + 50y — 164 = 0. 52. y = (x - 2) (x -f 3). 53. y2 = (x + 3) (4 - x). 54. y2 = (x + 3) (x - 4). 160 TLANE ANALYTIC GEOMETRY. 55. Find the equation of the ellipse in terms of a and b, when the origin is at the vertex, the major axis lying along OX. 56. Find the equation of the hyperbola in terms of a and b, when the origin is at the vertex, the major axis lying along OX. 57. Prove that in tlie ellipse or hyperbola the squares of the ordinates of any two points are to each other as the products of the segments of the transverse axis made by the feet of these ordinates. 58. Prove that in the ellipse or hyperbola, the semi-conju- gate axis is a mean proportional between AF and FA'. ' 59. Prove that in the ellipse or hyperbola, the semi-latus rectum is an harmonic mean between AF and FA'. 60. A point moves so that the difference of its distances from (a, b) and (a, — b) is always equal to 2c. Find its locus. 61. A point moves so that the sum of its distances from (a, b) and (a, — b) is always equal to 2c. Find its locus. 62. A line of constant length k moves with its extremities on the two axes of coordinates. Find the locus described by any point of the line. 63. Show that -^ Hj + r^ — r^ = 1, where k is an arbitrary X y parameter, represents ellipses confocal to -^ + fi ^ 1 when k'^ < b^ or k- > a^, and represents hyperbolas confocal to X V — + f-„ ^ 1 when k" > b" but < a^, a^ being considered greater a' b" than bl 64. Show that if the focus is taken upon the directrix, the conic section becomes one of the limiting cases. CHAPTER VII. TANGENT, NORMAL, AND POLAR. 75. We propose, in this chapter, to discuss three straight lines, which we have found to be of importance in the case of the circle : namely, the tangent, the normal, and the polar. The discussion will be almost exactly like that of the chapter on the circle, but we will proceed without direct reference to that chapter, both for the sake of the review of important properties, and also because of the greater generality in our treatment, 76. Tangent. Beginning with the tangent, we have the definition : A tangent at any point of a curve is the straight line ivhich the secant through that point and any second p)oint of the curve approaches as a limit, as the second point approaches the first along the curve. In accordance with this definition, the slope of a tangent may be found when the point of contact is known, by first finding the slope of the secant, and then passing to the limit. Let Pi(xi, yi) and P2(x2, ya), Fig. 71, be two neighboring points upon a conic, the equation of which is Ax'^ + By- + 2 Gx + 2 Fy + C = 0. Draw the secant PiPo and the tangent PjT, and let the angle RP,P2 = ^, and RPiT = <^. Then tan^^^y^:^- Xo — X, 162 PLANE ANALYTIC GEOMETRY. Y Fig. 71. As P2 approaches P^ y.2 approaches yi, Xg approaches Xi, and Si — Vi the fraction — approaches the indeterminate form -. X2 — xi ^^ Hence it is impossible to find the limiting value of tan from this fraction directly. But we can find another expres- sion for tan 6, by considering the fact that both Pj and Pg are on the conic. We have, namely, Axi^ + Byi^ + 2 Gxi + 2 Fyi + C = 0, (1) and Ax2-+ By2- + 2Gx2 + 2Fy2 + C==0. (2) By subtracting (1) from (2), we obtain A(x2^-Xi^)+B(y2^-yi^) + 2G(x2-x0 + 2F(y2-yi)=0, and, by factoring, (x2-xO[A(x2+xO + 2G] + (y2-yO[B(y2 + yi) + 2F] = 0. Whence y2-yi ^ A(x2 + xO + 2G xo-xi B(y2 + yO+2F TANGENT, NORMAL, AND POLAR. 163 Hence, we have t„„fl- A(x, + x,)+2G '''"^— B(y. + y,)+2F If now P2 approaches Pi, tan 6 approaches tan <^, and B(y. + yi)+2F''PP'°^^^^' By^ + F Hence, by the theory of limits, . ^ Axi + G '"^^ = -By7+7- The equation of the tangent is, therefore, by [13], § 33, But this equation may be placed in a much more conven- ient form. Let us first clear of fractions and transpose. There results Axix — Axi^ + By^y - By^^ + Gx — Gxi -f- Fy — Fy^ = 0. Whence, by adding (1), AxiX + ByiY + G (X + Xi) + F (y + yO + C = 0. [34] This equation of the tangent may be readily remembered by comparing it with the equation of the conic. It will be seen that, as in the case of the circle, ive have only to replace x^ arid y^ in the equation of the conic by XjX and \ji\j, respectively, and to replace 2x and 2y 6y x -(- Xj and y + yi, respectively, in order to obtain the equation of the tangent. The following are special cases : (1) Parabola, y- = 4px ; tangent, yiy = 2p (x + x,)- (2) Ellipse, ^. + ^2=1; tangent, ^ + M _ i. v\/ Yy\/\/ (3) Hyperbola, - — ^- = 1 ; tangent, Tir — xf = 1- a' b^ ' « ' a- b^ Ex. Find the equation of the tangent to the hyperbola 2x2 - 3y2 + 4x + G = 0, 164 PLANE ANALYTIC GEOMETRY. passing through the point (1, — 2). Since (1, — 2) satisfies the equation of the hyperbola, it is the point of contact. Hence the tangent is 2x (1) - 3y (- 2) + 2 (X + 1) + 6 = 0, or 2x + 3y + 4 = 0. For some purposes, it is convenient to be able to find a tangent with a given slope, the point of contact being un- known. In such cases the method used in § 18 and § 53 is sufficient to solve the problem. We call m, namely, the given slope. Then the equation of the tangent is of the form y = mx + c, where c is to be so determined that the points of intersection of the curve and the tangent shall be coincident. The student may verify the following results for special cases : (4) For the parabola y^ := 4px, there is one tangent with slope m ; namely, y = mx + ^ • x V (5) For the ellipse "i + f^ = Ij there are two tangents d. D with slope m ; namely, y = mx ± Va^m^ + b^. y2 y2 (6) For the hyperbola — — ui ^= !> there are two tangents a o with slope m ; namely, y = mx ± Va^m^ — b^ Ex. Find the tangents to the conic 2x2 + 3y2 — 4x-|-6y-|-4=0, which are parallel to the line 4x — 2y + 3 == 0. Here, m = 2. Hence the tangent is y = 2x + c. Substituting in the equation of the curve, we have, for the abscissas of the points of intersection, the equation 14x2 + (12c + 8) X + 3c2 + 6c + 4 =0. This equation has equal roots, if (12c + 8)2 = 56 (3c2 + 6c + 4), whence c = - 3 ± i V21. Hence the tangents are y=:2x— 3±i V21. TANGENT, NORMAL, AND POLAR. 165 77. The Normal. The normal to a curve at any point is the straight line pass- ing through that point and perpendicular to the tangent at that point. To find the equation of the normal, it is most convenient first to find the equation of the tangent. Then, if m is the slope of the tangent, the equation of the normal is y-yi = --(x-xi)- The following are special cases : (1) For the parabola y- =: 4px, the normal is yi / \ y-yi = -2^(x-xO. 9 9 X" y (2) For the ellipse ~^ + f^ = !> the normal is (3) For the hyperbola -^ — ^^ = 1, the normal is Ex. Find the equation of tlie normal to the parabola x2 + 2x + 3y - 9 = 0, at the point (— 2, 3). The equation of the tangent at this point is (- 2) X + (X — 2) + I (y + 3) - 9 = 0, or 2x - 3y + 13 = 0. Therefore, the equation of the normal is y-3 = - j!(x + 2), or 3x + 2y = 0. If the slope of the normal is given, the normal may be found as in the following example. Let it be required to find the normals to the conic 2x2 + 3y2 _ 4x + 6y + 4 = 0, which are parallel to 3x — 2y + = 0. 166 PLANE ANALYTIC GEOMETRY. Calling m the slope of the normal, we have, at once, m = |. If we can find the point where the normal cuts the curve, we can solve the problem. Let us call (xi, yi) this point. Then the equation of the tangent is, by [34], § 76, 2xix + 3yiy - 2 (x + Xi) + 3 (y + yi) + 4 = 0, whence, if mi is the slope of the tangent, _ 2x1-2 T, , 1 mi- — q., I q - But mi = ; oyi + 3 m 2xi - 2 _ 2 •'■ 3yi + 3~3' or xi — yi = 2. But (xi, yi) is on the conic ; hence 2xi2 + 3yi2 - 4xi + 6yi + 4 = 0. These two equations are sufficient to determine (xi, yi). We find xi = l±^V5, yi = - 1 ± ^ Vs. Hence there are two normals, their equations being y + l:F^V5=|(x-l:F| VB), or y = |x-|±xVV5. 78. Pole and Polar, If, through a fixed iwint P^ any number of secants he drawn to a conic section, and tangents to the latter he drawn at the points of intersection of each secant with the co7iic, then each pair of tangents will intersect upon a straight line, called the polar of Pj. Reciprocally Pi is called the pole of the sti'aight line. Let Ax2+ By2 + 2 Gx + 2 Fy + C = be the equation of the conic section, and (xi, yi) the coordi- nates of the point Pi. Then if Pi lies on the conic section, it follows from the definition that the polar of Pi is simply the tangent at Pi, as shown in Fig. 72. In this case, then, the equation of the polar is Axix + Byiy + G (x + Xi) + F(y + y,) + C = 0. The above case is unimportant ; the polar has real im- portance when the point Pi is not on the conic. In this TANGENT, NORMAL, AND POLAR. 167 FlO. 72. case we may draw for the ellipse either one of Figs. 73 and 74, and similar figures for the parabola and the hyperbola. Fig. 73. Take any secant PiN intersecting the conic at N and M, and let (xa, y^) and (xs, ys) be the unknown coordinates of N and M. The equations of the tangents at N and M are, respectively, by [34], § 76, Ax.,x + By,y + G (x + x,) + F (y + y,) + C = 0, (1) Ax,x + By,y + G (x + X3) + F (y + y,) + C - 0. (2) 168 PLANE ANALYTIC GEOMETRY. Fig. 74. The coordinates of R, the point of intersection of these tangents, satisfy both of these equations and hence satisfy their difference A(x2-x3)x + B(y2-y3)y + G(x2-X3) + F(y2-y3) = 0. (3) This equation contains the unknown coordinates (xg, y^) and (xg, y3),but does not contain the given coordinates (xi, yi), a defect which may be remedied by remembering that Pj, M, and N lie upon a straight line, and that therefore X 2 '^~ X 3 X J ~~~* X 2 y2 — ys yi — 72' Substituting in (3), we have A (xi - X2) x + B (yi - y^) y + G (x^ - x^) + F (y^ - y^) = 0, and adding (1), we have, finally, AxiX + Byiy + G (x + Xi) + F (y + yi) + C = 0. [35] This is an equation satisfied by the coordinates of R. It contains the coordinates of the given point Pi, and not the coordinates of M and N. Hence the result would be un- changed if we had used any other secant passing through Pj. Moreover, equation [35] is an equation of the first degree, TANGENT, NORMAL, AND POLAR. 169 and therefore represents a straight line. Hence the theorem at the beginning of the article is true. It is to be noticed that the form of the result is the same as that obtained when the point Pi lay on the conic. In other words, the equation of the 'polar is identical in form xvith the equation of the tangent. Hence, if the point (xj, yi) is given and we form equa- tion [35], it will not be known whether this is the tangent or the polar until we examine the position of the point (xj, yj). If (xi, yi) lies on the conic we have the tangent ; otherwise, we have the polar. Ex. Given the ellipse 9^4 ' and the point (4, — 5). The coordinates (4, — 5) do not satisfy the equation ,,+i = '- Therefore, the line is the polar of the point (4, — 5) 9 4 4x 5y 9 4=1 79. To find the Tangents from a Point not on the Conic Section. Let Pi(xi, yi) be a point not on the conic section. FlO. 75. 170 PLANE ANALYTIC GEOMETRY. Assuming that it is possible to draw tangents from Pi to the conic, let us call (xj, y^ the coordinates of the point of tan- gency, Q, of any tangent. If it is not possible to draw the tangents, that will be shown in the result by the fact that X2 and y2 have imaginary values. The equation of the tangent QPi is, by [34], § 76, Ax.,x+ By,y + G (x + x,) + F (y + y^) + C = 0. (1) Since this tangent passes through the point P^, the coordi- nates (xi, yi) satisfy the equation (1), and we have Ax^xi + By^yi + G (x^ + x,) + F (y^ + y^) + C = 0. (2) Now consider the polar of Pj. Its equation is, by [35], Axix + Byiy + G (x + xO+ F (y + yO + C = 0, (3) and equation (2) shows that the coordinates X2, yo satisfy (3), that is, that the point Q lies upon the polar of Pj. We see, therefore, that tangents from a point Pi to a conic touch the conic in the points where the latter is cut by the polar of Pi- Hence, to find the tangents from a 2>oint Pi to a conic, proceed as folio ivs : First, find the polar of Pi. llien, find the 2Joi?its of inter- section of the polar and the conic. Fhially, find the tangents at these points. Ex. Find the equations of the tangents to the ellipse 4x2 + y2 + 24x - 2y + 17 = 0, from the point (— |, ^-f). The polar of the point is 2x + y — 1 = 0. Solving the two equations, we find that the polar intersects the curve in the points (— 1, 3) and (— 2, 5). The tangents at these points are 4x + y + 1 =0, and X + y — 3 = 0. Since the polar is a straight line, it may have any one of three relations to the curve. TANGENT, NORMAL, AND POLAR. 171 (1) It may be tangent to it. (2) It may cut it in two real points. (3) It may not intersect it at all. The first case is possible only when the pole is itself on the conic ; the last two occur when the pole is not on the conic. In case (2), by applying the rule, we find two tangents from the pole. The polar is then the chord connecting the points of contact of the tangents from the pole or tlie so-called chord of contact. In case (3), by the application of the rule, we see that no real tangents can be drawn. From this follows the theorem : From a 2Joint not on a conic, either two or no tangents can he drawn to the conic. We may accordingly separate the points of a plane into two classes, those outside and those inside the conic ; the latter being those from which no tangent can be drawn. And now, summing up, we may say : According as a point Pi lies (1) without, (2) on, (3) within a conic, the polar of P^ is (1) the chord of contact, (2) the tangent, (3) a straight line not intersecting the conic. If it is questioned whether a point lies within or without a conic, the question may be answered by determining the posi- tion of the polar; unless, as is often the case, the question may be more quickly solved by inspection. 80. Properties of Poles and Polars. Two of the most important properties of polars are the fol- lowing : 1. If Pa lies upon the polar of Pi, then reciprocall// Pi lies tipon the polar of P^- 17,2 PLANE ANALYTIC GEOMETRY. Let the coordinates of Pi be (xi, yj), those of Pa be (xa, yo)- The polar of Pi is Axix + Byiy + G (x + x^) + F (y + y,) + C = 0, (1) and that of Pj is Ax,x+By,y + G (x + x,) + F (y + y,) + C = 0. (2) The condition that Pj shoukl lie on (1) is Axix,+ Byiy,+ G (x^ + x^) + F (y, + yO + C = ; but this is exactly the condition that Pi should lie on (2). This establishes the theorem. This theorem may be other- wise stated as follows : The polars of all i^oints lying in the same straight line pass through the same jjoint, namely the pole of that straight line. With the aid of this theorem, theorems similar to those in § 60 may be readily established and many others, the discus- sion of which lies outside the range of this book. 2. Any secant passing through Pi is divided harmonically by the conic and the j^olar of Pi. Fig. 76. Let PiN be any secant through Pi, and let M and N be the points where this line is cut by the conic, and Q the point where it is cut by the polar of Pi. We are to prove that the line MN is divided harmonically; i.e., that it is divided externally and internally in the same ratio. We must have TANGENT, NORMAL, AND POLAR. 173 then 1^=^. PiN QN' 1 f 11 t3^ 2.P1N.P1M whence follows P,Q = ^^ttt , ^ . - • Let us take the point Pi as the origin of coordinates and consider the axes of coordinates drawn parallel to the axes of the conic. Then the equation of the conic is Ax2+By2 + 2Gx + 2 Fy + C = 0, (1) and the equation of the polar of Pi, found by placing Xi = and yi = in [35], is Gx+Fy + C=:0. (2) Let us now introduce polar coordinates, using [7], § 10. Equations (1) and (2) become resj)ectively r=^ (A cos^ ^ + B sin^ ^) + r (2 G cos ^ + 2 F sin ^) + C = 0, (3) and r (G cos e + F sin ^) = — C. (4) The* roots of (3) are the distances PiM and PiN. Hence, by the theory of quadratic equations, C P,N. PiM = and P^N + PiM=- A cos='^+ B sin^^' 2 G cos e + 2 F sin 6 A cos^ ^ + B sin^ 6 The value of r in (4) is the distance PiQ ; hence p 0=r ~ ^ '^ G cos ^ + F sin e These results give, at once, pp,^ 2P.N.P.M '^ PiN + PiM' as was to be proved. This property is sometimes taken as the definition of the polar. 174 PLANE ANALYTIC GEOMETBY. EXAMPLES. Find the tangents and the normals to each of the following curves at the given points : X 1. 9x^ + 5y2 + 36x + 20y + ll=0; (-2,1). 2. 9x2 + 4y- + 36x + 24y = 0; (0,0). 3. X- - 3y2 + 6x + 12y - 9 = ; (0, 3). 4. y2 - 6y - 8x - 31 = ; (- 3, - 1). 5. y^ = 9x; (1,3). 6. x2 + 4y2 = 8; (2,-1). 7. 2x2-y=' = 14; (3,-2). \^^ 8. Find the tangent to the parabola y^ = 4x, which makes an angle of 45° with the axis of x. 9. Find the tangents to the conic 2x2 + 3y^ + 4x-6y-l=0, which make an angle tan~^ \ with the axis of x. 10. Find the tangents to the conic 4x^ + y- == 4, which are parallel to the straight line 2x — 4y 4-5=0. 11. Find the tangents to the conic x" — 3y^ — 4x + 3 =: 0, which are perpendicular to 3x + 2y — 1 = 0. 12. Find the tangents to the conic 7x^+8y^ = 56, which make an angle tan~^ 3 with the line x + y -f- 1 ^ 0. 13. Find a normal to the conic x^ + 2y2 + 5y + 2 = 0, mak- ing an angle tan~^ \ with the axis of x. 14. Find a normal to the conic 3x- — x + y = 0, parallel to the line 3x + 2y + 1 = 0. 15. Find a normal to the conic 4y- — x- -f 2x — 6y -f 20 = 0, perpendicular to the line 3y + 3x + 2 = 0. Find the polars of each of the following points with respect to the given conic, and also the points in which the polar intersects the conic : 16. (-3, 4); 4y2 + 16x-4y-3=0. 17. (0, 0) ; 3x2 _ 2y2 _ 6x - 8y - 6 = 0. TANGENT, NORMAL, AND POLAR. 175 18. (-|,i);| + f = l. 19. (3,-1); 2x- + y--8x + Gy + 11=0. Find the tangents drawn through the given points to the following conies : 20. 7x2 + 3y2 + 14x - 12y - 36 = ; ^4^ _y_^_ 21. y^=|x; (-1,-1). 22. 2x- - 6f = 21 ; (5 + V3, 2 + V3.) 23. y2 + 5x — lOy — 35 = 0; (3,1). 24. What is the slope of the tangent to the conic 3x2 -18x + 7y + 13 = through the point (1, 4) ? 25. Find the angle between the ellipse 4x2-|-9y2 = 13 and the circle x^ + y^ ^ 2 at any one of their points of intersec- tion. 26. At what angle does the straight line 2x — y + 3 = intersect the conic 3x- -(- y^ -|- 6x — 4y — 14 = ? x" v" X v" 27. Show that the curves — -. + ;^ = 1 and -; =^=1 cut lb i 4 5 each other at right angles and are confocal. 28. At what angles do the conies x^ — 3y2 — 4x — 6y — 5 = and x^ — 4x — 9y — 5 = intersect ? 29. Using the method of § 62, find a conic through the points of intersection of ^^7^ + ^ = 1 and - — j = 1, and also through the point (0, 6). What is the conic ? 30. Use the method of § 62 to find the equation of a para- bola which passes through the intersections of the hyperbola x^ — y2 =: 1 and the circle x-' + y^ — 4x — 1 == 0. 31. By the method of § 76, show that the tangent to the curve y = x^ at the point (xj, yj is y -|- 2yi = 3xi-x. 176 PLANE ANALYTIC GEOMETKY. 32. By the method of § 76, show that the tangent to y^ = x'' at the point (xi, yi) is 2yiy + yj^ = 3xf x. 33. By the method of § 76, show that the tangent to xy = 4 at the point (xi, yi) is Xiy + yix = 8. 34. Solve Ex. 33 by transforming the given equation to new axes bisecting the angles between the old axes, finding the tangent, and transforming back. 35. Show that for any conic section the polar of the focus is the directrix. 36. Where is the polar of the centre of an ellipse or hyper- bola with respect to that curve ? 37. If mi is the slope of the polar of a point P^ with x^ y^ respect to an ellipse —^-\-^^ = l and mj is the slope of the line joining Pj to the centre, show that mim2 =^ \- Find the SI similar relation for the hyperbola. CHAPTER VIII. THE PARABOLA : y2 = 4px. 81. We have seen in § 67 that, if we take the axis of the parabola as the axis of x and the line through the vertex per- pendicular to the axis of the parabola as the axis of y, the equation of the parabola assumes its simplest form, y- = 4px. This is the form of the equation we shall use in studying the parabola in this chapter. When the equation is in this form, the focus is tlie point (p, 0), and the equations of the axis and the directrix are respectively y = and x = — p. Also, according to § 78, the polar for the point (xj, yi) is yiy = 2p (x + Xi)/and the tan- gent at the point (xj, yi), if (xj, yi) is on the curve, is yiy = 2p (x + Xi). If, however, the tangent is determined by its slope m, its equation, by (4), § 76, is y ^ mx + 82. Constructions. Since the eccentricity of the parabola is always unity, it follows that every point of the parabola is equally distant from the focus and tlie directrix. From this fact we derive the two following methods of construction : (1) By compasses. Let F (Fig. 77) be the focus and DD' and FX be the directrix and the axis, respectively, of the parabola to be constructed. The vertex, V, will be on the line FX halfway from F to DD'. To determine other points of the parabola, draw a line, as AB, parallel to DD' and to the right of V. With F as a 178 PLANE ANALYTIC GEOMETRY. Fig. 77. centre and a radius equal to the distance from DD' to AB describe an arc of a circle. This arc will intersect AB at two points P' and P", which will be points of the parabola. Proceeding in this way we may determine as many "joints of the curve as we wish. (2) By a mechanical device. As before, F (Fig. 78) shall be the focus and DD'the direc- trix. ACB is a triangle, right angled at C, which moves with CB always coinciding with DD'. A thread of length equal to AC is fastened to the triangle at A and to the plane at F. Then, as the triangle slides along DD', the point of a pencil which always touches AC and keeps the thread taut will describe the parabola. THE I'AKABOLA. 179 Fig. 78. 83. Angular Properties of Tangent and Normal. 1. The tancjent viakes equal angles with the focal radius drawn to the iwint of contact and the axis of tlie -parabola. In Fig. 79, let PjT be tangent at the point Pi (xi, yi) to the parabola y^ ^=- 4px, of which the focus is F. We have then to prove that Z FPiT = Z FTPj. Now FPi= ^'^^^E^A^^iE^ = Vxi^— 2pxi + p- + yi^ = Vxi^ — 2pxi + p2 + 4pxi, (for yi" = 4pxi) = Vxi^ + 2pxi -I- p2 = Xi + p. 180 PLANE ANALYTIC GEOMETRY. Fig. 79. And TF = OF - OT = p — (— xi) = p + xi, for OT is the intercept of the tangent on OX and is found to be — Xi. Therefore, since FPi = TF, the triangle TFPi is isosceles and the angles FPiT and FTPi, opposite the equal sides, are equal. 2. The normal makes equal angles with the focal radius drawn to the jmmt tvhere the normal intersects the parabola and a line through the same point parallel to the axis of the jyarahola. If, in Fig. 79, PiN is the normal to the parabola at Pi and KKi is drawn through Pj parallel to OX, we are to prove ^ FPiN =:Z NPiKi, or that PiN bisects the angle FPiKi. Now ZKPiT = ZFTPi, since they are the alternate- interior angles of two parallel lines cut by a secant ; and we have just proved /_ FTPj = Z. FP^T. Therefore Z KPiT = Z FPiT, and TPj bisects the angle KPiF. It follows that PiN bisects the angle FPjKi, for a line, through the point of intersection of two lines, perpendicular THE PARABOLA. 181 to the bisector of one pair of the vertical angles formed by the two lines bisects the other pair. 84. Perpendicular Tangents meet on the Directrix. Fig. 80. As y^ = 4px is the equation of the parabola, the equation of its directrix is x = — p. Let the equation of any tangent be y = mx + £- (1) Then, if (2) is the tangent perpendicular to (1), its equation n 1 will be y = m'x + — „ where m' = , m' or y = pm. (2) If tangents (1) and (2) meet at (xj, y,), we find, by solving (1) and (2) simultaneously, that Xi = — p and yi = -^ — pm, values which satisfy the equation of the directrix. 182 PLANE ANALYTIC GEOMETRY. As (1) and (2) are any pair of perpendicular tangents, it follows that any two perpendicular tangents meet on the directrix, and the directrix is the locus of the point of inter- section of perpendicular tangents. 85. Perpendicular to Focal Chord. The perpendicular to any focal chord at the focus meets the tangent at the extremity of the focal chord on the directrix. Fig. 81. f = Apx (1) is the parabola, x = -p (2) is its directrix, and F (p, 0) is its focus. If Pi(xi, Yi) is any point of the parabola, the equation of the tangent through Pj will be, by (1), § 76, y,y = 2p(x + x0. ' (3) THE PARABOLA. 183 By [14], § 34, the equation of FPj will be y^U2-'-:ZJh ovy-y, = -^^(x-xO. (4) yi — xj — p Since FR is perpendicular to FPj its equation will be, by [18], § 36, y-0 = P^(x-p). (5) yi Solving (3) and (5) simultaneously, we find the coordinates of R to be — p and '^-^-^ — ^, so that R is a point of the yi directrix, as we set out to prove. This theorem gives us an easy method of constructing the tangent to the parabola when the point of contact is given. For we have only to draw the focal chord to the point, then draw through the focus a line perpendicular to this focal chord, and from the point where this perpendicular inter- sects the directrix draw a line to the given point of contact. This last line will be the required tangent. 86. Perpendicular from Focus to Tangent. The perpendicular from the focus to any tangent ivill meet it on the tangent at the vertex of the parabola, and will meet the directrix at the point where the latter is intersected by a line from the point of contact parallel to the axis of the parabola. y2 = 4px (1) is the parabola, x = -p (2) is its directrix, and F (p, 0) is the focus. The equation of the tangent at the vertex, which is the point (0, 0), will be 0y = 2p(x + 0) or x = 0.' (3) 184 PLANE ANALYTIC GEOMETRY. Fig. 82. If Pi(xi, yi) is any point of the parabola, the equation of the tangent through P^ will be yiy = 2p(x + xO. (4) Since FRS is perpendicular to the tangent, its equation will be ^ = -2p^'^-P> (5) If R is the point of intersection of (4) and (5), we find R to be the point ( 0, ^ ) and hence to be a point of (3), as was to be proved. If S is the point of intersection of (5) and (2), we find S to be the point ( — p, yi). Therefore, S is at the same distance from OX as is P^, and hence is on the line through Pj parallel to OX. THE PARABOLA. 185 It is readily showu that tlie distances RS and RF are both V. yi equal to \/p'^ + ^, so the tangent is perpendicular to FS at its middle point. And here we find a method of constructing the tangents to a parabola, passing through a point not on the parabola. Y Fig. 83. Let Pj be the point from which the tangents are to be drawn, and let F be the focus of the parabola. With P^ as a centre and a radius equal to PiF describe a circle, denoting by $2 and S3 the points at which it meets the directrix. From $2 and S3 draw lines parallel to OX, which will intersect the parabola at P2 and P3 respectively. Then P1P2 is one of the required tangents. For P1S2 = PiF, being radii of the same circle; and S2P2^FP2, since the eccentricity of a parabola is unity. Therefore, P1P2 is per- pendicular to FS2 at its middle point and hence coincides witli the tangent at P2. Similarly it can be proved that PiPj is a tangent to the ])arabola. 186 PLANE ANALYTIC GEOMETRY. Note. If Pi is so situated that tlie circle with radius PiF does not cut the directrix, the above construction fails. But it is evident that this case can occur only when Pi is a point within the parabola, and from such a point no tangent can be drawn to the parabola. 87. Diameter. The definition of a diameter which was given in § 61 for the circle is the definition of the diameter for all the conies. Hence the diameter of the 'parabola is the locus of the middle jyoints of a system of parallel chords. Let y2 = 4px (1) be the parabola, and let m be the common slope of the parallel chords. Then y = mx + b (2) will represent one of the chords, as PiPo, which intersects the parabola at the points Pi(xi, yi) and P2(x2, ys)- If THE PARABOLA. 187 PsC^s; ys) is the middle point of this chord, Substituting for x from (2) into (1), we have ^ ' y m /-iPy+ie-^=0, (4) the roots of which equation are yi and yj. .••yi + y. = ^, (5) whence, substituting in (3), we have y. = ?^. (6) As this is an expression for a coordinate of the middle point of one of the chords, independent of b and therefore the same for all chords, y = — [36] is an equation satisfied by the coordinates of the middle point of every one of the system of chords, i.e., by the coordinates of every point of the diameter, and is accordingly the equa- tion of the diameter. Note. The work here differs from that of § 61, in that ys alone is found, while there we found xj and ys both. The aim in both cases has been the same, — to find an equation in xs and ys which shall be inde- pendent of b, and hence be true for the middle point of every chord of the system. In the case of the circle the values of both xs and ys involved b, so that the only way in which we could obtain the desired equation was by eliminating b from the equations containing xs and ys. In the case of the parabola ys does not involve b, so that we have immediately an equation of the right form, and hence are not obliged to find Xs. 188 PLANE ANALYTIC GEOMETRY. It is to be noted that this diameter is parallel to the axis of the parabola and hence that all diameters of the parabola are parallel. We can now state the theorems of §§ 83, 86, in much better forms, as follows : The tangent makes equal angles with the focal chord and the diameter drawn through the point of contact. The normal makes equal angles with the focal chord and the diameter drawn through the point where the normal inter- sects the parabola. The perpendicular from the focus to any tangent will meet it on the tangent at the vertex of the parabola, and will meet the directrix at the point where the directrix is cut by the diameter drawn through the point of contact. Ex. 1. Find the diameter of the parabola y- = — 7x, bisecting chords parallel to the line x — y + 2 = 0. For the chords m = 1, and p = — |. 2(— 1) .-. the diameter is y = •* , or 2y + 7 = 0. Ex. 2. A diameter of the parabola y- = 6x passes through the point (2, — 1). What is its equation, and what is the slope of the chords bisected by it ? 3 If m is the slope of the chords bisected, the diameter is y = — • But m (2, — 1) is a point of the diameter. \, 3 3 .•. — 1 = — and m = — 3, and the diameter is y = — -, or y = — 1. m — 3 We could have written down this last immediately, for the diameter is parallel to OX, so that, if one of its points is distant — 1 from OX, all its points are distant — 1 from OX, and its equation is y = — 1. 88. Properties of the Diameter. The tangent at the extremitu of a diameter is imrallel to the chords bisected by the diameter. 2p If the equation of the diameter is y =:— ^, it will intersect m THE PARABOLA. 189 the parabola at the point (2' )• The tangent to the parabola at this point is — y = 2p ( x + 2 ) or y = mx + ^, the slope of which is m, or the slope of the chords bisected by the diameter. The, polar of any point of a diameter is pjarallel to the chords bisected by the diameter. "p\ -P x,, — ) is any point of the diameter y = --. my m Its polar will be 2p — !- y = 2p (x -}- Xi) or y = mx -|- inx, , m which line has its slope equal to m, the slope of the chords bisected by the diameter. It follows that the pole of every chord of the parabola is on the diameter wliich bisects the chord. As a chord of con- tact is a special case of a polar, the corresponding tangents meeting at the pole, the tangents at the extremities of any chord meet on the diameter ^vhich bisects the chord. 89. Parabola with Focus at the Origin. This form of equation of the parabola is readily found by transforming the equation y^ = 4px to F (p, 0) as a new origin, the directions of the axes not being changed. The formulas of transformation are x = x' + p and y = y', so tliat the trans- formed equation is, after the primes are dropped, y2 = 4px+4pl (1) It is evident that y^ = — 4piX-}--lpi^ is the equation of a parabola having its focus at the origin, but extending indefi- nitely toward the left, as it is the form (1) assumes if the sign of p is changed, p and pi being positive. 190 PLANE ANALYTIC GEOMETRY. Fig. 85. Any two parabolas having the same focus are said to be confocal, and the above is a special case of confocal parabolas. Moreover, if we define the angle between two intersecting curves as the angle between their respective tangents at their point of intersection, two such confocal parabolas as above will intersect at right angles. Solving y^ = 4px + 4p'^ and y- = — 4piX + 4pi2, we find these parabolas meet at the points (pi - p, 2 Vpip) and (pi — p, - 2 Vpip). The respective tangents at the point (pi — p, 2 Vpip) are by [34], § 76, 2 Vpip y = 2p (x + pi - p) + 4p2, and 2 Vpipy = — 2pi (x + pi — p) + 4pi2, THE PARABOLA. 191 are or Vpipy = px + ppi+ p2, and Vpipy = — pix + pi-+ pip. Their slopes are respectively -i/— and — \/ 5 so they perpendicular to each other, and hence the parabolas cross at right angles. In like manner the parabolas may be shown to cross at right angles at the other point of intersection, (Pi- p, -2Vpip). 90. Parabola Referred to a Diameter and the Correspond- ing Tangent as Axes. In this article we will find the equation of the parabola "with respect to a diameter as the axis of x and the tangent at the extremity of the diameter as the axis of y. Y' Fig. 86. 192 PLANE ANALYTIC GEOMETRY. 2p If the equation of the diameter O'X' is y = -^, the equation of the tangent O'Y' will be y — mx +— and the coordinates of 0' will be (^^o,— \ V m'' my If we take first a set of axes O'X' and O'Y" parallel to the original axes, we shall transform the equation y^ := 4px by the P 2p formulas x = -^ + x" and y = -^ + y", obtaining the equation ■' m To transform this last equation from axes O'X' and O'Y" to O'X' and O'Y' the formulas will be y' y'm for in formulas [21], § 45, ^ = and 6' = tau"^ m. The result of this substitution will be By[l],§3, FO' = P-(i±^\ Therefore if we denote FO' by p', our equation becomes, after dropping the primes, y^ = 4p'x. [37] It is to be noted that y'^ = 4px ahvays represents a parabola, the X axis being a diameter, the y axis a tangent, and the distance of the focus from the origin being one fourth the coefficient of x. THE PARABOLA. 193 91. Parabola in Polar Coordinates. For the polar equation of the parabola we will take the axis of the parabola as the initial line and the focus as the origin. In Fig. 79 draw PiM perpendicular to OX ; then FPi = p + Xi = p + OM = p + OF+ FM=2p+ FM. But in polar coordinates FPj = r and FM = r cos $, and we have r = 2p + r cos 6, whence we derive as our required equation of the parabola „_ 2p 1 — cos 6 EXAMPLES. [38] 1. Find the length of the chord of the parabola y^ + 2x=0 cut from the line 4x — 3y -|- 2 = 0. 2. If the focal distance of a point on a parabola y^ = 4px is n times the latus rectum of the parabola, what are its coor- dinates ? 3. Two straight lines are drawn throvigh the vertex of a parabola at right angles to each other and meeting the curve at P and Q. Show that the line PQ cuts the axis of the parabola in a fixed point. 4- Yi, y25 and yg are the ordinates of three points of the parabola y^ = 4px. Prove that the area of the triangle formed by joining them in succession is grCyi-ys) (y^-ya) (ys-yO- 5. Taking the definitions of subtangent and subnormal given for the circle, prove : (a) That the subtangent of any point of a parabola is twice the projection on the axis of the parabola of the line joining the point to the vertex. 194 PLANE ANALYTIC GEOMETRY. (b) That the subnormal always equals the distance from the focus to the directrix. 6. Find the equations of the tangent and the normal to the parabola y^ = 9x which pass through (4, 6). 7. Find the equation of a tangent to the parabola y^^ 6x which passes through (2, 4). v>^ 8. Find the equation of a tangent to the parabola y^ = 9x /^ perpendicular to the line 2x + y — 2 = 0. 9. A tangent to the parabola y^ ^ 7x makes an angle of 60° with the axis of the parabola. What are the coordinates of its point of contact ? 10. Find a normal to the parabola y^ = 4x which shall pass through (2, — 8). 11. Find a normal to the parabola y'-=;9x parallel to the line X + y — 1 = 0. 12. What are the slopes of tlie tangents drawn to the parabola y^ = x from the point ( — 1,2)? 13. Find the tangents and normals at tlie extremities of the latus rectum of the parabola y^ = 4px. 14. If perpendiculars be let fall on any tangent to a para- bola from two given points on the axis which are equidistant from the focus, the difference of their squares is constant. 15. If the chord PQ is a normal to the parabola at P, and the tangents at P and Q meet at T, show that PT is bisected by the directrix. 16. Find the coordinates of the point where the normal at (xi, yi) again meets the parabola y^ = 4px, and the length of the intercepted line. 17. M P is the ordinate of a point P of the parabola y^=: 4px. A straight line is drawn parallel to the axis and bisecting M P, which cuts the curve at Q. If MQ cuts the tangent at the vertex A at T, show that AT = f M P. 18. Show that the locus of the point of intersection of two tangents to a parabola, the ordinates of the points of contact of which are in a constant ratio, is a parabola. THE PARABOLA. 195 19. Show that perpendicular normals to the parabola y^ = 4px intersect on the curve y^ = px — 3p^. 20. Prove that the locus of the point of intersection of two tangents to a parabola is a straight line if the product of their slopes is constant. 21. Prove that the ordinate of the point of intersection of two tangents to a parabola is the arithmetical mean between the ordinates of the points of contact of the tangents. 22. If P, Q, and R are three points on a parabola, the ordi- nates of which are in geometrical progression, show that the tangents at P and R will meet on the ordinate of Q. 23. Prove that the area of the triangle formed by the axis of a parabola and the tangent and the normal of any point, is the same as the area of the rectangle forined by the directrix, the axis of the parabola, and the two lines through the point respectively i)arallel to the directrix and the axis. 24. Find the equation of the line through the vertex and the upper extremity of the latus rectum of the parabola y^ = 4px. 25. How far are the tangents at the extremities of the latus rectum of the parabola y^ + 3x := from the vertex of the parabola ? 26. The tangent at any point of the parabola y^ = 4px will meet the directrix and the latus rectum produced in two points equidistant from the focus. 27. Show that the normal at one extremity of the latus rectum of a parabola is parallel to the tangent at the other extremity of the latus rectum. 28. The normal at L, the upper end of the latus rectum, meets the parabola again at P. Show that the diameter on which the tangents at L and P intersect passes through the other extremity of the latus rectiim. 29. Show that the tangents at tlie ends of the latus rectum are twice as far from the focus as from the vertex. 196 PLANE ANALYTIC GEOMETRY. 30. Find the equation of the circle through the vertex, the focus, and the upper end of the latus rectum of the para- bola y2+ 12x =:0. 31. Find the equation of the circle through the vertex and the ends of the latus rectum of the parabola y^ = 4px. 32. Find the length of the perpendicular from the focus of the parabola y'^ = 4px to the tangent at any point (xi, yi) in terms of Xi and p. 33. Prove that the line joining the focus to the intersection of two tangents bisects the angle which their points of con- tact subtend at the focus. 34. If tangents are drawn at the ends of two focal radii, show that the angle between the radii is double the angle between the tangents. 35. Find the equation of the circle circumscribing the triangle formed by the axis of the parabola y2 = 3x, the tangent at the point (i-, 1), and the focal radius drawn to that point. 36. Show that any circle described on a focal chord as diameter is tangent to the directrix of the parabola. 37. Show that the circle described on any focal radius as diameter is tangent to the tangent at the vertex of the parabola. 38. A circle has its centre at the vertex A of a parabola y2=z4px, and the diameter of the circle is 3 AF. Show that the common chord bisects AF. 39. A diameter of the parabola 3y^ = 5x bisects chords per- pendicular to the line x + 3y + 1 == 0. What is its equation ? 40. A diameter of the parabola y^ + 7x = passes through the point (2, 1). Find its equation and the slope of the chords which it bisects. 41. y + 3 is a diameter of the parabola 2y^ + 3x = 0. What is the slope of the chords which it bisects ? 42. (2, — 5) is the middle point of a chord of the parabola y2 — 3x = 0. What is the equation of the chord ? THE PARABOLA. 197 43. Find the diameter of the parabola Sy'-' + 5x = which bisects chords parallel to the polar of (— 1, — 5). 44. Find the equation of the tangent to the parabola y2 — 5x = parallel to the chords bisected by the diameter y+4=:0. 45. Show that the tangents at the extremities of any chord of a parabola meet on the diameter which bisects that chord. 46. If tangents are drawn at the extremities of any focal chord of a parabola, show : (1) that the tangents will inter- sect on the directrix ; (2) that the tangents will be perpendic- ular to each other ; (3) that the straight line drawn from the point of intersection of the tangents to the focus will be per- pendicular to the focal chord. 47. Prove that the portion of the axis included between the polars of two points eqiials the projection on the axis of the line joining the points. 48. Find the locus of the feet of the perpendiculars from the focus to the normals of the parabola y'^ = 4px. 49. P is any point of a parabola, A the vertex, and through A a straight line is drawn perpendicular to the tangent at P. Find the locus of the point of intersection of this line with the diameter through P, and also the locus of the point of intersection of this line with the ordinate through P. 50. Two parabolas have the same axis, and tangents are drawn from points on the first to the second. Prove that the middle points of the chords of contact with the second lie on a parabola. 51. If the tangent to the parabola y^^4px meets the axis at T and the tangent at the vertex A at B, and the rectangle TABQ be completed, show that the locus of Q is the parabola y^ -[- px = 0. 52. A point moves so that its shortest distance from a fixed circle is equal to its distance from a fixed diameter of that circle. Find its locus. (Take tlie diameter of the circle as the axis of x and the centre of the circle at 0.) 198 PLANE ANALYTIC GEOMETRY. 53. If a straight line be drawn from the origin to any point Q of the line y = a, and if a point P be taken on this line such that its ordinate is equal to the abscissa of Q, show that the locus of P is a parabola. 54. Two equal parabolas have their axes parallel and a common tangent at their vertices, and straight lines are drawn parallel to the axes. Show that the locus of the middle points of the parts of the lines intercepted between the curves is an equal parabola. 55. Find the locus of the centre of a circle which shall always be tangent to a given circle and a given straight line. 56. Prove that the locus of the centre of a circle which always passes through a given point and is tangent to a given straight line is a parabola. p 57. If, in the triangle ABC, tan A tan iT = 2, and AB is A fixed, show that the locus of C is a parabola, the vertex of which is A and the focus B. (Take A as the origin and AB as the axis of x.) 58. The base of a triangle is 2b and the sum of the tan- gents of the angles at the base is s. Find the locus of the vertex. (Take the base of the triangle as the axis of x and the middle point of the base as origin.) 59. Find the eqiiation of the parabola referred to the tan- gents at the ends of the latus rectum as coordinate axes. CHAPTER IX. THE ELLIPSE : — + i- = 1. 92. We have seen, in § 68, chap. VI, that the equation of the ellipse takes its simplest fonu when the centre of the curve is at the origin of coordinates and the major axis of the curve coincides with the axis of x. The foci of the ellipse are then at the points (± ae, 0), and the . n n- • a , Va^ - b^ equations or the directrices are x = ± - , where e = We have further seen, § 78, chap. VII, that tlie polar of the • 1 1 • • ^iX . y,y . ^, point (xj, Yi) with respect to this curve is — - -\- \J, =" 1, the a" 0" polar coinciding with the tangent, if the point (xi, yi) is on the curve. Finally, if m is the slope of any tangent, the equation of the tangents with that slope may be written y = mx ± Va^m^ + bl [(5), § 76.] These simple equations will be used throughout this chapter. 93. To find the Foci when the Axes are given. Let the axes A A' and BB' (Fig. 87) be given both in length and position ; it is required to find the foci F and F'. By §68, CF' = ae = Va^-b-. But BC = b. Therefore, from the right triangle BCF', B F' = a. Similarly, B F = a. Hence we may give the simple rule : 200 PLANE ANALYTIC GEOMETKY. iVith either extremity of the minor axis as a centre, and with a radius equal to the semi-major axis, describe an arc of a circle. The arc will cut the major axis in the foci. 94. Focal Distances. The distances of a point on an ellipse from the foci are called its focal distances, and the lines joining the point to the foci are called the focal radii. The focal distances are easily expressed in terms of the abscissa of the point and the constants of the ellipse. For, consider the ellipse of Fig. 88, in which the directrices DM and D'M' have been drawn. Fig. 88. THE ELLIPSE. 201 By the definition of the ellipse, But MPi= DN = DC+CN, e Therefore, FPi = a + exi. Similarly, F'Pi = e. PjM', = e (CD'-CN), = ^(e-'^' = a — exi. 95. Sum of Focal Distances. From the results of the previous article follows at once a very simple property of the ellipse. For, if we add our two results, we obtain FPi+ F'Pi = 2a. That is : tJie sum of the focal distances of any jwhit on an ellipse is constant and equal to the major axis. Moreover, it is evident that this property is not true for any point not on the ellipse. This follows from the theorem of Geometry, that the sum of two lines drawn from the extremities of a straiglit line is less than the sum of two lines similarly drawn, but included by them. Hence we may say : The ellipse is the locus of a point the sum of the distances of which from two fixed points is constant. This is sometimes given as the definition of the ellipse, and the equation and properties of the ellipse deduced from it. Such a method of derivation, however, fails to bring out the relation of the ellipse to the other conic sections as shown by the definition of chapter VI. 202 PLANE ANALYTIC GEOMETRY. 96. Constructions. The last article forms the basis of two methods of con- structing an ellipse when the two axes are given, or when the foci and major axis are given. (1) By compasses. Find the foci, if they are not already given. Then take a line equal to the major axis and with compasses divide it into two parts, r and r'. With one focus as centre describe an arc of a circle of radius r, and with the other focus as centre an arc of radius r'. The intersections of the two arcs are points on the ellipse. Repeated applications of this process give as many points as are wished. (2) By a mechanical device. Find the foci by § 93, if they are not already given. Fix a pin at each focus and attach to each pin one end of a thread of length 2a. Place the point of a pencil in the loop of the thread and move it so as to keep the thread always taut. The pencil then traces an ellipse. 97. Angle between Tangent and Focal Radii. The tangent to an ellq^se snakes equal angles with the focal radii of the point of contact. Y Fig. 89. THE ELLIPSE. 203 Let TT' be a tangent, Pi(xi, yi) the point of tangency, and FPi and F'Pi the focal radii of Pi. To prove ZFPiT'^ZF'PiT. Draw the normal, PiR. By (2), § 77, its equation is Call CR its intercept on the axis of x ; then, placing y ^= 0, we find ro a-— b- CK = :; — Xi = e-Xi. a" Hence FR = FC + CR, = ae + e"Xi ; and RF'=CF'-CR, ae — e-x, Therefore, FR ae + e^xi a + exi FPj RF' ae — e% a — exj F'Pi Hence Z RPiF = Z RPiF' ; for, if a line through the vertex of a triangle divides the opposite side into segments proportional to the adjacent sides, it bisects the angle at the vertex. Therefore, finally, ZFPiT' = ZF'PiT. It may be noticed, in passing, that the value of CR shows tluit the normal does not pass through the centre of the ellipse, but cuts the major axis between the centre and the foot of the ordinate from the given point. It is, in fact, a distinct characteristic of the circle that the normal always passes through the centre. 98. Point of Intersection of a Pair of Perpendicular Tangents. Let Pi(xi, yi) be the point of intersection of a pair of per- pendicular tangents to the ellipse ; it is required to find the 204 PLANE ANALYTIC GEOMETRY. Fig. 90. locus of Pp The equation of any tangent to the ellipse may be written in the form y = mx it Va^m^ + b'^. If the tangent passes through P^, we shall have yi = mxi zt Va^m^ -\- b^, which is an equation from which the value of m may be com- puted in terms of Xi and yi. In fact, the equation reduces to the quadratic in m, (a^ — xf) m- + 2xiyim + b- — y^^ = 0, the two roots, trii and nria, of which correspond to the two tangents through P^. Now, by the theory of quadratic equations, 2xiyi iTii + rrio = — mim2 a^ — Xi b^ — Yi' a^ — x,^* THE ELLIPSE. 205 But, by hypothesis, the two tangents are perpendicular. Hence mirng =: — 1, and therefoi-e, a'^-xr~ ' or Xi^ + yi"= a^+ h'\ Hence the point Pi must lie upon the circle x^ + y- = a- + b-, and hence the theorem : The locus of the j^oint of intersection of a pair of 2)6vpendicvr lar tangents to an ellipse is a circle tvith its centre at the centre of the ellipse and its radius equal to the square root of the sum of the squares of the semi-axes of the elUjyse. 99. Perpendicular from the Focus upon any Tangent. Let y = mx ± Va^m=' + b^ (1) be any tangent to the ellipse. Then the equation of the per- pendicular from the right-liand focus upon it is, by [18], § 36, y = --(x-ae), or my -f- x = ae- (2) It is required to find the locus of the point of intersection of (1) and (2). If Xi, yi are the coordinates of the required point, then we have the two equations yi — mxi = ± Va^m''^ + b^, myi + xi = ae, out of which the parameter m must be eliminated. This elimination may be effected by squaring and adding, and then ^2 b2 substituting e^ ^= ^ — . We have, then, a (1 4- m-) Xi" + (1 + m-) yi'- = a''m- + b" + a^e", 206 PLANE ANALYTIC GEOMETRY. or (1 + m^) X,' + (1 + m^) y,' = (1 + m") a', whence, by dividing by (1 + m^), The same result would be obtained by using the left-hand focus (— ae, 0). Hence : The perpendicular from either focus upon any tangent intersects that tangent on the circle described with the centre of the ellipse as a centre and the major axis as a diameter. 100. Equation of a Diameter. We have already defined a diameter of a conic as the locus of the middle points of a system of parallel chords. To find the equation of a diameter, we proceed as in § 61. If nrii is the slope of the parallel chords, then y = rui X + c is the equation of one chord, which intersects the ellipse, we will say, in the points Pi(xi, yi) and P2(x2, yg). Y P, Fig. 91. If X3 and ya be the coordinates of the middle point of the chord, then xi -f- X2 X3 2 yi + y2 THE ELLIPSE. 207 But Xi and Xo are the roots of the equation (b^ + a-mr) x2 + 2a-miCx + a' (c- - b^) = ; so that x and hence Similarly, whence, by division, Xg a-irii This equation is satisfied by the middle point of any one of the parallel chords, since it does not involve c. Hence, y = :r- X [39] ■^ a-rrii ■- -^ is the equation of the required diameter. This equation shows that t/ie diameter is a straight line passi7i(jf through the centre of th e ellipse. In [39] iDi is the slope of the chords which the diameter bisects. If nrig is tlie slope of the diameter itself, then 4- X - 2a^miC 1 ^^2 - b-' + a^mi^' a-miC Xg — - b- + a'^rrii^' b^c ys — b'+a-mi^' y3. b^ or b- nriimo = r,. a" Out of this relation may be deduced the important fact that a diameter of an ellipse is in general not perpendicular to the chord it bisects. For, if it were, then mirrio would equal — 1. This is always true, if b = a ; but, in that case, the ellipse becomes a circle, and we are led back to the case discussed in chapter V. 208 PLANE ANALYTIC GEOMETRY. But if b is unequal to a, rnima is not equal to — 1, and the two lines are not perpendicular to each other. An exception occurs when the chords are parallel to one of the axes of the ellipse. In that case mi = or co, and the work of finding the diameter cannot be carried out exactly as above. It is, however, easy to prove that the diameter is then the axis to which the chords are perpendicular. Hence : The two axes of an ellipse are the only diameters perpen- dicular to the chords they bisect. Ex. Find a diameter of the ellipse ^ + 77 = 1, bisecting chords per- 25 9 pendicular to the line y + 3x + 7 = 0. The slope of the given line is — 3, therefore the slope of the parallel chords is ^. Substituting in the equation b2 mim2- — -^, we have im2 = - i^, or mg = - ||. . •. the diameter is y = - l|x, or 27x + 25y = 0. 101. Conjugate Diameters. We have seen that, if rrii represents the slope of a chord, the slope m2 of a diameter bisecting it is given by the relation a iDi Similarly, if nria is the slope of a chord and nrix the slope of a bisecting diameter, then b^ mi = ^ — . a^'roo Each of these conditions is equivalent to b^ mirrio = r,. a^ THE ELLIPSE. 209 Hence we may assert the proposition : If the slopes of two diameters are comiected by the relation ^1^2 = - -2' [40] then each diameter bisects all chords parallel to the other. Such diameters are called conjugate diameters. From the condition [40] follows : 1. The two axes are the only pair of conjugate diameters which are perpendicular to each other. The proof is the same as in the previous article. 2. If one of two conjugate diameters makes an acute angle with the axis of x, the other makes an obtuse angle with the axis of X. For, if mi is positive, m., is negative, since b- 171^1X12 = 7,. a- But a positive slope corresponds to an acute angle, and a negative slope to an obtuse angle. Hence the upper portions of conjugate diameters always lie on opposite sides of the minor axis ; for example, CP3 and CQ in Fig. 91. 102. The Tangent at the Extremity of any Diameter is Parallel to the Conjugate Diameter. Let P1P2 (Fig. 92) be any diameter, QiQo its conjugate diameter, and PiN the tangent at Pi. Then, if we call the coordinates of Pi, (xi, yi), the equation of CPi is and therefore its slope yi mi = ^. X, 210 PLANE ANALYTIC GEOMETRY. Fig. 92. To find the slope mo of CQi, we have, by substitution in [40], § 101, ll m2 = — b^ or The tangent at Pj is (Da __b a -Xl ^yi' x,x 3? 1 yiy. = 1, and its slope, therefore , is b^x, This is equal to mo. Hence the theorem is proved. 103. Supplemental Chords. Chords drawn from the same point on an ellipse to the extremities of any diameter are called supplemental chords. THE ELLIPSE. 211 We shall prove that diameters 2}cvallel to a pair of stqyjyle- mental chords are conjugate. To prove this we need simply to show that the slopes of the two chords satisfy the relation mim2 = — 7^- Fig. 93. Let Pi be any point on the ellipse, PoPs any diameter. Then P^Pa and P1P3 are supplemental chords. Let the coor- dinates of Pi be (xi, Yi) and those of P2, (x.,, y2). Then the coordinates of P3 are ( — X2, — ^2)^ since P3 is diametrically opj)osite to P2. Then, if mi and mg are the slopes of PiPo and P1P3, respectively, we have, by [2], § 3, yi — y2 mi — — m2 mim Xi X2 _ yi + y2 Xl + X2' ^ _yi^-y2^ Hence Xi^ — Xj" Now, since Pi and P2 are botli on the ellipse, Xi and — + a^ ^ b- a^ "^ b^ yi^_ 1, 212 PLANE ANALYTIC GEOMETRY. By subtraction, 3? ^ b^ ~"' or, otherwise written, yi^ — ^2^ _ b^ Comparing this with (1), we have, finally, b^ mitTia^ ^, a which proves the proposition. 104. Given the Extremity of any Diameter, to find the Extremities of the Conjugate Diameter. Let Pi(xi, Yi) (Fig. 92) be the extremity of a diameter P1P2 ; to find the points Qi and Q., at the extremities of the conju- gate diameters. The equation of CPi is y = ~x, (1) Xl and therefore the equation of QiQo is, by [39], § 100, y = -^- (2) a \/i This line meets the ellipse /2 -+y=i a^ ^ b^ -"' in points the abscissas of which are given by x_2 bV 2_. a^^aV "~ Solving, we have „2_ ^ yi a^i'+b^xi^' But since (xj, yi) is on the ellipse, b-xi^ + a'^yr = a'^b^. (3) THE ELLIPSE. 213 Hence, substituting in (3), „2 ^ yi whence x ^ ^r - yi. The corresponding values of y, found from (2), are b y = ±-x,. The coordinates of Qi and Q^ are, therefore, In solving numerical examples the student should follow the methods of this article and not use the results. For example, given the ellipse ^ + y" = 1 9 4 find the diameter through the point ('/, |), and the extremities of its conjugate diameter. The diameter through (y-, |) is 1 y = ti^^ '5' or y = ix, with slope mi = h. The slope m2 of the conjugate diameter is given by 2 "2 a > whence m2 = — |. The equation of the conjugate diameter is, therefore, y = - 'x- To find the intersection with the ellipse ^4.y!- 1 9 4~ ' we solve the two equations with the result 105. Lengths of Conjugate Diameters. We will denote the length of semi-diameter CPi (Fig. 92) by a', that of the semi-diameter CQi, conjugate to CPi, by b'. 214 PLANE ANALYTIC GEOMETRY. Let the coordinates of Pi be (xj, yi) ; then those of Qi are By[l],§3, Adding, a b b a a«=xr + y,», (1) b° = ^x.' + ^y.'- (2) But, since Pi is on the ellipse, a- ^ b--^ Hence a'--^ + b" = a^ + b^. We may state this result in the following theorem : The su7ii of the squares of two conjugate semi^diameters is constant and equal to the sum of the squares of the semi-axes. 106. Angle between Conjugate Diameters. Let Pi Pa and Q1Q2 (Fig- 92) be two conjugate diameters; required to find the angle ^ between them. Draw the tan- gent PiN, and the perpendicular from the centre upon it. As in the previous article, denote the length of CPi by a', and that of CQi by b'. Then, by § 102, PiN is parallel to Q1Q2 and, hence, the angle CPiN equals ^. Therefore, CN CN ^^"'^^CP^^V It remains to find the value of CN. Now the equation of the tangent PiN is THE ELLIPSE. 215 a- b" or, cleared of fractions, b'-XiX + a'Yiy = a'^b". By § 32, the length of the normal from the centre upon this line is which may be written Vb^Xi^ + aV ab zr b'^ But by (2j of the preceding article, the quantity under the radical sign is b'-. Therefore, CN=^- Hence sin ^ a'b' 107. Parallelogram on Conjugate Diameters. Consider the parallelognun GHKL formed by drawing tan- gents at the extremities of a pair of conjugate diameters, 216 PLANE ANALYTIC GEOMETRY. PiPo and QiQ,. Since (§102) GL and HK are parallel to Q1Q2 and KL and GH parallel to PiPo, and since, furthei-, C bisects PiPo and Q1Q2, it follows tliat tlie parallelograms PiCQoL, PiCQiG, P2HQ1C, and PgCQ.K are equal. Now, the area of CP1LQ2 is, by Trigonometry, CPi.CQo sin <^, = a'b' sin <^, = ab. (§ 106). Therefore, Area G K LH = 4ab. Hence : The area of the parallelogram formed by dratvlng tangents at the extremities of any pair of conjugate diameters is con- stant and equal to the rectangle formed by drawing tangents at the extremities of the axes. 108. Auxiliary Circle. Fig. 95. The circle described on the major axis of an ellipse as a diameter is called the major auxiliary circle of the ellipse. If, THE ELLIPSE. 217 now, any ordinate N Pi is prolonged until it cuts the auxiliary circle in a point R^, then P^ and Ri are called corresponding points. It is evident that the abscissas of corresponding points are the same, but the ordinates differ. There is, however, a very simple relation between the ordinates NPi and NRj, which may be found as follows. Call the coordinates of Pi (xi, yi) and those of Ri (xj, yg). Then, since Pi is on the ellipse a^ ^ b^ ' b we have Vi ^ ^ Va''' a and since Ri is on the circle x' + y- = a^ we have ya = Va^' — Xi" ; whence follows ^ = -. y, a That is, the ordinate of any 2^oint on the elVqjse is to the 07'di- nate of the conxspondiny p)o^^^^ ^f ^^^^ major auxiliary circle as b:a. 109. Eccentric Angle. I'he eccentric angle of a point on an ellipse is the angle made tvith the major axis by the line joining the centre to the corre- sponding point of the auxiliary circle. In Fig. 95, then, NCRi is the eccentric angle of the point Pi. The coordinates of a point are readily expressed in terms of its eccentric angle. For, in tlie right triangle NCRi, de- noting the Z NCRi l^>y ' we have CN^CRiCos <^. But CN = xi and CRi = a. .*. Xj = a cos <^. 218 Also, But, by § 108, PLANE ANALYTIC GEOMETRY. NRi=CRi sin <^ = a sin cos cf>' But mim^ = - K., by [40], § 101. 3," Therefore, sin <^ sin <^ _ _ cos (ft cos cji' or cos c}> cos ' + sin sin ' = 0. This may be written, by Trigonometry, cos (<^' — <^) = 0, and, therefore, ^' — <^ = -. Prom this follows the theorem : The eccentric angles of the extremities of a pair of conjugate diameters differ hy 90°. This proposition may be used to prove the propositions of §§ 104 and 105. For example (§ 104), if Xi, yi are given, we find by [41] 220 PLANE ANALYTIC GEOMETRY. Xi COS = P-- Then for Qi, X2 = a cos cji' ^ a cos (^ + 75) = — a sin ^ = — - yi y2 = b sin <^' ^ b sin [ ^ + ^ ] = b cos <^ = - Xi. Similarly, for Q^, X3 = a cos ^" = a cos ( (j> — -\ =^ a sin (f> = - yi, yg = b sin " := b sin ( (f> — -j= — b cos , b'- = xo^ + y2^ = a^ sin^ <^ + b" cos^ <^. By adding, remembering that sin- (f> -\- cos'-' ^ =: 1, we get a'2 + b'2 == a- + b-. 111. Ellipse referred to Conjugate Diameters as Axes. Let it be required to transform the equation of the ellipse a- b^ from the old axes OX and OY to new ones OX' and OY', the latter being a pair of conjugate diameters of the ellipse. Since the transformation is from rectangular axes to oblique axes having the same origin, we have, by [21], § 45, the formulas of transformation, x = x' cos ^ + y' cos 6', y ^ x' sin 6 -\- y' sin 6'. THE ELLIPSE. ■221 FlfJ. 97. Substituting in the equation of the ellipse, we have (x' cos ^ + y 'cos O'y (x' sin 6) + y' sin 6')- _ a^ + b^ ~^' which reduces to cos^ 6 , sin^ , „ , , , cos cos 6' , sin 6 sin 6' + 2x'y •; -, h = 1. , ,„ , ccs'^ 0' , sin^ e' We need now to see what special form the coefhcients assume owing to the hypothesis that CX' and CY' are con- jugate diameters. In the first place, the relation b^ is the same as b^ tan tan 0' = — which is equivalent to cos 6 cos + sm a sm a' b^ Hence the coefficient of x'y' vanishes. 0. 222 PLANE ANALYTIC GEOMETRY. Moreover, if we call CPi = a', as before, and take the coor- dinates of Pi, referred to the old axes, as Xi, y^, we have cos 6^-„ sin 6 = ^. a a' The coefficient of x'^ is, therefore, 1/xi^ yA a'2 i^a^"^ b-/ which equals simply — , since Pi is on the ellipse. Similarly, a the coefficient of y'- may be proved to be — . The equation of the ellipse referred to the new axes is, therefore, the primes on the coordinates being dropped. It will be noticed that this is an equation of the same form as the ori^rinal one. 112. Polar Equation of the Ellipse. To find the polar equation of the ellipse, the pole being at the centre. To transform from rectangular coordinates to polar coordi- nates, we have the relation [7], § 10, X ^ r cos Q, y = r sin Q. These values, substituted in a^ ^ b^ -^' l'' / ^cos=^^ , sin2^\ ' I , a' + b' y a=b' give r^ f — ^^ -f -^j- ] == 1, whence . — , „ „ . , , • 2 ^' b^ cos^ Q-\- z? sm^ Q THE p:llii\se. 223 a^— (a^— b^) cos^^' This is the required, equation. ■•■'•'= l-e-cs'e ' W3J EXAMPLES. 1. Given the ellipse 3x- + 5y- = 1 ; find the semi-axes, eccentricity, foci, directrices, and. latus rectum. 2. Find the equation of an ellipse which passes through the points (1, — 2) and (3, — 1), the centre of the ellipse being at the origin, and. the axes of tlie ellipse coinciding with the axes of coordinates. 3. Find the equation of an ellipse, the eccentricity of which is f and the foci of which lie at the points (0, ± 6). 4. Find the equation of an ellipse, the eccentricity of which is § and the latus rectum 10, the centre of the ellipse being at the origin, and the axes of tlie ellipse coinciding with the axes of coordinates. 5. Does the point (1, 3) lie inside, on, or outside, the ellipse 10x'+lly2=110? 6. Find the eccentricity and the equation of an ellipse, if the foci lie halfway between the centre and the vertices. 7. Find the eccentricity and the equation of the ellipse, if the latus rectum is ^ the minor axis. 8. Find the eccentricity, if the line connecting the positive extremities of the axes is parallel to the line joining the centre to the upper end of the left-hand latus rectum. 9. Find the equation of the tangent and normal to the ellipse 36x-+ 108y"^20 at a point the abscissa of wliieh is ^. 10. Find the lengths of the subtangent and the subnormal for the point (2, — 3) on the ellipse 4x- -j- 7y^ = 79. (See definition, § 56.) 224 PLANE ANALYTIC GEOMETRY. 11. Find the tangent and the normal to the ellipse at the upper extremity of the right-hand latus rectum. 12. Find the tangents to the ellipse 4x^ + 9y^ = 36 which are parallel to the line joining the positive extremities of the axes. 13. Find the tangents to the ellipse 2x^ + 5y^ = 10 which make an angle of 60° with the axis of x. 14. Find the point in which the polar of (4, 5) with respect to the ellipse — + =^ = 1 intersects the straight line joining the pole to the origin. 15. Find the equations of the tangent and normal at the upper extremity of the left-hand latus rectum of the ellipse x^ y2 — + r^=lj in terms of the major axis and the eccentricity. 16. If the normal in Ex. 15 passes through the upper end of the minor axis, show that the eccentricity is given by the equation e* + e^ — 1=0. o 9 x" V 17. Find a point on the ellipse — + f^ = 1? such that the tangent there is equally inclined to the two axes. 18. Find expressions for the subtangent and the subnormal for a point (xi, yi) on the ellipse "^ + ri = 1- cl D 19. Show that the point (xj, yi) lies inside, on, or outside, x^ y2 the ellipse — 2 + ui ^^ 1' according as |- + ^^-l<, = ,o,>0. 20. Prove analytically that if the normals at all points of an ellipse pass through the centre, the ellipse is a circle. 21. Find the coordinates of a point such that the tangent there is parallel to the line joining the positive extremities of the major and the minor axes. THE ELLIPSE. 225 22. Find the equation and the length of the perpendicular from the centre to any tangent to the ellipse 23. Show that the product of the perpendiculars from the foci upon any tangent equals the square of half the minor axis. 24. Show that if any ordinate M P be produced to meet the tangent at the end of the latus rectum through F in Q, then QM = FP. 25. Show that the perpendicular from the focus upon any tangent meets the line drawn from the centre to the point of contact on the corresponding directrix. 26. Prove that any choxd through the focus is perpen- dicular to the line joining its pole to the focus. 27. Find the locus of the middle points of a system of y2 y2 chords of the ellipse — + ^ = 1, which are parallel to the line 2x -f- 3y — 1 = 0. 28. In the ellipse 2x- + 5y^ = 3, find two conjugate diam- eters, one of which bisects the chord x + 3y + 2 ^ 0. 29. In the ellipse 2x^ + 3y^ = l, find two conjugate diam- eters, one of which passes through (^, |). 30. In the ellipse 4x2 _^ 25y^ = 100, find two conjugate diameters, one of which is perpendicular to 3x — 4y + 8 = 0. 31. Given the point (— 2, 3) on the ellipse x- + 2y=^ = 22, find the conjugate points. 32. In the ellipse 3x2 + 4y"^ = 12, g,-^(j ^]^g equation of a chord, the middle point of which is (2, — 1). 33. In the ellipse -,-\-h — '^, finfl the equations of two S" D conjugate diameters, one of which bisects the chord deter- mined by the upper end of the minor axis and the right-hand focus. 226 PLANE ANALYTIC GEOMETRY. 34. In any ellipse, show that the diameters parallel to the lines joining the extremities of the axes are conjugate. 35. Show that the tangents at the extremities of any chord meet on the diameter which bisects the chord. 36. If the tangent at the vertex A cut any two conjugate diameters produced in T and t, show that AT. At = — b^ 37. Show that the product of the focal distances of any point is equal to the square of half the corresponding conju- gate diameter. 38. If from the focus of an ellipse a perpendicular is drawn to a diameter, show that it will meet the conjugate diameter on the corresponding directrix. 39. Find the diameter of the ellipse 16^ 4 which is equal to its conjugate diameter. 40. Show that there can be only one pair of equal conju- gate diameters of the ellipse x^ y^ b b ^ + ^2 "" ^ ' namely, y = -x, y = --x. 41. Given the point (— 3, — 1) on the ellipse x^ + 3y2 = 12 ; find the corresponding point on the auxiliary circle, and the eccentric angle. x^ y2 42. In the ellipse - + j = 1, find the point of which the eccentric angle is 30°. 43. Show that the tangent to the ellipse at (xj, yj) and the tangent at the corresponding point of the auxiliary circle pass through the same point of the major axis. 44. Find the equation of the tangent at any point of an ellipse in terms of the eccentric angle at that point. 45. Show that the area of a triangle inscribed in an ellipse is ^ ab [sin (/8 — y) + sin (y — a) + sin (a — ^)], where a, (3, y are the eccentric angles of the vertices of the triangle. THE ELLIPSE. 227 46. Show that the perpendicular from either focus upon the tangent at any point of the auxiliary circle equals the focal distance of the corresponding point of the ellipse. 47. Show that the eccentric angles of the extremities of two equal conjugate diameters are -r and —-• 48. Find the equation of the ellipse 2x^ + 4y^= 12 referred to a pair of conjugate diameters as axes, one of which passes throiigh the point (2, 1). 49. Find the equation of the ellipse 2x^ + 4y' = 12 referred to its equal conjugate diameters as axes. 50. Show that the equation of any ellipse referred to its equal conjugate diameters as axes is x^ + y^= — - — . 51. If A and A' be the extremities of the major axis of an ellipse, T the point where the tangent at any point P meets AA', QR a line through T perpendicular to AA' and meeting AP in Q and A'P in R, show that QT = RT. 52. From a point P on an ellipse straight lines are drawn to A and A', and from A and A' straight lines are drawn per- pendicular to AP and A'P. Show that the locus of their point of intersection is an ellipse. 53. Find the locus of the points of intersection of normals at corresponding points of the ellipse and the auxiliary circle. 54. Find the locus of the vertices of a parallelogram formed on a pair of conjugate diameters. 55. Chords are drawn througli the end of the major axis of the ellipse ; find the locus of their middle points. 56. Find the locus of the middle points of a system of chords passing through a fixed point (xi, y^). 57. An ellipse moves so as to be always tangent to a pair of rectangular axes; show that the locus of its centre is a circle. 58. A point moves so that the sum of the squares of its distances from two intersecting straight lines is constant. 228 PLANE ANALYTIC GEOMETRY. Prove that the locus is an ellipse, and find its eccentricity in terms of the angle between the lines. 59. Two tangents to an ellipse are so drawn tliat the prod- uct of their slopes is constant. Show that the locus of their point of intersection is an ellipse or hyperbola according as the product is negative or positive. CHAPTER X. ^ = 1. 113. Conjugate and Equilateral Hyperbolas. We have seen in § 69 that the equation of the hyperbola takes its simplest form S-^.=i. (1) when the centre of the curve is taken as the origin of coordinates and the transverse axis of the curve as the VaM-^ axis of X. The eccentricity is e = ; the foci lie £1 at the points (± ae, 0) ; and the equations of the directrices are x ^ rb - . The polar of the point (x,, y,) is -^ — ^~ = 1, 6' ^ a'' b- this equation representing the tangent as special case of the polar when (xi, yi) is on the curve. If the slope m of a tan- gent is given, its equation is of the form, y = mx ± Va"^in- — b". Closely connected Avitli this hyperbola is the one the equa- tion of which is where a and b have the same numerical values as in the first equation. The hyperbolas (1) and (2) are so related that the transverse axis of each is the conjugate axis of the other; hence they are included between the same asj^mptotes y = ± - x. They lie, therefore, as in Fig. 98, the hyperbola 230 PLANE ANALYTIC GEOMETRY. (1) being composed of the right and left-hand branches, and the hyperbola (2) of the upper and lower branches. Such hyperbolas are said to be conjugate to each other, and in particular (1) is usually called the primary hyperbola and (2) the conjugate hyperbola * The eccentricity of the conjugate hyperbola is its foci lie at the points (0, ± be'); and the equations of its ,. . b directrices are y = ± — . e Evidently two conjugate hyperbolas are, in general, of quite different shapes, and could not be made to coincide. An exception occurs only when b ^ a. In tliat case, the equation of the primary hyperbola is * It would be just as correct to call (2) the primary hyperbola, and (1) the conjugate hyperbola. That hyperbola is the primary one which appears first in the discussion of a problem. THE HYPERBOLA. 231 The asymptotes are the lines y = ± x, which bisect the angles between the axes of coordinates, and are perpendicular to each other. Sucli an hyperbola is called an equilateral or rectangular hyperbola; the lirst name being derived from the equality of the axes; the second, from the angle between the asym})totes. The conjugate hyperbola is then — x^ + y^ = a.', which is equal to the primary hyperbola, but turned through an angle of 90°. In this chapter, we shall work with the primary hyperbola a- b- In those cases where the conjugate hyperbola is used, ex- plicit mention will be made of the fact. The special form of the result for the equilateral hyperbola may be obtained by placing b =: a, but we will omit this in all cases. The student will notice that much of the following work is similar to that of tlie preceding chapter. In nuiny cases tlie results differ only in a sign, corresponding to the fact that the equations of the ellipse and the hyperbola differ only in that way. 114. To find the Foci when the Axes are given. From § 69, the distance of the foci from tlie centre is ae, which equals Va^ + b^. This equals the lengtli of the semi- diagonal CH of the rectangle on the axes (Fig. 98). The distance of the foci of the conjugate hyperbola from the centre is be', which also equals Va- -|- b". Hence the following rule : From the centre of the hijperhohi describe a circle with a radius equal to the semi-diagonal of the rectangle on the axes. The circle will cut tlie transverse axis in the foci of the prlnnrri/ 232 PLANE ANALYTIC GEOMETRY. hyperbola, and the conjugate axis in the foci of the conjugate hyperbola. 115. Focal Distances. The definitions of focal distances and focal radii are the same as in the case of the ellipse. To investigate we proceed as in § 94. Y Let C be the centre of the hyperbola, F and F' the foci, and DM and D'M' the directrices. Take Pi a point on the right- hand branch and draw the ordinate NPi, the focal radii F'Pi and FPi, and the line P^MM' perpendicular to the directrices. Then by defijiition of the hyperbola FPi=re. MPi, , = e.(CN- CD), = e x, — Also, ^ ex, — a. F'Pi = e. M'P„ = 6. (CN + D'C), = e ( xi -f = exi -)- a. THE HYPERBOLA. 233 Similarly, for a point on the left-hand branch, we find F'Pi = — exi — a, and FPi = — exi + a. 116, Difference of Focal Distances. From the results of the last article, we find F'Pi-FPi = 2a for the right-hand branch, and FPi- F'Pi = 2a for the left-hand branch. Hence : The difference between the focal distances of any point on an hypevhola is constant and equal to the transverse axis. This property is true for no point not on the hyperbola, as may be readily proved by use of the axiom that a straight line is the shortest distance between two points. Hence we may say : The liyj^erhola is the locus of a j^oint, the difference of the distances of which from two fixed points is constant. Tliis is sometimes given as the definition of the hyperbola and is made the basis of the investigation of its equation and properties. 117. Constructions. The preceding article forms the basis of the following methods of constructing an hyperbola when the two axes, or the foci and the transverse axis, are given. (1) By means of compasses. If the foci are not given, find them as in § 114. Then, take a straight line equal to the transverse axis, and divide it externally in any two portions r and r'. With one focus as a centre and r as a radius, describe an arc of a circle ; with the other focus as a centre and r' as 234 PLANE ANALYTIC GEOMETRY. radius, describe another arc. The intersections of the two arcs are points on the hyperbola. In this way any number of points may be found and the curve drawn. (2) By a mechanical device. First find the foci, if they are not given. Then take a ruler longer than the transverse axis, and a shorter thread sucli that the difference in length of the ruler and the thread shall equal the transverse axis. Pivot one end of the ruler at one of the foci, and attach one end of the thread to tlie free end of the ruler and the other end of the thread to the unused focus. Insert a pencil in the loop of the thread and press it tightly against the ruler. As the ruler is revolved around its pivot, the pencil describes one branch of the hyperbola. By interchanging the positions of the ruler and the thread, the other branch of the hyperbola is obtained. 118. Angle between Tangent and Focal Radii. The tangent bisects the angle hetiveen the focul radii of the jioint of contact, and the normal makes eqital angles with these focal radii. Fig. 100. THE HYPERBOLA. 235 Let PiT be the tangent at any point Pi(xi, yi), P^R the normal, and FPi and F'Pi, the focal radii. We are to prove ZF'PiT=ZTPiF, and ZFPiR = ZRPiQ. The equation of the tangent is xix _ yiy ^ a^ b^ ' whence follows that CT, the intercept on the axis of x, a^ Therefore FT ^1 = F'C + CT, 1 ^' = ae H > Xi aexi + a^ Xi TF -:CF-CT, a^ = ae > Xl aexj — a" (§ 115) and Xl FT _ aexi + a- _ exi + a F'Pi TF aexi — a^ exj — a FPj Hence, by Geometry, ZF'PiT = ZTPiF. Also, PiR is by definition perpendicular to PjT. .-. ZFPiR = ZRP,Q. The value we have found for CT sliows that a tangent to the right- hand branch of the hyperbola cuts the transverse axis at the right of the centre, and a tangent to the left-hand branch cuts the transverse axis at the left of the centre. A reference to this fact will .sonietimes save the student from false constructions. 236 PLANE ANALYTIC GEOMETRY. From § 18, any straight line cuts a conic in no more than two points, and a point of tangency is counted as two coincident points. Hence the tangent to one branch does not cut the other branch. 119. Equation of a Diameter. As already defined, a diameter of a conic is the locus of the middle points of a system of parallel chords. It is to be noticed that parallel chords in an hyperbola may be drawn in two ways. They may connect points on the same branch or they may connect points on different branches, as shown by the two systems of parallel chords in Fig. 101. In either case, by using the method of § 61, we find that the equation of the diameter bisecting the system, the slope of which is rrii, is y = a-m X. [44] If m2 is tlie slope of the diameter, then the relation between m2 and rrii is mim2 = -:i. THE HYPERBOLA. 237 From this follows that the axes are the only diameters per- pendicular to the chords they bisect. For m^ cannot equal — — , unless mi = and m., = go, or vice versdi; nrii 120. Conjugate Diameters. As in the ellipse, conjugate diameters are such that each bisects all chords parallel to the other. It readily follows that the relation between their slopes is rriim., = -,• [45] a From this follow the theorems : 1. The two axes are the only i^air of conjugate diameters that are j^erpendicular to each other. 2. Two conjugate diameters make either both actite or both obtuse angles with the transverse axis. For the product of trii and nria is always positive, and hence the two factors have the same sign. Therefore the upper portions of two conjugate diameters lie upon the same side of the conjugate axis ; for example, see P1P2 and QiQ- in Fig. 101. 3. One diameter intersects the hyperbola, the other does not. For a line y ^ iriix intersects the hyjjerbola in points the abscissas of which are ab Vb^ — a'-rrii^ These values of x are real when irii is numerically less than -, and imaginary when mi is numerically greater than -. a a But of two conjugate diameters the slope of one is always numerically greater than - and the other less, by virtue of the a relation mim., = — . a' 238 PLANE ANALYTIC GEOMETRY. An exception occurs only when rrii = ± -. Then rria = ± - also, and the two diameters coincide with each other and with an asymptote, 4. A diameter- intersects either the primary Jtyperhola or the conjugate hyperbola. For as we have just seen, the diameter, y = mjx, intersects the primary hyperbola a^ b^ when irii is numerically less than -, but not when mi is a numerically greater than -. In the latter case, however, it a Avill intersect the conjugate hyperbola a^ ^ b^ ' for the abscissas of the intersections of y = miX and this curve are given by ab V— b^ + a-^m^ which is real when nrii is numerically greater than -. a 121. Propositions on Conjugate Diameters. The following propositions are similar to corresponding ones for the ellipse. We state them and leave the proof to the student. One point, however, needs a little explanation. We have seen that of two conjugate diameters, one intersects the hyperbola, the other does not. The one appears, therefore, indefinite in extent, while the other may be limited by the THE HYPERBOLA. 239 curve. We may overcome this want of uniformity by remem- bering that a diameter wliich does not intersect the primary hyperbola does intersect tlie conjugate liyperbola. Henca tve will agree to call the extreinities of any diameter the points in which it cuts either the i)riniary or the conjugate hijperhola ^ as the case may be. Moreover, we may remark tliat diameters conjugate with reference to the primary hyperbola are conjugate with refer- ence to the conjiigate hyperbola. For we may change the equation of the primary hyperbola into that of the conjugate hyperbola by writing — a" for a^ and — b^ for b^. Such a change, however, leaves the relation [45] unaltered. 1. The tangent at tlie extremity of any diameter' is parallel to the conjugate diameter. (§ 102) 2. Any pair of supplemental chords is 2)arallel to a pair of conjugate diameters. (§ 103) 3. If (>ii, yi) is the extremity of any diameter, then a b b^^'-r^ are the extremities of the conjugate diameter. (§ 104) 4. If a' and h' are the lengths of a pair of conjugate semi- diameters, then a'2 - b"-^ =■ a- - b\ (§ 105) 5. If (f) is the angle betiveen a 2)uir of conjugate diameters, sin<^ = ^,- (§106) 6. The area of the parallelogram formed by tangents at the extremities of a pair of conjugate diameters is constant and equal to the rectangle on the axes. (§ 107) 7. The equation of the hyperbola referred to a pair of conju- gate diameters as axes of coordinates is ^-^=1- [46] 240 PLANE ANALYTIC GEOMETRY. 122. Asymptotes. We have already noticed in § 69 that the two straight lines y b a ' which are readily constructed as the diagonals of the rect- angle on the axes, possess the peculiar property of having the coordinates of their points of intersection with the hyper- bola expressed as infinite quantities. These lines Ave call the asymptotes of the hyperbola. In general we may define an asymptote of a curve as a straight line of ivhich the distance from a point on the curve decreases below any assignable quantity as the jmint of the curve recedes indefinitely from the origin. We proceed to show that the asymptotes of the hyperbola satisfy this definition. For that purpose, let us take a point Pi(xi) Yi) on the upper, right-hand portion of the hyperbola, and denote its perpendicular distance from the neighboring asymptote by PiR. Fig. 102. If we write the equation of the asymptote in the form bx — ay = 0, _ bxi — ay i then, by § 32, PiR Va^ + b^ THE HYPERBOLA. 241 , since (xi yi) i« on the hyperbola, yi h -3 Vx;^-a^ ice, by sill )stituti = b(x on. PiR — Vxi'^ — a-) Va'^ + b- ba^ Va^ + b^ (xi + V xi- - a^) Now as Xi is allowed to increase indefinitely, the value of PjR decreases indefinitely, as was to be proved. We give in the following articles a few of the proi)erties of the asymptotes of the hyperbola. 123. The tangents (it the extremities of a pair of conjugate diameters intersect on the asymptotes. Fig. 103. Let Pi(xi, Yi) be an extremity of that one of tlie conjugate diameters which intersects the i)rimary hyperbola. The tan- gent PiT is Xix y^y 3== b^ 1. 242 PLANE ANALYTIC GKOMETRY. The extremity Qi of the conjugate diameter is and the tangent QiT to the conjugate hyperbola is ayi X I bxi y ^^ b a'^ a b" The coordinates of the point of intersection T of these two tangents are easily found to he y = yi + -xi- a But tliis is a point on the asymptote, since it satisfies the equation y = - x. * 124. The straight line connecting the extremities of a pair of conjugate diameters is parallel to one asymptote and bisected by the other. The straight line connecting the points Pi(xi, yi) and Qi[ - yi, - Xi ) (Fig. 103) has the equation y — Yi _ X — Xi b a yi- which reduces to y.-^x, x,-^y, y-yi = -- (x-x,). d. This line is evidently parallel to the asymptote y = x. 3. Moreover, the figure CPiTQ, is a parallelogram by § 121, 1, and CT is an asymptote by § 123. Therefore, QiPi is bisected by CT, since the diagonals of a parallelogram bisect each other. THE HYPERBOLA. 243 125. The portion of a tangent between the asi/mptotes is bisected by the point of tangency. For in Fig. 103, PiT = CQi and PiT' = CQo, being opposite sides of parallelograms. But hence PiT=P,T'. 126. The po)'tio?is of a chord between the hyperbola and its asymptotes are equal. Let RQ be any chord intersecting the hyperbola in S and M, and the asymptotes in R and Q ; to prove QM = SR. Draw the diameter CN to the middle point of SM, and at the point P where CN intersects the hyperbola draw the tan- gent TT'. Then by § 121, 1, TT' is parallel to SM. But TP = PT', by previous article, .-. QN = NR. But MN = NS, by construction, .-. QM = SR. 244 PLANE ANALYTIC GEOMETRY. 127. Equation of an Hyperbola Referred to its Asymp- totes as Axes of Co rdinates. We wish to transform the equation of the hyperbola ^ _ y: — 1 a' h'~ ' from the axes CX and CY to the axes CX' and CY', where CX' and CY' are the asymptotes. Fig. 105. We have, by [21], § 45, as the formulas of transformation X == x' cos 9 -\- y' cos $', y = x' sin ^ + y' sin 6', where d = XCX' and 0' = XCY'. Since CA = a and AH' = — b, it follows from Trigonometry that cos 6 Also, since = , sin Va^+b^' Va'+b^ CA = a and AH = b, cos u' = a . „, b , sin 6' = Va^ + b^ Va^ + b^ THE HYPERBOLA. 245 Hence the formulas of transformation become a = (x' + y'), Va- + b^ •va-+ b- Substituting in we have or, dropping primes, a^ b^ -^' 4x'y' _ a- + b' ^ -, xy==-^: — [47] Similarly, the equation of the conjugate hyperbola is a^+b^ ^y= — 4— In numerical equations, the right-band side of the equation [47] will appear as a single number and we shall have an equation of the form xy = c. To obtain the axes of the hyperbola, reference must be had to the angle between the coordinate axes. For example, given the hyperbola xy = 8, the angle between the coordinate axes being G0°. Since the transverse axis of the hyperbola bisects the angle between the coordinate axes, the angle between the transverse axis and an asymp- tote is 30°. Hence, ^ = tan :J0° = I V3. Al.so from [47], a2 + b2 _ ^ From these two ecjuations we find a = 2V0, b = 2 V2. 246 PLANE ANALYTIC GEOMETRY. 128. Polar Equation of the Hyperbola, the Pole being at the Centre. To obtain the polar equation, we have the formulas of transformation X = r cos ^, y = r sin 6. Substituting in a^ b^"-"' we obtain Whence, r2 a^b^ b'-' cos^ ^ - a^ sin^ 0' a^b^ (a^ + b'O cos- e -a^' r=- .. ^; e'cos^e-1 W" y EXAMPLES. 1. Given the hyperbola 3x^ — 5y- ^ 7, find the semi-axes, eccentricity, foci, directrices, asymptotes, and latus rectum, both of the given hyperbola and its conjugate. 2. Find the equation of the hyperbola which has the lines y = ± fx for its asymptotes and the points (± 4, 0) for its foci. 3. Find the equation of the hyperbola which has the points (0, zh VT) for foci and passes through the j^oint (2, — 1). 4. Find the equation of the hyperbola which passes through the points (— 2, 3) and (3, — 5), the centre of the curve being at the origin and the transverse axis coinciding with the axis of X. 5. Find the equation of the hyperbola with eccentricity 3, which passes through the point (2, 4), the centre of the hyper- bola being at the origin. THE HYPERBOLA. 247 6. Find the equilateral liyperbola which passes through (5, — 2), the centre being at the origin. 7. Find the equation of the hyperbola, the vertices of which lie halfway between the centre and the foci. 8. Does the point (4, 5) lie inside or outside the hyperbola 9x2 — 10y2 = 90? 9. Show that the point (xi, yi) lies within, on, or without the hyperbola —,— , :, = !, according as ^ — -j^ — 1 >, = , or < 0. a- b- a^ b- 10. If e and e' are the eccentricities of the primary and the conjugate hyperbolas, respectively, show that —^-] — ;;; = 1- 11. If the vertex lies two thirds of the distance from the centre to the focus, find the slopes of the asymptotes. 12. Express the angle between the asymptotes in terms of the eccentricity of the hyperbola. 13. Show that the eccentricity of an equilateral hyperbola is the ratio of the diagonal of a square to its side. 14. Show that in an equilateral hyperbola the distance of a point from the centre is a mean proportional between its focal distances. 15. Find the equations of the tangent and the normal to the hyperbola 9x^ — 4y^ = 36 at a point the abscissa of which is equal to its ordinate. X" V" 16. Find the tangents to the hyperbola q "~ ^ = ~ ^ which are parallel to the straight line 5y — 2x + 3 = 0. 17. Find the lengths of the subtangent and the subnormal for the point (3, 2) on the hyperbola x^ — 2y- = l; also the lengths of the portions of the tangent and the normal in- cluded between the curve and the transverse axis. 18. Show that the subtangent and tlie subnormal for the point (xj, Yi) on the hyperbola x2 y2 . ... x,'-a' , b^ — — —, = 1 are, respectively, — ■ and — Xi. a' b" Xi a 248 PLANE ANALYTIC GEOMETRY. 19. Find where the tangents from the foot of the directrix will meet the hyperbola, and what angles they will make with the transverse axis. 20. Prove that tangents to the conjugate hyperbola at the points where it is cut by the tangent at the vertex A of the primary hyperbola pass through the other vertex A'. 21. If any number of hyperbolas have the same transverse axis, show that tangents to the hyperbolas at points having the same abscissa all pass through the same point on the transverse axis. 22. Prove that an ellipse and hyperbola with the same foci cut each other at right angles. 23. Show that the portion of the normal between the axes is divided by the curve in the ratio a^ : bl 24. If the tangent at any point P on the hyperbola inter- sects the transverse axis at T, and if CP meets the tangent at the vertex A at R, show that RT is parallel to A P. 25. If a tangent to an hyperbola is intersected by the tan- gents at the vertices in the points Q and R, show that the circle described on QR as a diameter passes through the foci. 26. In the hyperbola 4x" — 2y^ = 7, find the equation of the conjugate diameters, one of which bisects the chord 3x - 2y + 3 = 0. 27. In the hj^perbola x- — 3y^= 6, find the equation of the chord the middle point of which is (4, 1). 28. Given the point (i, |) on the hyperbola 48x'-' — 4y^ = 3, find the point on the conjugate hyperbola at the extremity of the corresponding conjugate diameter. 29. Find the equation of the hyperbola 4x^ — 9y'-' = 36 referred to a pair of conjugate diameters as axes, one of which makes an angle 30° with the transverse axis. 30. Show that every diameter of an equilateral hyperbola is equal to its conjugate. 31. Sho\^that in an equilateral hyperbola conjugate diam- eters are equally inclined to the asymptotes. THE HYPERBOLA. 249 32. Show that in an equilateral hyperbola all diameters at right angles to each other are equal. 33. Show that the polar of any point on a diameter is par- allel to the conjugate diameter. 34. Show that if an ellipse and an hyperbola have the same axes in magnitude and position, then the asymptotes of the hyperbola coincide with tlie equal conjugate diameters of the ellipse. 35. Show that the area of the triangle made by any tangent and its intercepts on the asymptotes is constant. 36. Show that the focal distance of any point on the hyperbola is equal to the line drawn through the point par- allel to an asymptote to meet the directrix. 37. A perpendicular is drawn from a focus of an hyperbola to an asymptote ; show that its foot is at a distance a and b from tlie centre and the focus, respectively. 38. Show how an hyperbola may be graphically constructed by aid of § 126, if the asymptotes and one point of the curve are given. 39. If a circle be described from the focus of an hyperbola with radius equal to half the conjugate axis, it will touch the asymptotes in the points where they intersect the directrix. 40. If, from any point P of an hyperbola, PK is drawn parallel to the transverse axis, cutting the asymptotes in Q and R, then PQ . PR^a"; if PK is drawn parallel to the conjugate axis, then PQ . PR := — b^. 41. If two concentric equilateral hyperbolas be described, the axes of one being the asymptotes of the other, show that tliey will intersect at right angles. 42. Find the equation of 2x^ — 3y^ = 6 referred to its asymptotes as coordinate axes. 43. Find the semi-axes, eccentricity, and vertices of the hyperbola xy = 1, the axes being rectangular. 44. Find the semi-axes, eccentricity, and vertices of the hyperbola xy ^ — 12, tlie angle between the axes being 60°. 250 PLANE ANALYTIC GEOMETRY. 45. Show that xy + ax + by + c = is the general equa- tion of the hyperbola, when the axes of coordinates are parallel to the asymptotes. 46. Prove that the equation of the tangent to the hyperbola, referred to its asymptotes as axes, is Xiy + yiX = — ~ — , where (xi, yj) is the point of contact. 47. If the ordinate N P of an hyperbola be produced to Q, so that NQ = FP, find tlie locus of Q. 48. AOB and COD are two straight liiaes which bisect each other at right angles; find the locus of a point which moves so that PA. PB= PC. PD. 49. A straight line has its extremities on two fi.xed straight lines and makes with them a triangle of constant area ; find the locus of the middle point of the line. 50. Given a fixed line AB and a fixed point Q; from any point R in AB a perpendicular is drawn equal in length to RQ ; find the locus of the extremity of the perpendicular. 51. A chord PQ of an ellipse is perpendicular to the major axis ; PA and QA' are produced to meet in R ; show that the locus of R is an hyperbola having the same axes as the ellipse. 52. Show that the locus of the centres of all circles which touch two given circles is an hyperbola or an ellipse. 53. Find the locus of the intersection of a pair of perpen- dicular tangents to an hyperbola. 54. Find the locus of the foot of tlie perpendicular from either focus of an hyperbola to any tangent. CHAPTER XI. THE GENERAL EQUATION OF THE SECOND DEGREE. 129. Statement of Problem. The most general form of the equation of the second degree is Ax-+2 Hxy+ By2 + 2 Gx + 1> Fy + C=0, (1) where the coefficients A, H, B, G, F, C may have any values, positive or negative or zero, except that all three of the coefficients A, H, B shall not be zero at the same time. For, in that case, the equation would not be of the second but of the first degree. This equation evidently includes that treated in chapter VI, § 70, for the latter is obtained from (1) by placing H ^ 0. We propose now to examine the general equation, and to show that, whatever may be the values of the coefficients, the locus represented is one of the conic sections, under which name we include not only the circle, ellipse, hyperbola, and parabola, but also the limiting cases noticed in chapter VI. The method to be followed is to reduce the equation to its simplest form by a proper choice of new axes of coordinates. It will be found that the final form obtained depends upon certain hypotheses respecting the values of the coefficients, but that the possible results are limited to just those equations which we know represent conic sections. To simplify the Avork we shall assume for the present that the equation is referred to rectangular axes. The case of oblique axes will be liandled briefly in § 136. 252 PLANE ANALYTIC GEOMETRY. The general equation has already been partially handled in chapter III, § 41, and it was shown that, if ABC + 2 FGH- AF^-BG--CH2 = 0, the equation represents two straight lines. This fact will be proved anew in the course of our present work. Therefore we do not need to assume it, though it is well to bear it in mind. 130. To Remove Terms of the First Degree. We first proceed to see if it is possible to reduce the equa- tion so that the terms of the first degree in x and y shall dis- appear. For that purpose, let us move the axes parallel to themselves, by placing X — Xq -j- X , ,„, y = yo + y', ^^ Xo and Yo, the coordinates of the new origin, being left unde- termined for a moment. Substituting these quantities in (1), we have Ax'2 -f 2 HxV + By'2 -f 2(Axo + Hyo + G) x' + 2(Hx, + Byo + F)y'+Ax,^ + 2 Hxoyo+By„^ + 2 Gxo + 2 Fyo+C = 0. If now we can choose for Xq and yo such quantities that the coefficients of x' and y' shall vanish, we shall accomplish our object. We must have Axo+Hyo+G = 0, Hxo+Byo+F = 0. ^^ The solution of these equations is possible, when AB — H'- is not zero, giving as the necessary values of Xq and yo, _ HP- BG '°"AB-H^ ' M) _ HG- AF '^ ^ y"~AB-H'-^' THE GENERAL EQUATION. 253 If AB — H^ = 0, however, the equations (3) are contradic- tory, and hence no vahies of Xq and yo can be found, wliicli will make the coefficients of x' and y' vanish. There is one case even when AB — H- = 0, in which values of xq and yo can be found. This is the case in which the relations hold H~ B F ■ The equations (3) are then identical, and values of xo and yo which satisfy either will make the coefficients of x' and y' vanish. This case will be fully provided for later, and need not be discussed here. We will distinguish three cases, in the following discussion, namely : I, AB- H->0; II, AB- H2<0; III, AB - H'^^O. But before passing to them we will sum up the results of this paragraph as follows : The terms of the first degree may he removed from the general equation jnovided the quantity AB — W is not zero. The equa- tion theyi talces the form Ax'2+2Hx'y'+By'^ + (Axo^ + 2Hx,y,+ Byo^+2Gxo + 2Fyo + C)=0, whence it is seen that the coefficients of the terms of the second degree are unchanged, and the absolute term is ■ obtained by substituting the coordinates of the neiv origin in the original equation. 131. Case I. AB - H- > 0. We have just seen that by means of a cliange of axes the equation may be reduced to Ax'2 + 2 Hx'y' + By'- + C == 0, (5) where C = Ax,r + 2 Hx,y, + By,^ + 2 Gx^ + 2 Fy, + C. 254 PLANE ANALYTIC GEOMETRY. The value of C may be expressed in terms of the original coefficients as follows. Take the two equations (3), multiply the first one by Xq, the second by yo, and add them. There results Axo^ + 2 HxoYo + Byo^ + Gxo + Fyo = 0. Subtracting this sum from the value of C, we have C'=Gxo+Fyo+C, whence, finally, by substituting the values of Xq and yo, as given in (4), we have ^, ABC + 2 FGH-AF2-BG2-CH- ^^ AB^^hP ^^> The quantity in the numerator is called the Discriminant of the equation (1) and is usually denoted by the Greek letter A. If we represent the quantity AB — H^ by D, we have then Returning now to the equation (5), Ax'2 + 2 Hx'y'+By'2+C' = 0, we will next endeavor to remove the term 2 Hx'y'. For that purpose we turn the axes through an angle $, keeping the origin fixed and the axes rectangular. The three sets of axes we have now used lie as shown in Fig. 106, the figure being drawn on the assumption that Xo, yo? and 6 are all posi- tive. The formulas of transformation to the new axes are by [20], §44, x' = x" cos — y" sin 6, y' = x" sin 6 -\- y" cos 6. Substituting in (5), we have (A cos^ ^ + 2 H sin ^ cos e + B sin^ 6) x'"' + 2 {(B — A) cos 6 sin ^ + H (cos^ $ - sin^ 6)} x"y" + (A sin^ ^ - 2 H sin ^ cos e + B cos^ 6) y"- + C = 0. THE GENERAL EQUATION. 255 Fig. 106. We must now determine 6 so that the coefficient of x"y" shall vanish; i.e., so that 2 (B — A) cos ^ sin ^ + 2 H (cos^ d — sin^ 0) = 0. This equation is equivalent to 2 H cos 2 ^ + (B — A) sin 2^ = 0, 2 H whence or tan 2 e = 4 tan" - B' 2 H A-B (») Among the various values which 6 may have we shall agree always to choose that one which makes 2 6 lie between 0° and 180°. Hence 9 is always a positive acute angle. With this value of 9, our equation takes the form A'x"-+ B'y'"+C'==0, (9) where A' = A cos^ 9 -\- 2 H sin 9 cos. 9 -\- B sin^ 9, and B' = A sin^ ^ — 2 H sin ^ cos ^ + B cos" 9. These values of A' and B' may be expressed in terms of the original coefficients by use of (8). 256 PLANE ANALYTIC GEOMETRY. We have A' 1= A cos^ 6 + 2 H siu ^ cos ^ + B siu^ 0, .l±cos20.^ . ^ ^ p 1 - cos 2 g = A 2 h H sm 2 ^ B , = i [A + B + (A - B) cos 2 6 + 2 H sin 2 ^]. But since, by (8), 2 H tan 2 = A- B' it follows that ■ on 2 H ; „- A-B sm 2 ^ = - and cos 2 6 = V(A-B)- + 4H2 V(A-B)^ + 4H^ Hence A- = irA+B+ (A-B)- + 4Hn or A' = i[A + B + V(A-B)2 + 4 H^J. Similarly, B' = |(A + B - V(A- B)'^ + 4 H^. From these values we obtain the following relations : A'+B'=A+B, A'B'= AB- H^ 2 H (10) A'-B' = V(A- B)^ + 4 H2 = -^^— . ^ ^ sin 2 6 (11) J The first two of these relations give sufficient data to com- pute A' and B' from A and B ; any possible ambiguity in the result being removed by the last equation, which shows that A' — B' has the same sign as H, since sin 2 ^ is always positive. Moreover, since, in the case before us, AB — H^ is positive, it follows from the second relation that A' and B' have the same sign. We are now ready to see what locus is represented by the equation (5). An important difference arises according as THE GENERAL EQUATKJN. 257 C is or is not zero, or, what is the same thing [ since C = ^ according as A is or is not zero. (1) A < 0. We may then write the equation in the form A'x"-+B'y"' = -C' A' B' Since A' and B' have the same sign, the denominators must be either both positive or both negative. In the first case, the equation may be written a-^ b- and the locus is an ellipse, including the special case of a circle. In the second case, the equation may be written in ,12 a- b and the locus is an " imaginary ellipse." (2) A = 0. The equation then is A'x"-+ B'y'" = 0. A' and B', having the same sign, can both be made positive, and the equation may be written and the locus is ix point. Summing up briefly, we say : 258 PLANE ANALYTIC GEOMETRY. If AB — H">0, the general equation of the second degree may he reduced to one of the three following forms : - — I- ^- = 1 a^ ^ h"" ' a b a- b" The locus represented is a real elUptse (including the circle^, an imagiiiart/ ellipse, or a point It was proved in chapter III that if A = 0, the general equation rep- resented two straiglit lines. Here we have a case where A = 0, and the equation represents a point. The contradiction is only apparent. For we may factor into (7 + ^'^^)(7-f^-) = »- Each factor placed equal to zero gives an equation of the first degree and hence may be said to represent a straight line, but because of the V— 1 involved, no real point lies on either line except the point (0, 0). We may say either that the equation represents " two imaginary straight lines," or, if we confine ourselves to real quantities, that it represents a point. 132. Case II. AB - H" < 0. The work proceeds exactly as in Case I, and we find, as before, the simple equation A'x"-+ B'y"'+C' = 0, where A'. B', and C are defined by the same formulas as before. THE GENERAL EQUATION. 259 But here A' and B' have opposite signs, since A'B' = AB - H-, and AB - H' < 0. As before, we consider two cases. (1) A<0. We may then write the equation in the form which is either C' ' C A' or ^+^z=rl. The locus in either case is an hi/jjerhola. (2) A = 0. Tlie equation then is A'x"-+ B'y"2 = 0, which may be written a- b-^ Hence the locus consists of two intersecting straight lines. Summing up, we say : If AB — H'- < 0, the general equation of the second degree may be reduced to one of the three following forms : ^' _ y!! — 1 a^ b^ ~ ' a^ "^ b^ ~ ' a- b- The locus represented is either an hyperbola or two intersect- ing straight lines. 260 PLANE ANALYTIC GEOMETRY. 133. Case III. AB — H^ =^ 0. As we have seen in § 130 it is not possible to begin, as in Cases I and II, by removing the terms of the first degree. We must try some other method of procedure, and the follow- ing seems most convenient. Since, by hypothesis, AB — H^ = 0, the terms of the second degree in the general equation make a perfect square, and the equation may be written (VAx+ VBy)2 + 2 Gx + 2 Fy + C = 0, (12) where VB is taken with the same sign as H. We will now assume such axes, OX' and OY', that VAx + VBy = shall be the equation of OX' with respect to the original axes. The line perpendicular to OX', namely VBx — V Ay = 0, will then be OY'. Then, since x' and y' are the perpendiculars on OY' and OX' respectively, , VBx — VAy x' = , } VA+B VAx + VBy y =^ — • Va + b Solving these equations for x and y, we have _ VBx' + VAy^ -^1+^ I (13) _ -VAx' +VBy' ^~ Va+b A comparison of these formulas with [20], § 44, shows that Vb OX' makes an angle 6 with OX, where cos $=^ , — = and Va+b THE GENERAL EQUATION. 261 - Va sin B = , . Hence ^ is a neaatiue anale, and is acute VA+ B or obtuse according as VB is positive or negative. Substituting the values of x and y in (12), v^ have VA+ B VA+ B or y'-' + 2 G'x' + 2 F'y' + C == 0, (14) -, gVb-fVa GH-FA where G — F' = V(A-(-B)3 VA(A+B)3 G V A + F Vb _ GA+ FH V(A+B)3 ~ V A(A + Ef C A+ B The partial rationalization of G' and F' is effected by multiplying numerator and denominator by Va. This, of course, would not be possible if A = ; but in that case H = (since AB = H^), and hence the equation is already in the form we are seeking to obtain, and may be treated as in § 70. The further classification now depends upon the question whether the value of G' be zero or not. But this in turn depends upon A. For if we substitute H = VAB in the ex- pression for A given in § 131, we find A = 2 FG VAB-AF^-BG^ = - (G Vb - F VA)- = - (A + B)' G'-. Hence, if A < 0, G' < 0, and if A = 0, G' = 0. We examine each case separately. (1) A < 0. Then G' is not zero, and the equation (14) contains a term in x'. The equation may be handled, therefore, by tlie method of completing the square as in chapter VI, § 70. 262 PLANE ANALYTIC GEOMETRY. There results finally y"2 = 4px", which is the equation of a parabola. (2) A = 0. Then G' = 0, and the equation (14) contains no term in x', but is simply y'2 + 2 F'y'+C' = 0. By the theory of quadratic equations, this equation may be factored into (y'-a)(y'-^)=0, i where a and /3 are the roots of the equation. Hence the locus consists of two parallel straight lines, ivhich 7nay be real, coincident, or imaginary. Summing up, we say : If AB — H'^^^O, the general equation of the second degree may be reduced to one of the two follounng forms : y"2 = 4px", (y'-a)(y'-^)-0. The locus represented is, therefore, either a pjarahola or two parallel straight lines, which may be real, coincident, or imaginary. 134. Summary. The results of the previous articles are exhibited in the following table which gives the simplest form to which the general equation may be reduced under the various hypoth- eses. General Equation. Ax" + 2 Hxy + By2 + 2 Gx + 2 Fy + C = 0. A = ABC + 2 FGH- AF2-BG--CH^ D = AB- H^. THE GENERAL EQUATION. 263 A^o A = o D>0 Ellipse ¥''''2 m'^'I V + v = 1 a- b- Imaginary Ellipse z? b2 Point (Two imaginary inter- secting straight lines) a^ b- D <0 Hyperbola a- b- Two intersecting straight lines a- h- D = Pai-abola y"2 = 4px" Two parallel straight lines (real, coincident, or imaginary) (y' - a)(y' - iS) = 135. Rule for Handling Numerical Equations. Out of the foregoing discussion we may deduce tlie follow- ing working rule for the reduction of any given numerical equation. Compute AB — H^. I. 7/'AB-H2<0. 1. Substitute X = Xo + x', y = Yo + y', and determine Xq and yo so that the coefficients of x' ami y' shall vanish. Compute the absolute term C, by s^chstitutiny Xq and yo in the given equation. 264 PLANE ANALYTIC GEOMETRY. 2. Compute A' and E' from the relations A'+ B'=A+ B, A'B'= AB-H^, noting that A' — B' has the same sign as H. 3. The equation is noiv A'x"-+ B'y"'+C'=0, and the form of the locus is at once evident. 4. The position of O'X" is found by passing a line through 2 H (xo, yo) making the acute angle ^ tan"' ~ with OX. A — B II. If AB- H2 = 0. 1. Write the equation in the form ( VAx + VBy)' + 2 Gx -f 2 Fy + C = 0, VB being taken ivith the same sign as H, and. let y' ^ , -? Va + b VBx — VAv X ^ — — • Va + b 2. Solve these equations for x cmd y in terms of x' and y' and substitute in the given equation. 3. The equation is now in the form y'2 + 2 G'x' + 2 F'y' + C' = 0, and the further reduction is to be made by completing the square. 4. The p)ositio7i of OX' is found by turning OX through a negative angle, tan~^ ( — -=^ Ex 1 8x2 - 4xy + 5y2 _ 36x + 18y + 9 = 0. Here, A = 8, B = 5, H = — 2. ... AB- H-^ = 3G. THE GENERAL EQUATION. 265 (1) We place x = xo + x', y = yo + y', obtaining 8x'2 - 4x'y' + 5y'2 + (16xo - 4yo - 3G) x' + (- 4xo + lOyo + 18) y' + Sxq^ - 4xoyo + 5yo- - 36x0 + 18yo +9 = 0. If we place 16xo — 4yo — 36 = 0, -4xo+10yo+18 = 0, we find xo = 2, yo = ~ 1- Hence the reduced equation is 8x'2 - 4x'y' + 5y'2 - 36 = 0. (2) A' + B' = 13, A'B' = 36. From these equations follow A' = 4 or 9, B' = 9 or 4. But since H is negative, we take A' = 4, B' = 9. (3) The equation is now 4x"- + 9y"2 = 36. Y Fig. 107. 266 PLANE ANALYTIC GEOMETRY. This is an ellipse, of which the major axis along O'X" is 6, and the minor axis is 4. (4) The line O'X" passes through (2, — 1) and makes angle i tan-i (— f) = tan-i 2 with OX. Ex.2. 3Gx2 - 48xy + lGy2 + 52x - 200y - 39 = 0. A = 36, B = 16, H = - 24. .-. AB- H2 = 0. (1) We write and place (2) Then (6x - 4y)2 + 52x - 260y - 39 = 0, , _ 6x — 4y _ 3x — 2y ^ V52 ~ Vis ' - , — 4x — 6y — 2x — 3y X ^ ^rz — - ^ = ^ * V52 Vl3 - 2x' + 3y' Vl3 — 3x' — 2y' Substituting, we have, after reduction, ~4 y'2 + Vl3y' + Vl3x' - ? = 0. (3) Completing the square, * (,.f7=-Vir=(.-4), or finally, y"^ = — Vl3x", ,,_ , , Vl3 ,,_ , 4 where y — Y +—?;—? x = x ^^• 2 Vl3 (4) The axis of the parabola is parallel to OX', the equation of which is 3x - 2y = 0. The positive extremities of the axes are marked in Fig. 108. THE GENERAL EQUATION. 267 Y . / / / A \S" ^\ ^ l^ V / / x^ — ^ -K, {" X' Xy" \v' Fig. 108. 136. Oblique Coordinates. We have assumed, thus far, that the general equation is referred to rectangular coordinates. If, however, the equa- tion Ax*-^ + 2 Hxy + By" + 2 Gx + 2 Fy + C = has reference to oblique coordinates, it may be transformed to any conveniently chosen pair of rectangular coordinates. Formulas for this purpose are given in § 46, and it has been proved in § 48 that su.ch a transformation does not alter the degree of the equation. Therefore, the new equation is of the form A'x'-^ + 2 H'x'y' + B'y" + 2 G'x' + 2 F'y' + C = 0. 268 PLANE ANALYTIC GEOMETRY. This equation may now be investigated by the methods of this chapter. Hence we have the result : Any equation of the second degree, whether referred to rect- angular or to oblique coordinates, re2)resents one of the conic sections. 137. Conic through Given Points. The general equation of the conic section, Ax^ + 2 Hxy + By2 + 2 Gx + 2 Fy + C = 0, contains six coefficients. However, only five of these are essential, since we may divide the equation by any one of them without altering its meaning. For example, we may make the last coefficient C equal to 1, except in the case when the conic passes through the origin. It follows, therefore, that in general one and only one conic may he 2)assed thi'ongh any five points in the plane. For, if we substitute for x and y the coordinates of the given points, we shall have five linear equations out of which to find the values of the five essential constants. A more convenient method of procedure is as follows : Let us take any four of the given points and connect them by straight lines, so as to form a quadrilateral (Fig. 109). Let the equation of PiPg be Ajx + Bjy + Ci = 0, or, more shortly, Ui = 0, where Ui is simply an abbreviation for the left- hand member of the equation. Similarly, let the equation of P2P3 be U2 = 0, that of P3P4 be U3 = 0, and that of P4P1 be Form now the equation UiU3 + kU,U4 = 0, (1) where k is an undertermined factor. This equation is of the second degree in x and y ; therefore, it represents a conic section. Moreover, this conic section passes through P^ ; for THE GENERAL EQUATION. 269 Fig. lOD. the coordinates of Pi make Ui = and U4 = 0, and therefore satisfy equation (1). Similarly, this conic passes through Po, P3, and P4. If now we substitute in (1) the coordinates of P5, we determine the value of k which we must assume in order that the conic may pass through P5. We thus deter- mine the equation of a conic through the five given points. Ex. Let it be required to pass a conic tlirougli tlie points Pi (2, 3), P2(-l, 2), P3(-3, -1), P4(0, -4), P5(l, 1). The equation of PiPo is x — 3y + 7 = 0, that of P2P3 is 3x — 2y -I- 7 = 0, that of P3P4 is X + y + 4 = 0, and tliat of P4P1 is 7x — 2y — 8 = 0. We form the equation (x - 3y + 7) (x -1- y+ 4) + k (3x - 2y + 7) (7x - 2y - 8) = 0, and, substituting the coordinates of P5, find k=:l- Hence, the required conic is (x - 3y + 7) (X + y + 4) + ^ (3x - 2y + 7) (7x - 2y - 8) = 0, or 109x2 - I08xy -|- 8y- + l(>i»x - lOy - 108 = 0. 270 PLANE ANALYTIC GEOMETRY. It is evident that the above method is always applicable, if no three points lie in a straight line. Hence : Through any five points, no three of which lie in a straight line, one and only one, conic may he drawn. If three of the points lie in a straight line, the method is applicable, but it is evident that the conic must be one of the limiting cases, for it must consist of the straight line in which the three points lie, and the straight line connecting the other two points. If four or five of the points lie in a straight line, the method is not applicable. It is geometrically evident that in this case the problem is indeterminate. For the conic may consist of the straight line in which the four points lie, together with any line through the fifth point, if that is not on the line with the four, or any line whatever if the fifth point lies on a straight line with the four others. If it is required to determine a parabola, only four points are necessary. This follows from the fact that one rela- tion connecting the coefficients is always given ; namely, AB — H^ = 0. We may use here, also, the method involving the parameter k, if no three of the points are in the same straight line. We form, as before, the equation UiU3 + kU2U, = 0. We form then the equation AB — H- = out of the coeffi- cients of this equation. The result is a quadratic equation in k, and hence we will have two, one, or no real parabolas, ac- cording as the values of k are real, equal, or imaginary. It should be noticed that in this connection "parabola" may mean two parallel straight lines. Ex. Let it be required to pass a parabola through the points Pi (1,-1), P2(2, 3), Pa (2, -5), P4(5, 7). We find the equations of the following lines : P1P2, 4x - y - 5 = ; P.Pg, x - 2 = ; P3P4, 4x - y - 13 = 0; P4P1, 2x-y-3 = 0. THE GENERAL EQUATION. 271 The equation of the conic is, then, {4x - y - 5) (4x - y - 13) + k (x - 2) (2x - y - 3) = 0, or (16 + 2k) x-'^ + (- 8 - k) xy + y^ + (- 72 - 7k) x + (18 + 2k) y + 65 + Gk = 0. Since this is to be a parabola, we must have (4 + ^)'-(16 + 2k) = 0, whence follows k = or — 8. The two parabolas are, therefore, IGx-' - Bxy + y2 -72x + 18y + G5 = 0, f - 16x + 2y + 17 = 0. The second equation represents a true parabola ; the first factors into (4x - y - 5) (4x - y - 13) = and represents two parallel straight lines. EXAMPLES. Determine the nature and position of the following conies : 1. 4xy + 3y2— 8x-16y + 19 = 0. 2. x2 — 6xy + 9y" - 280x — 20 == 0. 3. 11x2 — 4xy + Uf — 26x + 32y + 59 = 0. 4. 5x2 _ 26xy -f 5y2 + lOx - 26y + 71 == 0. 5. 4xy + 6x — 8y + l=0. G. x^ - 2xy + y2 + 2x — 2y + 1 = 0. 7. 13x2 ^ loxy + 13y2 + 6x - 42y — 27 = 0. 8. x2 - 4xy - 2f - 14x -f 4y + 25 = 0. 9. 6x-' — 5xy — fiy-' — 46x — 9y + GO = 0. 10. 4x2 — 8xy + 4y2 + Gx - 8y + 1 = 0. 11. x2 + Gxy + 9y2 - Gx - 18y + 5 = 0. 12. 41x2 — 24xy + 34y2 — 188x + llGy + 19G = 0. 13. 31x2 - 24xy + 21y2 + 48x - 84y + 84 = 0. 14. Show that the conic represented by the general equa- tion is an equilateral hyperbola when A = — B. 15. Show that, if the general equation contains the term in xy and not more than one of the terms containing x2 or y-, the conic is an hyperbola. 272 PLANE ANALYTIC GEOMETRY. 16. Prove that the necessary and sufficient conditions that the general equation shoukl represent a circle are A = B, H = 0, provided the axes are rectangular. 17. Show that the asymptotes of the conic are parallel to the two straight lines Ax- 4- 2 Hxy+ By=^ = 0. 18. Show that if A and B have opposite signs the conic is an hyperbola. 19. Show that the tangent to the conic at the point (xj, yi) is Axix + H(xiy + yix) + By^y + G(x + Xi) + F(y + y^j + C = 0. Find the equations of the conies through the following points : 20. (3, 2), (- 2, - 3), a, - 3), (2. - 2), (|, - f ). • 21. (1,-2), (6, 3), (3,2), (2,1), (9, 2). 22. (0, a), (a, 0), (0, - a), (- a, 0), (a, a). 23. (1,1), (-1,5), (2,4), (0,3), (3,1). 24. Find the equation of a parabola through the four points (5,-4), (9,4), (6,-1), (51, -21). 25. Show that a parallelogram cannot be inscribed in a parabola. 26. Given the base, 2b, of a triangle, and the difference, a, of the angles at the base, find the locus of the vertex. 27. Given the base, 2b, of a triangle, and the difference, d, of the tangents of the angles at the base, find the locus of the vertex. 28. A straight line has its extremities on the coordinate axes and passes through a fixed point ; find the locus of its middle point. PART II. SOLID ANALYTIC GEOMETRY. CHAPTER I. THE POINT. 1. Rectangular Coordinates in Space. z N n' M X- / F //' / . — ^/. / 1 / 1 y 1 / / / • • / / L ^ ,/ M I Fig. 1. Let XX', YY', and ZZ' be three lines meeting at so that each is perpendicular to the other two. They will determine 274 SOLID ANALYTIC GEOMETRY. three plaues, each of which will be perpendicular to the other two, and which will be called the coordinate planes, the three lines XX', YY', and ZZ' being called the axes of x, y, and z, respectively, or the coordinate axes. Distances from the plane YOZ measured parallel to XX' shall be denoted by x, x being positive if measured in the direction OX, and negative if measured in the direction OX'. Similar meanings are assigned to y and z. There can then be but one set of simultaneous values of x, y, and z corre- sponding to any given point. For through the point only one plane can be passed parallel to the plane YOZ, and the distance of this plane from the plane YOZ, measured accord- ing to the rule above, will be the value of x corresponding to the point. Similarly, there will be but one value of y and one value of z corresponding to the point. On the other hand, a point is completely determined if simultaneous values of x, y, and z for the point are known. For example, let x = a, y = b, z — c. If x = a, the point is in a plane parallel to the plane YOZ, and a units distant from it. If y = b, the point is in a plane parallel to the plane XOZ, and b units from it. If z == c, the point is in a plane parallel to the plane XOY, and c units from it. These three planes meet in one and only one point, which has for coordinates x = a, y = b, z = c, or the point (a, b, c). Therefore we see that, given the point, there can be but one set of coordinates corresponding to it, and that, given the coordinates, there can be but one point corresponding to them. In Fig. 1, OL, ML', M'P, NN' all represent x; OM, NM', N'P, LL' all represent y; and ON, MM', L'P, LN' all repre- sent z. It follows that the point may be plotted in several ways, ars represented in Fig. 2 ; but the first method shown is the one most often chosen. If the axes are not mutually perpendicular to each other, the coordinates are called oblique, but the rules for their THE POINT. 275 Fig. 2. measurement and the plotting of points are the same as those just given for rectangular coordinates. 2. Polar Coordinates in Space. In § 1 we have determined tlie position of a i)oint by- means of its three distances from the three coordinate planes respectively. In this article we are to determine the posi- tion of the point by means of its distance from a fixed point and the direction in which this distance is measured. Let (Fig. 3) be the fixed point, or origin, and OX and OZ be two lines of reference at right angles to each other and determining a plane XOZ. The distance OP of the point from the origin shall be called r. The direction of OP will 276 SOLID ANALYTIC GEOMETRY. Fig. 3. be determined if tlie plane in which it lies and the angle it makes with OZ are known. The angle ZOP shall be called (f), and the diedral angle between the planes ZOX and ZOP shall be called 6. We now have the point completely deter- mined, its coordinates being (r, , 6). At first it may not be evident that in polar coordinates the point is determined by the intersection of three surfaces, yet such is the case. For, if is given, the point is in one of the planes radiating from OZ. If with the axis. These two surfaces intersect in a straight line radiating from 0. Finally, if r is given, the point is on the surface of a sphere having its centre at 0, and is the point where the spherical surface is pierced by the above line of intersection. (See Fig. 4.) <^ will vary from 0° to 180°, from 0° to 360°, and r can have all numerical values. If we wish to give r a negative value, we may do so as follows : As noted above, the coordi- nates <^ and $ determine a straight line radiating from 0, and r will be positive if the point is on this line, and negative if the point is on the extension of this line backward through the origin. 277 ^ Fig. 4. 3. Distance between Two Points Pi (xi, yi, Zj) and Pj (xj, I2, 22)- Z Draw Pi Ml and P2M2 parallel to OZ and meeting the plane XOY at Ml and M2, respectively, and MiLj and MgLg parallel 278 SOLID ANALYTIC GEOMETRY. to OY and meeting OX at Lj and Lj, respectively. Then OLi = xi, LiMi = yi, MiPi = Zj, etc. In the plane XOY draw- Mo R parallel to OX, and in the plane determined by the par- allel lines PiMi and P2M2 draw P2S parallel to M1M2, the line of intersection of this plane with the plane XOY. Then P1P2 and M1M2 are the respective hypotenuses of the right triangles PiSPg and MiRMo. Therefore Pi P2' = S P2' + S Pi = ivuvi;'+sp;' = RM2"+ RMr+SPi - (xi - X2)^ + (yi - y,f + (zi - Z2)^ for RM2 = X2 — Xi, RMi = yi — ya, and SPi = Zi — z^. Or, if PiP2 = d, d = ^y(Xl - x,y + (yi - y^f + (Zi - Z2/. [1] 4. To find the Coordinates of a Point P (x, y, z) which divides the Line joining Pi (xi, yi, Zi) and P2 (Xo, Y2, Z2) so that PiP : PPo = li : L. z Fig. 6. THE POINT. 279 To find X, draw PiNi, PN, and P2N2 parallel to OX and meeting the plane YOZ at Ni, N, and Ng, respectively. In the plane determined by the parallel lines PiNi and PgNo draw PiR and PS parallel to NiNN., the line of intersection of this plane with the plane YOZ. Then NiPi = Xj, NP = x, and N2P2 = Xo, so that RP = x — x^ and SPo = X2 — x. The triangles PiRP and PSPo are similar, since their sides are respectively parallel. PiP_ RP PP.,~SP.,' Therefore or ll X — Xi whence X = Xo — X 1 1X2 + I2X1 ll + I. Similarly, y = \ ^\ ' ^^^^^ ^ ^ "[^^^TJ^ * ^2] 5. Direction Cosines. The angle between any two lines is the angle between their respective parallels drawn from a common point. Hence any line in space may be regarded as making angles a, ft, and y, respectively, with the axes of x, y, and z, these being the angles made with the axes by the line through the origin parallel to the given line. Then cos a, cos j8, cos y are called the direction cosines of the line. If P (x, y, z). Fig. 7, is any point, and 0P = r, and the direction cosines of OP are cos a, cos y8, cos y, by Trigonom- etry OL == OP cos a, OM =: OP cos /?, and ON = OP cos y, or x = r cos a, y = r cos 13, z ^ r cos y. Squaring and adding, we have X" + y' + z" = r^ (cos^ a -\- cos'- /5 + cos'-' y). 280 SOLID ANALYTIC GEOMETRY. oy. Fig. 7. But by [1], x2 + y- + z2=r^; .•. cos^ a + cos'-^ P + cos- 7 = 1, [3] a fundamental relation between the three direction cosines of any line. The quantities, r, cos a, cos (3, cos y, may be regarded as the coordinates of a point, since they are sufficient to fix it uniquely. Any three quantities, I, m, and n, are proportional to the direction cosines of some line ; for if cos a cos (3 cos y I m n ' and the value of this constant ratio be denoted by r, we have cos a = Ir, cos /? ^ mr, cos y := nr, whence, by substituting in [3], we find V> + m^ + n= THE POINT. 281 SO that cos a cos /3 = ^ =^, Vl^+ m" -|- n- cos y ^ VP + m^ + n'^ 6. Projection. The projection of one line on a second line is the portion of the second line included between two planes passed through the respective ends of the first line perpendicular to the second. 1. The projection of one line on a second is the product of the first line hy the cosine of the angle between the lines. R M Fig. 8. Let AB be any line, and MN and RS be planes through A and B, respectively, perpendicular to the line C D and meeting CD at the points A' and B', respectively. Then A'B' is the projection of AB on CD, and if 6 denote the angle between AB and CD, we wish to prove A'B' = AB cos $. Erom A' draw A'E parallel to AB and meeting the plane RS at E. Then A'E erpials AB, since they are parallel lines included between parallel planes. Biit, by Trigonometry, A'B' = A'E cos ^ ; therefore A'B' ^ AB cos ^, as we wished to prove. 282 SOLID ANALYTIC GEOMETRY. If we define the projection of a broken line on a given straight line as the algebraic sum of the projections of its segments on that line, we shall have, as a second theorem : 2. The projections on any given line of a broken line and of a straight line joini7ig the ends of the broken line are the same. Fig. 9. For example, the projection of the broken line ABCDE on MN is A'B'+ B'C' + C'D'+ D'E', which equals A'E', and A'E' is the projection of the straight line AE on MN. 7. Angle between Two Lines. Let OM and ON (Fig. 10) be two lines having as direction cosines cos ai, cos ^i, cos yi, and cos ag, cos /?2, cos yo, re- spectively. Let angle MON be denoted by 6. Take P (xi, yi, z^) any point of OM, distant ri from 0. By 1, § 6, the projection of OP on ON is ri cos 6, and the projection of the broken line OLMP on ON is OL cos a., + LM cos /3.J + M P cos ya, or Xi cos aa + Yi cos ^2, + Zi cos yo. Therefore, by 2, § 6, Xi cos 6=Xi cos a2 + Yi cos fi^ + z^ cos y^, whence cos 6^~ cos a, + ^ cos /3o + — cos y^, 1-1 ' ri ^ ri ' THE POINT. 283 Fig. 10. or cos 9 = cos ai cos ao + cos Pi cos Pa + cos 7i cos y^, [4] since by § 5, — = cos aj, etc. As the angle between any two lines is the angle between their respective parallels drawn from a common point, [4] is obviously true in all cases. If ^ = 90°, cos ^ =; 0. Therefore, if two lines are perpen- dicular to each other, cos ai cos tto 4- cos Pi cos po -\- cos "Yi cos y., = 0. [5] If two lines are parallel, it is evident that ai = tto, pi = po, \i = V2- [6] 8. Transformation of Coordinates, the New Axes being respectively Parallel to the Old Axes. Let OX, OY, OZ denote the original axes, and 0'X\ O'Y', O'Z' denote the new axes, the coordinates of 0' with respect to the original axes being Xq, yo, Zo- 284 SOLID ANALYTIC GEOMETEY. Let P be any point in space, its original coordinates being X, y, z, its new being x', y', z'. From P draw a line parallel to OX, and hence parallel to O'X', and meeting the planes YOZ and Y'O'Z' at the points M and M', respectively. Then, by definition, MP = x and M'P = x'; also, since parallel lines included between parallel planes are equal, MM' = Xo. But by (1), § 1, Part I, MP=MM'+ M'P. Similarly, .-. X = Xo + X'. y = Yo + y'» z = Zo + z'. [7] 9. Transformation of Coordinates from One Set of Rect- angular Axes to a New Set of Rectangular Axes having the Same Origin. .^'X' .,-Hr7 'M' /L Fig. 11. In iFig. 11, OX, OY, and OZ are the original axes, and OX', OY', and OZ' are the new axes, their direction cosines with THE POINT. 285 respect to OX, OY, and OZ being respectively cos ai, cos fii, cos yi, cos a^, COS jSg, COS y^, and cos as, cos jSs, cos ys. Let P be any point in space, its old coordinates being X, y, z, its new being x', y', z'. From P draw PM parallel to OZ and meeting the plane XOY at M, and draw ML parallel to OY and meeting OX at L. Also draw PM' parallel to OZ' and meeting the plane X'OY' at M', and draw M'L' parallel to OY' and meeting OX' at L'. Then OL = x, LM == y, M P == z, OL' = x', L'M' = y', M'P = z'. Now the projection of OP on OX is OL or x, and by 2, § 6, this is the projection of the broken line OL'M'P on OX. The projection of OL' on OX is OL'cos ai, for the angle between OL' and OX is aj. The projection on OX of L'M' is L'M' cos aa, for L'M' is parallel to OY' and the angle between OY' and OX is aj. Similarly, the projection on OX of M'P is M'P cos as. Therefore, by § 6, the projection on OX of OL'M'P is OL' cos ai + L'M' cos a., + M'P cos as, so that finally X = x' cos tti + y' cos a. + z' cos a-^. Similarly, y = x' cos Pi + y' cos Po + z' cos ps, [8] z = x' cos "Yi + y' cos y., + z' cos ^3. In similar manner, we can deduce the following expressions for x', y', z' in tei'ms of x, y, and z : x' ^ X cos ai + y cos ^1 + z cos yi, y'= X cos as + y cos (So -\- z cos yo, z' := X cos a;5 + y cos jSs + z cos ys. As was to be expected, these formulas are of the same form as [8], for the two transformations are of the same kind. 286 SOLID ANALYTIC GEOMETRY. In using these formulas the student should bear in mind that the axes are perpendicular to each other, and hence that cos tti cos a2 + cos (3i cos ^2 + cos yi cos y2 = O5 ^ cos aa COS as -|- COS (S^ COS /Sg + COS y^ COS yg = 0, I COS ag COS tti + COS jSg COS ySi + COS yg COS y^ = ; J also that cos ai COS )8i + COS ag COS /Jg + COS ag COS /Sg := 0, '^ COS /?i COS yi + COS ^2 COS ya + COS jSg COS ys = 0, I COS yi COS tti + COS y2 COS a2 + COS yg COS ag = ; I also, by [3], that cos- ai + COS^ I3i + COS^ yi = 1, cos^ ttg + cos^ ^2 + cos^ y2 = 1; COS^ ag + COS^ ygg + COS^ yg = 1 ; j and COS^ ai + COS^ ao + COS- ag = 1,^ cos^ /3i + COS- 132 + cos^ /Sg = 1, J> COS^ yi + COS^ y2 + COS^ yg = 1. J 10. All transformations involving only rectangular co- ordinates can be made by the use of one or both of these sets of formulas, as the case may be ; and since both sets give the original coordinates as expressions of the first degree in terras of the new, it follows that the degree of an equation is not changed by any such transformation. The proof is like that given in § 48, Part I, for the coordinates of the point in the plane, and for tliat reason need not be given here, though the student is advised to write out the proof for himself. 11. Relation between Rectangular and Polar Coordinates. If the lines OX and OZ used in defining the polar coordi- nates of a point are respectively the axes OX and OZ of rectangular coordinates, it is not difficult to express the rectangular coordinates of any point in terms of its polar coordinates, and vice versa. THE POINT. 287 Fig. 12. For if, in Fig. 12, the polar coordinates of P (x, y, z) are r, (j>, and 0, and PM is drawn parallel to OZ, it is readily seen that OP = r, Z ZOP = , and Z XOM = 0, since Z XOM is the plane angle of the diedral angle between the planes XOZ and MONP. Therefore and also and ON = OP cos <^, OM = OP sin <^; LM = OM sin 0, OL = OM cos 0. By substitution of the respective values of OP, ON, OM, L^ and OL in these equations we get, as our desired equations. z = r cos (|), X = r sin a-, we can find no corresponding point on the locus, for y becomes imaginary. But if Xi"-^ + zi^ < a^, i.e., if (xj, Zj) is within the circumference of the circle with radius a which has its centre at the origin, we find two corresponding points of the locus. If Xi^ + z^^ = a^, i.e., if (xi, Zj) is on the circumference of the above-named circle, the two values of y coincide. Since, then, (xi, Zi) may be taken anywhere within or on the circle, it follows that the locus has extension in two dimensions. It has, however, no thickness, for the two values of y corresponding to (xi, z^) are isolated, and the points between them do not belong to the locus. Hence the locus is a surface. A reference to § 3 shows that this surface is in fact a sphere with radius a and centre at the origin. 4. It is evident that the above reasoning may be applied to any given equation. For we may give to x and z any two values (xi, Zj), thus obtaining for the locus extension in two dimensions. We compute, then, from the equation the cor- responding values of y, which may be n in number. There are, then, n points of the locus perpendicularly above or below the xz-plane, but these points are isolated and the space between them does not belong to the locus. Hence the locus has no thickness. Therefore, we may say : A single equation involving the coonllnates of a j^oint repre- sents a surface. There are apparent exceptions to the above tlieorem, if we demand that the surface shall have real existence. Thus, for example, x2 -f- y2 -|- 22 z= — 1 is satisfied by no real values of the coordinates. It is convenient in sucli cases, however, to speak of ' ' imaginary surfaces. ' ' 292 SOLID ANALYTIC GEOMETRY. Moreover, it may happen that the real coordinates which satisfy the equation may give points which lie upon a certain line, or are even isolated points. For example, the equation x2 + y2 = is satisfied in real coordinates only by the points (0, 0, z), which lie upon the axis of z ; while the equation X2 + y2 + z-^ = is satisfied, as far as real points go, only by (0, 0, 0). In such cases, it is still convenient to speak of a surface as represented by the equation, and to consider the part which may be actually constructed as the real part of that surface. The imaginary part is considered as made up of the points corresponding to sets of imaginary expressions in x, y, and z, which satisfy the equation. With this understanding the general dis- cussion (4) may be read without change, there being no distinction made between real and imaginary quantities. 13. Two or More Equations. Two simultaneous equations represent a line. When two simultaneous equations are given, it is under- stood tliat the locus consists of those points, the coordinates of which satisfy both equations. Now, any point, the coordi- nates of which satisfy either equation, must be on the surface represented by that equation. Hence, any point, the coordi- nates of which satisfy both equations at the same time, must be in each of the surfaces represented by those equations, i.e., in their line of intersection. Therefore, two equations repre- sent the line of intersection of the two surfaces respectively represented by the equations when taken separately. As in the previous article, it may happen that it is necessary to speak of "imaginary lines." This happens when the real parts of the two sur- faces represented by the two equations do not intersect. In exceptional cases, also, the line of intersection contains only isolated real points. Such would be the case if either surface were of the kind mentioned in the latter part of the fine type of the previous article. Three independent equations taken simultaneously determine a mimber of points. INTERPRETATION OF EQUATIONS. 293 For the sets of values of x, y, and z, which satisfy three simultaneous equations, are known, by Algebra, to be limited in number, provided the three equations are independent. Geometrically, these values correspond to the points which the three surfaces represented by the equations have in com- mon. These points may be found by noticing the points in which the line of intersection of the first two surfaces meets the third surface. It is not necessary that all or any of these points should be real. 14. Cylinders. If a given equation involves only two of the coordinates, it might appear to the student to represent a curve lying in the plane of those two coordinates. For example, the equa- tion x^ + y^ = a"'^ might appear to represent an equation in the plane XOY. Such an interpretation of the equation would be incorrect, for it amounts to restricting z to the value z = ; whereas, in fact, the value of z, corresponding to two values of X and y which satisfy the equation, may be anything what- ever. Hence, corresponding to a point of the circle in the plane XOY, there is an entire straight line" in space, parallel to OZ and lying on the surface represented by x- -|- y- = a^. Hence this surface is a right circular cylinder. In like manner, z^ = 4py is a parabolic cylinder, the ele- ments of which are parallel to OX. So, in general : An equation containing only two coordinates represents a cylindrical surface, the elements of ivhich are jparallel to the axis of the missing coordinate. If only one coordinate is present in the equation, as, for example, in x''' — (a + b) x -f ab = 0, the locus is a number of planes. The above equation may be written (x — a) (x — b) = 0, 294 PLANE ANALYTIC GEOMETRY. and hence by reasoning like that of § 40, Part I, it represents the planes X — a = and x — b ^ 0. It is evident that any equation involving only one coordinate can always be treated in this way. As a plane is a cylindrical surface of which the directrix is a straight line, we may say in general that any equation not contaiyiing all three of the coljrdmates rejivesents a cylindrical surface. If the axes are oblique, this class of equations will repre- sent cylindrical surfaces of which the elements are not per- pendicular to the plane of the directrix. 15. Surfaces of Revolution. A surface formed by the revolution of a plane curve about one of the coordinate axes is called a surface of revolution. The equation of- such a surface is readily formed from the equation of the plane curve which is rotated to form it. By a few examples we will show how to do this, and at the same time we will deduce a rule by which we can often tell by in- spection that a given equation represents a surface of revo- lution. 1. The circle x^ + z^=a^ Fig. 13, is rotated about OX as an axis, forming the sphere of radius a. If Pi is a point of the circle, then the equation of the circle may be written OL' + LPT = a-". As the circle is rotated about OX, OL remains fixed, but LPi takes the position LP. Hence we have for the point P, Ol' + LP' = a^. INTERPEETATION OF EQUATIONS. 295 Fig. 13. If, now, X, y, z are the coordinates of P, OL = x and LP = Vy^+ z^. Therefore the equation of the sphere is x^ + y^ + z" = a^- Here we have formed the equation of the sphere from tliat of the circle by replacing z by Vy^ + z^. 2. The hyperbola — — -" = 1, Fig. 14, is rotated about OX. The equation of the hyperbola is OL' _ Lp;' ^ 296 SOLID ANALYTIC GEOMETRY. Z Fig. 14. When Pi is rotated into the position P, OL is fixed and LPi becomes LP. OL' LP' _ -. But OL = x, and LP=: Vy" + z^. Therefore the required equation is x; _ y- + z- _ or — — 7^ — 7^ = 1. b- The change from the equation of the hyjjerbola to that of the surface is effected by replacing z by Vy" + z"'^. x'^ z^ 3. The hyperbola ^ — ri = !> Fig. 15, is rotated about OZ. cl D The equation of the hyperbola is a^ b'^ As Pi is rotated about OZ, ON remains fixed and NPi becomes N P. Therefore NP' ON' ^ -I INTERPRETATION OF EQUATIONS. 297 Fig. 15. But for the point P, ON = z and NP = Vx'' + f. Hence the equation of the surface is jj2 + y'-^ _ z^ _ or x^ y' z- a- a- b"' 1. The change from the equation of the hyperbola to that of the surface is made by replacing x by Vx- + y-. We thus see that to obtain the equation of a surface of revo- hition formed by rotating a plane curve about one of the coordi- nate axes, we leave the equation of the plane curve unchanged as far as the coordinate parallel to the axis of rotation is con- cerned, and replace the other coordinate by the square root of 298 SOLID ANALYTIC GEOMETRY. the sum of the squares of itself and the coordinate not present before. Conversely, if in any equation two coordinates appear only in the form of the square root of the sum of their squares or some algebraic expression of that quantity, the corresponding surface is one of revolution. The equation of the generating plane curve may be formed from the given equation by replac- ing the radical by either of the coordinates involved and leaving the third coordinate unchanged. The axis of rotation is then the axis of this third coordinate. 1. For example, y^ + z" = 4px is the same as (Vy^ + z-)''' = 4px, and is, therefore, the equation of the paraboloid of revolution formed by rotating the parabola y^ = 4px about OX as an axis. 2. x2 + y2 + z- + 2 Gx + 2 F Vy"-' + z- + C = is the same as x2 + (Vy-' + z-)' + 2 Gx + 2 F Vy- + z- + C = and is, therefore, the equation of the surface formed l)y rotat- ing the circle x^-}-z--|-2Gx-f2 Fz-|-C = around OX as an axis. EXAMPLES. 1. If U and V are expressions involving the coordinates X, y, and z, show that I U + kV ^ is the equation of a sur- face passing through all points common to the surfaces U = and V = 0, and meeting them at no other points. 2. If U and V are expressions involving the coordinates X, y, and z, show that UV = represents the two surfaces U = and V = 0. 3. Find the equation of a cylinder, the base of which is an ellipse and the axis of which is OX. INTERPRETATION OF EQUATIONS. 299 4. Find the equation of a right circular cylinder which is tangent to the plane XOZ and has its axis in the plane ZOY. 5. Find the equation of a right circular cylinder which is tangent to the planes XOZ and XOY. 6. The axis of a right circular cylinder is the line x = a, y = b, and the radius of its base is r. Find its equation. What kind of a surface is represented by each of the fol- lowing equations ? 7. y2-[-z2 + 4y-Gz = 0. 8. z- + 7x = 0. 9. a V + by — a-^b^ = 0. 10. aV-by-F a-b2 = 0. 11. 3x2 _^ (jxy _^ 3y. _^ 2x + 7 = 0. 12. 2z- + z — 21 = 0. 13. xy + 2 = 0. 14. Find the equation of the surface generated by revolv- ing the straight line x =: 3 about OX. 15. Find the equation of the cone generated by revolving the line y = mx about OX. IG. Find the equation of the surface generated by revolv- ing the parabola y- = 4px about OY. 17. Find the equation of the surface generated by revolv- ing a straight line around OX : (a) when tlie line is perpendicular to OX ; (b) when the line is parallel to OX ; (c) when the line makes an oblique angle with OX. 18. Find the equation of the ring surface formed by revolving about OX the circle x- + (y — b)- = a^, where a> b. 19. Find the equation of the surface formed by revolving ay2 = x° about OX. 20. Find the equation of the surface formed by revolving a^y = x" about OX. 300 SOLID ANALYTIC GEOMETRY. 21. Find the equation of the surface formed by revolving xi + yi := as about OX. 22. Find the equation of the surface formed by revolving (x2 + ff = 8? (x2 — f) about OX. Interpret the following equations : 23. x- + y- = al 24. z" + X- — 7y = 0. 25. (x^ + z2)l + y^ = ai 26. y* — 16x^ — 16z- = 0. 27. What kind of a line is represented by the two equa- tions, 3x — 5 = 0, 4y + 1 = ? 28. What kind of a line is represented by the two equa- tions, x'' -f z^ - 6y == 0, y + 2 = ? 29. What kind of a line is represented by the two equa- tions, x- - 9y - 36 = 0, X + 5 = ? CHAPTER III. THE PLANE AND THE STRAIGHT LINE. THE PLANE. 16. Any Equation of the First Degree Represents a Plane. The most general equation of the first degree has the form Ax + By + Cz + D = 0. [10] Since, by § 12, this equation represents a surface, we have only to prove that this surface is a plane. Let Pi(xi, Yi, Zi) and Po(x2, y2, Zg) be any two points upon the surface represented by [10]. Then, since the coordinates of a point upon a locus satisfy the equation of the locus, we have Axi+By. + Czi + D-O, (1) Ax2 + Byo + Czo + D == 0. (2) Let us join Pj and P2 by a straight line, and take a third point P3, which divides the distance PiPo, internally or exter- nally, in the ratio of li : l.j. By suitably varying this ratio we may take P3 any point on the line P1P2. By [2], § 4, the coordinates of P3 are _ I1X2 + Ui _ Iiy2 + l2 yi _ liZo + I2Z1 ""'" I1 + I2 '^'~ I1+I2 '"'" Ii + l2"' If we substitute these values in the left-hand side of [10], it becomes |-^ (Ax2 + By2 + Cz2 + D) ^- |-^^ (Axi + By, + Cz, + D). This is equal to zero by virtue of (1) and (2). Hence the point Pg, which is any point of the line joining P^ and Pj, 302 SOLID ANALYTIC GEOMETKY. lies on the surface. Therefore the surface is a plane, since a straight line joining any two of its points lies entirely upon it. 17. To Find the Equation of a Plane in Terms of the Length and the Direction Cosines of the Normal from the Origin. Z Fig. 16. Let ABC be the plane, OD the normal from the origin, p the length of the normal, and cos a, cos ^, cos y its direc- tion cosines. Take P any point in the plane, with the coor- dinates X — OL, y = LM, z == MP. By 2, § 6, the projection of the broken line OLMP upon OD equals the projection of the straight line OP upon OD. The projection of OP upon OD is OD itself, since the angle PDO is a right angle. The projection upon OD of OL is x cos a, that of LM is y cos /B, and that of M P is z cos y. Hence we have X cos a + y cos p + z cos "Y = p. This is called the normal equation of the plane. The general equation of the plane. Ax + By + Cz + D = 0, [11] THE PLANE. 303 can be readily put in the normal form [11 J. For, as it stands, A, B, C are evidently proportional to cos a, cos /?, cos y, and, therefore, by § 5, A „ B cos a= — , cos B = — , , Va^+b^+c- Va^+b^+c^ C cos y = , • Va^ + b^ + c- Hence the equation, Ax+By + Cz+D ^,^ V A^' + B^ + C^ corresponds to [11]. The ambiguity in regard to the sign of the radical is removed by noticing that, in [11], the absolute term, if transposed to the left-hand side of the equation, is negative. Hence the sign of the radical must be so taken as to make the absolute term of the general equation nega- tive. 18. To Find the Equation of a Plane in Terms of its Intercepts upon the Three Axes. Let a, b, c be the intercepts upon the axes of x, y, z, respec- tively. To find a, we put y = 0, z = in [10], and take the corresponding value of x. Hence D Similarly, b = — — ' D Equation [10] may be written _D ' _D ' _ D ABC 1 + 1 + 1 = '- M 304 SOLID ANALYTIC GEOMETRY. 19. To Find the Equation of a Plane which Passes through Three Given Points. Since the points are on a plane, their coordinates satisfy an equation of the form Ax+ By + Cz+ D=rO. We have, therefore, Axi+Byi + Czi+D = 0, Ax,+ By2+Cz2+D = 0, Ax3+By3+Cz3+D = 0. These three equations are in general sufficient to determine ABC — , — , jr, and the values of these quantities substituted in the general equation give the required result. Two possible exceptions deserve notice. ABC (1) It may happen that the three equations in — , — , — are contradictory. This would always happen, if the true value of D, at the outset unknown, were zero. In this case the BCD student should try other ratios, e.g., -r, -r, -r-. (2) It may happen that the three equations are not inde- pendent. There are, then, an infinite number of solutions, and it will be found that the three given points lie upon a straight line. 20. Angle between Two Planes. Let the two planes be Ax+ By+Cz+ D = 0, and A'x+B'y + C'z+D' = 0. The angle between them is equal to the angle between their normals. Therefore, by [4], § 7, if ^ be the angle, AA'+BB'+CC cos u^ Va- + B^ + C^ VA'2 + B'2 + C'^ THE PLANE. 305 If the planes are perpendicular, AA'+ BB'+CC' = 0. If the planes are parallel, A__B ^C A'~ B'~C'" 21. Perpendicular Distance from a Given Point to a Given Plane. Let the equation of the given plane (M) be in the normal form X cos a -[- y cos 13 -{- z cos y = p. Let Pi (xj, yi, Zi) be the given point, and pi its required distance from the plane. Pass a plane (Mi) through Pi par- allel to (M). Its equation will be X cos a + y cos /3 + z cos y = p + pi, when Pj is on the opposite side of (M) from the origin, and X cos a + y cos j8 + z cos y ^ p — pi, when Pi is on the same side of (M) as the origin. Since Pi is on (M,), we may substitute its coordinates and obtain Xi cos a + yi cos ^ + Zi cos y = p ± pi, or ± pi = Xi cos a + Yi cos /8 + Zi cos y — p. Hence our working rule will be : Put the equation of the plane in tJieforni X cos a 4" y cos i8 + z cos y — p = and substitute for x, y, and z the coordinates of the point. The resulting value of the left-hand member ivill he the distance of the p)oint from the plane, and will he jtositive if the p)oint and the origiii are on opposite sides of the 2'lane, and will he negative if the point and the origin are on the same side of the plane. 306 SOLID ANALYTIC GEOMETRY. THE STRAIGHT LINE. 22. Any Two Simultaneous Equations of the First Degree Represent a Straight Line. For each equation, taken alone, represents a plane. The values of x, y, z which satisfy both equations represent points which are common to both planes. Such points lie evidently upon the straight line which is the intersection of the two planes. Hence the most general equations of the straight line are Ax + By + Cz + D =0, ) A'x + B'y + C'z + D' = 0. i Li«5J These equations may be put into another form by first eliminating z and then y. There results y = qx + r, ( z== sx + t. j These equations contain only four independent parameters. Therefore four conditions are sufficient to fix a straight line in space. These four conditions may be taken as the coordi- nates of the points in which the straight line cuts two of the coordinate planes. 23. Equations of a Straight Line in Terms of its Direction Cosines and Any Known Point upon it. Let Pi (xi, yi, Zj) be a known point of the line, P (x, y, z) any point, and cos a, cos (3, cos y, the direction cosines. On Pi P as a diagonal construct a parallelopiped, the edges of which are parallel to the coordinate axes. Then PiL = x — xi, LM=y — yi, MP ^z — Zj. THE STRAIGHT LINE. 307 P y' R ,'' / M Fig. 17. From the riglit triangle LPiP we have PiL= PiPcos LPiP, or X — Xi ^ r cos a, if PiP is denoted by r. Similarly, y — yi = r cos /3, z — Zi := r cos y. Hence Xi Zi [14] cos a cos p cos "Y which are the required forms of the equations. The following special cases deserve notice : (1) One denominator zero, as cos a, for example. Omitting, for the time being, the fraction containing this denominator, we shall have left the equation yi_z — zi , which will cos ^ cos y represent one plane tlirough the line. TUit if cos a:=0, a is some odd multiple of 90°, and the line is perpendicular to 308 SOLID ANALYTIC GEOMETRY. OX, and is, accordingly, in a plane perpendicular to OX. Since, however, one point of the line is in the plane x = Xi, which plane is perpendicular to OX, the whole line must be in that plane. Therefore, the equation of the second plane through the line will be x = Xi. (2) If two denominators are zero, as, for example, cos a and cos ji, the line, according to case (1), will be in the planes X = xi and y = yi, and these equations will be its equations, the fraction in z being superfluous. It is evident that in this case the line will be parallel to OZ. 24. To Place the General Equations of the Straight Line in the Form [14], Prom Ax + By + Cz + D = 0, 1 A'x+ B'y + C'z+D' = 0, f ^^^ we obtain, as shown in § 22, y = qx + r,) . z = sx + t. I ("'^ Equating the values of x from these two equations, we have y — r z — t X = = ) q s X — y — r z — t or — - — = = 1 q s The coordinates (0, r, t) correspond exactly to (xi, yi, Zi) of [14]. Hence cos a cos /? cos y 1 q s and, therefore, by § 5, 1 cos a = — =^^= > Vl + q^ + s^ Vl + q^ +s^ THE STRAIGHT LINE. 309 Hence the equations X — y — r z — t 1 q s Vl + q' + s^ Vl + q' + s- Vl + q' + s^ are in the form [14]. The above method is to be employed if the problem is to find the direction cosines of a given line. 25. Equations of a Straight Line Passing through Two Given Points. A line through (xi, yi, Zi) has the equations X — xi ^ y — yi _ z — zi _ cos a COS (3 COS y If it also passes through (xo, yo, Zg), then X2— Xi _ y2 — yi _ Z2 — Zi _ cos a COS /3 COS y By division we get, after changing the signs of the denomi- nators, xi — X2 yi — y2 zi — Z2 ■- -■ as the required equations of the line. The following are the special cases to be noted : (1) One denominator zero, as Xj — Xo, for example. Omit- ting, for the time being, the fraction in x, we shall have left the equation y — yi _ z — zi ^ yi — y2 zi — Z2 which will represent one plane through the line. Now if Xi — X2 = 0, X2 = Xi, and two points of the line are at the same distance from the plane YOZ, and hence the line is in a plane parallel to the plane YOZ and distant Xj from it. Therefore the equation of the second plane through the line will be X =: Xx. 310 SOLID ANALYTIC GEOMETRY. (2) If two denominators are zero, as, for example, Xj — Xj and yi — ys, the line, according to case (1), will be in the planes x = x^ and y = yj, and these equations will be its equations, the fraction in z being superfluous. The line is obviously parallel to OZ. (3) If all three denominators are zero, the two points which are to determine the line are coincident, and hence no line can be determined. 26. Plane through a Given Line and Subject to One Other Condition. If the eqixations of the line are Ax+ By + Cz+ D =0, A'x + B'y + C'z + D' = 0, then I (Ax + By + Cz + D) + k (A'x + B'y + C'z + D') = (1) is the equation of some plane through the line, I and k being arbitrary constants. For it represents some plane, since it is of the first degree ; moreover, the line lies entirely upon this plane, for the coordinates of any point of the line will neces- sarily satisfy the equation of the plane. Hence, (1) represents any plane passed through the line, and a second condition can be imposed upon the plane by giving different values to I and k. We will illustrate this work by the two following numerical examples. Ex. 1. Find the equation of a plane through the line x+y + z=0 2x + y + 2z = l and the point (1, — 1, 1). Let I (x + y + z) + k (2x + y + 2z — 1) = be the required equation. Since the plane passes through the point (1, — 1, 1), the equation of the plane is satisfied by the coordinates of this point. .-. 1(1-1 + 1) + k(2- 1 + 2-1) = or I = -2k. . • . the equation of the plane will be - 2k (x + y + z) + k (2x + y + 2z - 1) = or y + 1 rr 0. THE STRAIGHT LINE. 311 Ex. 2. Find the equation of a plane tlirough the line x+y + z=0 2x + y + 2z= 1 and parallel to the plane 3x + 2y + oz + 2 = 0. If the required plane is I (x + y + z) + k (2x + y + 2z - 1) = 0, or (I + 2 k) x + (I + k) y + (I + 2 k) z - k = 0, , .on I + 2 k I + k I + 2 k ^ ^ I + 2 k I + k by }t 20, — ^ — — - _^ - = — - — . But — 3 — = -— is the only independent equation here, and from this equation I = k. .-. k(x + y + z) + k(2x + y + 2z-l) = 0, or 3x + 2y + 3z — 1 = is the required plane. EXAMPLES. 1. What is the equation of a plane 5 units distant from the origin and perpendicular to the line joining the origin to (1, 6, 5) ? 2. How far is the plane 3x + 4y — 12z + 10 = from the origin ? Find the direction cosines of a line perpendicular to this plane. 3. Find the equation of a plane through the three points (1, 2, 3), (— 1, — 2, — 3), and (4, — 2, 4). What will be its intercepts on the axes ? 4. What is the angle between the planes 3x + 5y — Gz + 1 = 0, 3x — r>y — ()z + 1=0? 5. How far is the point (1, 5, 0) from the ])lane 3x-4y + 12z+26 = 0? G. A line making angles 60°, 45°, and C0°, respectively, with the axes of x, y, and z passes througli point (1, — 3, 1). What are its equations ? 7. Find the equations of a line passing through the points (6, 2,-1) and (3, 4,-4). 312 SOLID ANALYTIC GEOMETRY. 8. What are the direction cosines of this last line ? 9. Find the equations of a line passing through the points (1, 5, — 3) and (2, 3, — 3). 10. Find the equation of a })lane passing through the point (2, — 3, 3), perpendicular to the line of Ex. 7. 11. Find the direction cosines of the line of intersection of the planes 3x + y — 6 = 7z, 2x — 3y + 4z = 7. 12. Find the equation of the plane perpendicular to the line joining (2, 4, 6) and ( — 6, — 4, — 2) at its middle point. 13. Find the point of intersection of the line X— 2 _ y+l _ z— 2 3 "" 6 ~ 1 with the plane 3x + 4y — 6z = 0. 14. Find the angle which the line 3x+2y + 7 = 2y + 4z+3 = makes with the plane x-l-y + 3z — 2^=0. 15. Find the equation of the locus of points at a distance + 3 from the plane x + y + zH-3 = 0. 16. Find the locus of points equally distant from the planes x + 2y + 3z + 4 = 0, x-2y-f 3z-5=0. 17. A plane is passed through (2, 6, — 7) parallel to the plane 3x — 4y + 12z + 17 = 0. Find its equation and dis- tance from the given plane. 18. Determine the equation of a plane through (1, — 3, 1) and (3,-1, 3) which shall be perpendicular to the plane 2x-3y + Gz — 14 = 0. 19. What are the equations of a line through (1, 3, — 5) parallel to the line y = 3x-14^^ 7x-2z = 17 \ ' THE STRAIGHT LINE. 313 20. Is the point (6, 2, 1) on the straight line determined by (2, 4, 3) and (- 1, 6, - 1) ? 21. Find the angle between the lines 2 = 3y-2j (z = 2x + 3 22. If the above lines meet, find their point of intersection. 23. Where is the point of intersection of the lines 3x — y — 3z — 8 = 0) ,( x + y — z = ' and ' -' X — y+ z + 2 = 0j |6x — Oy — 3z — lo = 0j' 24. What is the equation of the plane determined by the point (1, 3, — 1) and the line 2x + r>y - 6 = 3y + 7z + 1 = 25. What is the equation of the plane determined by the two lines S«- y_l3 = 0) <7x-;)y-22 = 0) y + 4z+ l = 0S'""M2x + 3z- 3 = oi- 20. Find the equation of a plane passing tlirough the line I X + 2y + 1 =: ( j y-3z + 4 = j and perpendicular to the plane x + y + z = 0. 27. Find tlie equations of the projection on the plane 2x + 6y — 3z + 7 = of the line 4x - 3y + 15 = I 3y + z - 1 = j • 28. Prove that the three planes x + 2y - 3z + 1 = 0, y + 5z-l = 0, 3x + 4y — 19z + 5 = pass through the same straight line. 29. A line is drawn through (1, 2, 3) in the direction a = 45°, p = 60°, y = 60°. Find the coordinates of a point upon this line distant 5 units from (1, 2, 3). 314 SOLID ANALYTIC GEOMETRY. 30. Find a point in the plane XOY equally distant from the origin and the points (1, 2, — 1) and (3, — 1, — 3). 31. Find a point in the plane x + y + z =0 equally distant from the points (0, 1, 0), (1, — 2, 1), and (2, 3, 1). 32. Find the centre of a sphere of radius 7 passing through (— 3, 1, — 9) and (1, — 5, 3) and tangent to the plane 2x + 3y + 6z - 23 = 0. 33. Find the distance of the point (4, 5, — 6) from the line (3x-2y + 7 = ( 2y + z - 3 = 0. 34. Find the distance between the lines determined by the points (1, 2, 1), (- 1, 2, - 1), and (2, 1, 3), (3, 1, 2). 35. Show that, if the lines qx + r ) ( y = q'x + r' ^ ' and ' intersect, z = sx + 't ) ( z = s'x + t' r— r' t — t' q — q s — s 36. If cos ai, cos ySi, cos y^ and cos ag, cos p^, cos yo are the respective direction cosines of any two lines, and cos a, cos yS, cos y are the direction cosines of a line perpendicular to the two given lines, show that cos a cos p cos )8i cos y2 — cos ^2. cos yi cos yi COS ag — COS yg COS a^ COS y cos y COS ai COS 182 — COS a-i COS /?! 37. A point moves so that the sum of its distances from any number of given planes is constant. Show that its locus is a plane. 38. Pi (xi, yi, zi) and P2 (xo, y^, z.,) are any two given points. Find the locus of a point Q of the plane XOY so that the angle P1QP2 is a right angle. THE STRAIGHT LINE. 315 39. If Pi (xi, yi, zj) and P2 (xa, y2, Zo) are any two given points, find the locus of a third point Q snch that PjQ and P2Q are equal, or are in a constant ratio. 40. Show analytically that the locus of points equally dis- tant from any three given points is a straight line perpendic- ular to the plane of the three points. (Take the three points in the plane z = 0.) 41. Show analytically that the planes bisecting the diedral angles of any triedral angle meet in a common line. 42. If the normal distance from the origin of the plane which makes the intercepts a, b, and c respectively on the axes of X, y, and z is p, prove that p2 a^ ^ b^ ^ c' 43. If A, B, and C of equation [10] approach the limit zero, show that the limiting form of the equation is D = and that the plane becomes infinitely distant from the origin. 44. What two planes are represented by the equation x^ + 4xy -f 3y^ + xz + yz -f X + 5y + 2z — 2 = 0? CHAPTER IV. EQUATIONS OF THE SECOND DEGREE. 27. The Cones : -, ± ^ ± -^ = 0. a- D- c^ If we take all the different combinations of the algebraic signs possible, we see that there are four surfaces, the equa- tions of which are respectively "^-f. 4-^ = "^-t-L^^o a^ b2"^c-^ ' a^ b^ c^' But the last three equations agree in having two terms of the same sign, the third term being of the opposite sign, so that we have essentially but two equations to discuss, i.e., i!+i;.+L;=o, a^ b- c^ 2 9 9 and ^ + ^ - ^ = 0. a^ b- c" a^ ^ b'^ ^ c' This equation can be satisfied in real coordinates by (0, 0, 0) only, for it may be written (-] +(f) +('") ^=^j '^^^^ ^^^^ sum of three squares cannot be zero unless each of them is zero. Therefore, this equation represents a single real point, — the origin. If imaginary values of the coordinates are con- EQUATIONS OF SECOND DEGREE. 317 sidered, we should say that the equation represents an imagi- nary cone, the proof being carried on as in II. „. ?_: + ^: _ ?!, = 0. a- b- c- It is evident that the origin is a point of this surface, since the equation is satisfied by (0, 0, 0). Moreover, if (xj, y^, Zj) is a point of this surface, ( — Xj, — yi, — Zj) will also satisfy the equation, and the surface is symmetrical about the origin as a centre. In fact, this surface is a cone with its vertex at the origin. For, if P^ (xi, yi, Zj) is any point of this surface, the coordinates of any point on the straight line through the origin and P^ are mxi, myi, mzi, where m is an abritrary factor. Substituting these coordinates in the equation, we which is true, since Pi (xj, yi, Zj) is on the surface. Hence the line through the origin and Pi lies entirely upon the sur- face, so that the surface is composed of straight lines passing through the origin and is, accordingly, a cone. We will complete the study of this surface by the method of sections made by planes parallel to the coordinate planes. If we let z = 0, the resulting equation in x and y will be the equation of the plane section of the cone made by the plane XOY, of which the equation is z = 0. This equation is x" f ~2~t" u2^^^' ^° ^^^^^ ^^^^^ section is, by § 71, Part I, a single di D point, the origin. If we let z = Zi, some fixed finite value, the resulting equa- tion in X and y will be the equation of the plane section made by the plane z = Zi. Por, by placing z z= Zi instead of z = 0, we have virtually transferred the plane XOY, parallel to itself, X^ y2 2,^ through the distance z,. This equation is— „-l-fT, r^O, a"* b" c" 318 SOLID ANALYTIC GEOMETRY. which may be written — —, + /^ \2 — 1- Hence the sec- c 7 y c tion made by a plane parallel to the plane XOY is an ellipse, which increases in magnitude as the cutting planes are taken farther away from the origin, since its semi-axes are "-^and^^\ . c c If z = — Zi, we shall get a section equal to that made by the plane z = Zi, so that sections made by planes parallel to the plane XOY, equally distant from the origin but on oppo- site sides of it, will be equal. Hence the cone is symmet- rical with respect to the plane XOY. If y == 0, the equation of the section made by the plane x^ z^ XOZ is —^ 2 ^ ^' which may be written a c / ya c y Therefore, by § 71, Part I, this section is the pair of straight X Z X z lines, = and — I — = 0, which intersect at the origin. a c a c If y = Yi, the resulting equation is — -(- ^-, ;, = 0? which a" b" c" z- x^ may be written — — : ; — : = 1. Hence the section ^cyiV /ayi- ^ b; (,b made by a plane parallel to the plane XOZ is an hyperbola having its transverse axis parallel to OZ. Since y = — yi gives an equal section, the cone is sym- metrical with res2:)ect to the plane XOZ. As X and y appear symmetrically in the equation, it follows from the above that the plane YOZ cuts the cone in a pair of straight lines intersecting at the origin, and planes parallel to the plane YOZ cut the cone in hyperbolas having their EQUATIONS OF SECOND DEGREE. 319 transverse axes parallel to OZ, and also that the cone is symmetrical with respect to tlie plane YOZ. We may now sum up our work as follows : TJt.is surface is a cone generated by a stra'ujht line passhuj throwjli the origin and following an ellipse as directrix* The term which has its algebraic sign unique will show which of the coordinate axes is the axis of the cone. Fig. 18. If b^ = a", the equation becomes have a circular cone. x^ + y^ -„ ^ 0, and we * This theorem might have been stated immediately after the finding of the first elliptic section, but we continued to investigate other sections because of the instructiveness of tlie method. 320 SOLID ANALYTIC GEOMETRY. X- y^ z^ , 28. The Central Quadrics : ^ ± :j^ ± ^^ = 1. The distinct types of equation here are ^ _ y! _ l' = 1 a-2 b2 ^,2-1- In the case of each of these equations it is evident that if it is satisfied by (xi, yi, Zi) it is also satisfied by (— Xi, — yi, — Zj). Hence the origin is a centre of each of these surfaces, for which reason they are called central quadrics. Proceeding to a separate discussion of each of these sur- faces we will take them up in the order in which their equa- tions are written above. a- b- c- If z = 0, the equation of the section made by the plane x^ v^ XOY is - + ;-:; = l. This section, then, is an ellipse with a'^ b" semi-axes a and b. If z = Zi, the resulting equation is 3^2-1-1,2+^2 -^o^a^+b-i c^' and there are three cases to consider : (1) If Zi < c, 1 ^ is positive, and the equation may be written x^ , y^ ^ (w-^y (w-«' This section is, accordingly, an ellipse which decreases in magnitude as z^ increases toward c as a limit. x^ y2 (2) If Zi =:= c, the equation becomes "i + r^ == 0, and the section is a single point on the axis of z. EQUATIONS OF SECOND DEGREE. 321 (3) If Zi > c, 1 ^ is negative, and the section is imaginary. Since z = — zj gives an equal section to tliat made by tlie plane z = z^, the surface is symmetrical with respect to the plane XOY. Therefore, this surface is symmetrical with respect to the plane XOY, is bounded by the planes z ^ — c and z = c, and all sections made by planes parallel to the plane XOY are ellipses, which decrease in magnitude as we go farther from the coordinate plane and finally reduce to the two points (0, 0, — c) and (0, 0, c). Since x, y, and z appear symmetrically in the equation we can say immediately that : (1) The surface is symmetrical with respect to the plane XOZ, is bounded by the planes y = — b and y = b, and all sections made by planes parallel to the plane XOZ are x^ z^ ellipses, of which — ^ -| — ^ = 1 is the largest, and which de- a c crease in magnitude, as we go from the coordinate plane, to the points (0, — b, 0) and (0, b, 0). (2) The surface is symmetrical with respect to the plane YOZ, is bounded by the planes x = — a and x = a, and all sections made by planes parallel to the plane YOZ are ellipses, of which f 2 -f -„ = 1 is the largest, and which decrease in ^ b c- magnitude, as we go from the coordinate plane, to the points (— a, 0, 0) and (a, 0, 0). This surface is an ellipsoid, of semi-axes a, b, and c, and is represented in Fig. 19 for the case in whicli a > b > c. x^ y^ + z^ If c = b, the equation becomes — + ^— r-^ — = 1, the equa- tion of the ellipsoid of revolution, called the prolate spheroid, formed by revolving an ellipse around its major axis, OX. 322 SOLID ANALYTIC GEOMETllY. Fig. 19, If b = a, the equation becomes ~- + — = 1, the equa- d, c tion of the ellipsoid of revolution, called the oblate spheroid, formed by revolving an ellipse around its minor axis, OZ. x^ + y^ + z^ If c = b ^ a, the equation becomes —^ = 1, or a x" -j- y^ + z^ ^ a^, which is the equation of the sphere. II. ^" . y_ _ ^ =, 1 a- ^ b^ c^ If z ^ 0, the equation of the section made by the plane x^ y2 XOY is -;; + r^=:l, so that this section is an ellipse with a'' b^ semi-axes a and b. If z := Zi, the resulting equation is a^~^ b^ = 1 or {^^^ (^V C" = 1. Hence the section made by a plane parallel to tlie plane XOY is an ellipse, which increases in magnitude without limit as we go from the coordinate plane. Since z := — Zj gives an equal section, the surface is sym- metrical with respect to the plane XOY. EQUATIONS OF SECOND DEGREE 323 If y ^ 0, the equation of the section made by the plane XOZ is — ^ = 1, so that this section is an hyperbola with dL c its transverse axis on OX. If y = y,, the resulting equation is —,-\-Tr, — -^ =^ Ij or a'' b- c- -^ z=^l — 7^, and there will be three cases to consider : a^ c^ b- y,^ (1) If Yi < b, 1 — r-^ is positive, and we may write the equation x'^ _ z- — r,\ 2 ^^ b^ which is the equation of an hyperbola with its transverse axis parallel to OX. (2) If Yi ^ b, the equation becomes 'Vi-0 (Wi- which is tlie equation of a pair of intersecting straight lines. (3) If equation (3) If yi > b, 1 — =j-^, is negative, and we may write the (W^-0^ (^V? ^-1 1, which is the equation of an hyperbola with its transverse axis parallel to OZ. Examining these three cases, we see that as y^ increases from 0, the vertices of the hyperbola come nearer together, till, when y^ = b, they coincide and the hyperbola reduces to a pair of intersecting straight lines, and that when yi > b, tlie axis of the hyperbola is turned through an angle of 90°, and the vertices go farther apart as y^ increases. Since y =^ — yj gives an equal section, the surface is symmetrical with respect to the plane XOZ. 324 SOLID ANALYTIC GEOMETRY. As X and y appear symmetrically in the equation, it is evi- dent that the surface is symmetrical with respect to the plane YOZ, and the plane sections made by planes parallel to the plane YOZ are : (1) hyperbolas with the transverse axis par- allel to OY, if the cutting plane is between the planes x = — a and X == a ; (2) a pair of intersecting straight lines for the planes x = — a and x = a ; (3) hyperbolas with the trans- verse axis parallel to OZ, if the cutting plane is outside the space bounded by the planes x ^ — a and x ^ a. In the above work we have seen that XOZ and YOZ, both of which pass through OZ, cut this surface in hyperbolas. The question naturally arises : Do all the planes through OZ cut this surface in hyperbolas ? If we rotate the axes about OZ as an axis through any angle $, the formulas of transformation will be X := x' cos — y' siu 9, y ^=x' sin ^ + y' cos 6, z = z'; for in [8], § 9, ai^e, ^1 = 90°-^, yi = 0°, a2 = 90° + ^, /82 = 0, 72 = 0°, as = 90°, /?3 = 90°, 73 = 0°. The equation of the surface becomes 'cos^ , sin^ J + 2x'y' cos 6 sin d f—^ i ) , ,, /sin^ 6 . cos^ ^\ r . + y'( -T^ + ^T^ )--2 = l• If y':=0, the equation of the section made by the plane X'OZ' is "" ' ^^ + ~b^ J ~ ^ ~ ' EQUATIONS OF SECOND DEGREE. 325 or V F^' which is the equation of au hyperbola. But as is given different values, the plane X'OZ' becomes any plane through 01, and, therefore, all sections of this surface made by planes through OZ are hyperbolas. The surface is an unparted hyperboloid or hyperboloid of one sheet, with OZ as its axis, and is represented in Fig. 20 for the case in which a > b > c. Fig. 20. If b =: a, the hyperboloid becomes one of revolution, OZ being the axis of revolution, and its equation is x' + y - _L=:i^ III. y b- 326 SOLID ANALYTIC GEOMETRY. If X = 0, the equation of the section made by the plane YOZ is f,_; H — ^ = — 1, so that this section is imaginary. If X = Xi, the resulting equation is xi^ _ r _ z a- b^ c and there will be three cases to consider a^ b^ c? ^ °'' b^ "^ c^ ~ a^ ' (1) If Xi a, -^ — 1 is positive and the equation may be 3." written . v^ z^ (WS \ ^2 c a Therefore, all these sections will be ellipses of increasing magnitude as the cutting planes are taken farther from the coordinate plane. Since x ^ — Xx gives an equal section, the surface is sym- metrical with respect to the plane YOZ. Therefore this surface, which is symmetrical with respect to the plane YOZ, cuts OX at the points (— a, 0, 0) and (a, 0, 0) and has no points between the planes x = — a and X = a, while outside the space bounded by these planes it increases in magnitude as we go from the coordinate plane, all the sections parallel to the plane YOZ being ellipses. If z ^ 0, the equation of the section made by the plane X^ y2 XOY is — — ^ == 1, so that this section is an hyperbola with a D its transverse axis on OX. EQUATIONS or SECOND DEGREE. 327 If z = Zi, the resulting equation may be written b" c- (aVl + f)' (bVl + l') i = l- Therefore, the section made by any plane parallel to the plane XOY is an hyperbola with its transverse axis parallel to OX. The surface is also symmetrical with respect to the plane XOY, for z = — Zi and z = Zi give the same result. Similarly, the surface is symmetrical with respect to the plane XOZ, all sections parallel to that plane being hyper- bolas with the transverse axis parallel to OX. It may also be shown that every plane through OX cuts this surface in an hyperbola with its transverse axis on OX. The surface is a biparted hyperboloid, or hyperboloid of two sheets, having OX as its axis (the axis always being determined by the term of unique sign), and consists of two separate parts extending indefinitely away from the planes X == — a and x = a. Fig. 21 is for the case when a > b > c. -f-r-x Fig. 21. If c = b, we have the hyperboloid of revolution, a^ b^ "" ' of which the axis of revolution is OX. 328 SOLID ANALYTIC GEOMETRY. 29. Relation between the Cone r^ + vi "~ ll- = and the HOC Hyperboloid a^ + v- ~ c^ ~ If in place of x, y, and z Ave use r, cos a, cos ji, cos y as the coordinates of a point, the equation of the cone becomes COS" a COS^ P _ cos- ;- x _ ., '^T '^ b^ ~^ ) ~ ^-'■^ and that of the hyperboloid becomes COS^ a COS^ ^ COS^ y" a- b^ c" 1 COS^ a COS- yS COS- y (2) If now r, COS a, cos /?, and cos y are the coordinates of a point coanmon to the two surfaces, r cannot be zero since the hyperboloid does not pass through the origin. Therefore (1) cannot be satisfied unless COS^ a COS^ /3 cos- y a- b c But then we have from (2) - = 0, whence r = oo. r Therefore the cone, i.e., any element of the cone, intersects the hyperboloid at infinity. x" y" z^ Hence we say that -^-\- tt^ 2 ^^ ^ '*'* ^^*^ asym/ptotic cone of the hyperboloid x" y- z" In the same way we may prove —, — ^ 2 ^= ^ to be the o 9 9 X y z asymptotic cone of the hyperboloid —^ — f^ ^ ^^ 1- EQUATIONS OF SECOND DEGREE. 329 30. The Unparted Hyperboloid is a Ruled Surface. If we put the term ~ on the right-hand side of the equa- tion, the equation of this hyperboloid becomes 9 9 o ^ _ £ _ 1 _ r which may be written in the form since each member is the difference between two squares. Now, if ki is any constant whatever, it is evident that the above equation may be regarded as the product of the two equations / w a c y b a c ki I b But these last two equations, being simultaneous equations of the first degree, are the equations of a straight line. This line lies entirely upon the surface of the hyperboloid, for the coordinates of any point which satisfy these equations will necessarily satisfy the equation of the hyperboloid. By giving ki different values we can get a whole system of lines lying entirely upon the surface of the hyperboloid. The par- ticular line of the system, which passes through any given point (xi, Yi, Zj), is found by substituting Xj, yj, Zi in either equation and thus determining k^ Now, a surface which may be generated by means of a straight line is called a ruled surface, the sim})lest examples being the cone and the cylin- der. It appears, therefore, that the iinjjarted hyperboloid is a mded surface. There is one other way in which the factors of the two members of the equation 330 SOLID ANALYTIC GEOMETRY. a-c a + n = 0-n + ^B i_£ = k /i + y may be paired, i.e., c " V ^ These equations represent a second system of straight lines lying entirely upon the surface of the hyperboloid. Hence it appears that through any point of the unparted hyperbo- loid, two straight lines, one of each of the above systems, may be drawn upon the surface. These lines are called the rectilinear generators, and every point of the surface may be regarded as being determined by the intersection of two of them. X V" 31. The Paraboloids : T2^u — 4pz. There are bvit two surfaces to be considered here : (1) the '> 2 X V elliptic paraboloid, -^ + ^ = 4pz, represented in Fig. 22 for the case in which a > b and p > ; (2) tlie hyperbolic parab- oloid, — — 7^ = 4pz, represented in Fig. 23 for the case m a' b- which a > b and p !> 0. The detailed study of these surfaces by the method of plane sections is left to the student, the work of § 28 serving as a guide. 32. It will be noticed that the equations of the cones and the central quadrics can all be represented by the single equa- tion ^ + fi + ~2 = K where k is either zero or unity, and a b c where a'^, b^. and c^ can be given either sign, according to the EQUATIONS OF SECOND DEGREE. Z 331 Fig. 22. surface we wish to study. Accordingly, we will use this equation in the work that follows, writing our formulas as general formulas in which the student must make the ap- propriate substitutions for a^, b^, c^, and k in any particular problem. Fig. 23. 332 SOLID ANALYTIC GEOMETKY. 33. Tangent Plane. The locus of the taiujent lines drawn to the surface at any 'point is a plane called the tangent plane, the point being called the point of tangency or the point of contact. Let Pi (xi, yi, Zi) be any point of the surface x^ v^ z^ a" b- c'' ^ Then, ^1j^Y1^^1 = ^, (2) Transforming to P^ as a new origin by formulas [7], § 8, we get, as the equation of the surface, ^'_l/'_l^_! . 2xix_' 2y^y' 2ziz' x^^ y,=^ zi_^_, a^"^ b^'"^ c^'"^ a2 +-b2 "^ c^ "^a^^ b=^"^ c^~ ' x'^ y'^ z" 2xix' 2yiy' 2ziz' by reason of (2). If, now, we let x' = r' cos a, y' = r' cos y8, and z' = r' cos y, (3) becomes ,2 /COS^ a . COS" y8 COS^ y\ + 2rY^^" + ^^^ + ^^)=0, (4) in which the values of r' are the distances from P^, the new origin, to the surface measured along the line having as its direction cosines cos a, cos /?, cos y. One value of r' is always zero, since the origin is a point of the surface. And if cos a, cos /?, cos y have such values that this line is any one of the tangent lines to the surface, both EQUATIONS OF SECOND DEGREE. 333 values of r' in the above equation must be zero, and this can happen only when Xi cos g yi cos /3 zi cos y _ a- b'' c" ^ '' Multiplying (5) by r' and then replacing r' cos a by x', etc., we have , , , f + ^ + f=0, (6) in which x', y', and z' will be the coordinates of any point on any one of the tangent lines to the surface at P^. This equa- tion is, accordingly, the equation of the locus of the tangent lines at the point Pi ; and since this equation is of the first degree, the locus is a plane. Transforming back to the original axes by the formulas x' =: X — Xi, y' =: y — yi, z' = z — Zi, we have j^O^^O _^ y> (y - yO ^ zj^z-z^) ^ a" b" c'^ a^ + b-. + c'^ - a^' + b^ + c'^ ^^■> Therefore, reducing by the aid of (2) we have, as the required equation of the tangent plane, ^+^^r-+t?=k. [16] 34. Normal Line. The normal line at any point of a surface is the 2)erpen- dicular to the tangent plane at that pioint. Now, the direction cosines of the normal to the tangent plane will be, by § 17, i^ yi fi a^ b z" where R = V7? + y^ + ^' R R R ' >' a^ ' b^ c" 334 SOLID ANALYTIC GEOMETRY. Therefore, by [14], § 23, the required equations of the normal line will be, after the denominators have been multi- plied by R, X — Xi ^ y — yi ^ z — Zi Xi Yi ^ [17] a' b' c' 35. Polar Plane. Let Pi (xi, yi, Zi) be any point from which secant lines are x^ y2 z^ drawn to the surface ^ + f^ + — ^ k. a^ h^ c We shall define the polar of Pi, ivith resj^ect to this surface, as the locus of the line of intersectio7i of the tivo tangent jjkmes to the surface at the points where it is cut by any one of the secant lines from Pi. Conversely, Pi will be called the pole of this locus. Let any line from Pi meet the surface at the points P2 (x2? y2j Z2) and P3 (xs, y3, Zg). The equations of the line P2P3 will be, by [15], §25, X — xo _ y — y2 _ z — Z2 .^. X2 — X3 y2 — ys Z2 — Z3 ' and since Pi is a point of this line, xi — X2 _ yi — y2 _ zi — Z2 X2 — X3 y2 — ys Z2 — Z3 The tangent planes at Pg and P3 are respectively X3X I ysy , Z3Z b ^ + ^^ + ^-k = 0. (2) (3) The equations (3) are the equations of a straight line, the locus of which is the polar of Pi. To find the equation of EQUATIONS OF SECOND DEGREE. 335 this locus it is necessary to eliminate from (3) the coordi- nates, X2, y2, Z2, X3, ys, Z3. This may be done by aid of (2), as follows : Subtracting the second of the equations (3) from the first, we have (X2 — XQ X (y2 - Ys) y , (Z-2 - Z3) z ^ ^ a^ b- 0- ' ^ ^ If we multiply tlie first term of this equation by — -, the second term by — ^, and the third term by ~ -, y2 — ys Z2 — Z3 these multipliers being equal by virtue of (2), we have (xi - xo) X (yi — ya) y , (zi - zo) z ^ ^^ 32 1^2 (,2 Finally, by adding to (5) the first of the equations (3), we as the equation of the locus of the intersection of the two tan- gent planes at P2 and P3. But this equation, which is that of a plane, is independent of Xj, y2, Z2 and X3, y3, Z3, and is, there- fore, true for every secant line from Pi. ■■■'^ + ^+^ = ^ [18] is the polar of the point Pi (xi, yi, Zi). 36. Diametral Plane. The locus of the middle 2)oints of a system, of parallel chords of the surface a- b- c- is a plane, called the diametral plane. 336 SOLID ANALYTIC GEOMETRY. Let cos a, cos /?, COS y be the direction cosines of each of the system of parallel chords, and let Pi(xi, y,, Zi) be the middle point of any one of them. Transforming to P^ as origin by formulas [7], § 8, we have the equation of the surface in the form v'2 v/'2 7'2 9y y' '^\/ \l' ^7 7' y ^ ,, - 7 ^ ^ + ^u2 + -' + ^ + ^+"-^ + ^ + T;^ + ^ = k- (1) a^ b'' c- a^ h" c a- b- c- ^ ^ If, now, we let x'==:r' cos a, y' = r' cos (3, z':=-r' cos y, (1) becomes /COS^a i^COS^/S COS^yX ^ ^^Xi COS a _|_ yi COS ^^ Zj COS y But if cos a, COS /8, COS y are the direction cosines of the chord bisected at Pi, the new origin, the two values of r' which satisfy this equation must be numerically equal and of opposite sign, since they are measured in opposite directions. That such may be the case, (2) must be a pure quadratic equation, so that Xi cos a yi cos j8 1^ Zi cos y a^ b" c" ^ ' But (xi, yi, Zi) may represent the middle point of any one of the system of chords. X cos a , y cos B , z cos V _ r.^-y •■■ ^;r- + V^ + ^^ = [19] is the required equation of the diametral plane, for we can now call this locus a plane, since it is proved to be a plane by the form of its equation. As (0, 0, 0) satisfies [19], it is evident tliat every diametral plane passes through the centre of the surface, and, conversely, it may be proved that every plane through the centre is a diametral plane. EQUATIONS OF SECOND DEGREE. 337 37. Diameters. Ani/ straiffht line through the centre of a surface is called a diameter. It follows that any two diauieters must neces- sarily determine a plane through the centre of the surface, which will be a diametral plane by § 36. Noiv, three diameters such that the plane determined by two of them bisects chords parallel to the third are called conjxujate diameters. We will now determine the necessary conditions that three diameters of direction cosines cos ai, cos ^i, cos yi, cos a^, cos /?o, cos yo) ^nd cos ag, cos ySs, COS yg, respectively, shall be conjugate diameters. By [19], § 36, X cos ai y cos ^1 z cos yi _ ^2 -^ b^ -^ c- ~ ' ^^ is the diametral plane bisecting chords parallel to the tirst diameter. If these three diameters are conjugate, the second diameter must lie in this plane, and hence must be perpendicuhu* to the normal to the plane. Writing down the condition that these two lines shall be perpendicular to each other, and reducing, we have cos tti cos a2 . cos /8i COS ySo I cos yi cos ya „ a'^ ^ b^^ ^ ? -"' ^"^ as the necessary condition that the second diameter sluill be in plane (1). Similarly, if the third diameter is in plane (1), cos ag cos tti cos /?g COS /3i , COS yg cos yi a- b"" c^ ^ In like manner, if the first and the third diameters lie in the plane bisecting chords parallel to the second diameter, we have 338 SOLID ANALYTIC GEOMETRY. COS ai COS az , COS /8i COS ^2 , COS yi COS ya r^ ^.^ COS a2 COS gg C OS /jg COS jgg COS ya COS yg _ ^i +- b^ + ^i -<^5 (^) and, if the first and the second diameters lie in the plane bisecting chords parallel to the third diameter, we have cos as cos ai cos 183 cos I3i . cos yg cos yi cos a2 COS as COS (32 COS ySg , cos y2 COS ys ^^ ^ b^ ^ ^^ ~ ^^ On reviewing this article we see that there are but three distinct necessary conditions for the three diameters to be conjugate, i.e., cos ai cos a2 , cos (3i COS ^2 , COS yi cos y2 „ I' + b^ "^ c' ~^' cos ttg cos Us , COS ^2 COS ySg COS y2 COS yg ^ ^^ + b^ "^ ^^ ~ ' 093 ttg COS tti COS /Ss COS ^1 COS y3 COS yi and that, if these conditions are satisfied, the diameters, taken in pairs, will determine the three diametral planes, X cos tti y cos /81 . z cos yi "^"^ b^~"^ c' ~ ' X COS a2 1^ y COS (3^ j^z cos y2 ^ a X cos 03 y cos ^3 z cos y3 _ .1- "^ b^ "^ c^ ~ ' each of which bisects all chords parallel to the line of inter- section of the other two. EQUATIONS OF SECOND DEGREE. 339 38. Auxiliary Line of the Ellipsoid. In § 109, Part I, we have seen that, if x and y are the coor- X" v"^ dinates of any point of the ellipse —^ -\-^-=^l^ then x = a cos ^ 3. D and y = b sin <^, tlie point being thus determined by means of its eccentric angle. Similarly, if x, y, z are the coordinates of any point of the ellipsoid -+^+-=1 we may place x ^ a cos A, y = b cos fx, z = c cos v, where cos^ A + cos*^ /x + cos- v ^ 1, this condition being necessary in order that the assumed values of x, y, z may satisfy the equation of the ellipsoid. The quantities cos X, cos jx, cos v are, then, the direction co- sines of some line, which we will call the mixiliarij line of the point (x, y, z). It is to be noticed that the auxiliary line is different for different points of the ellipsoid. The use of the auxiliary line will be found to aid materially in the solu- tion of some problems, as indicated in the two following : 1. The auxiliary lines corresponding to the extreviities of two conjugate diameters of an ellipsoid a7'e 2i&^2Jendicular to each other. Let the extremities of tlie diameters be (xj, yi, Zj) and (x2, y2, Z2), respectively, and the lengths of the semi-diameters be ri and r,, respectively. Then their direction cosines will be — , — , — , and — , — , —, respectively, and since the diam- •"i Tj r^ r2 fz ''2 eters are conjugate we have, by § 37, jLjx^ y^ z^J g (1) rifa ( a-^ b^ c- ^ ^ ^ But if cos Xi, cos /xj, cos vi and cos A2) cos fi^, cos v^ are the direction cosines of the corresponding auxiliary lines, 340 SOLID ANALYTIC GEOMETRY. xi = a COS Xi, Yi = b cos fx-i, etc., so that substituting in (1) and reducing we have cos Xi cos A2 + cos fjLi cos ju.2 + cos vi cos 1/2 = 0, (2) an equation proving the theorem stated. 2. The sum of the squares of three conjugate semi-diameters of an ellijjsoid is constant. Let ai, bi, and Ci be conjugate semi-diameters, the extremi- ties of which are, respectively, (x^, yi, z^), (xg, yz, z-i), and (xg, yg, Z3), or, if we use tlie auxiliary line, the extremities of which are, respectively, (a cos Ai, b cos /^i, c cos j/i), (a cos A25 b cos ju.2, c cos V2), and (a cos A3, b cos jUg, c cos V3). Then, a^^ -\- hi -\- Cj^ := a^ cos^ Ai + b- cos^ /xi -\- cr cos^ vi 4" a^ cos^ A2 + b' cos^ fXi -\- c^ cos^ v^ + a^ cos^ A3 -\- \y cos^ fi^ + c^ cos^ V3. But, by virtue of our first theorem, the three auxiliary lines are perpendicular each to the other two, so that cos Ai, cos A2, cos A3 may be regarded as the direction cosines of OX with respect to a new set of rectangular axes, whence cos^ Ai + cos^ Ao + cos^ A3 = 1. In like manner cos^ fxi + cos^ fxo -{■ cos^ ju.3 = 1, and COS^ Vi + COS^ V2 + COS^ V3 = 1. Therefore aj^ + bi^ + Ci" = a^ + b" + c^ and our proposition is proved. 39. Parallel Sections. Two conic sections are said to be similar and similarly ■placed, if the coefficients of the quadratic terms in the equa- tions of the two are the same or differ only by a constant multiplier. From this definition it is easy to show by aid of chapter XI, Part I, that the axes of the two conies are paral- lel and proportional in length, and hence that the eccentrici- ties are equal. In particular, any two circles are similar conies, as are also any two parabolas. EQUATIONS OF SECOND DEGilEE. 341 We shall now prove that all parallol lAane sections of any surface of the second degree are similar conies. The most general equation of the second degree is Ax^ + By- + Cz- + 2 Fyz + 2 Gzx + 2 Hxy + 2 Lx + 2 My + 2 Nz + D = 0.* Now, if we let z = 0, the equation of the section made by the plane XOY, is Ax- + By2 + 2 Hxy + 2 Lx + 2 My + D = 0. If we let z = Zi, the equation of the section made by a plane parallel to XOY is Ax'^+ By- + 2 Hxy + 2 Lx + 2 My + Cz^^-f 2 Fyzi+2 HziX + 2 Nzi+ D = 0. In these two equations the coefficients of x^, xy, and y^ are the same. Hence sections parallel to the plane XOY are similar. We may, however, take any plane as the plane XOY, by a proper transformation of coordinates, without altering the degree of the equation. Hence the theorem is true universally. 40. Circular Sections. The problem of finding when a plane section of a central quadric will be a circle reduces, by § 39, to the problem of finding when a section made by a plane passing through the centre will be a circle. We will solve this problem for the ellipsoid K+j'.+^=i. (1) a b- c- * It may be shown that this equation represents one of the cones, the central quadrics, or the paraboloids, mentioned in the text, except in the cases in which the equation may be separated into two linear equations, the locus being then two planes. 342 SOLID ANALYTIC GEOMETRY. Now, all points which are a constant distance r from the origin lie upon the sphere t+t+'-.=l. (2) r r r If such points lie also upon the ellipsoid (1), then, by Ex. 1, chapter II, they lie upon the surface which passes through the origin. This equation (3) repre- sents, in general, a cone, thus showing that if points on the ellipsoid at a constant distance from the centre be joined to the origin, the joining lines form a cone, except in excep- tional cases. These exceptional cases occur when the equa- tion (3) may be factored into two linear factors, thus showing the corresponding surface to consist of two planes. But such factoring is possible only when r has such values that one of the quantities -^ -j, r^ ^, -^ -2 is zero. Accordingly, if a > b > c, let us place r = b ; equation (3) then becomes b- aO ^ U' b'-' 7.-rA=^, which factors into W^-i + ^i-i-o, Vs-^'-^Vi-^==»' and b^ a'^ ^c^ b^ thus giving two planes through the origin, which cut the ellipsoid in circular sections. These planes are determined by the axis of y and the semi-diameters in the plane XOZ which are equal to b. If Ave had made either of the other two coefficients zero, the equation (3) could not have been factored into real factors, EQUATIONS OF SECOND DEGREE. 343 and the reason is not far to seek. If we had placed r = a, thus making the coefficient of x^ zero, we shoukl have been required to draw in tlie phme YOZ a semi-diameter greater than b, which is the greatest semi-diameter in that plane ; on the other hand, if we had made r ^ c, thus making the coefficient z^ zero, we should have been required to draw in the plane XOY a semi-diameter less than b, which is the least semi-diameter in that plane. Treating the equation of the unparted hyperboloid i^ _L y' _ e! = 1 a^ "^ b^ c^ in the same way, we find that the planes cut the surface in circular sections, if a > b, and that conse- quently the planes pass through the semi-diameter a. If we apply this same method to the biparted hyperboloid and if b > c, the only way in which we can reduce this equa- tion to the equation of two planes is by letting r^ = — b^, when the equation becomes These planes will not cut the hyperboloid at all, but the })lanes parallel to them, Avhich do intersect the hyperboloid, will intersect it in circles. 41. Umbilic. If the plane cutting the ellipsoid in a circle is moved away from the centre, the radius of the circle will decrease as the 344 SOLID ANALYTIC GEOMETRY. plane approaches a position of tangency to the surface. The point of tangency thus obtained as the limit of a system of circular sections is called an umbilic. It is characterized by the fact that a plane indefinitely near the tangent plane and parallel to it cuts the surface in a small circle. X^ y2 jQ, Ex. Find the umbilics of the ellipsoid 77^ + tt + t =i- do y 4 Equation (3), § 40, is 2p cos yi = a- b - X cos a.2 , y cos B2 n and —^^ \- —^ - -P t'os y. = 0, 350 SOLID ANALYTIC GEOMETRY. of the parabolid -i,-\- h = 4pz, are such that a" b" COS tti cos ag , cos /Si COS (32 ^ a^ + b^ ~ ' then each phine bisects all chords parallel to the other. 49. Find the umbilics of the hyperboloid x^ y^ z^ 4 ~ 9 ~3G~ ■ FOEMULAS. (Rectangular coordiuates are used unless otherwise stated.) PART I. PLANE ANALYTIC GEOMETRY. Distance between two points : d= V(xi-x,)^ + (yi-y2)^. [1], §3. Slope of line through two points : ^ = yL_y^. [2], §3. Xi X2 Point of internal division : I1X.2+ I2X1 _ Iiy2 + l-iYi I. + I2 ' ^ I1 + I2 ' [3], §4. Point of bisection : x = li±^,y = >4X.. [4], §4. Point of external division : I 1X2— I2X1 _ Iiy2— l2yi '^^V=V''^^2- ™'§*- Area of triangle : Area PiPgPs^ i{(xiy2 — Xoy,) + (xoya — x^y^) + (x:,yi-x,y3)}. [6], §5. 352 PLANE ANALYTIC GEOMETRY. Relations between rectangular and polar coordinates : X = r cos ^, y = r sin 6. [7], § 10. r = Vx^ + f, = tan- 1 ^ • [8], § 10. Equations of the straight line : y = mx-|-b. [9], §21. ^ + ^ = 1. [10], §22. X cos a + y sin a= p. [11], §23. r cos (^ — a) = p. [12], §31. Line through a point with a given slope : y-y,= m(x-xi). [13], §33. Line through two points : yj-yi _ X — Xi yi — y2 xj — X2 [14], §34. Angle between two lines : tan /? = ^^ ^ • [15], § 35. Parallelism : trii ^ nria. [16]? § 35. Perpendicularity : rrii = [17], § 35. Line through a point perpendicular to a given line : y-yi = --(x-xO. [18], §36. Translation of origin : x = Xo+x', y = yo + y'. [19], §43. Rotation of rectangular axes through an angle 6 : X = X' cos «-/ sine ^ y = x' sm ^-|- y' COS ^. FORMULAS. 353 Transformation from rectangular to oblique axes with the same origin : x = x'cos^ + y'cos^', y = x' sin ^ + y' sin 6'. ^ ^' ^ Transformation from oblique to oblique axes with the same origin : _ x' sin (co — ^) + y' sin (o) — 0') sin CO ' ,.,,,.„ [22], §46. X sm y + y sin 6 y— __/ . ^ sin w Equations of the circle : (x-d)-+('y-e)2 = rl [23], § 49. x^' + y^'=r. [24], §49. x2 + y2 + 2 Gx + 2 Fy + C = 0. [26], § 50. Tangent to circle : xix + yiy + G (x + xO + F (y + y,) + C = 0. [26], § 54. Polar with respect to the circle : xix + y.y + G (x + xO + F (y + y,) + C = 0. [27], § 59. Diameter of circle : y-e = -^(x-d). [28],§61. Polar equation of the circle : r + rf — 2rri cos {6 - 6,) = a'^. [29], § 65. Equation of conic section : (x-2p)2+y^-eV. [30], § 66. Simplest equation of the parabola : r = 4px. [31], §67. 354 PLANE ANALYTIC GEOMETRY. Simplest equation of the ellipse : ^, + ^■=1. [32], §68. Simplest equation of the hyperbola : |-^, = 1. [33], §69. Tangent to conic section : Axix + Byiy + G (x + x^) + F (y + yO + C = 0. [34], § 76. Polar with respect to the conic section : Axix + Byiy + G (x + xO + F (y + Yi) + C = 0. [35], § 78. Diameter of parabola : y = ^- ' [36], §87. Parabola referred to diameter and tangent as axes : y2 = 4p'x. [37], §90. Polar equation of the parabola : 2p r = ' • 1 — cos 6 Diameter of ellipse : [38], § 91. y = - ^ x. [39], § 100. Condition for conjugate diameters of ellipse n^^rr^, = --,' [40], § 101. a Equations of the ellipse in terms of the eccentric angle : = a cos <^, y == b sin <^, [41], § 109. X FORMULAS. 355 Ellipse referred to conjugate diameters as axes : ^^ + ^. = 1- [42], §111. Polar equation of the ellipse : ''=^ ^— ^^- [43], §112. 1 — 6' COS'' 6 L J' o Diameter of hyperbola : y = ^ X. [44], §119. Condition for conjugate diameters of hyperbola : mim2 = -2. [45], §120. Hyperbola referred to conjugate diameters as axes : ^,1-^, = 1. [46], §121. Hyperbola referred to the asymptotes as axes : a- + b' xy = ^ . [47], § 127. Polar equation of the hyperbola : '' = -2 ?^ T • [48], § 128. 6'' COS^ 6—1 L J^ o Keduction of the general equation of the second degree, when AB- H'<0: To remove the terms of the first degree, transfer the origin to (xq, yo), where Axo+Hy„+G-0, Hxo+ By„+ F=0. 356 PLANE ANALYTIC GEOMETRY. To remove the xy term, rotate the axes through an angle 6, where 2 H « = *»--' a3^; the coefficients A', B', C of the resulting equation A'x"^+ B'y"- + C' = will be determined from the equations C = Ax,/ + 2 Hx„yo + By,2 + 2 Gx„ + 2 Fy., + C, A'+ B'= A+ B, A'B'= AB- H-, where A'— B' shall have the same sign as H. Reduction of the general equation of the second degree, when AB- H2=0. To remove the xy term, rotate the axes through such an angle that the equation of OX' shall be VAx+ VBy = 0, ,_ VBx — VAy ' ~ Va+~b~' ,_ VAx+VBy ' ~ Va+b ■ and The reduction may now be finished by the method of completing the squares. FOEMULAS. 357 PART II. SOLID ANALYTIC GEOMETRY. Distance between two points : d-V(xi-x,)^+(y,-y,)^+(zi-Z2)^. [1], §3. Point of division : I1X0 + I2X1 l,yo+loyi liz„+l,z, Direction cosines : COS^a+COS^^ + COS'^y^l. [3], § 5. Angle between tw^o lines : COS 6 = COS ttj cos a^ + COS ^j COS (3^ + COS yi COS y2- [4], § 7. Perpendicularity : COS ai cos a2 + COS /3i COS ^2 + COS yj COS y^ = 0. [5], § 7. Parallelism : ai = a2, /3i = (32, 71 = 72- [6], § 7. Translation of origin : X = Xo + x', y = yo + y', z = Zq + z'- [7], § 8. Rotation of rectangular axes : X ^ x' cos tti + y' cos a2 + z' cos as, y = x' cos /?i + y' cos /?2 + z' cos /Sj, [8], § 9. Z = x' cos yi + y' COS yo + z' COS ys- Relations between rectangular and polar coordinates : z = r cos (f>, X = r sin <^ cos 6, [9], § 11. y = r sin <^ sin 6. 358 SOLID ANALYTIC GEOMETRY. Equations of the plane : Ax+By + Cz+D = 0. [10], §16. X cos a + y cos /5 + z cos y = p. [11], § 17. - + ^ + - = 1- [12],§18. a b c L j^ o Equations of the straight line : Ax + By + Cz + D = 0, A'x+ B'y + C'z+ D' = 0. yi [13], § 22. [14], §23. [15], §25. §27. §28. §28. X" y z" Biparted Hyperboloid : ^ ~ r^ 2 — 1- § 28. 2 2 Elliptic Paraboloid : ^ + f-, = 4pz. §31. a^ h- x^ f Hyperbolic Paraboloid : ~i~h^ 4pz. § 31. a b cos a cos /? cos y ' x — Xi _ y — yi_ z — Zi Xi- - X2 yi — y2 Zi — Z2 Equations of the second degree : Cone : 9 9 Ellipsoid TTT^T^Ol.fr:>^^ TTttv 9 9 9 a^ ^ b^ ^ c^ Tangent plane : ^ + f+7»=''- [16], §33. FORMULAS. 369 Normal line : _i Zi Zi d? b^ c^ Polar plane : f + ^^ + f = l'. [18], §35. Diametral plane : [19], §36. X cos a , y cos y8 z cos y PART I. CHAPTER I. 1. 2V4T, |. 2. V73, -|. 3. 2V5, -|; Vri, -|; V26, -5. 4. Vl3, I; V4T, f; 7 ^^, 1. 7. 2 (a + b + Va^Tb^). 10. (V-, V). 11. (- -¥, - i)- 12. (0, I). 13. (f , V), (V, ¥)• 14- (24, 18). 15. (- 5, 21). 16. (13,27). 17. (-11, -16). 18. (7,27). 19. 4, V37, Vei. 20. V37, 6. 21. Vl9, ^V3. 22. 1. 23. 31. 24. 18. 25.5:3. 26. 11^, 2J. 27. 3V5 + 2 V3 +V26 - 5V3 + V53 - 14 V3. 30. Vl9, 3V3, V9T; fV3. 31. V1O-3V34- V34-I5V3 + ^^; ^(18-5V3). 35. (-f, 1)^ 36. (±|V2, =Fl^)- 37. (±|fV34, ±i|V34). 38. (f, 3f). CHAPTER II. 31. y = 4x. 32. 6x - 3y + 14 = 0, 6x - y + 10 = 0. 33. 3x — 7y + 2 = 0. 34. 5x + 4y + 24 = 0, 5x + 4y - 26 = 0. 35. x2 + y2 + 2x + 4y-20 = 0. 36. y2-8x + 16 = 0. 37. r = tan 6. 38. r = a^. 39. x2 - 3y2 - 8y + 16 = 0. 40. 5x2 + 5y2 _ 50x - 46y + 22 = 0. 41. 5x2 _ 4y2 + 24x + 16y - 52 = 0. 42. (- f , \). 43. (23, - 16). 44. (- 1, 0). 45. (0, 0). 46. (tV, H), (-3, -3)._ 47. (3, -3±2V6). 48. (-3^5,4). 51. .(5±W5, 20t4V5-^ 52. (- ffVio, /,V^). 53. (-H, -3f). 56. (±5, ±4), (±4, ±6). 57. (- 2, 1), (- H, - If). 58. (± 5, 0). 59. ± 7 VlO. 362 PLANE ANALYTIC GEOMETRY. 60. ±7. 61. f. 62. ±iV6. 63. ±fV2. 66. (IxV, Hi)- 67. (± 15, zf 5). _ 68. (12^, 0). 69. (2i, 21), (-11, If). 70. (±V5, ±2V5). 71. (4, 3±V2T). a 13. h 14. 2, 15 i-O 16. 0, 17. -3, 18. 00, 19. y = X + 3. 20. 2y = (-jQ- V2)x - 4. 21. x + y = 0. 22. 2x + (V3 + l)y =4V2. 23. x Vg + 2y - 10 V2 = 0. 25. a = - i, b = i, p = CHAPTER IIL b P m cos a sin a -1, 1 Vio' 3, 3 Vio' 1 Vio -5, 10 V29' 2» 5 V29' 2 V29 2> 0, 2, 0, 3 -2, h |V5, 4. W5. CO,. 3, CO, -1, 0. h h 0, 0, 1. 2V13 + 6V3' - 3 . 1 L, sin a 2V13 + 6V3 V13 + 6V3 28. -%= • 29. ^ VlO. 30. Vl3. 31. ^\ VlO. V37 32. 3x + 4y - 20 = 0, 3x + 4y + 10 = 0. 33. 2x + y + 2±2V5 = 0. 34. 2x + y - 3 =: 0, x - 2y + 4 = 0. 35. 4x - 3y - 3 = 0. 36. x - y - 3 = ; (3, 0), (0, - 3). q 17 20 37. 2x + 5y + 17 = 0; -^, -iL, ^- 38. f. V29 V29 V29 39. 3x - 2y - 4 = 0. 40. llx - 3y + 18 = 0. 41. 2x — y — 7 = 0. 42. X + 1 = 8. 43. x - y + 2 = 0. 44. tan~^ if 45. tan~^ |. 46. 5x + 3y + 4 = 0. 47. X — 2y = 0. 48. 30x + 24y — 35 = 0. 49. 4x - 14y + 5 = 0. 50. 5x - 27y + 5 = 0, 15x - 23y + 15 = 0. 51. 21x + y + 101 = 0, 19x - 9y + 131 = 0. 52. 3x - 28y - 37 = 0, 27x - 8y - 89 = 0. 53. 7x - 24y + 17 = 0, x - 1 = 0. 54. 8x - 7y - 59 = 0, 8x + 7y + 11 = 0. 55. x + y = 0. 56. 29x + 32y + 15 = 0. 57. 4x - 4y - 9 = 0. ANSWERS. 363 58. 161x - 4(;y — 88 = 0. 59. 4x + 3y - 1 7 = 0. 60. x-y-l = 0. 61. 2x-5y-41=0. 62. X + y - 9 = 0. 63. 8x - y - 17 = 0. 64. 25x - 3y + 3 = 0, 31x + 78y - 190 = 0, 37x - 122y + 10 = 0. 65. 3. 66. 99x-27y + 61 = 0, 21x + 77y- 121 =0. 67. (1 ± V3)x + (2 ± V2)y + 5=0. 68. (15 ± 4 ViO)x + (5 qi 3 Vl7))y = 0. 71. VSS, -^L • V58 72. (2, 0), (12, 0), (5, 7) ; tan"^ |, 45°, tan"' §. 73. f Vil, |, § VtI. 74. 2x - y — 8 = 0, llx - lOy - 17 = 0, x — 5y - 22 = 0. 9 45 45 75 ' — 1 — — • V5 V22I V2G 76. 2x + 4y - 13 = 0, 20x + 22y - 51 = 0, 5x + y + 7 = 0. 77. (— 2^5,, 4^8). 78. V29 + 2 Vn + 5, 13. 79. llx + 6y-87 = 0, x-.3y-2 = 0, X -7 =0. 80. (7, |). 81. 2x + 5y - 8 = 0, X - 4y - 1 = 0, 4x - .3y - 10 = 0. 82. 8x + 2y - 55 = 0, 3x + 4y — 32 = 0, 5x — 2y - 30 = 0. 83. e = c. 86. ((V6-V2)a, I). 87. (^^ a, | W^ a, ^J^V 88. 12x - 7y + 8 = 0, Vl93. 89. (- 14^, 9^). 91. (- 2 j-\, 2^1). 102. X + y = 0, 2x + y - 1 = 0. 103. 3x - y + 3 = 0, x + 3y + 1 = 0. 108. tan~^ 2 VH-^ - AB . ' ' A+ B CHAPTER IV. 1. 3x + 4y = 0. 2. X- + y2 = 10. 3. y2 = x'' - x. 4. y2 = 8x. 5. xy = — 8 or xy = 8. 7. 3xy = 1. 8. 2x2 + 2y2 = 25. 9. x2 = 4y. 10. 4x2 — 6y2 = 3. 11. 12x2 + 16y2 = 51. 12. 3x2 + 4y = 0. 13. xy = 0. 14. 3x3 + ^y'^ - ^- ^6- ■■ ^"^ (^ ~ '^) ~ p- 17. r2 sin 2 61 = 14. 18. r-^ = -, , ,^"^" . -^ ■ a — (a — b-) cos- 6 2 a sin2^ 19. r = 8a (cos ^ + sin 6). 20. r = -— • 21. r = a tan ^. ^ cos 6 22. x + y=10^^. 23. x =4V3. 24. x- + y'- - ay = 0. 25. (x2 + y2 + ax)2 = a2(x2 + y-). 26. (x2 + y2)3 = 4 a2x-y2. 27. (x2 + y2)2 = a2 (x2 - y2). 28. (x-' + y2)3 = a2 (x-' - y2 + 2xy)2. 22. 364 PLANE ANALYTIC GEOMETRY. CHAPTER V. 5. x2 + y2 + 2x — 12y + 32 = 0. 6. x2 + y2 - 4x + 6y + 8 = ; x2 + y^ - Ox + 2y + 5 = 0. 7. x2 + y2 + 3x — y - 10 = 0. 8. x^ + y^ = a^ + b^. 9. x2 + y2 + 4x — 2y = 0. 10. x'^ + y^ ± 2ax = 0. 11. x2 -f- y2 ± 2ax ± 2ay + a^ = 0. 12. 25x-' + 25y2 =^ 49. 13. 5x2 + 5y2 + lox - 20y -|- 24 = 0. 14. 13x2 + i.3y2 _ 7yx + 52y + 151 = 0. 15. x2 + y2 + 4x + 4y + 4 = ; 9x2 + 9y2 + 12x - 12y + 4 = 0. 16. (3, 1); Vn. 17. (-2, 5); 2. 18. (2, -3); 2V3- 19. (1, -1); iVl41- 20. (I, - I); Vt. 21. (i, - 1) ; 0. /_G^ _FN VG'^+F2-CA. 23_ 6^_5 +3^0. \ A A / A 25. x2 + y2 - 19x - 17 y = 0. 26. 3x2 + ,3y2 _ 5x + y _ 12 = 0. 27. 3x2 + 3y2 + 2x + 2y — 10 = 0. 28. x2 + y2 = a2 + b2. 29. 19x2 + 10y2 — 65x — 55y — 50 = 0. 30. x2 + y2 - 8x - 9 = 0. 31. x2 + y2 - 3x + 4y - 31 = 0. 32. 2x2 + 2y2 + 2x - y - 15 = 0. 33. x2 + y2 — 3x — lly — 40 = 0. 34. x2 + y2 - 8x - 20y + 31 = ; x2 + y2 + lOx + lOy - 35 = 0. 35. 5x2 4. j-jyi _ 22x — 6y - 39 = ; 5 x2 + 5y2 - 8x - 4y — 61 = 0. 36. x2 + y2 + 2x - 14y + 41 = 0. 37. x + y = 0. 38. 2 (a - a') X + 2 (b - b') y = a2 - a'2 + b2 - b'2. 40. x2 + y2 + 4x — 6y — 36 = ; x2 + y2 - 80x + 50y + 1000 = 0. 41. 3x2 + 3y2 + 5x _ 5y _ 20 = ; 40x2 + 40y2 _ 400x + 520y + 2429 = 0. 42. 3x2 + 3y2 + lly - 17 = 0. 43. 3^2 + 3y2 + 4x — 4y — 19 := 0. 44. x2 + y2-2x + 4y — 20 = 0. 45. yV3 = x±4V3. 46. y = xV3 + 2V3 + 7; y=xV3+2V3-l. 47. 6x — 8y + 49 = ; 6x — 8y — 71 = 0. 48. x - 2y ± 5 = 0. 49. 3x + 2y + 1 ± 3 Vl3 = 0. 50. 3x - 4y ± 20 = 0. 51. 0, — f 52. y -e= m(x — cl)± rVl +m2. 53. y + F = m (x + G) ± V(G2 + F2- C) (1 + m2). 54. 2x - 3y — 13 = ; 3x + 2y = 0. 55. y + 3 = ; x - 2 = 0. 56. 3x - 4y + 50 = ; 4x + 3y + 25 = 0. 57. X — 2y = 0; 2x — y — 3 = 0. 58. 4x + 2y Vs - 12 - 3V5 = 0. 59. 2x - y + 2 = 0. 60. Vl7. 61. V6 ; 0. 62. lOx - 7y + 4 = 0. 66. i^^, i). 68. ^ • ANSWERS. 365 70. I Vs. 71. X + 3 y - 10 = ; 2 V39. 72. x + 2y + 3 = 0. 73. 12x + 12y V3 - 3 V3 + 2 = 0. 74. x - y + 6 = 0. 75. 6x + 4y - 1 = 0. 76. 5x + 4y - 1 = ; *. 77. 21x - 12y + 1 = 0. 78. 14x - 20y + 5 = ; lOx + 7y + 1 = 0. 79. (^, 2^5)- 80. (If, -If). 81. 13x2 + i3y2 + 24x — Gy — 184 = 0. 89. A concentric circle. a2 92. x- + y'" = X ■ ^"^^ ^ circle through 0. 94. A straight line perpendicular to the line joining the point to the centre of the circle. 98. A circle. CHAPTER VI. 1. 5x2 + 9y2 _ 72x + 144 = 0. 2. 3x2 _ y2 _ lOx - 25 = 0. 3. x2 - 4y + 4 =: 0. 4. 8x2 + 9y2 _ igx - .3Gy + 45 = 0. 5. 25x2 _ iiy2 _ i50x + 272y + 589 = 0. 6. y2 — 4x - Oy — 3 = 0. 7. x2 = 4y. 8. 3x2 + gy = q. 9. y2 + 12x = 0. 10. y2 — IGx = 0. 11. x2 + 8y - 10 = 0. 12. y2 + 20x - (iy - 51 = 0. 13. x2 — 2x - 4y + 5 == 0. 14. x2 + 9y2 = 9. 15. 4x2 + 9y2 = 1. 16. x2 + 4y2 - 2x - IGy + IG = 0. 17. 4x2 + y2 + 24x + 20 = 0. 18. 25x2 + G4y2 + lOOx + 128y + IGO = 0. 19. lGx2 + 25y2 = 400. 20. x2 + 4y2 = 12, 21. 5x2 + 9y2 _ 20x - 25 = 0. 22. 9x2 + 25y2 = 225. 23. ^ + ^ = 1. 3t) 169 24. 24x2 + 25y2 = 210. 25. 9x2 + 8y2 = 209- 26. 30x2 _ y2 =r 36. 27. 9y2 - 10x2 = 144. 28. 9x2 _ 64y2 = 4. 29. 4x2 _ y2 _ gx - 4y — 4 = 0. 30. 2x2 — 3y2 - 8x + 18y -18 = 0. 31. 40x2 _ 0y2 = (J40. 32. 2x2 - 5y2 = 70. 33. 5x2 _ 4y2 = 125. 34. 7x2 - 3y2 + 20 = 0. 35. 25x2 - 9y2 = IG. 36. 9y2 - x2 = 35. 37. (1) (0,0) ; (2) y = ; (3) 7, 2 ; _(4) (± 7, 0) ; (5) f V5 ; (6) (± 3V5, 0); (7) x = ± f| Vs. 38. (1) (0, 0) ; (2) X = ; (3) ^^, V|; (4) (0, ± Vi) ; (5) Vi; (6) (0, ± V^); (7) y = ± Vj. 39. (1) (0, 0) ; (2) (0, - |) ; (3) x = ; (4) 4y - 7 = 0. 40. (1) (0, 0) ; (2) X = ; (3) 4, 3 ; (4) (0, ± 3) ; (5) ^ ; (6) (0, ± 5) ; (7) 5y ± 9 = ; (8) 4y ± 3x = 0. 366 PLANE ANALYTIC GEOMETRY. 41. (1) (0, 0) ; (2) (4, 0) ; (3) y = ; (4) X + 4 = 0. 42. (1) (0,0); (2) y = 0; (3) Vs, V2; (4) (± Vs, 0) ; (5) Vf ; (6) (± V5, 0) ; (7) X = ± I V5 ; (8) y V3 ± x ^ = 0. 43. (1) (0, 0) ; (2) (0, V) ; (3) x = ; (4) 4y + 15 = 0. 44. (1) {- i, f) ; (2) y = f ; (3) 1, V2 ; (4) {h f), (- f , |) ; (5) Vs ; (6) (-i±V3,f); (7) x=-i±iV3; (8) y - J = ± V2 ( x+ i). 45. (1) (2, - 1) ; (2) (2, |) ; (3) x = 2 ; (4) 4y + 15 = 0. 46. (1) (1,3); (2) x = l; (3) 1, VS; (4) (1, 3±V3); (5) Vf ; (6) (1, 3 ± \^) ; (7) 2y = 6 ± 3 V2. 47. (1) (- h i) ; (2) (- i|, 1) ; (3) 2y - 1 - ; j4) 24x - 1 = 0. 48. (1) (- 3, - 2) ; (2) y + 2 = ; (3) Vl^, i V26 ;_ (4) (-3±Vl3^-2); (5) iV2; (6) . (- 3 ± i V26, -2); (7) X + 3 ± V26 = 0. 49. (1) (I, -i); (2) 2x-3 = 0;j3) iV2, iV3; (4) (|^-i±iV3); (5) i VlO ; (6) (3, - i ± ^ VSO) ; (7) 15y + 5 ± VSO = ; (8) 6y + 2 = ± V6(2x-3). 50. (1) (1,^2); (2) y=j-2; (3) V?, VS; (4) (1 ± V?, -2); (5) V-V°- ; (6) (1 ± VlO, - 2) ; (7) lOx = 10 ± 7 VTo ; (8) y + 2 = ± Vf (x-1). 51. (1) (2,-1); (2) y = -l; (3) 5,3; (4) (7,-1), (-3,-1); (5) f ; (6) (6, - 1), (- 2, - 1) ; (7) 4x = 33, 4x + 17 = 0. 52. (1) (i, - -\5-) ; (2) (i, - 0) ; (3) 2x = 1 ; (4) 2y + 13 = 0. 53. (1) (i,0); (3) I; (5) 0. 54. (1) a, 0); (2) y = 0; (3) |, |; (4) (-3, 0), (4, 0); (5) V2; (6) (i ± I V2, 0) ; (7) 4x = 2 ± 7 V2 ; (8) 2y = ± (2x - 1). 56. ^-y' = ±^. a'^ b^ a _, (x — a)2 , y- , , CHAPTER VII. 1. y - 1 = 0, X + 2 = 0. 2. 3x + 2y = 0, 2x - 3y = 0. 3. X — y + 3 = 0, x + y-3 = 0. 4. x+y + 4 = 0, x-y + 2=0. 5. 3x - 2y + 3 = 0, 2x + 3y - 11 = 0. 6. x - 2y — 4 = 0, 2x + y - 3 = 0. 7. 3x + y - 7 = 0, X — 3y - 9 = 0. 8. x - y + 1 = 0. 55. x2 y2 _ 2x _ a2 + b2 ~ * a " 60. -«^>g = Mb>c, 62. Ellipse with centre at origin. ANSWERS. 367 9. X - 3y + 4 ± V2I = 0. 10. X - 2y ± Vl7 = 0. 11. 2x — 3y — 5 = 0, 2x — 3y -3 = 0. 12. 2x — y ± V39 = 0, X — 2y ± 6 = 0. 13. x-2y-3 = 0, x-2y-2 = 0. 14. 81x + 54y - 7 = 0. 15. X - y + 6 = 0, 2x - 2y - 13 = 0. 16. 8x + 14y - 36 - 0. 17. 3x+4y + 6 = 0; (2, -3), (f, - |). 18. 4x-3y + 24 = 0. 19. x + y - 2 = ; (3, - 1). 20. 7x + 12y - 72 = 0, 14x - 9y - 23 = 0. 21. 9x - 4y + 2 = 0, 3x + 4y + 6 = 0. 22. 2(5 + V3)x - 5(2+ V3)y =21. 24. 12_±2_^ _ 25. tan \v 26. tan ^ |, tan ^ f f. 28. tan ^^, tan '— • 29. 132x2 + 65y2 - 2340 = 0. 30. y2 = 2x. CHAPTER VIII. 1. 3i. 2. (p(4n -1), ±2p V4n - 1). 6. 3x - 4y + 12 = 0, 4x + 3y - 34 = 0. 7. X - 2y + 6 = 0, 3x — 2y + 2 = 0. 8. x - 2y + 9 = 0. 9- (i^^a^/^). 10. 2x-y-12 = 0. 11. 4x + 4y-27 = 0. 12. -l±iV5. 13. X — y + p = 0, X + y + p = ; x + y — 3p = 0, x — y - 3p = 0. 16. / (xi + 2p)2 _ 4p(xi + 2p) \ 4p^(xi + p)^ \ xi ' yi /' xi 24. y-2x = 0. 25. fV2. 30. x^ + y-^ + 3x - 6y = 0. 31. x2 + y2 — 5px = 0. 32. Vp (p + Xi). 35. 36x2 + 36y2 - 15x - 26y - 9 = 0. 39. 18y - 5 = 0. 40. y - 1 = ; — |. 41. i. 42. 3x + lOy +44=0. 43. y + 5 = 0. 44. 5x + 8y + 16 = 0. 48. y2 = p{x — p). 49. x + 2p = 0, x» = py-. 52. x2 — 2ay — a2 = 0. 55. Parabola. 58. Parabola. 59. (x — y)2 = 4 V2p(x + y) — 8p2, or, symmetrically, x^ + y- = a^. CHAPTER IX. 1- Vi, V|; Vf ; (± V^, 0); x = ± V| ; fVs. 2. 3x2 + 8y2 = 35. 3. 100x2 + 30y2 = 2025. 4. 5x2 + 9y2 = 405. 6. i; 3x2 + 4y2 = Sa*. 7. iV3 ; x2 + 4y2 = a2. 8. iV2. 368 PLANE ANALYTIC GEOMETRY. 9. 18x ± 12 Vl5y - 29 = 0, 6 Vl5x q: 9y - 2 Vl5 = 0. 10. 7|, If 11. x + yV3-3 = 0, 3x-yV3-l=0. 12. 2x + .3y ± 6 V2 = 0. 13. VSx - y ± Vl7 = 0. 14. (j\\, j%Oj). 15. ex - y + a = 0, x + ey + ae^ = 0. 17. ( ± ' ± . ) . 18. ' - xi. \ Va2 + b2 Va2 + b2/ xi a^ a b \ / a b ^^•(V2' V2)' ( V2' V2) 22. a2yix - b2xiy = 0, , ^ ' = ■ 27. 2x - 3y = 0. Vb*xi2 + a4yi2 28. X + 3y = 0, 6x - 6y = 0. 29. x - 4y = 0, 8x + 3y = 0. 30. 4x + 3y = 0, 3x - 25y = 0. 31. (3 V2, ^/2), (- 3 ^A, - V2). 32. 3x - 2y - 8 = 0. 33. ay - bex = 0, aey + bx = 0. 39. x±2y = 0. 41. (—3, - Vi), 210°. 42. (|V3, 1). 48. 4x2 + 5y2 = 20. 49. 2x2 + 2y2 = 9. 53. x2 + y2 = (a + b)2. 54. ^ + J't = 2. 55. b2x2 + a2y2 - ab^x = 0. a^ b^ -' 56. -'+g-^-^ = 0. 58. e = Jl-tan2f, ifa<^- a2 b2 a2 b2 \ 2 2 CHAPTER X. 2. 52x2 -117y2 = 570. 3. ^^^ + ^^ = 1. 1 + V29 6 - V29 4. 10x2 - 5y2 = 19. 5. 8x2 _ y2 - ig or - x2 + 8y2 = 124, 6. x2-y2==21. 7. 3x2-y2 = 3a2. 11. ±iV5. 12. cos-^^-^^- 15. 9x -4y ± 0V5 = 0, 4V5x + 9V5y ±78 = 0. 16. 2x - 5y ± 8 = 0. 17. 2|, H ; 3J, 2^. 19. (ae, ± -\ tan-i (± e). 26. 4x - .3y = 0, 3x - 2y = 0. 27. 4x-.3y-13 = 0. 28. (J V3, V3), {- J^/3, - V3). 2^- £-43 = ^- ^2. 4xy = 5. 43. a = b = V2; e = V2; (1, 1), (-1, -1). 44. a = e, b = 2V3, e = 2; (2V3, — 2V3), (-2V3, 2V3). 47. Straight line. 48. Equilateral hyperbola. ANSWERS. 369 49. Hyperbola. 50. Equilateral hyperbola. 53. x2 + y2 = a2 — b^. 54. x^ + y2 = a^. CHAPTER XI. 1. Hyperbola, 4x"2 — y"-^ = 1 ; centre, (1, 2) ; slope of axis, 2. 2. Parabola, ( y' ;== ) — ■p= ( x' ;= ) ; slope of axis, i. V Vio/ VioV 7V10/ 3. Imaginary ellipse. 4. Hyperbola, 4x"2 — 9y"2 = 33 ; centre, (— 1, 0) ; slope of axis, 1. 5. Hyperbola, 2x"2 — 2y"2 = — 13 5 centre, (2, — ?,) ; slope of axis, 1. 6. Two coincident straight lines. 7. Ellipse, 9x"2 + 4y"2 = 36 ; centre, (-1,2); slope of axis, 1. 8. Two intersecting straight lines. 9. Two intersecting straight lines. 10. Parabola, ( y' H 7= | = ;= ( x' '-^ ] ; slope of axis, — 1. V 8V2/ 4V2\ I6V2/ 11. Two parallel straight lines. 12. Ellipse, x"2 + 2y"2 = 2 ; centre, (2, — 1) ; slope of axis, f . 13. Point. 20. 2x2 + Sxy + y2 + 12x - 13y - 50 = 0. 21. x2 + y2 — 12x + 4y + 15 = 0. 22. x2 - xy + y2 - a2 = 0. 23. (2x + y - 3) (3x + y - 10) = 0. 24. x2 - 2xy + y2 — 22x + 18y + 101 = 0, 529x2 _ 5or,xy + 121y2 - 7126x + .3402y + 23957 = 0. 26. Hyperbola, centre at middle of base. 27. Hyperbola, centre at middle of base. 28. Hyperbola, asymptotes parallel to coordinate axes. PART II. CHAPTER I. 1. Vl49. 2. (3, 4, -0), (4, 5, -8). 4. (4, - 17, 21). 5. (5, - 6, 5). 6. (3, - 2, 1), (- |, |, V)- 7. Vl4 ;(-!,- 1, 8). 8. iV206; (i^s-ff, il). 9- cos ' Vi. 10. — -i — .± — • 11. -|=, -4-, -7=- 12. 7t. 13. I, |V3, f. V29 V29 V29 14. (13, cos"' xV. tan~' 4). 3T0 SOLID ANALYTIC GEOMETRY. CHAPTER IL 4. x2 + y2 ± 2ay = 0. 14. X - 3 = 0. 16. y* = 16p2 (x2 + z2). 17. (a) X - a = ; (b) y2 + z2 = a2 ; (c) y2 + z2 - (mx + b)2 = 5. y2 + z2 ± 2ay ± 2az + a2 = 0. 15. m2x2 — y2 — z2 = 0. -1 CHAPTER III. 1. X + 6y + 5z = ± 5 V62. 2 3. 7x + 4y - 5z = ; 0. 4. cos' 6. X — l^ -' _ =z — 1. 7. — 7r-=- V2 3 3 -2 3 yj22 V22 \^ 10. 3x - 2y + 3z - 21 = 0. 11. 12. X + y + z = 0. 5. 9. z + 1 -2 3 9. 2x + y - 7 = 0, z + 3 = 0. 17 26 11 '1086 13 /2 8 1_1 6 4\ )86 VIO86 14. sin~^ V671 15. X + y + z + 3 + 3 V3 = 0. 16. 4y + 9 = ; 2x + 6z - 1 = 0. 17. 3x - 4y + 12z + 102 = ; f|. 18. 9x - 4y - 5z - 16 = 0. 19. 1 _ y-3 _ z + 5 ~ 6 ~ 2 6 7 23. (-1, -2, -3). 21. cos -1 9 2V2r 24. 6x-18y-77z-29 = 0. 25. 2x + 4y + 17z + 1 = 0. 26. 2x + 7y - 9z + 14 = 0. 27. 2x + 6y — 3z + 7 = 0, 12x - 3y + 2z + 43 = 0. 29. (1 + I V2, f, V). 30. (V, - tV. 0). 31. (-\^ - f , - -1/). 32. (- 1, - 2, - 3). 33. VlO. 34. 1. 38. Circle. 39. Plane, sphere. 44. X + 3y + z - 1 = 0, X + y + 2 = 0. CHAPTER IV. 2. 3x2 + 3y2 - z2 = 0. 3. 2x2 + 3y2 - 6z2 = 0. 4. 2x2 + 3y2 + 3z2 = 1 ; 2x2 + 3y2 + 2z2 = 1. 14. ^±^ = 2p,z + ,,. 24. xix + yiy + ziz = a^. ANSWERS. 371 „„ /-a2A -a2B -a2C\ ^„ /-a2A - b2B -c2C\ 34. X cos a + y cos /3 + z cos 7 = 0. ^^ xcosa , ycosiS „ ^ 35. r— ± =^-7— -^ — 2pcos7 = 0. di^ b- a2A b2B -c2C 36 Va4A2 + b4B2 + c*C2 Va''A2+ b^B^+ciC^ Va''A2 + b*B2 + 0*02. 49. ( ± —^ . ± — =' 0)- V V13 V13 / UC SOUTHERN REGIONAL LIBRARY FACILITY "■"■■"T'l!ll'l|l| HM i|i' i|ifi!| I !: |i' 1 1 !' ' I I ! il