> u t I '' * 1 ^ '• ^ /,y(y SigSMM^g^M SMmSMS^I 1886. 3-«) 'PilHaxa avBV votjfidreov woqiog- voTjfiara av£v \yriiiar(ov ovdh. E. R. WriGiMT. " In the world there is nothitig great but man. In man there is nothisg great but mind. 1 i I If m University of California • Berkeley The Theodore P. Hill Collection Early American Mathematics Books 6s. ' It /. ELEMENTS OF ANALYTIC GEOMETRY. BY ARTHUR SHERBURNE HARDY, Ph.D., Professor op Mathematics in Dartmouth College. -»o>»ioo- BOSTON, U.S.A.: PUBLISHED BY GIXN & COMPANY. ■ 1889. Copyright, 1888, Bt ARTHUR SHERBURNE HARDY. All Rights Reserved. Typography by J. 8. Cusuing & Co., Boston, U.S.A. Presswork by Ginn & Co., Boston, U.S.A. PEEFACE. Although writing a text-book for the use of beginners following a short course, the teudencj' of an author is to sacrifice the practical value of the treatise to completeness, generalization, and scientific presentation. I have endeavored to avoid this error, which renders many works unsuitable for the class-room, however valuable they may be for reference, and yet to encourage the habit of generalization. To this end I have attempted to shun the difficulties involved in introducing the beginner to the conies, before he is familiar with their forms, through the discussion of the general equa- tion ; and at the same time to secure to him the advantages of a general analysis of the equation of the second degree. The teacher will observe the same effort to cultivate the power of general reasoning, which it is one of the objects of Analytic Geometry to promote, in the preliminary con- struction of loci, a process too often left in the form of a merely mechanical construction of points by substitution in the equation. In passing from Geometry to Analytic Geometry, the student should see that, while the field of operations is extended, the subject matter is essentially the same ; and that what is fundamentally new is the method, the lines and surfaces of Geometry being replaced by their equations. His chief difficulties are : IV PREFACE. First. A thorough uuderstanding of the device by which this substitution is effected ; hence considerable attention has been paid to this simple matter. Second. The acquisition of an independent use of the new method as an instrument of researcli ; hence the inser- tion of problems illustrative of tlie analytic, as distinguished from the geometric, method of proof. The function of numerical examples — tl)at is, of examples consisting of a mere substitution of numerical values for the general con- stants — is simply that of testing the student's knowledge of the nomenclature. The real example in Analytic Geometry is the application of the method to the discovery of geomet- rical properties and forms. The polar system has been freely used. It is not briefly explained and subsequently abandoned without application ; nor is it applied redundantly to what has been already treated by the rectilinear system. It is used as one of two methods, each of which has its advantages, the selection of one or the other in any given case being governed by its adaptability to the demonstration or problem in hand. The time allotted to the courses in Analytic Geometry for which it is hoped this treatise will be found adapted has determined the exclusion of certain topics, and has limited the chapters on Solid Geometry to the elements necessary to the student in the subsequent study of Analytic Mechanics. ■ ARTHUR SHERBURNE HARDY. Hanover, N.II., Oct. 6, 1888. CONTENTS. Part I. — Plane Analytic Geometry. CHAPTER I. — COORDINATE SYSTEMS. Section I. — The Point. The Rectilinear System. ART. PAGE. 1. Position of a point in a plane 1 2. Definitions . , 2 3. Construction of a point 2 4. Definitions 2 5. Equations of a point. Examples 3 6. Division of a line. Examples 4 7. Distance between two points. Examples 5 The Polar System. 8. Position of a point in a plane 6 9. Signs of the polar coordinates 7 10. Construction of a point 7 11. Equations of a point 7 12. Definitions. Examples 8 13. Distance between two points. Examples 8 Section TT. — The Line. The Rectilinear System. 14. Loci, and their equations 10 15. Distinctions between Analj-tic Geometry, Geometry, and Algebra 11 16. Quantities of Analytic Geometry 12 17. Construction of loci. Examples 13 VI CONTENTS. The Polar System. ART. PAGE. 18. Polar equations of loci 21 19. Construction of polar equations 22 20. General notation 2i Section III. — Relation between the Rectilinear and Polar Systems. 21. Transformation of coordinates 25 Eectilinear Transformations. 22. Formulas for passing from ;iny rectilinear system to any other . . 27 Polar Transformations. 23. Formulas for passing from any rectilinear to any polar system . . 30 24. Formulae for passing from any polar to any rectilinear system . . 31 CHAPTER II, — EQUATION OP THE FIRST DEGREE. THE STRAIGHT LINE. Section IV. — The Rectilinear System. Equations of the Straight Line. 25. General equation of the first degree 35 26. Common forms 36 27. Derivation of the common forms from the general form 38 28. Illustrations 39 29. Discussion of the common forms -40 30. Construction of a straight line from its equation. Examples ... 43 31. Equation of a straight lino through a given point 45 32. Equation of a straight line through two given points. Examples 45 Plane Angles. 33. Angle betiveen two straight lines 47 34. Equation of a line making a given angle witli a given line 48 35. Conditions of parallelism and perpendicularity. Examples.... 49 Intersections. 36. Intersection of loci. Examples 51 37. Lines through the intersections of loci , . 53 CONTENTS. Vll Distances between Points and Lines, and Angle-Bisectors. ART. PAGE. 38. Distance of a point from a line 55 39. Another method. Examples 56 40. Equation of the angle-bisector. Examples 58 Section V. — The Polar System. 41. Derivation of polar from rectangular equations 60 42. Polar equation of a straight line. Normal form. Examples... 60 Section VI. — Applications. 43. Recapitulation 63 44. Properties of rectilinear figures 64 CHAPTER III. — EQUATION OF THE SECOND DEGREE, THE CONIC SECTIONS. Section VIT. — Common Equations of the Conic Sections. 45. The Conic Sections 71 The Circle. 46. Definitions 72 47. General equation of the circle 72 48. The equation of every circle some form of y^'J^x:^-\-Dy + Ex + F=0 73 49. Every equation of the form ^^ 4. ^2 ^ £)j ^_ £j. _^ j^_ q ^Ij^, gqug. tion of a circle 74 50. To determine the radius and centre 75 51. Concentric circles. Examples 75 52. Polar equation of the circle 77 The Ellipse. 53. Definitions 78 54. Central equation 78 55. Definitions 80 66. Common form of central equation 80 57. Length of focal radii 81 58. Polar equation 81 59. The ratio 82 Viii CONTENTS. ART. PAGE. GO. Geometrical construction of the ellipse 83 01. The circle a particular case 85 02. Varieties of the ellipse. Examples 85 The Hyperbola. 63. Definitions 87 64. Central equation 87 65. Definitions 89 66. Common form of central equation 80 67. Length of focal radii 90 68. Polar equation 91 69. The ratio 9^ 70. Geometrical construction of the hyperbola 93 71. The equilateral and conjugate hyperbolas 95 72. Varieties of the hyperbola. Examples 95 73. The Parabola. Definitions 98 74. Common equation 98 75. Polar equation 99 76. Geometrical construction of the parabola. Examples 100 Section VIII. — General Equation of the Conic Sections. 77. Definitions 102 78. General equation of the conies. Examples 102 79. Every equation of the second degree the equation of a conic. . . . 103 80. Determination of species. Examples 105 81. The equation ^^2+ Cx2 + Dj/ + £'.r+jP=0 represents all species. 106 82. Definitions 108 83. Centres 108 84. The equation Ay^+ Cx^+F-0 represents all ellipses and hyper- bolas 108 85. Varieties of the parabola 109 86. Definitions HI 87. Locus of middle points of parallel chords Ill 88. Tangents at vertices of a diameter parallel to tlie cliords bisected by that diameter 114 89. Definitions 114 90. Conjugate diameters of ellipse 114 01. Every straight line through the centre of an hyperbola meets the hyperbola or its conjugate 115 CONTENTS. IX ART. PAGE. 92. Definitions 110 93. Conjugate diameters of the hyperbola 110 Construction of Conies from their Equations. 94. By comparison with the general equation. E.xamples 117 95. By transformation of a.xes. E.xamples 119 96. By conjugate diameters. Examples 122 97. Construction where i> = 0. PLxamples 125 General Theorems. 98. Conic section through five points. Examples 126 99. Intersection of conies. Examples 128 100. Definitions 131 101. Conies, similar and similarly placed 131 102. Condition for two straight lines 132 Section IX. — Tangents and Normals. 103. Definitions 134 104. Equations of tangent and secant. Examples 134 105. Problems and E.xamples 137 106. Chord of contact 141 107. Equation of the normal. Examples 142 108. Definitions 143 109. Subtangent and subnormal. Geometrical constructions 143 Section X. — -Oblique Axes. Conjugate Diameters. 110. Equations of ellipse and hyperbola 150 111. Ordinates to conjugate diameters of ellipse 151 112. Same for hyperbola 152 113. Parameter a tliird proportional to the axes 152 1 14. Circumscribed circle 152 115. Sum of the squares of conjugate diameters constant 152 116. Difference of the squares of conjugate diameters constant 153 117. Eectangle of the focal radii 154 118. Circumscribed parallelogram 154 119. Equal conjugate diameters 155 Supplemental Chords, 120. Definitions 156 121. Property of supplemental chords. Geometrical constructions. . 156 CONTENTS. Parabola referred to Oblique Axes. ART. PAGE. 122. Equation of the parabola 157 123. Property of ordinates 159 Asymptotes, 124. Equation of the hyperbola 159 125. Property of the secant. Geometrical construction 161 126. Property of the tangent. Geometrical construction 161 127. Tangents meet on the asymptotes 162 CHAPTER IV. — LOCI. 128. Classification of loci 164 Section XI, — Loci of the First and Second Order, 129. Examples 166 Section XTI. — Higher Plane Loci. 130. 1. The cardioid. Trisection of the angle 176 2. The conchoid. Trisection of the angle 177 3. Tlie cissoid. Duplication of the cube 179 4. The lemniscate 181 5. The witch 182 6-8. Examples 183 Section XITL — Transcendental Curves. 131. 1. The logarithmic curve 185 2. The cycloid 185 3-10, The circular functions 187 11, Spiral of Archimedes 189 12, Reciprocal spiral ; 190 13. The lituus 191 14. Logarithmic spiral 192 CONTENTS. XI Part IL — Solid Analytic Geometry. CHAPTER v.— THE POINT, STRAIGHT LINE, AND PLANE. Section XIY. — Introductory Theorems. ART. PAGE. 132. Definitions. Projections of lines 195 133. Length of the projection of a line on a line and plane 196 134. Projection of a broken line 197 Section XV. — The Point. 1.35. Position of a point 198 130. Equations of a point. Examples 199 137. Distance between two points 200 138. Polar coordinates of a point 201 139. Relations between polar and rectangular coordinates 201 140. Direction-angles and -cosines 202 141. Relation between direction-cosines. Examples 202 142. Angle between straight lines 203 Section XVI. — The Plane. 143. General equation of a surface 205 144. Equation of a plane 206 145. Intercept form. Examples 207 146. Normal form. Examples 208 147. Equation of a plane through three points. Examples 209 148. Angle between planes. Examples 210 149. Traces of a plane 213 Section XVII. — The Straight Line. 150. Equations of a straight line /. 214 151. Symmetrical forms ". 210 152. Reduction to symmetrical form. Examples 216 153. Equations of a line through two points 218 154. Angle between straight lines. Examples and problems 219 Xll CONTENTS. CHAPTER VI. — SURFACES OP REVOLUTION, CONIC SECTIONS, AND THE HELIX. Section XVIII. — Surfaces of Revolution. ART. PAGE. 155. Definitions 222 156. General equation of a surface of revolution 222 157. The sphere 223 158. The prolate spheroid 223 159. The oblate spheroid ... 224 160. Tlie paraboloid 224 161. Tlie liyperboloid of two nappes 224 162. The hyperboloid of one nappe 225 163. The cylinder 225 164. The cone 226 Section XIX. — The Conic Sections, 165. General equation of the plane section of a cone, and its discus- sion 227 Section XX. — The Helix. 166. Definitions 229 167. Equations of the helix 229 PART I. PLAICE Al^ALYTIC GEOMETEY. CHAPTER T. COORDINATE SYSTEMS. 3j«<0<^ SECTION I. —THE POINT. THE RECTILINEAR SYSTEM. 1. Position of a point in a plane. The usual method of locating a point on the earth's surface is by its latitude and longitude, reckoned respectively from the equator and some assumed meridian. A similar method serves to fix the position of a poiut in a plane. Thus: if X'X, Y'Y, be any two assumed straight lines intersecting at 0, the position of a point P' in their plane, with reference to these lines, will be known when the pn / pi distancesmP^iPf, are known, these / J" J distances being measured parallel to ,w / /o ml ° *- A—/ 4 -^ — -x, Y' Y and X'X. If, however, only / / / the numerical lengths of onPK 7iP\ ^,^/ / / are known, the point may occupy /y' any one of four positions, P\ P", Kig i. P™, P"'. This ambiguity will dis- appear if we distinguish distances laid off above and below X'X, parallel to I"' Y, as positive and negative, respectively ; and, in like manner, as positive and negative, respectively, those laid off to the right and left of Y'Y, parallel to X'X. Hence, the position of 0719/ point in the plane of X'X and Y'Yicill be com- pletely determined tvith reference to these lines when its distances from them, measured parallel to them, are given in magnitude and sign. Z ANALYTIC GEOMETEY. 2. Defs. The fixed lines X'X, Y'Y, are called the axes of reference; their intersection. 0, the origin; the distances mF\ nF\ the coordinates of the point P' ; and to distinguish these co- ordinates, nP^ is called the abscissa, and mP' the ordinate of P\ 3. Construction of a point. Since the coordinates of a point, when given, fix its position with reference to the axes, and since (Fig. 1) Oin = nP\ On. = mP\ to determine the position of a point whose coordinates are given we have simply to lav off the given abscissa from along X'X in the direction indicated by its sign, and at its extremity on a parallel to F'F, the given ordinate, above or below X'X according as it is positive or negative. This determination of the position of a point is called the construction of the point. 4. The axes of reference are always lettered as* in Fig. 1, and hence are often designated as the axes of X and 5", the former being usually taken horizontal. For brevity they will frequently be spoken of as X and Y simpl}-. The abscissa of a point, being always a distance parallel to the axis of X, is always represented b}^ the letter x ; and for a like reason the ordinate is always represented by the letter y ; hereafter, there- fore, these letters will always represent distances parallel to the axes of reference. As indicating the directions in which abscissas and ordinates are laid off, the axes are also distin- guished as the axis of abscissas {X'X). and the axis of ordi- nates {Y'Y). The angles XOY, YOX', X'OY', Y'OX, are known as the first, second, third, and fourth, angles, respectively. It is evident that so long as the angle XOY is not zero, it may have any value whatever. When /p' a right angle, the system of reference i is called a rectangular system; other- Q j,„ ^ wise, an oblique system. As nothing in general, is gained by assuming oblique axes, the axes will hereafter be supposed Y' rectangular, unless mention is made to the Fig. 2. contrary ; the abscissa and ordinate of a Y n X^ THE POINT. 6 point will thus be (Fig. 2) the perpendicular distances of the point from the axes. 5. Equations of a point. It is now evident that we ma}' designate the position of a point by giving its coordinates in the form of equations. Thus, .r = 2, ?/ = 3, designate a point in the first angle, distant 3 units from the axis of X, and 2 units from the axis of Y. These equations are called the equations of the point. But it is more usual to adopt the notation (2, 3) to designate the point, the abscissa being always loritten first. Examples. 1. What are the signs of the coordinates of all points in the first angle? In the second? In the third? In the fourth ? Where are all points situated whose ordinate^are zero? Whose ordinates are equal and have the same sign? Whose abscissas are zero? What are the coordinates of the origin ? 2. Construct the following points : (2,4); (3,-2); ( — 6, -2) ; (-4, 3) ; (2, 0) ; (0, -2) ; (-2, 0) ; (0, 2) ; (0, 0). 3. Construct the triangle whose vertices are (4, 5), (4, —5), ( — 4, 5). What kind of a triangle is it, and what are the directions of its sides ? 4. The side of a square is a, and its centre is taken as the origin, the axes being parallel to its sides. What are the coor- dinates of the vertices? What, when the axes coincide with the diagonals, the origin being still at the centre? 5. An isosceles triangle, whose base is h and altitude a, has its base coincident with X. What are the coordinates of its vertices when the origin is (1) at the centre of the l)ase? (2) at the left hand extremity of the base ? 6. Construct and name the figures whose vertices are (1). (a, a), («,-«), (-ff, -«), {-a, a). (2). {a,b), (a, -b), (-a, -6), (-«,&).'■ (3). (a,&), (o, -&), (-a, -6), {-a,c). (4). (o, &), {c,d), (-e, d), (-/,&). ANALYTIC GEOMETIIY. 6. To find the coordinates of a point ichich divides the straight line joining two given j^oints in a given ratio. Let P', P", be the given points, x', y\ and a*", 3/", their coordinates, and P a third point {x, y), dividing P'P" so that P'P : PP" :: m:n. Then, from similar triangles, Y P" (1) P'P BP PQ QP" n Bnt P'E =x-x',PQ = x"- x, RP=y- y', QP"= y"— y. Substituting these values, Fig. 3. X — x' _y — y' _ m x"—x y" — y n Solving these equations for x and y, we obtain x = mx" -\-nx' m + n my" -\-ny' m -j- n (2) For the middle point of P'P", m = n. Hence the coordinates of the middle point of a line joining two given points are X — x"-\-x' 2 ' y = y"+y' 2 (3) If the line is cut externally, we have from (1) and Fig. 4, X — x' _ y — y' X — x" y — y" m n whence Kii,'. 4. X mx"—nx' m — ?i y my" — ny' m — n (4) Oblique Axes. Tlie above formulas hold good for oblique .ixcs, since the triangles remain similar whatever the angle XO Y. Examples. 1 Find tlie coordinates of the middle point of the line joining (5, 3) and (3, 9); also (5, 8) and ( — 5, —8). Ans. (4, G); (0, 0). THE POINT. O 2. Find the coordinates of the middle points of the sides of the triangle whose vertices are (G, 2), (8, 4), (10, 12). Ans. (7, 3); (9, 8): (8, 7).' ^'3. The line joining ( — 2, —3) and ( — 4, 5) is trisected. Find the coordinates of the point of trisection nearest ( — 2, —3). 154. The line whose extremities are (2,4) and (G, —8) is divided in the ratio 3:2. Find the two points of division ful- filling the condition. > /22 1G\ /1 8 4 Qo. Find the two points on the line joining (2, 4) and (6, 3) twice as far from (2, 4) as from (G, 3). Ans. (10, 2); f— , - G. The vertices of a triangle are ( — 4, —3), (6, 1), (4, 11). Find the coordinates of the points of trisection, farthest from the vertices, of the lines joining the vertices and the middle points of the opposite sides. A71S. (2, 3). 7. To find the distance between two given j^oints. Let P', P", be the given points, x', y', and x'\ y", being their coordinates. Then FT" = VP'B' + BP"- ; or, representing P'P" bv rf, d = V{x"-x'r + {ij"-!j'y. (1) If one of the points, as P", is at the ori- gin, its coordinates will be zero. Hence the distance of any point P' from the origin is d = ^x" + y". (2) Fig. 5. |5 Oblique Axes. In this case the triangle P' RP" will not be a right triangle. Let ^ = inclination of the axes. Then P'P"= y/PIt^ + RP'"^-2P'P . HP" COS P' HP", or, since P'PP" = ISO^ - 3, d= V(a;"-a;')^+(2/"-2/')- + 2(a;"-a;') {y"-y')coBp, (3) which, when ^ = 90^, reduces to (1), since cos 90° =0. 6 ANAXYTIC GEOIVLETRY. Examples. 1. Find tlie distance between (2, 4) and (5,8). As the quantities under the radical sign are squares, it is immaterial whether we substitute (2,4) for {x', i/') and (5, 8) for {■'-",>/"), or vice versa. Thus (/ = V'(2-5)--2 + (4-8)^ = V(5-2)--2+(8-4)''2 = 5. 2. Find the distances between the following points : (—2, —4) and (-.'), -8); (7, -1) and (-G. 1); (7, 2) and ( — 7, -2). ^r^TT" Alls. .') ; Vl73 ; 2 V53. 3. Find the distance of (G, —8) from the origin. Ans. 10. 4. Find the lengths of the sides of the triangle whose vertices are (4, 8), (1, 4), (-4. -8). j^^s. 5; 13; 8V5. 5. Find the lengths of the sides of the triauole whose vertices are (4, 5), (4, -o), (-4, 5). Ajis. 10; 2 V4l ; 8. THE POLAR SYSTEM. 8. Position of a point in a plane. The position of a point on the earth's surface is often designated by its distance and direction from some other point ; as when A is said to be 25 miles northeast of B. In a similar way the position of a point in a plane may be designated. Thus : if OA be an}' assumed straight line through a fixed point 0, yP' the position of a point P' in the plane /^ AOP', with reference to 0. will be / ' ^ knoAvn when the angle AOP' and the / distance OP' are known. The fixed P" Fig..;. line OA is called the Polar Axis; the fixed point, O, the Pole; the angle AOJ" and distance OP', the Polar Coordinates, OP' being the radius vector and AOP' the vectorial angle. The ratlins vector will always be represented by the letter ?•, and the vectorial angle by the letter &. u THE POINT. 7 9. Signs of the polar coordinates. If the vectorial angle be always l:iid off above OA (Fig. 6) to the left, as iu trigonom- etry, the position of every point in the plane may be desig- nated without ambiguity, and were this the only consideration there would be no necessity for any convention as to signs. But as r and 6 often occur in the course of analytic investiga- tions with negative as well as positive signs, it is necessary to adopt some convention for the interpretation of the negative sign. For this purpose the vectorial angle is regarded positive when laid off above OA to the left, and negative when laid off below OA to the right ; while the radius vector is considered positive when laid off from O towards the end of the arc measur- ing the vectorial angle, and negative when laid oft" in the oppo- site direction. Thus (Fig. 6), for the angle AOP', OF' is the positive, and OP" the negative, direction of r. 10. Construction of a point. Since the coordinates r and 0, when given, fix the position of a point, to determine its position we have only to hi}" oft" the given value of above or below OA (Fig. 6) according as is positive or negative, and on the line through and the end of the measuring arc the given value of r, towards or away from the end of the measuring arc as the sign of r is positive or negative. This determina- tion of the position of a point is called the construction of the 2Mint. 11. Equations of a point. It is now evident that we may designate the position of a point bj' giving its coordinates in the form of equations. Thus (Fig. G), ?- = 4, ^ = 60°, locate a point P' distant 4 units from on a line inclined at -|- 60° to OA; while r = — 4, ^ = 00°, locate a point P" on the same line, but on the opposite side of 0. These equations are called the jwlar equations of a jyoint; but it is more usual to adopt the notation (r, ^), writing the radius vector first. Thus, the above points would be (4, GO") and (—4, G0°), respectively. 8 ANALYTIC GEOMETRY. 12. This system of reference is called the Polar System, aud that previously described, whether the axes be oblique or rec- tangular, the Rectilinear System. It will be observed that in each s^'stem tico things serve as the bases of reference ; in the rectilinear, the axes of X and Y\ in the polar, the pole and polar axis. Also that in each system tivo quantities are suffi- cient to refer the point ; in the rectilinear system, the abscissa and ordinate ; in the polar, the radius vector and vectorial angle. Again, that while in the rectilinear system a given point can have but one set of coordinates, in the polar S3^stem it may have an infinite number of sets. Thus (Fig. 6), P' may be designated as follows: (4,60°), (-4, -120°), (-4, 240°), (4, -300°), (4, 420°), etc. This fact, however, gives rise to no ambiguity in the position of P', for no one set of polar coordinates can locate more than one point. The rectilinear aud polar systems of reference, together with a third called the trilinear, are those in most common use. The two former only will be employed in this treatise. Examples. 1. Construct the following points: (o, 90°); (5,270°); (-3, 120°); (-G, -180°). 2. What are the coordinates of the pole? AYhat are all possible values of for points on the polar axis? 3. Construct the points (0, 45°) ; fo, -\ ; (4,0°) ; (-4,0°). 4. Give three sets of polar coordinates locating (10, 90°). 5. Construct (s.^^Y, f-S, ^) ; fs, ^) ^ (§' " 'f 6. The side of a square is 5V2, its centre at the pole, and sides parallel and perpendicular to the polar axis. AVhat are the coordinates of its vertices? P" 13. To find the distance between two I ^^p' (jiven points. Let P', P", be the given points, r\ 0\ and r", 0", their coordinates, ^ and d the required distance. Then, from / Fi^'. 7. the triangle P'OP", THE POINT. P'P"= VOP'--f- OP"'- 2 OF. OP" COS P'OP", or f?= Vr'2 + r"2-2/r''"cos (6^"— 6'). (1) If oue of the points, as P'\ is at the origin, d — OP— r' . Examples. 1. Fiud the distance between the points (3,^ and ( 4. -'^ o Since cos 6 = cos (— 6) , it is immaterial which of the two given points is designated as (c', 6') . Thus d= V9+ 16-24 cos 00°= VlO^ 9-24 cos (-60") = VlS. Observe, also, that if 6''— 6' > 90°, the cosine will be negative and the last term positive, as it should be, for then the triangle will be obtuse-angled at (Fig. 7). 2. Find the distances between the following points : (3, 60°) and (4, 150°) ; (5, 0°) and (5,-180°) ; (^2,-) and (1, 0°) ; (10, 30°) and (-10, - 150°) ; (G, G0°) and (0, 0°) Arts. 5 ; 10 ; 1 ; ; 6. 10 ANALYTIC GEOxMETKV, SECTION II.— THE LINE. THE RECTILINEAR SYSTEM. 14. Loci and their equations. Every line, straight or curved, may he regarded as generated by the motioi of a j'oint. The kind of line generated will depend upon the law which governs the motion of the generating point. Thus, a circle may be traced by a moving point, the law which governs its motion being that it shall always remain at a given distance (the radius) from a fixed point (the centre). If the origin be taken at the centre of the circle, P being any point of the circle, x, y, the coordinates of P, and 0P= B, the radius, then 0P-= Om- + mI'^, or X- + y- = 7?-, is true for every position of P while generating the circle. This equation is the algebraic expression of the law which governs P's motion, and is called the equation of the circle ; and, in genei-al, the equation of a. line is the algebraic expressitm of the law ivhich governs the motion of its generating point. Again, the relation x- + y' = R' is true for no point within or without the circle, ])ut is true for every point on the circle ; it thus expresses the relation between the coordinates of all points of the circle, and of no other jioints ; hence, in general, the equation of a line is the algebraic exjjression of the relation ivhich exists between the coordinates of any and every point of the line. Evidently if a point moves at random, without any governing law, the line it traces can have no equation ; for the latter is THE LINE. . 11 the algebraic expression of a law, and when the point moves at random, none such exists. The above equation, x- + y- = B^, being the equation of a circle whose radius is li, the circle is said to be the locus of the equation; i.e.. translating the word locus literally, it is the 2)lare in which the point, moving under the law expressed by the equation, is always found ; and, in general, the locus of an equation is the 2)ath of <( poiid so moving that its coordinates always fulfil the relation e.vpressed by the equation. It follows that if a point lies on a locus, the coordinates of the point must satisfy the equation of the locus. Thus, if the radius of the above circle be ."j, x*- + 7- = 25, and the points (3, 4) , (0, — 5) , ( — 4, — 3) , are all points on the circle because their coordinates satisfy the equation ; but (2, 4), (— 4, 4), are not on the circle. Hence, to ascertain tvhether a, given point ties on a given locus, substitute its coordinates in the equation of the locus and see whether they satit2, we have, for x= 2, ^ = 0, locating P\ x = 3, y = ±'2, loc. P" and P", X = 4:, y = ± 2 V2, loc. P'' and 7^^, Fis. n. X 5, 7/ = ±2V3, etc. x=G, y=±-i, etc., every value of x > 2 locating two points Fig. 12. 16 ANALYTIC GEOMETRY. The above method of constructiug a locus b}' points is a purely mechanical one. The greater the number of points located, the more accurate the construction of the locus. A simple inspection of the equation will, however, often indicate the general form and position of the locus. Thus, in the above example, every value of X gives two values of ?/ numerically equal with opposite signs, and the locus is therefore made up of pairs of points equidistant from X'X, or the axis of X is an axis of symmetry ; and, in general, tchenever the equation contains the square only of either variable, the other axis is an axis of symmetry. Thus y-=^x is symmetrical with reference to X\ y-\-of=2 is sym- metrical with respect to Y; while x~-{-'ir=2o, c(r—y-=i, 9. r^+ 16.^^=144, are symmetrical with respect to both coordi- nate axes. Again : since the ordinate of every point on the axis of X is zero, if the locus has any point on the axis of X it will be found by making y = ; and for a like reason if it has any point on the axis of Y, it will be found by making .x = ; and, in general, to find lohere a locus crosses or touches either axis, make the other variable zero in its equation. Thus, in the above example, to find where the locus of y-=- 4.r — 8 crosses X, make 2/ = 0, whence a;=2=0/". Making x=0, y is imaginary, showing that the locus does not meet the axis of Y. The dis- tances from the origin to the 2^oints tchere a locus meets the axes are called the intercepts of the locus. They are distinguished as the X-intercept and the 5''-intercept. Thus, the X-intercept of 7/2= 4 a; _ 8 is 0P'=2. 7. 25?/-+ 9x-=225. We observe that the locus is symmet- rical with res[)ect to both axes, flaking ?/ = 0, we find x= ±5, or 0^1 and OA' are the X-intercepts ; making x = 0, y= ±'d, or OB and OB' are the I'-intercepts. Solving the equation in succession for x and y, we have 2/ = ± f V25 - ic2, a- = ± f Vi) - y-. From the value of y we see that x cannot be numerically greater than ±5, otherwise y is imaginary ; hence no point of the curve THE LINE. 17 lies to the right of A or to the left of A' ; that is, x — ±i) gives the limits of the curve iu the direction of X, and these values are the roots of the eqxiation obtained by puttincj the quantity under the radical sign equal to zero. The reason for this is plain : y is real when 25 — x^ is positive, and imagiuar}' when 25 — 0."^ is negative ; hence the limiting values of y correspond to 25 — ar = 0, since in passing through zero 25 — x- changes sign. For a like reason, placing 9— 2/" = 0, y = ±S are the limits of the curve in the direction of Y. And, iu general, whenever the equation of the locus is of the second degree loith respect to one of the variables, if ice solve it for that variable, and place the radical equal to zero, the roots of this equation are Fia. 13, the limits in the direction of the other axis. (Thus, in Example 6, the equation is of the second degree with respect to y ; solved for y, the radical placed equal to zero gives 4 a; —8 =0, or x—2. Beyond this limit the curve extends indefinitely in the direction of X.) We have now determined the intercepts, symmetry, and limits, of the locus, and so have a general knowledge of its form and position. Points may now be constructed as before. Thus, for a; = 3, or - 3, y = ±i^, locating P' , P°, P"', and P'^, X = 4, or - 4, y = ± f , locating P^', P'\ P"S P^, etc. 18 ANALYTIC GEOMETRY. 8. 16^" — 9;r- = — 144. Making x = 0, y is imaginary; hence the locus does not meet the axis of Y. Making y = 0, x = ±i, or OA and 0A\ the X-intercepts. The curve is symmetrical with respect to both axes. Solving for x, .T = ± I V?/- + 9 ; but y- + 9 cannot change sign, or, otherwise, y- -\-i) = gives imaginary values for y, hence there are no limits in the direc- tion of Y, the curve extending indefiuitely in that direction. Solving for y, 2/ = ±f Va;--16. Placing a;-— IG = 0, the limits in the direction of X are seen to be + 4 and — 4. Having found the limits, it is always neces- sary^ to see whether the locus lies within or without the limits. In this case x cannot be numerically less than ± 4, and the curve therefore lies without the limits. Having thus deter- mined the general features of the locus, we proceed to construct a few points. For x=±o, ?/=±f, locating P', P", P™, P^, x = ± 6, y = ± f V5, locating P^, P", P'", P^'", etc. A curve of this kind, com- posed of two separate branches, Fig. 14. is said to be discontinuous. 9. x- -i-y- — Sx — 4y — !) —0. Making .t =0, ?/ =5, and —1, giving the intercepts OB, OB'. For y=0, x=4:± V2T= 4 ±4.6 nearly, or 8.G and — .6 for the intercepts OA, OA'. Solving for X, we have whence X = 4±^-y' + Ay + 2l. (1) THE LINE. 19 Now, every value of y gives two values for x of the form a; = 4 ±p, and thus locates two points distant p (the radical) from a line parallel to I'' and 4 units from it. Thus, for y — (J, x = 4 ± 3, locating P' and P", each distant 3 units from DD', DD' being parallel to Y and 4 units from it. Solving for y, we have whence y=2 ± V-.^•^ + 8a;+9, (2) from Avhich we see the locus is also symmetrical with respect to CC\ parallel to X and 2 units above it ; and, in general, whenever the equation, all its terms being transposed to the first member, is of the form Aa? 4- -B.« + etc. ivith respect to either variable, if the coefficient of the square be made positive unity, then half the coefficient of the first poiver, ivith its sign changed, will be the distance from the other axis of a line of symmetry parallel to that axis. Thus, v? -\- y^ —IQy -\- -i — Q is symmetri- cal with respect to Y, and also with respect to a line parallel to X and 5 units above it; xr -\-2x -\-y' — ^y = is symmet- rical with respect to two lines, one parallel to Y at a distance 1 to its left, the other parallel to X at a distance f above it. To find the limits along X, put the radical in (2) equal to zero, whence x 9 and — 1. Values of y are imaginary for a; > 9 or < — 1 , and the locus lies within these limits. For the limits along Y, {\) gives y = 1 and — 3 , or no point of the locus lies above -f- 7 or below —'3. Having now determined the intercepts, limits, and symmetry, we may construct a few points. For »= 4, ?/ = 7, or — 3, a; = 8, y = b, or — 1, \d' Fig. 15. P-, and P^^, p\ and P^'S etc 20 AJS"ALYTIC GEOMETRY. 10. Show that y- — Gy -i-x- —16 = 0, is symmetrical with respect to Y, and a line parallel to X, 3 units above it ; that its limits along X are ±5, and along Y, +8 and —2. Determine its intercepts, and construct. 11. ?/- — 10.T + x'^= 0. Determine the lines of symmetry, intercepts, limits, and construct. 12. af — 6x -\-9 -j-y- -}-10y = 0. Lines of symmetry are —5 and 3 from XandT^, respectively. Limits along X are 8 and —2 ; along Y, and —10. Intercepts on Y are —1 and — 9 ; on X, + 3. Construct. 13. ?/--2ar + 12a; -22 = 0. Show that the locus has no limits in the direction of X, lies wholly outside the limits ± 2 in the direction of I", has X and a parallel to Y distant + 3 units from it for lines of symmetry, and ± V22 for F-inter- cepts. Construct. 14. y"=^9x. This locus is symmetrical with respect to X, is without limits along I", has x = for a limit along X, lying wholly in the first and fourth angles. Construct. Observe that if x = 0, y = 0, and converseh' , or the intercepts are zero on both axes, and hence the locus passes through the origin. Otherwise, the coordinates of the origin satisf}^ the equation, and the origin is therefore a point of the locus. Evidently this cannot be the case when the equation contains an absolute term. Hence, in general, tvhenever the equation of the locus contains no absolute term, the locus j^^^sses through the origin. Thus, cc^ -}-y- — lOy = 0, x* — y^ ■{- 3x = pass through the origin. 15. xy = 10. Solving for y, y= — ■• By assigning values to x, and deriving those of y, we may construct the locus b}' points. But the student should endeavor in all cases to de- termine the general features of the locus by an inspection of its equation. In this instance we observe that there is no line of symmetry parallel to either axis, as the equation con- tains the square of neither variable ; also, that y is positive THE LINE. 21 when X is positive, and negative when x is negative, and tliere- fore the curve lies wholly in the first and third angles. Again, when x = 0, ^ = cc , and as X increases y dimin- ishes, but becomes zero only when .r = oo . In the first angle, then, the locus lies as in the figure, continnally ap- proaching the axes as — X changes, but touching neither within a finite distance from the origin. A line to ivldcli a cnrve thus continually ap- proaches, hilt does not touch icithin a finite distance Is called an asymptote. In the third angle, x being negative and decreasing algebraically, y increases algebraically, becoming zero, liowever, only when a; = — go. The axes are thus asymptotes to both branches. Constructing a few points, we have, for x= ±1, y= ±10, P\ P", a;_= ± 2, ?/= ± 5, x= ±5, y= ±2, x^ ± 10, .y= ± 1, etc. Fig. 16. P"S P'\ p\ P", / o THE POLAR SYSTEM. 18. Polar equations of loci. We have seen that the eqna- tion of a locus is the algebraic expression of the law governing the motion of the point which traces the locus, and that the quantities in terms of which this law is expressed are the coor- dinates of the moving point and certain constants. Nothing in this statement restricts us to tlie use of any particular system of 22 ANALYTIC GEOMETRY. coordiuates. Thus, if the law which controls the moving point is that it shall always remain at a given distance from a given point, the line traced will be a circle. C being the fixed point and CP = R the radius or constant distance, if we assume OA as the polar axis, and the pole at a distance from C equal to the radius, ^"^' ^ ' ■ OP = r and A OP =6 will be the polar coordiuates of P the moving point ; and since OPB will be a right angle for every position of P while tracing the circle, OP — — = cos BOP, or r=2R cos 6 is true for every position of P on the circle, but is true for no point within or without the circle. It is therefore the expression of the relation existing between the coordinates of any and every point of the circle, and is therefore the polar equation of the circle. And, conversely, the circle is the path of a point so moving that its polar coordiuates satisfy the above equation ; hence the circle is the locus of the equation. 19. Construction of polar equations. In a polar equation, the varial)les which correspond to x and y of the rectilinear sys- tem are r and 6, and by assuming values for one and deriving the corresponding values of the other from the equation, we may construct as many points of tlie locus as we desire. It is obviously convenient to make ^, the vectorial angle, the inde- pendent variable, and derive the values of r. Examples. 1. r=5. This ecjuation is independent of 0, that is, /• = 5 = rt constant, for all values of 0. It is tlien evi- dently the equation of a circle whose radius is 5, the pole being at tlie centre. We have seen (Art. 14) that tlic corresponding rectangular equation of the circle is or -\-y- = R-. Tlie student will observe the comparative simplicity of tlie polar form r = R, and will thus see that in many cases it might be preferable to THE LINE. 23 use the polar rather thau the rectangular equation of a locus because of its simpler form. 2. r= 10 cos ^. As in the case of rectangular equations, the student should endeavor to obtain a general idea of the form and position of the locus from its equation, rather than to con- struct the locus mechanically by points. In the present case we see that when 6 — 0°, cos $ has its greatest value, and there- fore also r ; that as 6 increases, cos 0, and therefore also, ?•, diminishes, becoming zero when 6 = 90°. That as 6 increases from 90° to 180°, r is negative and increasing numerically, becoming — 10 when 9 = 180°, the same numerical value which it had for^ = 0°. Constructing a few points, we have, for ^= 0°, e= oO°, e= Go°, 0= 90°, ^=120°, ^=l.-)0°, ^=180°, As when 6 = {f° the radius vector coincides with the polar axis, P' is constructed l)y making OP' = 10. Laying off AOP'' = 30°, and OP'' = 5 V3, P" is (5 VS, 30°) . 6 = 90°, gives r = 0, and locates the pole. P'^' and P^'are constructed in the same way, but the values of r when > 90° being negative are laid off awav from the end of the measurinsj arc. If 6 increases from 180° to 360°, the values of r are repeated (numerically), so that the entire locus is traced for values of 6 from 0° to 180°. As 0P'= 10, and r= lOcos^ is true for all positions of P, OP^'P', OP^S^'i etc., is always a right angle, and the locus is therefore a circle whose radius is W. ■■^ The above loci, and those of Art. 17, are constructed simply to familiarize the student with the meaning of the terms loci of equations, and, conversely, equations of loci. A clear concep- tion of these terms, and of a coordinate system as a device for 24 ANALYTIC GEOMETRY. representing lines by equations, is fundamental to the subject. In Chapter II we shall begin the systematic study of loci by means of their equations, commencing with the simplest, namely, the straight line. 20. General notation. Any equation of a locus referred to a rectilinear system of axes may be represented b}' the equation f(^x, y) = 0, read ' function x and y = 0,' this being a general form for what the equation of the locus becomes when all its terms are transferred to the first member. In such an equation, X and y are said to be implicit functions of each other. If the equation of the locus is solved for one of the variables, as y, the corresponding general form will be y =f{x), read ' y a function of X.' In such an equation, the way in which y depends upon x being fully indicated by the solution of the equation, y is said to be an explicit function of x. The primary object of Algebra is the transformation of implicit into explicit functions, and /(.T, y) = may be written y =f{x) whenever the former can be solved for y. Similarly /(/•, 6) = 0, and r =f(0)y are general forms for the equation of any locus referred to a polar system. KELATION RECTILINEAR AND POLAR SYSTEMS. 25 SECTION III. RELATION BETWEEN THE RECTILINEAR AND POLAR SYSTEMS. 21. Transformation of coordinates. It is evident that the coordinates of a point and the form of the equation of a locus will depend upon the system of reference chosen and its posi- tion. Thus, the coordinates of P (Fig. 19) referred to the oblique system XxOiTi are Oi?ni and vHjP; referred to the rectangular system XOY^ they are Om and mP\ while if the polar system O2A is employed they are O^P and AO2P. Again, we have seen that the equation of a circle referred to rectangular axes through the centre (Fig. 20) [sx^-\-if = R^ (Art. 14) , but if it is referred to the system XiOiYi its equation is (x^ —m)- + (^1 — ?i)- = R^ Y Fig. 20. (Ai*t. 16), the subscripts being used to distinguish the coordi- nates of the two systems. Again, in Art. 18, we found the polar equation of a circle to be r = 2 i2 cos when the pole was on the circle and a diameter was taken for the polar axis ; while the polar equation, when the pole was at the centre, was found in Art. 19, Ex. 1, to be t= R. It is thus clear that the form of the equation of any locus will vary with the system of reference chosen, and, from the above 26 ANAI.YTIC GEO:srETEY. illustrations, that one form ma}' be simpler than another. It is therefore desirable to be able to pass from one system to another. This passage from one system of reference to another is called Transformation of coordinates. As this transformation is of frequent use, it is important that the student should thoroughly understand its object and nature. The problem may be thus stated : Having given the equation of a locus referred to one sj^stem of reference (as the equation of the circle (a;i— ???)-+ (yi — n)- = R- referred to the axes XyOiYi)^ to find its equation when referred to any other system (as the parallel system XOl", to which when the same circle is referred its equation is x--\-y- = Br). Tlie object of this transformation is to obtain a simpler equation of the same locus ; the method will consist in finding values for the coordinates iKj, ?/i, in terms of the coordinates x and y, and substituting these values in the given equation ; the resulting equation will then be a relation between the new coordinates, and therefore the equation of the locus referred to the new axes. In the same way, having given the equation of a locus in terms of x and ?/, we pass to the polar equation of the same locus by substituting for x and y their values in terms of r and 9 ; the resulting equation will then be independent of x and ?/, and, being a relation between r and true for all points of the locus, is its polar equation. The problem thus reduces to : Tlie coordinates of any point P with respect to one system of reference being known, to find its coordinates icith resjyect to any other system. The system to which the transformation is made is called the new system ; that from which we pass, the primitive system. The three following cases will be considered : (A). To pass from any rectilinear system to any other recti- linear system. (J5). To pass from any rectilinear system to any polar system. (C). To pass from any polar system to any rectilinear B3'stem. TRANSFORMATION OF AXES. 27 RECTILINEAR TRANSFORMATIONS. 22. Formuke for jX(ss/»(/ from any rectilinear system to another. Let XOY be the primitive system, ^ being the iucliuation of the axes, and P any point whose primitive coordinates are Om = X, mP=y. Let XiOi\\ be the new system, its position being given by the coordi- ,-^- /I'j nates of its origin , OA — .r„, AOi — 2/o» and the angles y, yi, which its axes make with the primitive axis of X, the coor- dinates of P referred to the new system being OiWi = x^ and miP= 2/i- Draw OiB and niiC parallel to OX, and m^D parallel to r. Then Fig-2i. Om = OA + OiD + m,C. But Om = x, 0A = a'o, OyD : Oitn^ : : sin Oim^D : sin OiDnii, whence Oimi sin Oim^D _ Xi sin (^ — y) ^ ~ sin OiDnii "" sin /3 and whence Substituting these values. MiC : miP : : sin rUiPC : sin niiCP, 2/i sin ( ^ - yi) sin /8 CC — ^(j "P Xi sin (/? - y) + yi sin (^ - yQ sin )8 Again, mP= AOi + Dm^+CP. But ?»P= y, AOi = 2/o, I>?"i : OiWi : : sin DOim{ sin OiDmi, a-, sin y whence i^mj = — -. — ^ ' sin 13 28 ANALYTIC GEOMETRY. and whence CP: m^P: : sin CmiP: sin m-^CP^ CP = Vi sin yi sin ji Substituting these vakies, y = yo + Hence Xi sin y -f jji sin yj sin)8 , a;iSin(/3-y) + ?/iSin(^-yi) .risiny+?/iSinyi .,. are the required formulae. The following special cases may arise : First. To pass from any system to a jKiraUel one. In this case (Fig. 22) y = 0, yi = /S, and the general formulae (1) become a; = .T„ + .^i, y = yo + yi, (2) which are independent of /3 and apply to all parallel axes, oblique or rectangular. A "' Fig. 23. Second. To pass from, rectangular to oblique axes. In this case (Fig. 23) ^=90° ; and since sin(90°-vl) = cos^l, sin (90° — y), and sin (90° — y,), become cos y and cos yj, or the general formulae become n- = a-o + a:i cosy + .'/, cos yi, ?/ = ?/„ + -''i sin y + ^/i sin yj. (3) TRANSFORMATION OF AXES. 29 Third. To pass from one rectangular system to another. lu this case (Fig. 24) ^ = 90°, y, = 90° + y ; and since sin (90° + ^) =cos^, sinyi = sin (90° -|- y) = cosy, sin {ft — y) = cosy, sin(/3-yi) =sin(90°-90°-y) =sin-y= -siny, and the aeneral formulae become a; = (t'o + .^•lCOSy — 2/1 siny, y = y^ + Xj sin y + t/i cos y . r (4) A m Fis. 24. niA Fig. 25. Fourth. To pass from oblique to rectangular axes. In this case (Fig. 25) y^ = 90° + y ; and hence sin(/3-yi) = sin[/3-(90° + y)] = sin-[90°-(/3-y)] = -sin [90° - (^ - y)] = -cos(/3 - y), and the general formulae become aJi sin (/3 - y) - y-^ cos (/3 - y) ~] X — X(f-\- y = yo-\- sin^ Xi sin y + t/i cos y sin/3 (5) The student will observe that the special formulae, like the general ones, may be deduced directly from the accompanying figures. 30 AISTALYTIC GEO^METEY. If the new origin coincides with the primitive origin, Xq and 2/0 ill the above formulae become zero. Hence, To 2MSS from one oblique set to another, X z= ^^ ^^^^^ - 7) + Ih sip (/?- yO ^ .r, siny + .Visinyi ,^. sin/8 "^ bin/3 ^ ^ To pass from a rectangular to an oblique set, ic = .Ti cos 7 + 2/1 cos yi, y = iCi sin y + ?/i sin yi. (7) To pass from one rectangular set to another, X — Xi cos y — 2/1 sin y, y "= ^i sin y + i/i cos y. (8) To 2>ass from an oblique to a rectangular set, ^. ^ g-i sin (^ - y) - yi cos (^ - y) ^ a;i sin y + 2/1 cos y .g. sin/3 ^ sin/3 ^ ^ The student will observe that none of the above formulje involve higher powers of the new than of the primitive coor- dinates, and therefore that when these values of x and y are substituted in any equation, the transformed equation will always be of the same degree with respect to the variables as the primitive equation ; that is, the transformation from one rectilinear system to another affects the form but not the degree of the equation. POLAR TRANSFORMATIONS. 23. Formulcp. for jxi.ssing from any rectilinear system to any polar system. Should the primitive system be oblique, we may first pass to a rectangular system with the same origin by equa- tions (9) of Art. 22 ; the problem then consists in passing from an}' rectangular to any polar system. Let XOY be the primitive system, and P any point whose coordinates are x = Om, y = 7nP. Let 0, be the pole, its coor- dinates being OA = Xo, AOi = yoi and let the polar axis make an angle a with the primitive axis of X. Then OiP=r, and EELATIOJSr RECTILINEAK AND POLAR SYSTEMS. 31 = AoOiP, or AiOiP, according as the polar axis lies above or below O^X^, drawn parallel to OX, i.e. according as a is posi- tive or negative. Hence, in general, XiOiP=^±a. Now Om=OJ^+0,Z), in which Om = X, OA = Xq, 0,D= OrPcos DO,P =r cos {0± a). Hence x = Xq + rcos(^± a) ; similarly, (1) o^-i y = >/o+'''s\n(0±a). ^ If the polar axis is ^7ara??eZ to the axis of X, a = 0, and the general formulae become X = Xq + r cos 0, y = yQ-j-r sin 0. (2) If the pole coincides ivith the origin, Xq = y^ = 0, and x = rcos{6±a). y = rsm(0±a). (3) If the pole is at the origin and the polar axis coincident ivith X, a = 0, a^o = ?/o = 0, and a; = ?-cos^, , y=rsmO. (4) 24. Formulae for passing from any polar system to any rectilinear system. From Equations (1) of Art. 23, X — x^) = r COS (^ ± a) , y — y^^ = r sin (^ ± a) . Squaring, and adding, and substituting the resulting value of ?*, we have, since sin''^4 + cos-^l= 1, r= V(.c-.r„)^ + (?/-?/o)-', COS(^ ± a) = sin {$ ± a) .v-yo ^{x-x,Y + {y-y,y ^ (1) 32 ANALYTIC GEOMETRY. If the polar axis is parallel to X, a = 0, and r = V(a; — .To)-+ {y — 2/o)^ cos ^ = ~ ^^ sin e = ^~-^° V(a;-Xo)2+(2/-?/o)' . If the new origin is at the pole, o-y = 2/0 = 0? and K2) r = Va-' + /, cos (g ± g) = -^^ . sm(g±a^= -^ .(3) If the neio origin is at the pole and the neio axis of X coincides with the x>olar axis, a = 0, a-y = ?/o = 0, and r = Va?+y, cos^ = — . sin^ = ^ (4) Var4-2/- Vx^+y- By means of Equations (4) we may pass from any polar system to a rectangular system with the origin at the pole and axis of X coincident with the polar axis ; then, by Equations (3) of Art. 22, to any oblique system. Examples. 1. Transform v/ — .t — 4 = (Ex. l,Art. 17) to a new set of parallel axes, the new origin being at (0, 4). The formulae for passing from any rectilinear system to any parallel one being x=Xf,+ x^, >J = 1/0 + ^i, in which 3-„=0, and j/o=4, the values of the primitive coordinates in terms of the new are, in this case, x = x^, y = 4: + ijy Substituting these values in the given equation, we have yj — .r, = for the transformed equation. As the subscripts are only used to distinguish the two sets of coordinates, they may be omitted after the transformation is effected. By referring to Fig. 10, Art. 17, the student will see that the new origin is at P', a point of the locus, and that there- fore the transformed equation should have no absolute term. 2. Transform ?/ + 0; — 1 = (YjX. 2, Art. 17) to a new set of parallel axes, the new origin being at (1, 0). Ans. y -\-x=0. 3. Transform 3?/ — 2a; + 4 = to parallel axes, the new origin being ( — 4, —7). Ans. 'Sy—2x — = 0. 4. Transform ?/- = 4.i; — 8 (Ex. G, Art. 17) to parallel axes, the new origin being at (2, 0), that is, at P', Fig. 12. Ans. y' = 4:X. RELATION RECTILINEAR AND POLAR SYSTEMS. 33 5. Transform y-=ix — 8 to anew set of rectauguhir axes with the same origin, the new axis of X making an angle of — 90° witli the primitive axis of X. The forinulaa are x = x\ cos y — y^ sin y, y — 3\ sin y + i/i cos y, in which y = — 90°. They become, then, x= i/^, y = — x^ - Substituting and omitting subscripts, a:^ = 4 y — 8. 6. Transform x" -\- if ~8x - 4.y -b = (Ex. 9, Art. 17) to a new set of parallel axes, the new origin being at (4, 2), that is, at Oi (Fig. 15) . Ans. x- + 2/"= 25. 7. Transform .T^= 10 (Ex. 15, Art. 17) to rectangular axes with the same origin, the new axis of X making an angle of 45° with the primitive axis of X. The formulae are x = x^ cos y — jji sin 7, y = x-^ sin 7 + ^1 cos 7, in which 7 = 4.5°, and they become x = Vl (x^ — y^), y = \/| (.r^ + y^) . Substituting these in xy = 10, and omitting subscripts, .r- — y- = 20. 8. Transform .t- + ?/- = 25 to a polar system, the pole being at (—5, 0), and the polar axis coincident with X. The formulas are x = x^ + r cos d, y = ^o + '" ^i" ^> wliich for Xq = — 6, y^ — 0, become x = — 5 + r cos Q, y = r sin 6. Substituting these in x"^ + y'^ = 25, we obtain r = 10 cos 6. 9. Transform (.r + ?/-)" =(r{x- — y") to polar coordinates, the pole being at the origin and the polar axis coincident with X. Ans. r^ = a^cos2 0. 10. Transform the following equations, the origin and the pole being coincident, as also the axis of X and the polar axis, r = 20cos^ ; xy = a\ ^^^^_ _^, ^ ^, _ 20^- =0 ; 7-^ - ^^- sin 2 e 1 1 . Having the distance between two given points in a rec- tangular system, d = Vix" - x')- -{-{y" - y')- (Art. 7) , to find the polar formula for the distance, when the pole is at the origin and the polar axis coincident with X. Substituting x' = r' cos 0', y^ = r' sin 6', x" = r" cos 0", y" = r" sin 6", d= V(i-" cos e" — r' cos e'Y + (?•" sin 6" — r' sin d'Y =. Vr"%cos^e"+sin^e")+r'\cos^e'+sin^e')—2r'r '(cos 0"cos e'+sin9"sine') = Vri-^+r"-^-2 r'r" cos id"-e'), 34 ANALYTIC GEOMETRY. wliich is the formula of Art. 13, which was there derived directly from the figure. 12. Under the same conditions find the polar coordinates of the point midway between two given points, having given its 1 r + x' + x" y'+y" rectangular coordmates ■ , -l — —^ — 2 2 ^^^^ r' cos ^'4-/-" cos ^" r'sin^' + /-"sin^" 2 ' 2 ' OlS. The distance between two points referred to a rectangu- lar system is d= ■\/{x" —x'y-{- {y" — y')'- Find the distance for an oblique system with the same origin, the new axis of X being coincident with the i)rimitive axis of X, and the new axis of Y making an angle ft with it. The formulae are x = Xi cosy + Vi cosyj, ?/ = ajjsiny + ^/jsinyj, which for y = 0, yj = /3, become x = x^ -{-yiCosfS, y = yis'mfi. Substituting these values, and dropping the subscripts, d= -Vix" -hy" cos 13 — x' — y' cos l3y-\-{y" sin (3 — I/' sin fSy = V[(.t" - x') + {y" - y') cos^]^^ + [(//" - y') sin/3]- = V(a;" - x'Y + {y" - ?/)- + 2{x" - x') {y" - y') cos/?, a result already obtained in Art. 7. CHAPTER II. EQUATION OF THE FIRST DEaREE. THE STRAIGHT LINE. -ooJl^JrJOO- SECTION IV. — THE RECTILINEAR SYSTEM. EQUATIONS OF THE STRAIGHT LINE. 25. Every equation of the first degree between tivo variables is the equation of a straight line. Every such equatiou may be put under the form Ax-{-By-\-C=0, (1) in which A and B are the collected coefficients of x and y, and C is the sum of the absolute terms. Let P', P", P'", be three points on the locus of this equation, whose abscissas in order of magnitude are x', x'\ x'". Then, from (1), their ordinates y', y'\ y'", will also be in order of magnitude. As these three points are on the locus, their coordinates must satisfy its equation ; hence Ax'+By' + C=0, Ax" + By"+C=0, Ax'" + By'" + C = ; whence, by subtraction, Aix"- x') + B{y"-y') = 0, A (x'"- x') +B{y"'-y') = 0. Fig. 27. 36 ANALYTIC GEOMETRY. Equating the values of A, y"'-y' . x"'-x' y"-y' . (2) Let P'Q be drawn parallel to OX. Then, from (2), P"'Q ^P"R P'Q PR Hence the triangles P"'QP\ P"MP', are similar, and P" is on the straight line P'P". In the same manner it ma}'^ be shown that every other point of the locus is on the same straight line P'P'". The locus is therefore a straight line. The expression " the line Ax + By + C= " will frequently be used for brevit}^, meaning "the line whose equation is Ax + By + C=0." ^ Oblique Axes. The above demonstration depends only upon tbe similarity of the triangles and is therefore equally true for an oblique system. 26. Common forms of the equation of a straight line. There are three common ways of determining the position of a straight line 3IN with reference to the axes. First, by its intercepts OR, OQ; second, by its y-intercept OQ, and the angle XRQ which the line makes with the axis of X (always measured, as in Trigonom- etry, from OX to the left) ; thiixl, by the length of the perpendicular OD let fall from the origin on the ~^ line, and the angle XOD which this perpendicular makes with tlie axis of X. In each case the posi- tion of the line is evidently com- pletely determined. We are now to find the equation of the line when given in each of these three different ways. First. Let P be any point of the line, x, y, its coordinates, and OR = a, OQ = b, the given intercepts. Then Fig. 28. THE RECTILINEAR SYSTEM. 37 QO : OR : : PL : LR, or h : a : : y : a - x, whence hx-\-ay— ab, or. dividing by a6, ah Second. Draw OS parallel to MX, and let tan XRQ = m. Then LP = SP - SL. But SP= OQ = b, SL = tanSOL . OL = tan ORP . OL = -t&nXRQ . 0L= - mx. Hence y = mx -\- b. (2) The tangent of the angle which the line makes with the axis of X is called the slope. Third. Draw LK parallel to MN, and PT parallel to OD. LetXOZ) = a and OZ>=i). Then 0K+TP=0D. But 0K= OL cosLOK^ x cosa, TP= LP sin TLP = y sina. Hence x cosu + y sina = p. (3) All these equations are, as they should be, of the first degree (Art. 25). Other forms of the equation of a straight line might be found by assuming other constants to fix its position, and such forms will be given later. The reason for employing more than one is that one form is often more convenient than another for the solution of certain problems. Equation (1) is called the intercept, Equation (2) the slope, and Equation (3) the normal form, while the general equation Ax -{-By + (7= is called the general form. The student will observe that Art. 25 is an illustration of the general problem : Given the equation, to determine the locus ; Avhile this article illustrates the inverse problem : Given the law of the moving point {straight line) and the position of the locus (by the constants), to determine its equation. In the latter case, the problem always consists in finding a relation between X and y true for every point of the locus, and expressing this relation in the form of an equation. Whenever we have sue- i^ 38 ANALYTIC GE0:METRY. ceeded in establishing this equation, we have the equation of the locus, whatever the constants involved. Oblique Axes. Whatever the angle A'Or(Fig. 28), the triangles QOR and PLR are similar; hence the intercept form applies without change to oblique axes. For the slope form we have, as before, LP = HP— SL, in which LP=y, SP= h; but SL : LO : : sin SOL : sin LSO. Let u = XO Y= the inclination of the axes, and A= XRQ, the angle made by MN with A". Then SL : ar : : sin A : siu (A — OQ^fi. Then OD = OK + 7'P = X cos a + y cos /3 =i>. When AOr= 90^, DOQ is the complement of XOD, that is of a, cos^ = sin a, and the equation reduces to (3). 27. Derivation of the common forms from the general form. Since the equation of every straight line i.s of the general form Ax-\-Bi) + C=0, it must evidently be possible to derive the common forms from the general form, and to express the par- ticular constants a, Z>, ?/i, ^>, cosa, sin a, in terms of the general constants A, B, C. First. TJie intercept form. Assuming Ax + By + C=Q, transposing C to the second member, and dividing the equation by — C, i'.c, making the second member positive unitv, we have X y A B 1, which is the required intercept forui, the intercepts being a = - ^ h=-^- A ' b' Second. The slope form. Solving the general form for y^ we have A C y= -T>^'- B B THE RECTILINEAR SYSTEM. 39 A which is the required slope form, in which m. = — ^, /j= as before. Third. The normal form. This form requires that the second member {p) should be positive, as no convention has been made for the signs of a distance except as that distance is laid off on the axes ; and also that the sum of the squares of the coefficients of x and y should be positive unity, since C0S"a + sin'a = 1. Let R be the factor which will transform the general to the normal form. It nmst fulfil the condition {RA)- + {RB)- = \. Hence R = — • Introducing this factor and transposing V.1- + B' C, we have Ax Bu _ -C V A' + B' V22 + B' VA' 4- B' in which A . B -C COSa= , SUl (I = — , p — V^-1- + B' V^l- + B' VA' + B' To make the second member (p) positive, of the two signs of -VA--\-B^ we must evidently take the opposite one of C. 28. In tlie preceding article we have found the values of «, &, 711, p, cos a, and sin a, in terms of ^^^1, B, and C. Cut it is unnecessary for the student to burden his memory with these relations. Thus, suppose we have given the straight line 3a; — 4^ + 10 = 0, and its intercepts are required ; we have only to put the equation in the intercept form, i.e., transpose the abso- lute term to the second member and then divide by —10, giving -^+^=1, 1 ' 10 3" 4 and the intercepts are seen lo be a — — ^^, h=i^. A still simpler way of determining the intercepts is to make y and x successively zero (Art. 17, Ex. (>) . Thus, for .); = 0, y=h=^^ ; 40 ANALYTIC GEOMETRY. and for y = 0, .'c = « = — y. Again, suppose the slope is required. We then put the equation under the slope form by solving it for ?/, obtaining and the slope is seen to be m = |, the y-intercept being b = J^"-, as before. Final! v, if the distance of the line from the orig-in (p) is required, we put the equation nnder the normal form by dividing it b}' V-/1' -\- B' = 5, transposing the absolute term to the second member and changing the signs throughout to make the second member positive, thus obtaining the distance from the origin to the line being 2, and a lying between 90° and 180° since its cosine is negative and sine posi- tive. The exact value of a would be found from the tables, being the angle whose cosine is — |, or sine is ^. 29. Discussion of the common forms. First. The interaqit form. This form is X , y . . , . , (J J C - + - = 1 , m which a = , o — • a b A B If a and b are both jwsitive, the line occupies the position M^N^ (Fig. 30) , both intercepts being laid off in the positive directions of the axes. If a and b are both negative, the line occupies the position J/"iV", botli intercepts being laid off in the negative directions of the axes. In like manner when a is positive and b negative, the line lies as does il/"'jV'", and «7ie?i a is negative and b jyositive, as does J/'^-Y'^'. We observe, also, that when C and B, as also C and A, of the general form have like signs, the intercepts are negative, and when they have unlike signs tlie intercepts are positive. Ifa = / = b is the equation of a parallel to X at a distance b from it, above or below according as b is positive or negative. Notice that when THE RECTILINEAR SYSTEM. 41 a = 00, -4 = 0, and the geueral form is independent of x. Sim- ilarly, if b = , in which in. This form is B' B If m is positive, the line makes an acute angle with X and cuts Y above or below the origin according as b is positive or negative. If m is negative, the line makes an obtuse angle with X. We thus have y = — mx + h, JPN', y = -mx-b, 2P'N'\ y= mx-b, J/'"^"S y= mx + b, M'^N'-". If m = 0, the line is parallel to X, and y = 6 is its equation, as already seen. If m — co, the line must be parallel to Y, since the angle whose tangent is 00 is 90°. The equation then becomes ?/ = x .r + 6. To interpret this form, we observe that, as the line is parallel to A , C „ the con- Fig. 30. Y, b must also be x , and hence in m = — ^, 6 = B B' The ditions m= x, & = x, will both be fulfilled when J3 = (J general form then becomes Ax+C='d, or x = 7 = «r as before. If b = 0, we have y = mx, or Ax -]- By = 0, as before^ p: 42 ANALYTIC GEOMETRY. the line passiug through the origin. The form y = mx is the most convenient for lines passing tlirough the origin, the value of m fixing the inclination of the line to X. Tuna). Tlie 7iormal form. This form is '^ , <^, a; cos a -\-y sin a =j>, in Nvhich A . B C cos a = — , sin a = ., 2'> ^ — — V-42 + B- V^" + B' -v/^- + B' the sign of V^-l- + B- being such as to make p positive. If both cosa and sina are positive, a lies between 0^ and 90° (3/'iV'). If both are negative, a lies between 180° and 270° ( J/"iV^') . If cos a is 2)ositive and sin a is negative, a lies between 270° and 360° (7W'"iV"'), and ifcosais negative and siiia posi- tive, between 90° and 180° {J^P^'N^'). If p = 0, the line passes through the origin, and its inclination is known when sina and cos a. are known, its equation taking the form x cosa + y sina = 0, or — 3r=z=r H = 0, or Ax -j-By —0, as .before. VA' + B- VA' + B' If a = 0° or 180°, sina=0, and the equation becomes cosa V^ + # ^ ^ as before, the line being parallel to 1'. //' a = 90° or 270°, cosa=0, and y = —^ — =h, in like manner, the line being sin a parallel to X. OnLHiL'E Axes. Tlie intercept form beiug the same, tlie discussion above given applies equally to oblique axes. The slope form is ?/= ' x + b (Art. 26). Since sin A is alwaj's positive, the sin(uj— A) Bign of the coefl'icient of x clcpomls upon that of 8in(w — A), anil will be positive or nega- tive as (u> or (Art. '2n) is similar to the 1} 7} above, the equation taking the forms x = ' — , y = — i — , when the lino is i>arallel to the * "^ cosa •' C08/S ' fczes, i.e., when ^ = 9(P and o=00\ respectively. THE RECTILINEAIl SYSTEM. 43 30. To construct a straight line, having given its equation. If the equation is given in one of the three common forms, we may construct the line b}' means of the given constants. For example, ■ ■■ (1) (2) (3) 8 4 — 1_2 ~5"' are the intercept, slope, and normal forms, respectively, of 4a; + 3y— 12 = 0. From the first, make OR = 3, OQ = A, and QR is the line. From the second, make OQ = 4 and draw QR-, making an angle with X, whose tangent is 4 3' This angle may be constructed without the tables by making QN=2>, NP' = A, these lines being parallel to the axes, laying off QN to the right or left of Q, as the angle is acute or obtuse, j.e., as wi is plus or minus. From the third lay off XOD = angle whose cosine is |- (or sine f ) , make OD = ^, and draw QR perpendicular to OD. Since the line is a straight line, it ma}' evidently be con- structed by constructing an}' two of its points. Thus, for x = -3, ?/ = 8, (P'), and for x=G, y = -4, (P"). But the points most easily constructed are those in which the line crosses the axes. Thus, in any of the above forms, for x= 0, 2/ = 4, (Q), and for y=0, ic = 3, (R). Hence, practically, whatever the form in which the equation is given, to construct a straight line from its equation., construct its intersections with the axes. Oblique Axes. To construct the line, make x and y in succession equal to zero, and determine the intercepts. Examples. 1. Construct the line whose equation is x—y=2. Making y = 0, we have x = 2, the intercept on X; making x = 0, we have ?/ = — 2, the intercept on Y. is the required line. A line through the points thus found 44 ANALYTIC GEOMETRY. 2. Construct the following lines : ^ — 2.i' + G = 0; x -\- y = 7 ; Sy + x — [) = {); 3a; + o^ + 15 = 0. 3. Construct the line // + 2 x = 0. Since this equation has no absohite term tlie line passes through thu origin. In such a case construct any second point; thus .r = 1 gives y = — 2. Then join (1, — 2) with the origin. 4. Construct the lines y = x; y = — x; 2y — ox=0. 5. AVhat angle does y + Sx—l — make with X ? Putting the equation under tlie slope form, ^ = — 3 .( + 7, and the angle is the angle wliose tangent is —3. 6. What is the distance of 6 a; + 8 .?/ + 11 = i) from the origin ? Dividing by — V^"^ + B^ = — 10, \vc liave — -i x — ^y = \\, and ;j = ii = the required distance. 7. Find the intercepts, slope, distance from the origin, and angle made by the perpendicular from the origin on the line with X, of the line '2x+ly — 'd = 0. Making x~0, y = b=^; making y = 0, x = a=?^. Solving for y, y = — 'i-x+ 2, .-. ?u — — ?. The normal form is 2x . 7i/ 9 9 , -17 1 ■— = , .•./; = , and a = sni ^ — — \/53 \/53 \/53 v'SS VSS 8. Find a, 6, m, jh and a in the following lines : 3 >/ — 4 a; + 25 = ; 7.i- — // = ; y-\-x — o = 0. 9. The intercepts of a line are 6, 3 ; write its equation. Ans. x + 2?/ — G = 0. 10. A line makes an angle of 45° with X, and cuts i'at — 2 from the origin ; write its equation. Ans. y — x— 2. 11. The distance of a line from the origin is (!, and the per- pendicular upon it from the origin makes an angle of G0° with X; write the equation of the line. Ans. V3y + .'c=12. 12. Same as Ex. 11. when the angle is 120°. 13. AVrite the equations of p;u:illels to A", one 4 above, and one 10 below it. THE RECTILINEAIl SYSTEM. 45 14. Write Ihe equations of the sides and diagonals of a square whose side is 10, the sides being parallel to the axes and the centre at the origin. 15. What are the equations of the axes? Since x = a is a parallel to Y at a distance = a, if a = the line coin- cides with y. Hence t = is the equation of Y. Similarly y = is the equation of X. 16. Determine which of the points (2, 3), (1, —3), (_2, 7), ( — 3; 11) are on the line 2i/ + 7.i-— 1 = 0. 17. Find the length of the portion of the line 4a; + 3?/ = 24 included between the axes. Ans. 10. r^ 18. The line y = mx passes through {x\ y')\ find the value of m, 31. Equation of a straight live x>assing through a given point. Let {x\ y') be the given point. Since the required line is a straight line, its equation will be of the form (1) Ax-\-By + C= 0, and since it passes through the point (x\ y'), we have Ax'+By' + C=0. Subtracting this from (1), A{x-x')+B{y-y')=0. ..,y-y'=-^{x-x'). But -|=m. Hence V — y' = "^ {^ — ^') •> (2) being a relation between x and y in terms of the given constants x\ y', is the required equation. An infinite number of lines may be drawn through a given point ; hence the line is not determined unless its slope m is also given. Thus, the line through (1, —4), making an angle of 45° with X, is y-\-i=l(x — l), or y = x—o. Oblique Axes. Th2 above equation applies to oblique axes, understanding that „,=_jiE^ — f*/ sin (u> — A) '^ 32. Equation of a straight line passing through two given points. Let (x',y'), {x",y") be the given points. The required 46 ANALYTIC GEOMETRY. equation will be of the form (1) Ax + Bj/ -\- C=0, since the Hue is a straight line, and must be satisfied for the co- ordinates of the given points; hence (2) Ax' -\- By' -\- C =0, (3) A'c"+%"+C=0. Subtracting (2) from (1), and (3) from (2), we have A{x-x') + B{y-y') = 0, A{x'-x")-\- B (y'-y")=0. Transposing, and dividing, which is a relation between x and y in terms of the given constants, and hence the equation required, the coefBcient of x, v'— v" ^^ — ^, being the slope (Art. 27). Thus the line passing through (1, 2) and (-3, 4) is y - •> = ^^^ (x-l), or 1+3 2y + x—b = 0. It is immaterial which point is designated 4 — 2 as (x',y'). Thus, y - 4 = — ^^— (.-« + 3), or 2?/ + .^- - 5 = 0, as before. Oblique Axes. Nothing in the above reasoning being dependent upon the inclina- tion of the axes, the equation is the same if the axes are oblique, only the coellicient •^ ~y is then the ratio of the sines of the angles which the line makes with X and Y. x'-x" Examples. 1. "Write tlie equation of a line through (2, 4) having the slope 5. Ans. y — 5 .^• -j- 6 = 0. 2. "Write the equation of a line through (2, 3) and (1, —2). In place of using Eq. (4), Art. 32, as there illustrated, it is quite as expeditious to determine the constants of any one of the three common forms directly. TIius, the form ij — mx + h, satisfied in succession for the two points gives 3 = 2 »« + h, and — 2 - m + h. Subtracting, we obtain m = 5. Substituting this value of m in either of the above, we find b = — 7. Hence y = 5 .r — 7. 3. Find the equations of the sides of the triangle whose vertices are (4,8), (1, 4), ( — 1, —8). Ans. Sy — 4x-H = 0; T)?/ - 12.x-- 8 = ; y-2x = 0. THE EECTILINEAll SYSTEM. 4r 4. Find the equations of the raedials of the triangle of Ex. 3, Ans. ll?/-2U;c-8 = 0; y-4x=0; 13y - 28x- - 8 = 0. 5. AVrite the equation of a line through (2, 5) and the origin. We may use the equation of a line tlirough two points, making one of the points (0, 0) ; or tlie slope form */ = mx (since b= 0), which, satisfieci for (2, 5), gives m = 5, and therefore y = | x. 6. Write the equations of tlie following lines : (1) through ( — 7, 1), making an angle 45° with X. (2) through (2, -1), and (-3, 4). (3) through ( — 1, —7), and the origin. (4) through (-G, -3), parallel to X. (5) through ( — 1, 2), parallel to Y. Ans. y — x — 8 = ; >/ + x — 'i = ; y—7x; ?/ = — 3 ; x = — 1. PLANE ANGLES. 33. To find the angle included bettvemi tivo given straight lines^ The slope form is best adapted to this problem. Let y = mx + 6, y = m'x -\- b', be the two given lines ; m and m' ai-e the tangents of the angles XRP=X, and XQP = X\ Then, if c = tan RPQ = tany, from Trig- onometry we have tan A' — tan A tany= or c = 1 + tan A tan A' m' — 7n 1 + mm' (1) Thus the tangent of the angle between Fig. 32. , and the angle may A O y= 4.T + 7 and ?/= 2a; — 1 is c = — , ...... .^v. ....j,- be found from a table of natural tangents. It is immaterial whether we substitute 4 for m' and 2 for m, or vice versa; in the latter case c = = , the difference in sign being due to the fact that the tangents of the supplementary angles EPQ 48 ANALYTIC GEOMETRY. and QPS which the Hues make with each other are numerically equal with opposite signs. We thus obtain the acute or obtuse angle, according as the sign of the result is positive or negative. 34. To find the equation of a straight line mcikincj a given angle v:ith a given line. Let y = mx + h be the given line, y = m'x + h' the required line, and c = tangent of the given angle. Then in the relation c = m' — m 1 + '^nm'' c and m are known. Solving for m', we have , m -4- c m'= 1 — mc Hence the requu'ed equation is m 4-c y = x + b'. (1) 1 — mc Since an infinite number of straight lines may be drawn making a given angle with a given line, b' is undetermined. We are then at libertN' to impose another condition upon the line, as that it shall pass through a given point. The equation of a line through a given point is y — y' — m' (x — x') , in which m' may have any value (Art. 31). Substituting the value found above, y-y m + c {X - X') (2) 1— mc is the equation of a straight line passing through a given 2wint and making a given angle loith a given line. Thus, the line through (2, 4), making an angle 45° with ?/= 2.^—4, \y is2/-4=^±i(.r-2), or ?/ = -3.v+10. Fig. 33. 1-2 ^ Constructing, MN is the given line ?/ = 2.^ — 4; P' the given point (2,4); and P'Q the hue ?/ = -3.x'+10. The student will observe that P'R makes :in angle 135° with MN, the angle be- ing measured, as always, from the line to the left ; and that, therefore, to obtain the equation of P'R wo should make c = tan 135° = — 1 . THE RECTILINEAR SYSTEM. 49 35. Conditions that two lines shall he parallel,, or perpendicu- lar,, to each other. First. If two lines are parallel,, their included angle is zero. All ' /JIT Hence (Art. 33) c= = 0, or m== m' ; th;it is, tivo lines 1+mm are paralld whenever, their equations being solved fur y. the coefficients of x are equal. This follows obviously from the fact that parallel lines make equal anoles with X, and hence the tangents of these angles, m,m\ must be equal. Thus, ?/=2a; + 4, y = 2x — 7, y — 2x = 0, are all parallels. Cor. The equation of a line passing through a given point (x', y') jyarallel to a given line y = mx + b, is (Art. 31) y-y' ^m {x-x'). (1) Second. If the lines are perp>endicular to each other, their included angle is 90°, and hence c- = = oo, 1 + 7nm' or l+mm' = 0, .•.m'= ; m that is, tivo lines are perpendicular to each other ivhenever, their equations being solved for y, the coeffi^cients of x are negative recip- rocals of each other . Thus, are all perpendicular to y = f a; + 7. Cor. The equation of a line passing through a given point (a-', y') perpendicular to a given line y = mx + b is y-y'= {x-x'). (2) m The equations m = m\ ra' = , are not the equations of m lines, for they contain no variables. Since they involve only constants (which serve to fix the position of the lines in ques- tion) , they express conditions imposed upon the position of the lines. Such equations are called equations of condition. 60 ANALYTIC GEOMETRY. Examples. 1. Find the angles between the lines ?/= —a; + 2, ?/ = 3« + 7; y = ^x—l, ?/ = -f-a; + 4; y = 2x-o, i/=2x+l; ?/ = i a; - 3, y = -2x + 0: 3 ?/ + 4a- + 1 = 0, 2.v + .<; + 5 = 0. Ans. c=2; c = -.^ ; 0°; 90°; c = i.^ 2. Write the equation of a line making an angle whose tangent is 3 with ?/ = — 3x + 4. Ans. y = 6. 3. Write the equations of lines making angles of 45° and 135° with 2y — x + d = Q. J^y^s_ y = Zx + h\ y = — \x-^h. 4. Write the equation of a line through (— 3, 7) making an angle whose tangent is V3 with 2y — a; + 1 = 0. Ans. (2 - V3) y - (1 + 2 V3) a; - 17 + V3 = 0. 5. Write the equations of two parallels to y = -| x-\-l ', also io'dy-\-lx = Q. ^ 6. Write the equation of a parallel to 3?/ — 4 a.' =2 through (1,2). Ans. Zy- 4.x -2 = 0. 7. Write the equations of two perpendiculars to y = — hx -\- 4 ; also to y — x + 4 = 0. 8. Write the equation of a line through (7,-1) perpen- dicular to ?/ = — 4a; + 1 ; also through (7, — 1) perpendicular to dy-2x = 0. Aiis. 4iy-a; + ll = 0; 22/ + 3.r- 19 = 0. 9. Write the equations of lines through (1, 3) making angles of 0°, 90°, and 45° with X. jins. y = 'i ; x=l; y- x-2 = 0. 10. Write the equation of a line through (5, 3) parallel to the line whose intercepts are 3, 2. ^ns, 3?/ + 2 a; — 19 = 0. 11. Write the equation of a line through (2, 3) perpendicular to the line joining (2, 1) with (—2,5). Ans. yz=x + l. 12. The vertices of a triangle are (—1,-1), (— 3, 5), (7, 11). Write the equations of its altitudes. yins. 3?/ -.0-26 = 0; 3?/ + 5.^- + 8 = ; 3// + 2.«;- 9 = 0. 1 THE IlECTILlNEAll SYSTEM. 51 13. Write the equations of the perpeadiculais erected at the middle points of the sides of tlie triangle of Ex. 12. Ans. 3y-x-S = 0; 3 ?/ + o.u - 34 = ; oy + 2x — 2i=0. 14. Prove that Ax + B>i -j- C = is perpendicnlar to A'x + B'y + C = if AA' + BB' = 0. 1.0. Prove that Ax + By -f C = is parallel to A'x + B'y + C = if AB' - A'B = 0. 16 Prove that the angle between Ax + By -f- C'= and A'x + B'y + C = is given by the relation tan y = -T- J-r to J ;' AA' + BB' 17. Write the equation of a straight line perpendicular to Ax-\-By+ C=0 and making an intercept a on the axis of X. 18. Write the equation of a line perpendicular to y = mx + b and at a distance d from the origin. 19. Write the equation of a line parallel to y = mx -\-h and at a distance d from the origin. 20. Prove that if the equations of two straight lines differ only in their absolute terms, the lines are parallel. INTERSECTIONS. / J 36. Intersection of loci. The point of intersection of two straight lines is the point common to both. But if a point lies on a given straight line, its coordinates must satisfy the equa- tion of the line ; hence the coordinates of the point of intersection must satisfy the equations of each line. Conversely, to find the point of intersection of two straight lines, combine their equations and find the set of values of the coordinates ichich satisfies them both. The above reasoning is obviously entirely general. AVhatever the loci under consideration, if the}' have a common point, or points, the coordinates of these points must satisfy both equa- tions. Hence, in general, to find the intersections of any two loci, combine their equations. 52 ANALYTIC GEOMETRY. Since the number of sets of values of x and ?/, obtained by making the equations simultaneous, is equal to the product of the numbers indicating the degrees of the equations, this prod- uct also indicates the possible number of intersections. If, for example, the equations are of the second degree, their loci may intersect in four points, but no more ; and as some of the val- ues of X and y thus obtained may be imaginary, the number of real intersections may be less than four. And, in general, the greatest possible number of intersections of two loci whose equa- tions are of the j>th and qili degrees, respectively, will be pq, and the number of real intersections will be the number of sets of coordinates, satisfying both equations, in which x and y are both real. Since all equations of straight lines are of the first degree, but one set of values of x and y can be found satisfying any two such equations, or two straight lines can intersect in but one point. If two straight lines are parallel, they cannot intersect, and the combination of their equations will give an impossible result. Thus, x-\-y = 4: and x + y=7 are parallels. Com- bining, we obtain 3 = 0., Hence non-intersection is shown by the occurrence of impossible or imaginary results. Otherwise, equating the values of .r, 0?/ = 3, or 7/ = |-=oo, showing that the lines intersect only at an infinite distance. Examples. Find the intersection of the following lines : 1. 2?/ — 3a;- 7 = 0, and 2i/+ .T- 10 = 0. Ans. (|,-V-)' 2. .T-f-2?/ — r) = 0, aud2.r + // — 7 = 0. Ans. (3,1). 3. 2/ — .r 4- 1 = 0, and y -f x -f 1 = 0. Ans. (0, - I ) . 4. 6^0; -t- 6 y — 1 = 0, and x-\-y = 4. 5. iB + y =0, and .^'— ?/ = 0. 6. Find the vertices of the triangle whose sides are 5y-12a;-8 = 0, 3?/ - 4.x- - 8 = 0, y-'2x = 0. Ans. (1,4), (-4, -S), (4,8). THE RECTILlNEAIt SYSTEM. 53 7. Show that y + 3x- - 1 = 0, ?/ + 2;i; + 7 = 0, y - .r+ 31 = 0, meet in a poiut. 8. Show that the medials of the triaugle of Ex. 4, Art. 32, meet iu a poiut. 9. Show that the altitudes of the triangle of Ex. 12, Art. 35, meet in a point. 10. Show that the perpendiculars erected at the middle points of the sides of the triangle of Ex. 13, Art. 35, meet in a point. Examples on the intersection of curves are reserved until the student is familiar with the equations of the curves ; but he will observe that the process is the same, whatever the degree of the equations or the system of reference : to find the intersec- tions of any lines, combine their equations. 37. Lines through the intersection of loci. Let (1 ) Ax + By + C = 0, (2) A'x + B'y + C = 0, be the equations of any two straight lines, aud 7i any arbitrary con- stant ; tlieu is (3) Ax + By + C + k {A'x -j- B'y + C') = the equation of a straight line through their intersection. For the values of x and y which satisfy (1) aud (2), evidently satisfy (3) also, hence (3) passes through the poiut of intersection of (1) and (2). Moreover (3) is of the first degree, hence the equation of a straight line. Note. This reasoning is entirely independent of the form and the degree of the equations. Hence if a = 0, ;3 = 0, be the equations of ani/ tiro loci, a and fi representing any functions of x and y, and k he any arbitrary constant, a + kfi = passes through all their points of intersection. So long as Ic is arbitrary, (3) will represent any straight line through the intersection of (1) and (2), and as h may have any value, it may be determined so that (3) shall fulfil any reasonable condition. Thus : 54 ANALYTIC GEOMETRY. First. To find the equation of a straight line passing tliroiigh the intersection of two given straight lines and also through a given point. Let (1) and (2) be the two given lines and {x',y') the given point. Then (3) is a straight line through their inter- section. Since this line is to pass through (x',y'), we have A.v' + By'-}-C-\-k {A'x'-{-B'y' + C') = 0, in which everytliiug is known hut A;. Determining Jc from this equation and substi- tuting its A'alue in (3), we have the required equation. Second. To find the equation of a straight line passing through the intersection of two given straight lines, and parallel (or per- pendictdar) to a given line. Let (1) and (2) be the given lines and y = mx + b the line to which (3) is to be parallel (or per- pendicular). Solve (3) for y and place the coefficient of x equal to m [or ] ; from this equation determine Ji and si;b- stitute its value in (3). The resulting equation will be the line required. Examples. Write the equations of the following lines : 1. Through the intersection of .T + 2?/ - 5 = aud y - 3.i- -f H = 0, and the point (6, 4). Substituting .r = G, // = 4, in x + 2 y — 5 + k (;/ — o .r + 8) = 0, we find ^ = f . Hence the required line is ?/ — x + 2 = 0. 2. Through the intersection of 2 a; + ?/— 7 = and x +2y — o = 0, parallel to G x — 3 y + 5 = 0. We have 2x + ;/ — 1 + k (.r + 2 y — 5) = 0. Solving for //, ^ 1 + 2 /.• "^ 1 + 2 A.-" Solving the parallel for y, ?/ = 2t+ J. Hence — - — -—=2. .-. A- = — <, and the required line is 2 x — y — [>■-= 0. "*" 3. Tlirough the intersection of 2x + ?/ - 7 = nnd y — x — l = 0, perpendicular to Sx + 3// — 1 = 0. Aus. y — x— I = 0. THE RECTILINEAR SYSTEM. 55 4. Through the intersection of y — x — l = and ?/ — 2 a; + 1 = 0, parallel to ?/ = 4 x + 7. Ans. y = 4:X — 5. 5. Through the iuterseetion of y = Sx -\- 14 and y = x-}-(}, making an angle of 45° with y = 2x. Ayis. y = — Sx— 10. 6. The line y=mx + h passes through the intersection of y = m'x + jy with ?/ = vi"x + h". Find the value of m. DISTANCES BETWEEN POINTS AND LINES, AND ANGLE-BISECTORS. 38. To find the distance of a given X)oint from a given straight line. Let a- cos a + ?/ sin a=p be the given line and (x',y') the given point. Through the given point, P', draw ST parallel to the given line MN. The perpendiculars OQ, OR, from the origin on these lines, coincide ; therefore a is the same for both (Art. 26), and the equations of the parallels will differ only in the lengths of the perpendicnlars. Hence, if OR^=p\ the equation of ST will be X cos a + y sin a - and since P' is on ST, x' cos a 4- ?/' sin a P\ \^^ P- Now DP' = QR is the difference be- tween 2^' ^ud 2'>i hence the required dis- tance is Z) = x' cos a + y' sin a. —p. lint this is simply what the equation of the given line becomes when j^ is transposed to the first member and x\ y', substituted for x, y. Hence, to. find the distance of a given point from a given line, Fig. 34. (1) 56 ANALYTIC GEOMETllY. put the equation of the given line under the normal form, trans- pose the absolute term to the first member, and siibstitute the co- ordinates of the given i^oint. Since to put Ax + By + C = under the normal form we divide by ^A- + J3-, we have jy ^ Ax' + By' 4- C As only the distance DP' is required, it is not necessary to attend to the sign of ^A--\-B-. If, however, we follow the general rule of signs for putting the general under the normal n form (Art. 27), the last term of (1), — . ivill always he VA + B' negative, since, when transposed, it must equal -\-p. Now if we make x' and y' zero, (1) will be the distance of the origin C from the line = — 1:^:2=^= ; hence the origin is always con- VA + B' ° ^ sidered as being on the negative side of the line. Whenever, then, for any given point, (1) is negative, the point is on the same side of the line as the origin. Thus, suppose the equation of MN is 2.T -f- ?/ — 2 =0. Dividing by Vo, the normal form is 2x V 2 — ^ + -^ r. =0. Substituting the coordinates of P', (|, f), Vo Vo Vo 9 3. 4- -3. _ 9 '}]_ D = ^ ' 2 ^2 1 _ -'3 _ i)p\ Vo v'o This being positive, P' is on the opposite side of tlie line from the origin. But, substituting the coordinates of P", (— f, 2), 9 3. I o 9 Q D = ' - = — = P " D' V5 Vo This being negative, P" is on the same side of the line as the origin. Were the axes oblique, tlie equation of the given line being a;co8a + y cob j3— p = (Art. 26), as the reasoning above is independent of p, x'cos a + j/'cosp —p would be the required distance. 39. The distance from a given point to a given line may also be found as follows : Write the equation of a line through the THE RECTILINEAR SYSTEM. 57 given point perpendicular to the given line ; find the intersection of the perpendicular and the given line ; then find the distance from the given point to this intersection by the formula d = V(a;' - x"y-{- (y' - y"y. Thus, to find the distance from (8, 1) to 3. r — 42/4-5 = 0; the perpendicular through (8, 1) to dx — ■iy-{-o = is y —1 = —^{x—8); combining this with dx — iy-{-o = 0, we have for the point of intersection (5, 6). Hence d = V(8-5)2+ (1-5)2=5. This method is usually less expeditious than that of Art. 38. Examples. Determine the length of the perpendicular from the point to the line in the following cases, ascertaining in each case whether the point and the origin are on the same or opposite sides of the line. 1. 3a; 4- -1^-2 = 0, (2, 7). Alls. -V2- ; on the opposite side from the origin. 2. 3.^• — 42/4-5 = 0, (8, 1). Ans. 5 ; on the side of the origin. 3. 4ic-3^-G = 0, (1, -1). Ans. i ; on the opposite side from the origin. 4. Sx + iy + 2 = 0, (2,4). Ans. ^ ; on the side of the origin. 5. y—'2x+ 1=0, (-1, -3). Ans. 0. 6. Find the lengths of the altitudes of the triangle whose sides are Ax-Sy + 8 = 0, 12.r- 5?/ 4-8 = 0, 2x~y = 0. Ans. The vertices are (1, 4), ( — 4, —8), (4, 8), and the ,.., , 2 16 16 altitudes , — , — V5 5' 13 7. Find the length of the altitudes of the triangle whose vertices are (1,2), (-2,0), (6, -1). . 19 19 19 Ans. ■ — r=, — :=? Vl3 V65 V34 58 ANALYTIC GEOMETEY. 8. Find the area of the triangle whose vertices are (2, 3), (-1,4), (6,5). The line througli (—1, 4) and (G, o) is :•■ — 7 // + 20 = ; its normal forni. r 7 ;/ '^n 1 n IS — '- + -!^ -^ = 0. The distance of (2, 3) from this side is =-^. V50 Vso VSU v^ The length of the line joining (— 1, 4) with (0, 5) is VdO. Hence the area = ). ( -^ X V50 ) = 5. ' VV5U / 9. Find the area of the triangle whose sides are 2.r + ?/— 7 = 0, y-x-\=0, x + 2y-o = 0. Ans. The vertices are (3, 1), (2, 3), (1, 2), and area f. 10. Find the distance between the parallels y =2x — i), y = 2x + 8. The line 3/ = 2 a- — 6 crosses Y at (0, —0); the distance of this point 14 from ?/ = 2 .r + 8 is — ^/5 11. Find the distance between the parallels y=3 X, y = ox—10. Ans. VlO. "40. To Jjnd the equation of a line bisecting the angle beticeen tico given lines. Let a; cos a +?/sinu — ^^ =0, (1) xcosa' -\-y sina' —]>' = <\ (2) be the two given lines. Then {x cos a' + y sin a' — p') -f- h: {x cos a + 2/ sin a — p) = (3) is a straight line throngh their intersection. Now the quantities in the parentheses are the distances of anv point (.r, y) from the lines (2) and (1) (Art. 38). Thus, if J/JV, 31' y, be the lines given bj' (1) and (2). then (3) is the equation of so?»e line VP through their intersection V, nnd the parentheses are the distances i'D\ PD, of any of its points from Pigg. M'N' and 3IN. Now if A" = -1, THE 1IECTIL[NEAR SYSTEM. 59 PD' = PD from (3), and (3) will be the equation of the line bisecting the angle MVN'. AVhen a point P is on the same side of a line as the origin, we have seen that the perpendicular PD is negative (Art. 38). For the angle MVN\ P is on the same side of both lines that the origin is, and hence both perpendiculars must be negative, that is, have the same sign, or A:= — 1. For the angle N'VN^ Q is on the same side of one line that the origin is, but on the opposite side from the origin in the case of the other line ; one perpen- dicular must therefore be negative, and the other positive, that is, have opposite signs, or ^'=1. Hence, to bisect the angle betiveen tivo given lines, put their equations under the normal form, and subtract or add them according as the origin does or does not lie ivithin the angle to be bisected. Examples. 1. Find the bisector of the angle between 12 a; + 5 ?/ — 2 = and 3 .^' — 4 y/ + ''^ = ^1 in which the origin lies. Ans. %^ri6^71y — 4W=0. 2. Find the bisectors of the angles between 2 a; -^'y + 8 = 0, x + 2y — 3 = 0. Ans. 3.T + 3 ?/ + 5 = 0; cc — y+11 =0. 3. Find the bisectors of the angles between 2.^•4-2/ + 8 = 0, and?y = 0. Ans. 2 ;r + (1 ± V5)//+ 8 = 0. 4. "Write the equations of the bisectors of the angles between the axes y = 0, x = 0. Ans. y ± x = 0. 5. Of what line would Eq. (3), Art. 40, be the equation if k = 2? if A- = n ? ^ / ' 6 - 60 ANALYTIC GEOMETRY. SECTION v.— THE POLAR SYSTEM. 41. Derivation of polar from rectangular equations. When the pole is taken at the origin and the polar axis is coincident with the axis of X, auy rectangular equation of a straight line may be transformed into the corresponding polar equation (that is, the polar equation expressed in terras of the same constants) by means of the relations a; = r cos ^, y = ?• sin ^ (Art. 23). The simplest and most useful of the polar equations is the normal form. 42. Normal polar equation of the straight line. The normal rectangular form being x cos a+^sin a=p, substituting a;=rcos 6 and y = r sin 6, we have r(cos cosa + sin sina) =p, whence r = P Discussion. If 6=0°, r = cos {0 — a) P P cos ( — a) cos a the point Q where MN crosses the polar axf| (1) OQ, locating If = a, r = P cosO° ~-^^' g^^'i^g the point D. If $>a and increasing, 6 — a is increasing. cos(^ — a) decreases, and hence r increases till 6* = a + 00°, when cos(^ - a) = cos 90° = and r = x , as it should be, since r is then parallel to 3IN and must be produced infinitely to meet the line. When $>a + 00°, ^-a>yO°, and r is negative, showing that it must be produced backwards, or away from the end of the measuring arc, to inect MN, and remains negative \ \ 'V^" Fig. 36. THE POLAR SYSTEM. 61 till = a-{- 270°, or 6 — a = 270°, when r = co agaiu, and is par- allel to J/iV^. For ^ = 360°, r = j^ — - = ^^=0Q. The cos (—a) COS a entire line is traced for values of 6 between 0° and 180°, for which latter value of 0, r = ,, ,,,,o : = = OQ. cos (180 —u) cos a If MN is perpendicular to the polar axis and lies on the right P of the pole, a = 0°, and the equation becomes r = -^ ; if on ^ ^ cos 6 the left of the pole, a = 180°, and the equation becomes p 2^ ~P r = cos (^-180°) cos - (180° - 6^) cos^ p P Hence r = ± ;; is the equation of all perpendiculars to the cos 6 ^ M A polar axis, the negative sign applying to those which lie on the left of the pole. If the line is parallel to the polar axis and above it, a — 90 and the equation becomes P P P . '' ^ cos (6 - yO°) ^ cos -(90°-^) ^ sln^ ' if below the polar axis, a = 270°, and P P P r = cos (^-270°) cos-(270°-^) sin ^ Hence r = ± —. — - is the equation of all parallels to the polar axis, the negative sign applying to those which lie below the pole. If the line passes through the p)ole, p) ~ 0, and r = 0, except when 6 = 90° + a, in which case r = - ; that is, r is zero for all values of 6 except when the radius vector coincides with the line, when r may evidently have any value. Examples. 1. Write the polar equation of a line whose dis- tance from the pole is 5, the perpendicular being inclined 45" to 62 ANALYTIC GEOMETRY. the polar axis. Find the intercept on the axis, and the values of for which r is infinite. ^"^- '■= ra — T^A ' 5\/2; 135°; 315°. cos {6 — 45 ) ' ' 2. Write the polar equations of lines for which p = 2, a= 60° ; p = 10, a = 120° ; and find their intercepts. 2 8 3. Construct r — cos(e-30°) cos (^-(30°) 4 5 4. Construct r = ± ; r = ± cos sin 6^ 5. Write the polar equations of the sides of a square whose centre is at the pole and side 10, one side being parallel to the axis. 3 — 5 G. Find the rectangular equations of r = — ^ — ; r 7. Find the rectangular equation of r = cos 6 sin 9 cos {0 — 45°) Ans. x-\-y — 9V'2 = 0. 8. Find the polar equation ofSa; — 4?/4-l = 0. Qx 4 V 4- 1 If the normal form is required, -^^—^ — = 0, whence p = h aii*l — 5 a — cos"* -|, which may be found from the tables. Then substitute p and a in r = f- — ■ — If the normal form is not specified, substituting directly cos(0 — a) , the values of :c — r cos 6 and y = r sin 6, we have r = 4 sin — 3 cos d 9. Find the polar equation of y = 3 x-^'2. Ans. r- sin ^ — 3 cos B <^"l- ^^ ' APPLICATIONS. 63 SECTION VI.— APPLICATIONS. 43. Recapitulation. The foregoing formulfe and equations relating to points and straight lines constitute the elementary tools, as it were, of analytic research on the properties of recti- linear figures. The student must remember that it is not the object of Analytic Geometry to produce these equations and formulas, but to investigate the properties of loci by means of them. While, therefore, familiarity with these expressions is indispensable, a mastery of analytic geometry implies a knowl- edge of their use in the discovery of geometrical truths ; that is, the mastery of a method of research. The more important of these expressions are here collected as a review exercise. The student should memorize them, and be able to explain the x' 4- x" meaning of all the quantities involved. Thus, x = — — — , v'+v" ^ y = - — 7-=^, are the equations of a point midway between two given points, in which x, y, are the coordinates of the required middle point, and x', t/', a;", y", the coordinates of the given points. aj = ^-±^, y = t±JL. Equation (3), Art. 6. f\2 d = Vr'-+ r"-- 2 r'r" cos {$" - 0') . ' ' x = Xa-{-Xi, y = y^, + yi. " a;=>-cos^, y = rsmO^ " Ax-\-By+C=0. " O y=imx + h. " (1), - 7. (1)^ " 13. (2), " 22. (4), " 23, (1), " 25, (1), " 26. (2), " 26. 64 ANALYTIC GEOMETRY. X cosa + y sin a =2>- Equation (3) , Art. 26. y — y' = vi{x — x') . " y-y' = ^~^„ {^-^')- " X' — X c = m' — m 1 + mm' (( y-y = , (x - x'). « 1 — mc , 1 m = m , m = • m' y — y' =: m (x — x') . *' y-y' = — {x — x'). " m ^,^ Ax'+By'-\-C _ ■Va'+ B' acosa' + y sina'— p'± (ajcosa + y sina— p) = 0. r= ^ " (1), cos(^-a) ^ ' (2>, '• 31 (4), " 32 (1). " 33 (2), " 34 " 35 (1), " 35 (2), " 35 (1). 38. 40. 42. PROPERTIES OF RECTILINEAR FIGURES. 44. 1. The diagonals of a square are j^erpendicular to each other. Take two adjacent sides for the axes. Then, if a= side, the vertices are (0, 0), (a, 0), (a, a), (0, a), and the equations of the diagonals are y = x, y = — x + a, in which m = ^ (Art. 35). '"^ 2. The line joining the middle points of two sides of a triangle is parallel to the third side. Take the third side for the axis of X, and the origin at its left-hand extremity. Tlien (0, 0), (o, 0), (&, c) are the ver- tices, [-, Hi [ " , -), the middle points, and 2/=| is the line joining them. APPLICATIONS. 65 3. Tlie diagonals of a parallelog)xim bisect each other With the axes as in the figure, let the side OB = a, the altitudi' mD=b, and Om = c. Then the coordinates of C are (a 4- c, &) , and the middle point of OC is '^ The coordinates of B are (a, 0), of D, (c, h), and of the middle ''a + c 6' 2 ' 2. Fii?. 37. point of BB, 4. T/te straight lines joining the middle points of the opposite sides of any quadrilateral bisect each other. Let CO, 0), (a, 0), (b, c), {d, e) be the vertices 0, -B, C, D, in order. Then the middle point of each line is ''a-\-b -{-d c-\- e"^ 5. Prove that the middle point of the line joining the middle points of the diagonals of any quadrilateral is the point of inter- section of the lines of Ex. 4. 6. The lines joining the middle points of the adjacent sides of a parcdlelogram form a parallelogram. With the notation of Ex. 3, the slope of the lines joining the middle points of DC and BC, DO and OB, is ; hence these lines are parallel. 7. The middle point of the hypothenuse of any right-angled triangle is equally distant from the vertices. Take the axes coincident with the sides. 8. Prove that if A, B, C, be squares on the sides of a right- angled triangle OJiQ, and OT is perpendicular to RQ, then RS, QP, and OT meet in a point. With the axes as in the figure, let c = OQ, d = OB, the sides. Then the coordinates of S and B are (c, — c), {— d, 0),. and the equation of SB is ^./ 66 ANALYTIC GEOMETRY. X — cd 4- d c + d .Similarly the equation of QP is y = ^^ x — c. The equation of RQ is ~ — \- -^ = d J —d — c 1, and of OT, perpendicular to it, y = -x. Substituting this value of y in the equations of BS and QP, the values of x are found to be the same ; hence OT intersects them both at the same point. Fig. 38. 9. The altitudes of a triangle meet in a point. Take the axes as in the figure, and let AB = c, C being given as (x', y') . Then the altitude through C is X = X' (1) r_ y The equation of BC is y — y'= —^ — (x — x') , and that of the altitude through A is Z/ = c — X' -X. y (2) The equation of AC is ?/ = ^a;, and of the altitude through B 13 y = -^(x-c). (3) y Combining (2) and (3) to find their intersection, we obtain x = x\ which satisfies (1). Hence (1), (2), and (3) meet in a point. APPLICATIONS. 67 10. The perpendiculars erected at the middle points of the sides of a triamjle meet in a point. „< The equation of AC (Fig. 30) is 7/ = -^ a;, and that of the perpendicular to AC through B' is y' x' / x'\ , . . The equation of BC is y — y' = —^ — (a; — x') , and that of X — c x' the perpendicular to BC through A' is y' c — x'f c + x'\ .r,\ 2 y The perpendicular to AB at C is X = -. (3) 2 ^ ^ Combining (1) and (2), eliminating y, we have .'k = — Hence (1) and (2) intersect on (3). 11. The medials of a, triangle meet in a point. The middle points A', B', C (Fig. 39), are ^c + x' y'\ fx' y'\ (c 2 '2;' w^r V2'^'' and the equations of the medials are y = -y^x, AA', (1) c + X _ y' y = -r^(x-c), BB\ (2) X — 2c y^ y\2x-c) ^ CC". (3) 2a;'- 1 Combining (2) and (3), we find they intersect in f — ^^—, - \ 3 3 and these values satisfy (1) ; hence (1), (2), and (3) meet in a point. 68 ANALYTIC GEOMETRY. 12. To find the general analytic condition that three straight lines may meet in a point. Let (1) y = m'x + b', (2) y = m"x + h'", (3) y = m"'x+h"' be the three lines. The intersection of (1) and (2) is jj"-h' m'b"-b'm" X = ■ , y ■ m'—m" m'—m" But these must satisfy (3) ; hence m' b" - m" b' + m"'b' - m'b'" + m" b'" - m'" b" = 0. 13. Shoiv that 11?/ — 20.r-8 = 0, y-4x = 0, 13y - 28.x- — 8 = 0, meet in a point. 14. To find an expression for the area of a triangle in terms of the coordinates of its vertices. Let {x\y'), {x",y"), (.r'", ?/'"), be the vertices. The equation of a line through the first two is y—y'—^ — ^(x — x'), or .^ — a;' (y"—y')x-i-(x'—x")y-\-y'x"—y"x'=0. Hence the perpen- dicular distance from this side to (x'", y'") is (y>'_ y<) x"'+ (x'-x") y"'+y'x"- y"x' ' ■\/(y'-y"y + {x'-x"y ~' But the denominator of this expression is the distance between (x', y') and {x", y"). Hence the area = Ibase X altitude = i [.t' (?/'"-//") +-'^""(?/'-2/"') +«"'(?/"-y)]- 15. Find the area of the triangle tvhose vertices are (2, 3), (-1,4), (6,5). 16. Fiiid the equation of a straight line passing through a given point and diriding the line joining tivo given points in a given ratio. Substitute in y — y'= •-, — ^, (x — x') for .r", y'\ the values of x—x' APPLICATIONS. 69 X and y in Equation (2), Art. 6, and for x', i/', the coordinates h, k, of the given point, and we have ^ m{x"-h) + n{x'-h)^ ' 17. The bisectors of the interior angles of a triangle meet in a point. Let the equations of the sides of the triangle be iKcosa' -\- y aiua' —j)' =0, (1) a; cosa" +// sina" — y)" = 0, (2) a:cosa"'+?/sina"'-iV"=0, (3) and let the origin be tvithin the triangle. Then the origin lies within each of the three angles to be bisected, and the equations of the bisectors (Art. 40) are xcosa' +i|/siua' — ^/ — (x cosn" -\- y s'ma" — j)") = 0, (4) cccosa" + y/sina" - p" - {x COSa'" + y nina'" - p'") =0 , (5) a;cosa"'+?/ sina'" — //"— (.^ cosa' -\- y sina — p') =0. (6) But values of x and y which satisfy any two of these equations also satisfy the third ; hence these three lines meet in a point. / 18. The lines tvhich pass through the vertices of a triangle and bisect the angles supplemental to those of the triangle meet in a point. F'or brevity, represent by a = 0, /3 = 0, y = 0, Equations (1), (2), and (3) of Ex. 17. Then a + /3 = 0, (3 + y^O, y + a = are the bisectors required. -^ 19. The bisectors of any two exterior angles of a triangle and of the third interior angle meet in a point. The bisector of the exterior angle of (1) and (2), Ex. 17, is a + ^ = 0, and of (2) and (3) is /? + y = 0, and the bisector of the interior angle of (1) and (3) is a — y = 0. Subtracting the second of these equations from the first, we have the third. 70 ANALYTIC GEOMETRY. 20. Tlie bisectors of the angles betioeen the bisectors of perpen- diculars are the perpendiculars themselves. Let y = mx + 6, y = x + b', be the perpendiculars. Their normal forms are y — mx — & _ n my + x — mb' _ ^ vr+w"- ' Vi + m' and their bisectors are y — mx — b± {my + x—mb') = 0. The normal forms of these latter are (1 +m)y + (l —m)x—(mb'+b) _^ V(l+m)2 + (l-m)2 (1 - m) ?/-(!+ m) X + {mb'— b) _ q ^{l+my + {l-my and their IMsectors are (1 + m) ?/ + (1 — vi) X — {mb'+h) ±[(1 - ra) z/ - (1 + m) x + {mb' - 6)] = 0, or ?/ — mx — & = 0, and my + x — mb'= 0, which are the given perpendiculars. 21. To find the condition that the three poiyits {x\ y'), {x", y"), {x"\ y'"), shall be coUinear, i.e., lie on the same straight line. 22. Prove that the line ivhich divides two sides of a triangle proportionally is parallel to the third side. ^ 9, > '■ CHAPTER III. EQUATION OF TEE SECOND DEaREE. THE CONIC SECTIONS. o-oXXoo SECTION VII. —COMMON EQUATIONS OF THE CONIC SECTIONS. 45. The Conic Sections. It has been shown in the previous chapter that every complete equation of \hQ first degree between X and 2/? Ax -f- Bij -f- 0== 0, and the various forms which such an equation may assume owing to a change in tlie values or signs of the arbitrary constants A^ B, C, is the equation of a straight line. In the present chapter it will be shown that every equation of the second degree between x and y, Ay^ -\- Bxy + Cx- -{- Dy -\- Ex + F — 0^ and the various forms it may assume when diilerent values and signs ai'e given to the arbitrary constants A, B, C, D, E, F, represents some one of a family of loci called the Conic Sections. These loci, which for brevity may be designated Conies, are so named because every section of the surface of a right cone with a circular base by a plane is one of this family. They may all be traced by a point so moving that the ratio of its distances from a fixed point and a fixed straight line re- mains constant, the particular locus traced depending upon the value of this constant. Since all the loci of this family may thus be generated b}' a pouit moving under a single law, it will evidently be possible to express this law in a single equation, and to derive the particular cases from this general equation 72 ANALYTIC GEOMETRY. by assigning the corresponding value to the ratio. The proof of the foregoing statements and the discussion of the general equation is, however, greatly facilitated by a knowledge of the forms and elementary properties of these loci ; we shall therefore first determine their equations separately from some of their properties with a view to the discovery of their forms, reserving the discussion of the general equation until the student has thus become familiar with the various loci which it represents. THE CIRCLE. 46. Defs. The path of a point so moving that its distance from a fixed point remains constant is a circle. The constant distance is the radius, the fixed point the centre. 47. General equation of the circle. Let (m, n) be the centre C\ R the radius, and P any point of the circle. From the right-angled triangle PCM, CP- = CM- + MP\ or {y-ny + {x-mY=R\ (1) which is the required equation. Hence, to tvrite the equation of any circle ichose X>osition and radius are knoiun, substitute the given values of m, n, and li, in the above equation. Thus, the equation of the circle whose radius is 6 and centre is (6, -2), is (y + 2)-+{x-6y=36, or /+a;- + 4.v-12.T+4 = 0. By assigning different values to m and n, we may derive the equation of a circle in any position from the general equation (1). Two of these derived equations are of frequent use and should be memorized. First: when the centre is at the origin, in which case m = 0, n = 0, ;>nd ( 1 ) becomes Fig. 40 y- -t- .r = R; (-0 Fiir. -11. CO:SDION EQUATIONS OF CONIC SECTIONS. 73 called the central equation of the ch-cle. Second : when the origin is at the left-hand extremity of any diameter assumed as the axis of X, in which case m = M, v(, = 0, and (1) becomes y^^-IRx-x". (3) Thus, the central equation of the circle whose radius is 6 is / + a" = 36, and when referred as in Fig. 42, y- = \2x-x^. The above forms may be obtained directly from the correspond- ing figures. The student will observe that, by transposition, either (2) or (3) shows that PM^ = AM.MA', a well-known property of the circle from which these equations might have been established. Fig. 42. 48. The equation of every circle is some form of the equation y^^^+By + Ex + F=0. (1) Expanding the general equation of the circle (y-ny-i-{x-my-=R^ (2) we have y- + xr — 2 ny — 2 mx + vr -{-n" — R- = 0, (3 ) which is of the same form as (1). The first two terms of (3) are independent of m, ?;, and R, so that no change in the position or magnitude of the circle can affect these terms. Every equation of a circle, therefore, tvill contain the squares of x and y loith equal coefficients and like signs. The remaining terms will vary with the radius and position of the circle. Thus E=0, when m = 0, and y~ -\-x^-\- Dy + F=0, is the equation of all circles whose centres are on Y\ i)= 0, when n = 0, and y^ -{-X? -\- Ex+F = () applies to all circles whose centres are on X; if both m and n are zero, than ^ = 0, D=0, and we have the central form y- + xr = R-, the centre being at the origin ; if the origin is on the curve, then the equation can have no 74 ANALYTIC GEOMETRY. absolute term, or i^ = 0, and (1) becomes y^ ^ x^ ^ By + Ex =0 ; if B, E, and F are all zero, we have y^ ^ a^ = 0, which is true only for.T^O, ?/=0, the origin, the circle becoming a point; this may be regarded as the limiting case of the central form as the radius diminishes indefinitely. ^ 49. Conversely, every equation of the form f-^ .^2+ Dy + Ex + F={), (1) which is not ivij^ossible, is the equation of a circle. Adding to both members of (1) the squares of half the co- efficients of a; and y, we obtain f + By + ^ + x^ + Ex + ^ = ^ ^^- F, 4 4 4 4 or (^y + ^'Y+(^a; + |J=i(i)^'+£'^-4F), (2) which is of the same form as {y-ny+{x-my = R\ (3) and in which, therefore, -f = 'N -f = *'i. \{D''+E--^F) = R\ (4) If D-+ E- >4:F, then li- is positive, E is real, and the equa- tion represents a circle whose centre is [ — ~, — — Y and whose , V 2 2 7 radius is J V/>^'-t- E'-iF. If D'+ E'=-iF, then i?- is zero, and the equation becomes | ?/ +— j -f- Li- _(- _y=o, which is F T) satisfied only for x = - , .?/=-^-, or the circle becomes a point, namely, the centre, which may be regarded as a circle whose radius is zero. If D-+E''<4 F, then E- is negative, E is imaginary, and the equation is impossible since the sum of two squares cannot be negative. We have thus three cases, in which A' is real, zero, or imaginary, and for brevity and COMMON EQUATIONS OF CONIC SECTIONS. ib uniformity of expression we may say that evenj equation of the form y--{-x"-\-Di/ +Ex+F=0 is the equation of a circle, real or imatjinary . The equation ay'-\- ax--\-dy +ex ■\-f=Vi may be reduced to the form of (1), and is therefore the most general form which the equation of a circle can assume. 50. To determine the centre and radius of a circle whose equation is given. When the equation is given in the form (^ — «)"+ (x — m)"=E-, the centre {m, n) and radius B may, of course, be determined by inspection. If given in the form y^-{- a;--f- Dy + Ex + F=0, we may put it under the above form by adding to both members the squares of half the coefficients of x and ?/, as in Art. 49. Otherwise, by equations (4), Art. 49, the coordinates of the centre are half the coefficients ofx and y loith their signs changed, and the radius R = .V ^^U- + £'- — 4 F. Thus , given y''-^x'-Ay+2x + l = 0, m = -l, ?i = 2, i? =Wl6+4-4 = 2. 51. The equations of concentric circles differ only in their absolute terms. Since the values of m( = — — j and ?if = — --j are inde- pendent of F, and i? = i V-D-+-E-— 4i^, if in the equation of any circle F changes, D and E remaining the same, the circle retains its position but changes its size. Hence circles are concentric ivhose equations differ only in their absolute terms. Examples. 1. Write the equation of the circle whose radius is 7 and centre at (0, 8). Ans. (7/-8)-+(.^•-0)^=49, ov y-+ x^-\i:,y +lb = Q. 2. Write the equations of the following circles: Centre at ( — 1, —4), radius 2; Centre at (0, 0), radius 9; Centre at (5, 0), radius 5; Centre at ( — o, 5), radius 5. 76 ANALYTIC GEOMETRY. 3. Write the equation of a circle whose centre is (6, 8), passing through the origin. 4. Which of the following equations are those of circles? x'+ 2f-+ 8.7; -4^ +7 = ; .«2+ y- + xy + x + // -1 = ; a-^- 2/-+ X- -22/ + 4 = 0; /+ ,;--4y- 8.« + 1=0 ; x'^^'if+x-Sy=(); 2.r+2/+4.r-3^+7 = 0; a-2 + 2?/-4.r-l = 0; f ?/-+ f x2-2a; =0. 5. Write the equations of a circle whose radius is 6 when (1) both axes are tangent to the circle; (2) when X' X is a tangent and y y passes through the centre (two eases). 6. Determine the position and radius of the following circles : ?/2+ar — 82/ + 4.T— 5 = 0. Ans. ( — 2, 4), 5. 2/2_|_.^^_|.107/-4a;-7 = 0. Ans. (2, -5), 6. y'+x--\-\Qy+Ax-'20 = (), Ans. (-2, -5), 7. ^2^.^^_2?/+G.^•=0. Ans. (-3, 1), VlO. y'i^x'+Zy--x-^^=0. Ans. (|, -|), 4. /+a;2+42/-2.T + r> = 0. Ayis. (1, -2), 0. 362/-+36rr-24?/-36.t;-i;31 = 0. Ans. (.^, i),2. ?/+ 3'-'+ .'/ + .« - 1 = 0. Ans. ( - 1-, -1) , I V(3. /+ a^+ 7/ + .i; + 1 = 0. ^ws. ( -i, -i) , i V ^. y-+x-- 2/ - .i- + 4 = 0. .4»s. ( 1. , I) , ^ V^a4. 7. Write the equation of the circle whose centre is at the origin, and which touches the line 3a;— 4_y +25 = 0. Putting the equation of the line under the normal form, 3a: , 4y . o which must equal /.'. Tloncc y-+a-2=25. 8. Write the cfiuation of the circle whose centre is (0, 0), and which touches the line 3.a- + //— = 0. 9. Wri1(> the ('((natiou of tlu' circle whose centre is (2, 3), and wiiicli tunchos 3.t; + 4 v + 12 = 0. COMMON EQUATIONS OF CONIC SECTIONS. 77 10. Prove that the sum of the equations of auy number of circles is the equation of a circle. 11. Prove that if the equation of a straight line be added to the equation of a circle, the sum is the equation of a circle. 52. Polar equation of the circle. /-v -- The general equation of the circle being {^y — nf-\-{x—mf=^R-, let the pole be taken at the origin and the polar axis coincident with X. Then, if ?•', B' (Fig. 43), are the polar coordinates of the centre C, and ?•, ^, those of any point P, from the formulaa for transformation, Eq. (4), Art. 23, we have .'C = rcos^, y=-r sin ^, m = r' cos 6', n = r' sin $', which being substituted in the above equation, there results, after reduction, r-2n-'cos(^-^')=^'-'' ..12 (1) Fig. 43. Fig. 44. From this equation we may derive that of the circle in any given position by assigning the proper values to r' and 0'. Thus, if the centre is at the pole, r'= 0, and (1) becomes r=E, (2) which is true for all values of 0. (See Ex. 1, Art. 19.) If the pole is on the curve, and the polar axis a diameter, ^' = 0, r'=B, and (1) becomes r=2Rcose, (3) which, being true for all positions of P (Fig. 44), shows that OPA\ or the angle inscribed in a semi-circle, is a right angle. (See Ex. 2, Art. 19.) 78 ANALYTIC GEOMETRY. Discussion of Equation (1). Solving the equation for r, we have r = r' coB(9-e') ± -y^Jf^-r'-Bin^ie-d'), which gives two values of r for every value of 9, locating two points P and Pj, so long as Ji->r'-sin-(9~e'), or iJ> r' 8in(e — S')- If Iioints is constant is called an ellipse. The two fixed points are called the foci, the point midway between them the centre, and the lines joining any point of the ellipse with the foci the focal radii. 54. Central equation of the ellipse. Let F, F' be the foci, the centre and origin, the axis of X being coincident with FF', P any point of the ellipse, FF'— 2c, and 2a the constant sum. Then FP + F'P=2a. But F FP = ^P3P+ ME^ = V(x + c)^+ y\ F'P= -^F^I\P+MP-^ ^(x-cy+y'. Hence Vix + c)'+ y'+V{x - cy+ ?/= 2a. Transposing the second term to the second member, and squaring, (X + c)2+ / = 4 a^- 4 a V {x-cf+f' + {x - cf+ y\ or ex — '/,- = — a V(.r — c^+y'^ COMMON EQUATIONS OF CONIC SECTIONS. 79 Squaring asrain, aV+ (a-- c-)x':= a\a-- c"). (1) Discussion of the Equation. Since only the squares of the variables enter the equation, the ellipse is symmetrical with respect to both axes. Making y = 0, the X-intercepts are ± a. Take OA==OA'=a, then ^^r=2a=the constant sum, and as the sum of two sides of a triangle is greater than the third side, PF+ PF'=2a>FF'= '2c, or A and A' are icithout the -X Fig. 45. foci. Making x = 0, the T-intercepts are ± Vo^— o", which are real, since a > c, and locate the points B, B'. Solving the equation for y, y = ± -^ {a'- c'){ct'- x"), a which is imaginary if x > a numerically, and therefore the curve lies wholly within the limits A and A' along X. Solving for x, X = *^^^f^ r r which is imaginary if 3 — 72 > 1, or y > Vfr— c- numerically, or the curve lies ivholly idthin the limits B and B' along T. The form of the ellipse is best observed by the following mechani- cal construction: Take a string whose length is AA'=2a, fix its extremities at F and F', place a pencil point against the 80 ANALYTIC GEOMETRY. string, keeping the string stretched ; as the pencil is moved it will trace the ellipse, for in all its positions FP-\- PF'= 2 a. 55. Defs. AA' is called the transverse axis of the ellipse, BB' the conjugate axis, A and A' the vertices, FA and FA' (or F'A and F'A') the focal distances, the double ordinate through either focus, as GG\ the parameter, and the distance from the / Ff)\ focus to the centre divided by the semi-transverse axis ( — i the eccentricity. As referred to an origin at its centre and axes of reference coincident with those of the ellipse, Eq. (1), Art. 54, is called the central equation of the ellipse. 56. Common form of the central equation. Representing the FO c eccentricity by e, we have e — = -, .-. c = ae, which substi- •^ -^ AO a tuted in Eq. (1), Art. 54, gives another form of the central equation, in terms of the eccen- tricity. Representing the conjugate axis BB' by 2&, 26 = 2Va^— C-, .-. a^—c'^=lr, which substituted in Eq. (1), Art. 54, gives ay+6-.r=fr&-, (2) the equation of the ellipse in terms of the semi-axes, and called the common form of the central equation. Cor. 1 . Since e = - and a > c, the eccentricity of the ellipse is a alioays less than unity. CoR. 2. Since c-= cfi—b', e = - = , the eccentricity . . ^ ., . a a in terms of the semi-axes. Cor. 3. Since e = -, c = oe, the distance of either focus from the centre. Note. The Btiident will observe Uiat Uie form of the ellipse will vary with a, b, c, Bnd e, and therefore that the constants in the equation of a locus may serve to determine Us form as well as its magnitude and position (Art. 16). COMMON EQUATIONS OF CONIC SECTIONS. 81 57. Length of the focal radii. F being any point of tlie ellipse (Fig. 45), FP2= F]\P^MP^= {ae + a;)-+ f (Art. 50, Cor. 3) ^{ae + x)-+{a--x'){l-e-) (Art. 5G, Eq. 1) = (r -\- 2 aex +e-a;" = (a + ex)- ; or FP = « + ex. But FP +FP =2 a, . • . F'P = 2 a - (a + ex) = a- ex. Hence, the focal radii to any point ivliose abscissa is x are a ± ex. 58. Polar equation of the ellipse. Let the pole be taken at the left-hand focus, and the polar axis coincident with the transverse axis. We shall obtain the equation directly from the figure, this being easier than to transform the central equation. From the triangle FPF', P being any point of the curve, F'P' = FP- + FF'- -2FP.FF' cos F'FP. But FP=r, F'FP= 6, FF' = 2ae, and F'P = 2a-FP=^2a-r. Making these substitutions, we obtain ail-e-) 1 — e cos 6 (1) Discussion of the Equation. "When ^=0°, r = a{l +e) = FA'; when $ = 180° , r = a{l — e) = FA ; hence the focal distances are 0(1 ±e). t~ '^'■' 82 ANALYTIC GEOMETRY. AVhen = 90% r = a ( 1 - e-) = a/l - '^\ = a^^^^ = - ; \ a-J a^ a hence the parameter GG' = 2a(l — e-), or ^^• a When » = i^'i^i3 = 008^^;^= COS" ^ — i r= r-i FB r ae- r .-. r = a=FB. This is also evident from the right-angled tri- angle FOB, in which FO = c, OB — b, and therefore FB = V c^TP = a, since a- — c^ = &-. Hence ^/ie distance from either focus to the extremity of the conjugate axis is equal to the semi-transverse axis. Therefore, to find the foci lohen the axes are given, with the extremity of the conjugate axis as a centre and the semi-trans- verse axis as a raditis describe an arc; it ivill cut the transverse axis in the foci. Representing the parameter GG' = 2a (I — e^) by 2^9, the polar equation (1) may be written }• = P 1 —e cos 6 (2) 59. The ratio. The ellipse can be traced by a point so mov- ing that the ratio of its distances from a fixed x>oint and a fixed straight line is constant. From the polar equation r = :, p., we have '■ 1 — e cos d r = p -\- er cos 6, e . or FP = J) -\-i FM (Fig. 46) . Take FS such that FS = -, or J) = eF/lS, and draw DD' perpendicular to FS. Then FP=e (FS + FM) = eSM= ePQ, FP PQ being parallel to MS. But e is a constant • hence — — is a constant. ^ The fixed line DD' is called the Directrix. COMMON EQUATIONS OF CONIC SECTIONS. 83 Cou. 1 . The ratio is equal to the eccentricity and is always less than unity. IF CoK. 2. Since ^ is a point of the curve, - — ^ = e, . cr -4^ a(\—e) ,. . ~o\ .'.AS= =-^ ^ (Art. o8). e e For tlie same reason - — = = VC) ; e = — ; 'Ip = 4. COJNOIOX EQUATIONS OF CONIC SECTIONS. 87 3. Find the axes, focal distances, and distance of the direc- trix from tlie centre of 6//-+ ^.r = 108. Ans. a=G; b = SV^ ; 3V2(V2 ± 1); gV^. 4. Write the equation of the ellipse whose axes are 18 and 10. Ans. 8\y-+26x' = 2025. 5. AVrite the equation of the ellipse whose eccentricity is | and transverse axis 10. Ans. 9 >/--{- ox' = 125. G. The conjugate axis of an elUpse is 4 and its lesser focal distance 1. Find its eccentricity and equation. Ans. f ; 257/'+ IGx- = 100. 7. Construct geometrically the eUipse in the following cases : (a) e=s. Observe the size is undetermined. (b) f = i ; distance from focus to directrix =10. (c) a = 5, h = 4. 8. The eccenti'icity of an ellipse being — -, what is the angle between the focal tangents ? ^^ Ans. 60°. 9. Find the focal distances, conjugate axis, parameter, and position of the directrix, of tlie ellipse r = -• ^ 16-9cos6' Ans. 4, 11 ; f Vt ? 2? 4^i from centre. 10. AYrite the polar equation of the ellipse whose axes are 18 -^^8- Ans. r= ''' 9 — V65 cos^ THE HYPERBOLA. 63. Defs. The path of a poi)d so moving that the difference of its distances from tico fixed points is constant is called an hyperbola. The two fixed points are called the foci, the point midway between them the centre, and the lines joining any point of the hyperbola with the foci the focal radii. 64. Central equation of the hyperbola. Let F, F', be the foci, the centre and origin, the axis of X 88 ANALYTIC GEOMETKY. being coincident with FF\ P any point of the hyperbola, FF' = 2 c, and 2 a the constant difference. Then FP-F'P= 2 a. But FP = Vi^J/- + MP' = V (x- + c) - + f, F'P = ■VF'3P+ MP- = V {X -cy+ y\ Hence V(a; 4-c)^+r— V(.i- — c)-+ ?/- = 2a. Transposing the second term to tlie second member, and squarmg, ^^^ ^ ^_^,,^ ^^^, ^ ^ ^^,^ ^ ^^ ^______,^ (.^_e)2+ ^.^ or ex — a- — a V (if — c) ^+ y^- Squaring again, ay 4- (a2 - C-) .r-' = (r{a' -c'). ' ( 1 ) Discussion of the equation. Since only the squares of the variables enter the equation, the hyperbola is symmetrical with respect to both axes. Making y = 0, the X-intercepts are ± o. Take OA=OA'=a, then yL4'=2cf, the constant diflerence. and as the difference between two sides of a triangle is less than the third side, PF - PF' = 2a< FF' = 2r, or A and A' ar e betzceen the foci. Making x = 0, the F-intercepts are ±Va'— o-, and are imaginary, as a < c ; hence the curve does not cross the axis of Y. Solving the equation for »/, 1 wliich, since cC-—(? is negative, is imaginary for all values of a: < a numerically, and the curve lies icholly tdthoiit the limits A and A' along X, extending to ± cc. Solving for x, ±a^ii — r_ \ a^- c- which is real for all values of y since a-— c- is negative, or the curve has no limits in the direction of Y. The form of the hyper- bola is best observed by the following mechanical construction : take a ruler of any lengtli Fl. (ixinu- one extremity at F, and a string F'PI, shorter than the ruler by .Ll'= 2a, one of whose COMMON EQUATIONS OF CONIC SECTIONS. 89 ends is attached to the farther extremity / of the ruler, the otlicr at the focus F'. Press the string against tlic ruler l)^' a pencil, as at P, keeping the string stretched, the ruler turning Fig. 48. about its fixed end F. As the pencil moves it will trace the hyperbola, for in all its positions FPI = F' PI -\- 2 a ; or, sub- tracting PI from each number, FP = F'P + 2 a, . • . FP - F'P = 2 a. 65. Defs. AA' is called the transverse axis of the hyper- bola, A and A' the vertices, FA and Fxi' (or FA' and F'A) the focal distances, the double ordinate through either focus, as (?(?', the parameter, and the distance from the focus to the /F0\ centre divided by the semi-transverse axis f |, the eccentric- •^ \AOj ity. Equation (1), Art. 64, is called the central equation of the hyperbola. 66. Common form of the central equation. The hyperbola does not cross the axis of Y, and does not therefore determine by its intercepts a conjugate axis, as in the case of the ellipse. Its equation, however, will assume a form similar to that of the ellipse if we represent the viirnerical value of V«" — c' by 6, laying off OB = OB' = b (Fig. 4.S) for a conjugate axis. We thus 90 ANALYTIC GEOIVIETRY. have «-— c- = — b'-, minus because a<:,c, aud e = -, .-. c = ae, e being the eccentricity. Substituting c = ae in Eq. (1), Art. 64, we have or, since o > a, and therefore e = - > 1, a the central equation in terms of the eccentricit\'. Substituting a^—c- — — b-, in the same equation, we obtain cr y- —h-!xr = — tr 6", ( 2 ) the common form of the central equation of the liyperbola. Note. The student will observe that Equations (1) and (2) diflter from the corre- sponding equations of the ellipse (Art. 56) only in the value of e and the sign ofb'-; also, that while .r=0, in Eq. (2), gives 2/= ^ V— li'-' ^f imaginary quantity (as it should be, since the curve does not cross Y), its numerical value is the semi-conjugate axis, as in the case of the ellipse. Cou. 1. Since ^= , aud e>o, tlie eccentricity of the luiper- a bola is cdicays greater than unity. Cor. 2. Since a^— c- = — ir, e = ~ = — — , the eccentricity in terms of the semi-axes. Cor. 3. Since e = -, c = ae= ■\/a:-'+b- = AB (Fig. 48), or a the distance from the focus to the centre is the distance from either vertex to the extremity of the conjugate axis. Hence, to find the foci ichen the axes are given, v:ith as a centre and AB as a radius, describe an arc; it iviJl cut the transverse axis in the foci. 67. Length of the focal radii. P l)eing any point of the hyperbola (Fig. 48), FP'' = FM- + MP- = {ae + x)-+y- (Art. G6, Cor. 3) = (ae + a;)2+(ar^-fr)(e'- (Art. GG, Eq. (1)) =e'^x'^+2aex + a^ = {ex + af ; or FP = ex -\- a . COMMON EQUATIONS OF CONIC SECTIONS. 91 But F'P = Fr-'2 a = ex-\-a—2a = ex - <(. Hence the focal radii to any point ivhose abscissa is x are ex ± a. 68. Polar equation of the hyperbola. Let the pole be taken at the left-hand focus, and the polar axis coincident with the transverse axis. From tlie triangle FPF', P being an}- point of the curve, F'P^ =FP^+ FF''—2FP . FF' cos F'FP. But FP=r, F'FP = 9, FF'=^2ae, and F'P=FP-2a = r-2a. Making these substitutions, we obtain a(e'—l) r = e cos ^ — 1 (1) Discussion of the equation'. When ^=0°, r=a{e-{-l)=FA'; when 6 = 180°, r = —a{e — l) = FA ; or the focal distances are a {e ± 1), numerically. 1 = b' a ; or the u Wlien ^ = 90°, r = - a (e--\) = - a parameter, GG', is 2a (e' — l), or ^^-, numerically. As 6 increases from 0°, cos (9 diminishes and r increases, tracing the branch ^1' P, r becoming infinity when e cos ^ = 1 , Fig, 49. 92 ANALYTIC GEOMETRY. or ^ = cos"^-- When ecos^^l, r is negative, and the branch e G'A is traced, in the direction G'A, r being FA when ^ = 180°. "When passes 180°, cos^ is negative and /• remains negative, tracing the branch AG, and becomes infinity again when ecos^ = l, or ^=cos^^- in the fourth angle; after which, e ecosO is greater than unity, /• is positive and ti'aces the branch LA'. Representing the parameter GG' — 2a{e- — l) by 2p-, the polar equation (1) may be written P e cos — 1 (2) 69. The ratio. The hyperbola can be traced by a xioint so moving that the ratio of its distances from a fixed point and a fixed straight line is constant. From tlie polar equation of the hyperbola, r= -^ , we ecos^ — 1 / have r = ercose-p, or (Fig. 49) FP=eF3I-p. Take FS P such that FS = , or p — eFS, and draw DD' perpendicular to FS. Then FP=e {FM- FS) = e SM= e PQ, PQ being par- FP . allel to MS. But e is a constant ; hence — - is a constant. Pil The fixed line DD' is called the directrix. Cou. 1. The ratio is equal to the eccentricity, and is always greater than unity. Cor. 2. Since A is a point of the curve, ^^=e, .•.^5 = -i^= "^'^-^^ (Art. 68V AS e e For the same reason ^^e, .-. A'S = — = 'ii^-tD. Hence A' S e e the distances f roni the vertices to the directrix are —^ — ~ — ^« COMMON EQUATIONS OF CONIC SECTIONS. 93 Cor. 3. FS = FA +AS =a (e - 1) + ^ ^^ ~ ^ ^ = ^ C^'-^) ^ e e the distance from the focus to the directrix. Cor. 4. OS = 0F- FS = ae - "'^^'~^^ =-= the distance e (' from the centre to the directrix. / ,-) y > ^ 70. Geometrical construction of the hyperbola when the ratio is given. Let e = -, in which A; > s, be the given ratio. Take FS = s, s draw GG' perpendicular to FS, and make FG = FG' = k. Draw SG and SG', and between these lines produced draw an}- parallel to GG', as N'L'. With i^ as a centre and M'N' as a radius, describe an arc, cutting the parallel in P' and P" ; these are points of the hyperbola. To prove that P is a point of the hyperbola, through S draw DD' perpendicular to SF, and P'Q' perpendicular to DD' . Then, from similar triangles, N'M':M'S:'.GF'.FS\ but N'M' = FP', M'S = P'Q' ; hence FP'-.P'Q':: GF:FS, or FP' GF P'Q' FS Since iViJij > MiS b}' construction, the arc described with FPi=M\Ni, as a radius will determine Pj, P^, on the right of DD', which may be proved to be points of the hyperbola as above ; and in the same manner any number of points may be constructed. It is evident from the construction that SG and SG' can have but one point each in common with the curve ; for this reason they are called the focal tangents. Since GF>FS, their in- cluded angle, is greater than 90°, the distance FS, taken to represent ^. simply determines the scale of the construction ; but if e varies, the angle G'SG will vary, and the hyperbola will differ in shape as well as size. 94 ANALYTIC GEOMETRY. Cor. 1. Since ^1 is a point of the curve, — — = e. But ^—=e,..AF=AK. Similarly, A'K'=A'F. Hence, to AS find the focal tangents ivhen the axes are given, first determine Fig. 50. the focus (Art. 66, Cor. 3), thou make AK=AF and A'K'= A'F. KK' will be the focal tangent, and its intersec- tion with the axis, S^ a point of the directrix. COMMON EQUATIONS OF OONIC SECTIONS. 95 Cor. 2. Since the curve is symmetrical with respect to its axes, aud CF' = CF, there is another directrix ou the right of the centre at the same distance from it as DD'. 71. The equilateral, and the conjugate hyperbola. When the axes of an hyperbola are equal, it is said to he equi- lateral. ]M:iking a = h m the common form of tlie central equa- tion a-y- — b"x- = — a~b~, we have y--x- = -a\ (1) for the equation of the equilateral hyperbola. Tivo hyperbolas are said to be conjugate to each other when the transverse and conjugate axes of the one are the conjugate and transverse axes of the other. If, in deducing the equation of the h3perbola, the transverse axis had been assumed coinci- dent with y, tlie equation of the hyperbola in this position would have been a-x^ — b^y^ = — a-b'', as this supposition simply amounts to interchanging x and y. Interchanging now a and ?>, this becomes ^^,^, _ ^,^ ^ ^^,^, . ^g) or, the central equations of conjugate hyperbolas differ only in the sign of the absolute term. Conjugate hyperbolas are distinguished as the X- and the T- hyperbola, each taking its name from the coordinate axis on which its transverse axis lies, and the equation of either may be derived from that of the other by changing the signs ofa^ and W. Cor. 1 . The eccentricity of the ^-hyperbola is — — Cor. 2. Since the distance of the foci of an hyperbola from the centre is the distance between the extremities of the axes (Art. 66, Cor. 3), the foxir foci of a pair of conjugate hyperbolas are equidistarit from the ceyitre. 72. Varieties of the hyperbola. Every equation of the form J/+ar + F=0 (1) 96 ANALYTIC GEOMETRY. is the central equation of an hyperbola, if A and C have unlike signs and neither is zero. First. Let F be positive. Then Ai/ — Cxr = —F, which can be reduced to the form a-y- — Irx- = — d-b'-, as in the case F of the ellipse (Art. 62), by introducing the factor B = -— ; p rp pi2 IP I Jp whence ~y- x^ = -, in which a = ^ — -, and b =\l — —- C A AC \ C y A numerically. If A — C, the axes are equal and the hyperbola is equilateral. Second. If i<'is negative, (1) becomes Ay^ — Cx^' = F, which is the conjugate hyperbola, since it differs from Ay- — Cxr = — F only in the sign of the absolute term (Art 71). fn Third. If F = {), (1) becomes At/ — Cxr = 0, or y=± \-7X, which is the equation of two straight lines through the origin making supplementary angles with X. In this case the axes are zero. There are then four varieties of the hyperbola, in which the axes are unequal, equal, interchanged, and zero; corresponding to the X-, equilateral, Y-hyperbola, and a pair of intersecting straight lines through the origin. Examples. 1. What are the axes of the hyperbola 9y--4.x- = -Ui? Multiplying by the factor R = -— = 4, the equation becomes AC 36 3/2 - 16 x^ = — 576, which is of the form a^ y^ — b- x- = — a- b- ; tlie axes are therefore 12 and 8. Or, directly, making y = and x = in succession, wo have numerically .r = a = 6, ?/ = 6 = 4, .-. 2 a = 12, 2 6 =r 8. 2. Find the axes, eccentricity, and parameter of 3?/2_2.r2 = -18. Ans. 6, 2V6; iVT5; 4. COMMON EQUATIONS OF CONIC SECTIONS. 97 3. Find the axes, focal distances, and position of the direc- trix of ?/-a-2 = —81. y Ans. a = b=0 ; 0( V2 ± 1) ; ~r: from centre. V2 4. Write the equation of the hyperbola whose axes are 18 and 10. Ans. 81 ?/2- 25 a- = -2025. 5. AVrite the equation of the hyperbola whose eccentricity is |- and transverse axis 10. Ans. D y- — 1 xr = — 175. G. The conjugate axis of an hyperbola is 4 and its lesser focal distance 1. Find its eccentricity and write its equation. Ans. |; 9?/--16.'>r = -36. 7. Construct the following hyperbolas : (a) e = f . Observe the size is undetermined. (^b) e = f ; distance from focus to directrix = 8. (o) a =8, b = 6. 8. Construct a pair of conjugate hyperbolas whose axes are 12 and 8. 9. Write the equations of the hyperbolas conjugate to those of Exs. 2 and 3, and determine their eccentricities and directrices. Ans. < 3 r- 2 X- = 18 ; ^;^- ; 2 J- from centr 9 i/ — af=81 ; V2 ; —7= from centre. V2 e. 10. The eccentricity of an hyperbola being V3, what is the angle between the focal tangents ? Ans. 120°. 11. Find the focal distances, conjugate axis, parameter and 6 directrix of the hyperbola r = Vl5 cos^-3 Ans. —^ ; 2 Ve ; 4 ; 3^ - from centre. Vl5q:3 ^5 12. Write the polar equation of the hyperbola whose axes are 8 and 6. ^9 Ans. r = -. 5 cos 0—4: 98 ANALYTIC GEOMETRY. THE PARABOLA. 73. Defs. The path of a point so moving that its distance from a fixed p)oiHt is always equal to its distance from a fixed straight line is called a parabola. The fixed poiut is the focus, the fixed straight Hue the directrix, and the line joining any point of the parabola with the focus, the focal radius. 74. Equation of the parabola. Let F be the focus, DD' the directrix. Draw iSi^ perpendicu- lar to DD' and let /SF = 2i- By definitiou, the middle poiut of SF is a point of the curve. Let be the origin and the axis of X coincident with OF. Then, P being any point of the curve, and PQ perpendicular to DD', PF= PQ, or P <-'^' Squaring and reducing, + r = x- + y- = 2X)X. (1) Discussion of the equation. Solving for ?/, we have y = ± V2pa;, or y has two numerically equal and increasing values for positive increasing values of x, but is imaginary when X is negative ; hence the curve lies wholly to the right of F, extends to infinity in the first and fourth angles, and is symmetrical with respect to X. The form of the parabola may be observed from the following mechanical construc- tion : take a ruler of any length QI, and a string, FPI equal in length to the rnler. Fix one end of the string at the focus, the other at the extremity / of the ruler, and, keeping the string pressed against the rnler at P by a pencil, slide the ruler along the directrix parallel to SF; the pencil will trace the curve, for in all its positions PQ = PF. Fig. 51. COISOrON EQUATIONS OF CONIC SECTIONS. 99 OX is called the axis of tlie parabola, the vertex, and the double ordinate GG' through the focus the parameter. CoK. 1. Substituting .r= OF = ^ in Eq. (1), we have y=^FG =p, or the jyarameter GG' = 2p = the coefficient of x in the equation of the curve. Also OS = 0F= -}j2^ = \GG' ; or SF=FG=FG'=2y. Cor. 2. FP= QP= SO + 0M= ^p + x; or the length of the focal radius to any point where abscissa is x is x + ^p. 75. Polar equation of the parabola. Let the pole be taken at the focus, and the polar axis coinci- dent with the axis of the parabola. The formuke for transfor- mation from rectangular axes at to the polar system, are p X = .!•„ + r cos 6 — -r + r cos 6., y = ?/,, -f- r sin 6= r sin 6. Substituting these values in the equation y^ = 2px, we have, r^sm-6=2p(^^-{-rcos6\ or ?*^(1 — cos- 6) = })- + 2pr cos 0. Transposing, r- = ?•- cos'^ -j- 22??' cos^ -f p^ _ ^j. cos6 +py. Extracting the root of each member, ?• = - 2' ovr = —^—^' (1) 1 — cos 6 vers 6 Let the student discuss the equation. Observe that the equation 1 — ecos0 is the general polar equation of the ellipse, circle, hyperbola, and parabola, when the pole is at the focus ; taking the forms 100 ANALYTIC GEOMETRY. r = P 1 — e cos 6 for the ellipse (Art. 58), that is, when el; and r = - ^ parabola, when e = 1 . Q 76. Geometrical construction of the parabola, the focus and directrix, or the i^arameter, being given. Lay off SF=2^ = J the parameter, or the given distance between the focus and the directrix. Draw GG' perpendicular to SF, and make FG = FG' = SF. Draw SG and SG\ and any chord N'L' perpendicular to SF. With i^ as a centre and J/'A^' as a radius describe an arc cutting the chord in P' and P". These are points of the parabola. To prove that P' is a point of the parabola, join P with P, and draw P'Q' parallel and DD' perpendicular to SF. Then N'M' ^GF^ P'F 31' S FS P'Q'' since the triangles GFS and N'3f'S are similar, and P'F= N'M', P'Q' = M'S, In the same way any number of points may D A t / /. w p' G / \ F w \ G X X \^\ p" D' \ ^^~-~ r \ Fig. 52. by construction be found. As in the case of the ellipse and the hyperbola. SN' and SL' have evidently but one point each in common with the curve, and are called the focal tangents; and as SF= FG =FG', the focal tangents of the para])ola make an angle of 90° with each other. — = -^-^ is called the ratio, and, evidently, the ratio FS P'Q' of all parabolas is unity. COMMON EQUATIONS OF CONIC SECTIONS. 101 The distance SF, taken to represent p, determines the scale to which the parabohi is constructed. Had a distance twice that of tlie figure been taken, the construction of the same parab- ola to the new scale would have been equivalent to the con- struction of a parabola whose parameter was 2 (2p) to the original scale. Hence, para^o^o.s, like circles, differ only in size. Examples. 1. Construct the parabola whose parameter is 10, and write its equation. 2. Construct the parabola the distance of whose vertex from its focus is 2. 3. Write the polar equations of the parabolas of Exs. 1 and 2. 4 4. The polar equation of a i^arabola is r = Write .^ , \ \. ^ 1-cos^ its rectangular equation. Ans. y- = 8x. / 0/ 102 ANALYTFC GEOMETRY. SECTION VIII. — GENERAL EQUATIONS OF THE CONIC SECTIONS. 77. Defs. A conic is the locus of a jwint so moving that the ratio of its distances from a fixed point and a fixed straight line is constant. This constant is called the ratio, the fixed point the focus, the fixed line the directrix, and the [)eipendicular to the directrix through the focus the axis of the conic. 78. General equation of the conies. Let P be any point of the conic, (»i, n) the focus jP, DD' the directrix, its equation being x'lcosa + ?/i sin a — p = 0, the sub- scripts being used to distinguish the coordinates of the directrix from those of the conic. Then FS, perpendicular to Z>Z>', is the axis. Join F with P, draw PQ perpendicu- lar to DI)\ and let e = the constant PF ratio. Then = e, or PQ PF- = e-Pq\ But PF= VJy - ny + {x - m.y (Art. 7) ; and PQ = X cos a + 2/ sin a —^) (Art. 38) ; hence (y — ny + {x — my = e-{xcosa + ysma—2)y (1) is the required equation, in which e determines the species, and m, ?i, a, and p, the position of the conic. Examples. 1. Write the equation of an ellipse whose centre is (1, 2), tran«;verse axis is 6, eccentricity axis parallel to X. V5 and transverse GENERAL EQUATIONS OF CONIC SECTIONS. 108 o — 180°, e. — — - ; .-. cos a — — 1, sin a -= 0, p = " _ — 1, m = \— Vo, n — 2. Substituting in Eq. (1), or i* .'/■- + i ■'■- — :5< ) .'/ — 8 .«• + 4 = 0. 2. Write the equation of a, parabola whose axis is parallel to X, vertex is at ( — 3, —2), aud parameter is 9. Ans. / + 4?/ -9a; -23-0. 3. Write the equation of an elli[)se whose eccentricity is — -•> V3 centre is (1,1), transverse axis 2V3, the latter being inclined at an angle 135° with X. m = 1 -, w = 1 -] -, j) = 3, e = — -, a = 135°, V2 V2 V3 and the equation is 5 f + 2 .17/ + 5 :r2 - 12 // - 12 ,f = 0. 4. Write the equation of a circle whose radius is o, the axes being tangent to the circle. m = 71 = 5; If- + X- - 10 // - 10 .r +25 = 0. 5. The centre of an ellipse is ( — |, 4), its eccentricity f, and its transverse axis = J^^, and is parallel to X ; write its equation. Ans. 9 ?/2 + 5 a;- - 72 ?/ -f 1 2 a; + 144 = 0. 79. Every complete equation of the second degree between x and y, and all its forms, is the equation of a conic; and, con- versel}', the equation of every conic is some form of the equation of the second degree. Expanding the general equation of the conies, Art. 78, we have (1 — e-sin-a)?/^ — 2e-sina cosa xy + (1 — ^-cos^a) a? ] -f (2e-psina — 2?;)?/+(2e^pcosa— 2m).T+7jr+n^— e-jr=0. J The complete equation of the second degree between a? and y, Af' + B.ry + av- + Dy + Ex-\-F=0, (2) 104 ANALYTIC GEOMETRY. is of the same form, but the coefficieats of corresponding terms are not necessarily the same, since any equation may be multi- plied or divided by any factor without affecting the qualit}^ Making these coefficients, therefore, equal, by dividing each equation by its absolute term, and designating the resulting coefficients of (2) by ^1', B\ etc.. we have e- sm-a rfi- + n^ — e^p^ — 2e- sin a cos a 9 O = A\ = B\ C. = D\ (3) m-+ ir 1 O 9 1— e- cos- a vr-\- ir— e-jr 'le-p sin a — 2 » 9,9 9 9 m--\- n- — e-p' 2 e-p cos a — 2 m _ , ,, 9,9 9 9 »r+ n- — e'p- From these five equations the values of the five constants A', B', C, etc., may always be determined when a, m, n, p, and e are given ; and as the latter are arbitrary, such values may be assigned to them, that is, the locus ma}' be assumed of such species and in such position, as to give A\ B\ C", etc., any and every possible set of values. Conversely, the values of a, ?)i, n, p, and e, can always be found from the above equa- tions when those of A\ B\ C, etc., are given ; that is, a conic of some species and position corresponds to any and every set of values which may be assigned to A', B', C", etc. Hence, every equation of a conic is some one of the forms assumed by the general equation of the second degree^ and every form of sxich equation is the equation of some conic. The axes were assumed rectangular, have been (Art. 7) IlatI tlicy been obliqnc, the distance FP would y/[y-ny+(,x — my + 2{y-n)(x-m) cos /S, and the distance PQ would have been (Art. 38) X COB a + y co»p' - p, in which p if the given inclination of the axes, .Tnd |3' the angle made by PQ with Y. The equation P F- = t'-PQ- Wduld, therefore, have involveil the Kainc arl)itrary constants, GENEEAL EQUATIONS OF CONIC SECTIONS. 105 and no others. Passing now to rectangular axes, since this trausforraation involves no new arbitrary constants, and cannot affect the degree of the equation, therefore the above reasoning is entirely general. 80. To determine the species of a conic from its equation. Forming 5'-— 4J.'C" from Eq. (3), Art. 79, we have _ 4 e^ sin^g cos^g — 4(1 — e- slir'a) {I — e- cos^g) m--\-n-— e-iry _ 4e^ sin^g cos-g — 4+46^ cos^a + 4e^ siii'a — 4e^ siu^g cos^g nr-\- n-— e-p-y ^ 4(e--l) (m^ + n-— e-p'^'Y Now the locus will be an ellipse, a parabola, or an hyperbola, according as e is less than, equal to, or greater than, unity. But, since the denominator of the above fraction is a square, and the sign of the fraction is thus that of its numerator, when e< 1 the first member is negative, when e = 1 it is zero, and when e > 1 it is positive. Hence the conic will be an ellipse, parabola, or hyperbola, according as B'~ — 4^'C" is negative, zero, or positive. To apply this test it is unnecessary to reduce the given equa- tion to the form'lfl^' A'y-+B'xy+C'x'^D'ii +E'x + 1 = 0; for if B''—4:A'C' be negative, zero, or positive, then will {KB')-~ 4 {KA') {KG') = K-{B"~ 4 A'C) • also be negative, zero, or positive. Hence, iv7iatever' the co- efficients, Ay- + Bxy + Cxr -\- Dy + Ex + F = is the equation of an ellipse, parabola, or hyperbola, according as B-—AAC is negative, zero, or positive. Examples. Determine the species of the following conies : (1) y-— bxy + 6ic^ — 14ic + by-\-A = 0, an hyperbola; (2) ?/-- 8 xy + 2b a? + 6y - 2.^- -f 49 = 0, an ellipse; (3) 3 ^2^ ioxy-\-Ax-—%y=Q, an ellipse ; ^/j: '9^, 106 ANALYTIC GEOMETRY. (4:) ?/^4- 2 a;?/ + x^— 7 + 1 = 0, a parabola ; (5) y-— 1 + 3 .T = (.V — ?/)-, CO) hyperbola; (6) ?/-= 4 (a; — 1 ) , a j^arabola; (7) 4:a;/y —16 = 0, a?t hyi^erbola. 81. T/ie equation Ay' + Cx^+Dy +E^F= represents all species of the conic sections. The general equation of the conies is Ay-+Bxy + Cx'+Dy +Ex+F= 0. Passing to any rectangular axes with the sarae origin by the formulae (Art. 22, Eq. (8)), X — .i'l cos y — v/i sin y, y = x^ sin y + 2/i cos y, we have, after omitting the subscripts, A {x^ sin^y + 2xy siny cosy + ?/ cos^y) + jB (a^ cos y sin y + xy cos' y — xy sin- y — ?/- sin y cos y) + C (x? cos^y — 2 xy cosy sin y + y- sin-y) + other terms not involving xy. The term containing oyy is [ 2 A sin y cos y + -B (cos^ y — sin- y ) — 2 C sin y cos y] xy, or ' [(yl— C) sin2y -f iJcos2y]a;y, which will be zero if {A-C) sin2y4-Bcos2y = 0, or if tan2y = -^I-^; (1) Now tan2y can have any and every value from 4-^ to — x, hence a value can always be found for y which will satisfy (1) whatever the values of A, 7?, and C\ that is, whatever the species of the conic. To (iud this value of y, we have (Art. 79) GENERAL EQUATIONS OF CONIC SECTIONS. lUT B tail 2 Y = = -. — 7^ = F 2e- sinu eosa o , 7, — = tan 2 a, 1 — e- siu-a — (1 — e- cos-u) and since tau2a = tan (180°+ 2a), (1) will be satisfied when- ever 2y=2a or 180"+ 2a; that is, wheny = a or 90°+ a, or whenever the axis of the conic is parallel to either axis of refer- ence. Hence every equation of the form Ay- + Cx-^ By +Ex + F = () is the equation of a conic ivhose axis is parallel to one of the axes of reference, and, since, B-~AAC = — 4:AC when B=0, the conic will be an ellipse, hyperbola, or j^rabola, according as A and C have like signs, unlike signs, or either is zero (Art. 80) . Thus, whatever the signs or values of D, E, and F, Ay-+Cx-+Dy-\-Ex-\-F=0 (2) represents an ellipse whose axes are parallel to the axes of reference ; ^^o_ Q^2_^j)y _^^j. +^^ q (3) represents an hyperbola whose axes are parallel to the axes of reference ; ^^o _^ j)y _^ ^^ + F={), or Cx^ -\-Dy + Ex + F=0. represents a parabola whose axis is parallel to X, or Y, respectively. Cor. rN^'hen referred to the new axes the coefficients of the square are A (cos^y + siu'-y) =A, C (cos-y + sin-y) = O, or the coefficients of x' cindr~tuaWe/ to PQ, a' remains the same, but h' differs. Eliminating then b' by substituting its value from the first in the second of the above equations, we obtain y = c a'A x=^. b- a'a? X, (3) which is a relation between the coordinates of the middle points of all chords parallel to PQ ; it is therefore the equation of a line through these middle points. Being of the first degree it is a straight line, and having no absolute term it passes through the origin, which is the centre. Hence, the locus of the middle points of parallel chords to the ellipse, or hyperbola, is a straight line throngh the centre. CoK. If A = C, or the locus is a circle, (3) becomes 1 2/= --^^ a' which is perpendicular to y = a'x + b'. Second. The p)arabola. Let 2/" = 2jXT be the parabola, and y = a'x Jf- b' any chord PQ. Combining as before, 2a'b'-2p _ ly^ a'' a" or, x"^ + qx = r, whence, in the same manner the coordinates of the middle point M are x = 9 2' u'q , , or, replacing q by its value, a'b' — » p "" a''' a' from which we see that the abscissa x of the middle point varies with b\ but that the ordinate y is constant if a' is constant ; that is, if the chords vlxq ptarallel. Hence, the loctis of the middle points of parallel chords to the parabola is a straight line parcdlel to X. Fig. 56. 114 ANALYTIC GEOMETRY. The studeut will observe that if a diameter be defiued as a chord through the centre, the diameters of the parabola are necessarily parallel as the centre is infiniteh- distant. The extremities of any diameter are called its vertices. 88. The tangents at the vertices of a diameter are parallel to the chords bisected by that diameter. Since the diameter TT (Figs. 54, 55, 56) bisects all chords parallel to PQ, as J/ approaches T (or T'), P and Q approach each other, and J/P, J/Q, remaining equal, must vanish together. Hence, when 31 coincides with T (or T'), PQ will have but one point in common with the curve, or is a tangent. 89. Def. One diameter is said to be conjugate to another hen it is parallel to the tangents at the vertices of the latter. ic 90. Conjugate diameters of the ellipse. Let KK' (Fig. 54) be drawn parallel to the tangent at T, that is, parallel to PQ. Its equation will he y = a'x. The equation of TT' is y= a"x=-— x (Art. 87, Eq. 3). 7 2 « "" Ilcnce o'a" = ^ is the relation which must exist between the a- slopes of a diameter and the chords which it bisects. But this relation is satisfied for KK' and the chords PQ', etc., parallel to TT. Hence, if one diameter is conjugate to another, the latter is conjugate to the former, and the tangents at the vertices of conjugate diameters form a ^xirallelogram. a'a"=-^l (1) a^ is called the equation of condition for conjugate diameters to the ellipse. Since the rectangle of their slopes is negative, the tangents of the angles which they make with X have opposite signs ; hence, if one diameter makes an acute angle with the transverse axis, the other will make an obtuse angle, or conju- gate diameters to the ellipse lie on opposite sides of the conjugate axis. GENERAL EQUATIONS OF CONIC SECTIONS. 115 Cor. If a = b, (1) becomes a'= , or conjugate diam- a" eters to the circle are at right angles to each other. 91. Every straight line through the centre of an hyperhola^ except the diagonals of the parallelogram on the axes, meets the hyperbola or the conjugate hyperbola. Let y= a'x (1) be any straight line through the centre, ay - Vx" = - a'b^ (2) the equation of the X-hyperbola, and (Art. 71) ay _ 6V = a-b' (3) that of the I"-hyperbola. Combining (1) in succession with (2) and (3), we have a-b' o _ a-b~ b' - d'a /-V,'2 (4) XT — a'a" - b^ (5) Now if a' <,-■, X is real in (4) and imaginary in (5), and the Fig. 57. 116 ANALYTIC GEOMETllY. line intersects the X-hyperbola, as TT'. If a' > —, x is imag> (Jj inary iu (4) and real in (5), and the line intersects the F-hyper- bola, as KK'. If a' = ± , both values of x are infinity. In this case (1) becomes y—± -.r, the equations of CS and CS', a the diagonals of the rectangle on the axes, neither of which meet either hyperbola within a finite distance. 92. Defs. The diagonals of the rectangle on the axes of a pair of conjugate hyperbolas are called the asymptotes. Their equations being y =±-x, if a — h their included angle is 90°, a and the hyperbola is said to be rectangular ; or, n^hen an hyper- bola is rectangular it is also equilateral (Art. 71). 93. Conjugate diameters of the hyperbola. Of ttvo conjugate diameters, one meets the X-, the other the Y-hyperbola. Let TT' be any diameter bisecting a system of parallel Ki),'. .-18. GENERAL EQUATIONS OF CONIC SECTIONS. 117 chords of which PQ is oue. Draw KK' parallel to PQ^ that is, to the tangents at the vertices of TT' ; it is then conjugate to TT'. Being parallel to PQ, its equation is ?/ = a'.t', and that of TT' is y = a"x = -^x (Eq. 3, Art. 87). Hence a'a" = — is the relation which must exist between the slopes of a- , a diameter and the chords which it bisects. If a' < , a" must 7 7 2 « evidently be > -, since their product = — , and conversely ; or, h " ^'' '^''" since - is the slope of the asymptote, if one diameter' intersects a the X-liyperhola, its conjugate will intersect the Y-hyperhola, and conversely. Again ; since the equation of the "1^- hyperbola is derived from that of the X-hyperbola by changing the signs of a^ and If (Art. 71), a'(7" = — is also the relation which must exist cr between the slopes of any diameter of the I'-hyperbola and the chords which it bisects. But this relation is satisfied for KK' and the chords P'Q', etc., parallel to TT' \ hence TT' is parallel to the tangents at K and K' , or is conjugate to KK' ; hence, if one diameter is conjugate to another, the latter is conju- gate to the former, and, as in the case of the ellipse, the tangents at the vertices of conjugate cjiaineters form a jKirallelogram. The equation a'a" — — a- is called the equation of condition for conjugate diameter's to the hyperbola. Since a'a" is positive, the angles which two conju- gate diameters to an hyperbola make with the transverse axis are both acute, or both obtuse, or the diameters lie on the same side of the conjugate axis. CONSTRUCTION OF CONICS FROM THEIR EQUATIONS. 94. First jMethod. By comparison ivith the general equa- tion. Make the coefficients of like terms in the given and the 118 ANALYTIC GEOMETRY. general equation equal by dividing each equation by the co- efficient of the same term. Equating the resulting coefficients of corresponding terms, we have five equations from which a, m, n, e, and p, may be determined. This method is tedious and of little practical value except as e = 1, or some of the co- efficients are zero. Example. 1. ?/2 + 4?/ + 4.i; + 4 = 0. Since B' — 4 AC = 0, the conic is a parabola, and therefore e= 1. The coefficient of y- being unity, divide the general equation (Eq. 1, Art. 79) by the coefficient of i/, 1— e^sin^a, and we have, after making e=l, V \ P -^ Q i ^ F j ^ s D' Fig. 59. — 2 sin a cos a 1 — sin-u = 0, (1) 1 — COS" a 1 — sin- a = 0, (2) 2j:»sintt — 2n — 4 ('^) 1 — sin- a ^1 2/? cos a— 2 m 1 «r-sin-a = 4, (4) m- + 11- — p- 1 ■ ■> = 4. (^) 1 sm-a From (1), -2sinacosa=0; .-.a must be 0°, 90°, 180°, or 270°. From (2), cosa = ± 1 ; hence a cannot be 90° or 270°, and is either 0° or 18U°. In either case (3) gives n = — 2. Substituting cosa = ±l in (4), we have ± 2]) — 2m = 4, or 111 = ±p — 2, according as a is 0° or 180°. From (;')), since 7i = — 2, m'^ = p'^\ or, substituting the above values of w, j) = ±l. Bnt\2^ is always positive; taking, therefore, the upper sign, a = 0°. Finally, from (4). making cosa=l and p=l, we have m = —\. Tlie values of the constants are thus : r' = 1 , a = 0°, m = — 1 , ?« = — 2, j) = 1 . To construct these results, lay off OQ =p=\ to the right, since a = 0°, and draw the directrix Z)7)' perpendicular to X. Construct F^ ( — 1, —2), and through F draw FS perpendicular to DD'. GENEKAL E(^)UAT10N8 OF CONIC SECTIONS. 119 Having thus the focus aud cHrectrix, the parabola may be cou- sti'ucted as in Art. TG. 95. Second Method. Bii transformation of axes. If B- — 4:AC is not zero, the conic is an ellipse or hyperbola, imd . ^ 2AE -BD ^ ^ 2 CD -BE are the coordinates of the centre (Art. 6-i). Transferring to parallel axes with (a^o, y^) as a new origin, we have the equation of the ellipse, or hyperbola, referred to its centre. If the term Bxy is not present in the primitive equation, the result of this transformation is the central equation of the ellipse, or hyperbola. If, however, this term is present, we must transfer again to new axes with the same origin, the angle between the new and primitive T> axes of X beiuo- determined by the condition tau2Y = ^ ^ A-C (Art. 81). If B' — ■XAC= 0, the conic is a parabola. Trans- fer first to new axes with the same origin, the new axes of X being subject to the condition tan2y = — ; then to parallel axes M'hose origin is (Art. 8o) _ I^-±AF _-D '"'- 4AE ' '^"-yz' the resulting equation will be the equation of the parabola referred to its vertex and axis. Examples. 1 . 5 if- -\-2xy -{- ox- — 12 y — V2x=0. B'-4AC=-96, hence the conic is an ellipse and has a centre. The coordinates of the centre are ^^ 2AE-BD ^^ ^ 2CD-BE ^ ' B'-4:AC ' "^^ B--4AC and the fornmhTe of transformation are X = .(•„ + X, = 1 + a'l, y = y, + y, = 1 + ?/,. 120 ANALYTIC GEOMETRY. -4. Substituting these in the given equation, omitting subscripts, we have . , , ^ , c -^ . ^ r> o/ 4- 2 x'^ + 5 .1" — 12 = 0. To obtain the central equation we must have -B tan 2 y = • = — GO y = - 4.7 A-C and the formulae of transformation are X = .Ci cos 7 - ?/i sin y = V^ {x^ + y^ , y = .Ti sin y + 2/1 cos y = Vi (2/1 - x^ ) . Substituting these values in 5?/- + 2a;?/4-5x'- — 12 = 0, and omitting subscripts, we obtain 3?/^ -f 2x-^= 6. The axes are therefore 2 V3 and 2 V2, and the eccentricit}^ 1 V3* To construct the ellipse, construct ( 1 , 1 ) , the new origin Oi, and draw OjXi, Oi ^'i, the parallel axes. Draw OiX, making the angle Xi Oi X2 = — 45°, and OxY., perpendicular to it. On O1X2 lay off 0i.l=0i.4'= V3, and on OxY^i OiB= 0^0= V2. AA' and OjB are the axes of the ellipse ; the focus may be found as in Art. 58, and the ellipse constructed as in Arts. 54 and 60. The curve may be traced with approximate accuracy by determining the intercepts on the axes. Thus, from 5y^ + 2a;i/ 4-5ar — 12y — 12.i; = 0, x — Q gives ?/ = and ^- ( and ^) ; .v = gives x = and ^ {0 and 9) • In the same way from 5 y' + 2xy +oxr — 12 = 0, 0,S=OiT= V/, OiU= OiF= V-U. Tln-ough the points thus found trace the curve. Fig. 60. 2. y^-2xy + x^ + 8x-lC> = conic is a parabola. tan2y = jr--iAC=0, hence the B A-C .— X, y = 45°, and the GENERAL EQUATIONS OF CONIC SECTIONS. 121 ^0^ 2A V2. formulae of Iransformation are x = V^ {x — ?/) , y = Vi {x + y) , tiud the transformed equation 2y- — 4 V2 2/ + 4 V2 .i- — 16 =0. From the latter, .r,, = ^^ — = , Transferring to parallel axes with the orio;in ( , V2 ), VV2 / we find ?/- = — 2 V2 x. To construct the parabola, draw the axes Y^OX^^ making XOXi = 45°. On these axes construct the vertex , V2 ], or Oi, Fig. 61. V2 and draw the parallel axes Xo Oi Y^. We may now con- struct the parabola whose parameter is 2 V2 as in Arts. 74 or 76, or determine the inter- cepts and trace the curve approximatively. OQ = — 4 + 4 V2, OQ'=-4-4V2, OR = OR' = i, 0*S=2V2, Or = V2 + ViO, 0T'= V2- VIO. 3. y~ — x;- + y — X + 2 = 0. B- — 4AC = 4, hence the conic is an hyperbola, its axes being parallel to the axes of reference, since B = 0. •2 ' yo = — i- Passing to parallel axes whose ori- gin is (-1, — I), we have y~ — a;- = — 2, an equilateral hyperbola whose axes are 2 a/2, eccentricit}' is V2, and cutting the primitive axis of X at 1 (Q) ^^^ Y' O' Fig. 62. 122 ANALYTIC GEOMETRY. 4. ?/2 -f 2 V.'5 .rji - X- - 64 = (). B' — 4:AC=\^, hence the conic is an hyperbola referred to its centre ( Art. 83 ) . tan2y=-V3, .•.2y=-(;0^ or y = -30°. Transferring to new axes such that -^ XOX, = - 30°, the equation becomes i-/2_x2=32, the equilateral F-hyperbola whose axes are 8 V2, cutting the primitive axis of I" at ±8 (Qand Q'). 5. y" - Axy + fx^ + 2?/ - 2a- + 3 = 0. 6. |ar + 4.T?/ + 3r-3 = 0. 96. Third Method. By conjugate diameters. The general equation of a conic being Ay--\- Bxy + Cx''+ Dy -^Ex + F^O, solving for y, we have Bx + D Cx-+Ex + F y "1 -, — .'/ — : » A whence Bx+D y=- 1 ■lA ±^^-^{B'-4.AC)x;'+2{BD--2AE)x+D'-4.AF. First. Construct the line QR whose equation is y— Bx+I) 2A Every value of x locating a point M on this line locates two points P' and P" of the locus, equally distant from Qli and on opposite sides of it, this distance being the radical in the value of ?/. Hence QR bisects a system of chords parallel to Y and is a diameter. GENERAL EQUATIONS OF CONIC SECTIONS. 123 Second. Values of x which reuder the radical zero give the same values for y for both the locus and the diameter ; hence the values of x found from the equation (B-- 4: AC) x--\- 2 (BD - 2 AE) x + D'-AAF= determine the points where the conic cuts QR ; that is, the Fig. 64. vertices T, T', of the diameter. This equation being a quadratic, there will be two such points except when B'- — 4:AC= 0, in which case the conic is a parabola and there will be but one vertex. If the conic is an ellipse it lies wholly between Tand T" ; if an hyperbola, w4iolly without these points. In either case the half sum of the above values of x determines the centre C, and the corresponding values of // locate K and K', the vertices of the conjugate diameter. Having thus the circumscribing parallelogram (Arts. 90, 93), a few other points may be con- structed, especially the intercepts on the axes, and the curve sketched with sufficient accuracy. Examples. 1 . 4 ?/- + 4 x>/ -{-5xr — 8y — 28 x + 24 B--^AC=-64, and the conic is an ellipse. Solving for y we find = 0. y = ^ (2 - x) ± V- X-+ 6x-o. 124 ANALYTIC GEOMETUY. Construct the diameter y = ^(2-x), QB. Placing — 0^ + 6a; — 5 = 0, we find x= 5 and 1, whence 7/ =i (2 — if) = — f and ^. or (5,— f) and (1, i) are the vertices T and T. The abscissa Fig. 65. of C is ^ (5 -f- 1) = 3, whence, from the equation of the conic 2/ = 4 and — f, locating K and 7i'. The circumscribing paral- lelogram may now be drawn. Making ?/ = we find the X- ' — Intermediate points uui}^ be found intercepts = o if necessary ; thus, for x=A, y = —l ± V3, locating P' and P". Trace the curve through the points tlius found, tangent to the circumscribing parallelogram at K, K', T and T'. 2. 2/2 + 2a;2/ + ar' + 2.v-7.c-8 = 0. GENERAL EQUATIONS OF CONIC SECTIONS. 125 B' — 4AC= 0, and the conic is a parabola. Solving for y, tj = - (.f+l)± V9a'+9. Y Construct the diameter QR, y = -{x + \). The radical gives but one value of x = — l, for which y = 0, lo- cating tlie vertex T. The X- intercepts are 8, — 1, and the F-intercepts 2, —4. Interme- diate points may also be found ; thus for x = S, y — 2 and — 10 (P' and P"). Trace the curve through these points and tangent at T to a parallel to Y 3. y^ + 2xy-2x:^-4:y-x+l0=0. B--4.AC= 12, .-. the conic is an hj'perbola. Fisr. 66. y = - {x - 2) ± Vdx--3x-G. The diameter is y = — (x — 2), its vertices are (2, 0) and ( — 1, 3). The X-intercepts are 2, —4, the F-intercepts being imaginary. 4. y^ + 2xy-^3x^-4:x = 0. 5. y^ — 2 xy -\- X- — y -\- 2 X — 1 = 0. 6. y--2xy-\-x''-\-x=0. Fig. 67 97. When the equation of the conic does not contain the term involving xy, the axes of the conic are parallel to the axes of reference, and its position may be determined by the prin- ciples of Art. 17. If the squares of both variables are present, it is an ellipse or an hyperbola according as their signs are like or unlike ; if these coefficients are equal in magnitude and sign, it is a circle ; if numerically equal and of opposite 126 ANALYTIC GEOMETRY. signs, an equilateral hyperbola. Solving the equation for either variable, as y, values of x which render the radical part of y zero give the extremities of the axis parallel to X, and the algebraic difference of these values is the length of this axis ; the half sum of these values of x is the abscissa of the centre, and the corresponding values of y determine the vertices of the axis parallel to Y^ then* algebraic difference being its length. If the term containing x is lacking, the centre is on Y; if the term containing y is absent, the centre is on X. If the equation involves the square of but one variable, the conic is a parabola whose axis is parallel to the other axis of reference, and coincides with it when the the first power of the variable whose square enters the equation is lacking. The vertex is found by solving the equation for the variable which enters as a square and placing the radical part equal to zero ; this equation determines the limit, i.e., the vertex. Examples. 1. 9y/_|-4.T- - .36?/ - 8.^• + 4 = 0. A and C have like signs, .*. the conic is an ellipse. Solving for ?/, 2/ = 2 ± ^ V— 4.X- + 8X- + 32. The limits along X are found from — 4a^ + 8a; + 32 = to be 4 and — 2, and the axis parallel 4 — 2 to X is therefore 6. The abscissa of the centre is ■ = 1, 2 ' and the corresponding values of y are 4, 0, or the axis parallel to Y'\s 4. Hence the locus is an ellipse whose centre is (1, 2) and axes 6 and 4, its transverse axis being parallel to X. 2. ?/- + 4 jy — 6.^ — 14 = 0. The locus is a pai*abola, its axis being parallel to X. Solving for y, y = — 2 i VGa; -j- 18 ; hence its vertex is (—3, — 2). 3. 4/ + .i^+ 16?/-4.^•+lG = 0. 4. 92/--4.r2-36y + 24a;- 36 = 0. GENERAL THEOREMS. 98. Tlirough any jive points in a plane one conic may be made to jjuss. GENERAL EQUATIONS OF CONIC SECTIONS. 127 Let (.fi, y,)^ (%' ?/-'), (-^3, ys), (■>',. /a), (a-55 .Vs), be the five given points. Dividing the general equation of a conic by the coefficient of any of its terms, and distinguishing the new coefficients by accents, we have A'f--{-B'xy+C'x''-\-D'y + E'x + 1 = 0. (1) Substituting in succession the coordinates of the given points, since the conic is to pass through them, we have A'yf + B'x.y, + C'x'^ + D'y, + E'x, + 1=0,^ A '2/2- + B'x.y. -f C'av + D'y, + E'x^ +1=0, Ahii + B%y, + C'x^ + D'y, + E'x, + 1=0, } (2) A'y,' + B%y, + C%' + D'y, + E'x, + 1 = 0, A'y-J' + B'.^5?/5 + C 'av + 1>'^, + E'x, + 1 = 0, in which A', B', C", Z>', and E', are the only unknown quanti- ties, and from which their values may be determined by elimi- nation. Since these equations are of the first degree, each of these quantities has but one value. Substituting in (1) the values of A', B', etc., found from (2), the resulting equation will be that of the conic passing through the five given points. If one of the points is the origin, one of the equations (2) would be 1 = 0, which is impossible. In such a case divide the general equation by any coefficient except the absolute term. This results from the fact that the equation sought can have no absolute term. Examples. 1 . Find the equation of the conic passing througli (4,4), (4,-4), (9,6), (9,-6), (0,0). Since the conic is to pass through the origin, F= 0. Dividing the general equation by A, we liave f- + B'xi/ + C'x'-2 + D'y + E'x = 0. Substituting the coordinates of the remaining points, 16+16i?'+ 16C" + 4Z'' + 4£' = 0. (1) 16-l6B'+l6C'-iD' + 'iE' = 0. (2) 36 + biB'+8\C'+6D' + 9E' = 0. (3) 36-54jB' + 81 C"-0Z>'+9£' = 0. (4) 128 ANALYTIC GEOMETRY. From (1) and (2), and (3) and (4), by addition, 32+ 32C'+ 8^' = 0. (5) 72 + 162 C" +ISE' = 0. (6) Eliminating E' between these we find C" = 0, which in (5) gives E' = — i. Substituting these values in (1) and (3), we have 54 B' + 6 Z>' = 0, whence B' = 0, D' = 0. The required equation is therefore y~-—4:X. 2. Fiud the equatiou of the conic passing through (—1, 2), (-hi)^ (-1,1), (-1, i), (-4,8). Ans. i/ + x^ + 2xy -\-3x — y -{-iz= 0. 3. Find the equation of the conic passing through (5, 0), (0, 5), (-5, 0), (0, -5), (0, 0). In this case F = 0, since one of the points is the origin. Divide the general equation by B, otherwise the term Bxi/ will disappear for every substitution, and B will be undetermined. Ans. The Axes. 4. Througli how many points may the conic be made to pass? 5. Find the equation of the circle circumscribed about the triangle whose vertices are (;5, 1), (2, 3), (1, 2). Ans. 3?/- + 3a;' -11?/ -13a; + 20 = 0. 6. Find the equation of a circle through the origin and mak- ing intercepts a and h on the axes. 99. Two conies can intersect in hut four points. The coordinates of the points of intersection of two conies will be found by combiniug their equations (Art. 36). But we know from Algebra that the elimination of one unknown quan- tity from two quadratic equations gives rise, in general, to an equation of the fouith degree. This equation will have four roots ; there will therefore be four sets of coordinates, all four of which may be real, two real and two imaginary (since imag- GENERAL EQUATIONS OF CONIC SECTIONS. 129 inary roots enter in pairs), or all four imaginary, and in any case, since equal roots occur in pairs, these four sets may reduce to two. When two sets of values reduce to one, that is, are equal, two of the points of intersection become coincident and the conies are said to touch each other at that 'point. Hence two conies can touch each other at but two points. The several cases are illustrated in the figure. 1 and 2 inter- sect in four points, all four sets of values of x and y being real ; Fig. 68. 1 and 3 intersect each other in two points and touch each other in one, two sets of values being equal ; 1 and 4 touch each other in two points ; 1 and 5 have no points in common, the roots being all imaginary ; while 1 and 6 intersect in two points, two sets of values being real and two imaginary. If the conies are circles, the simplest way of combining their equations is by subtraction. Thus, let (Art. 48) f^x' + D!/ + Ex-hF=0, (1) jr- + a-- + D'u + E'x + F' = 0, (2) be the given circles. Subtracting, 130 ANALYTIC GEOMETRY. {D-D')y + {E-E')x+{F-r) = ^, (3) from which we may find the value of either variable in terms of the otlier, and substituting it in either (1) or (2) find that ol the other. Cor. 1. Since (3) is of the first degree, two circles can inter- sect each other in but two points, and hence can touch each other but in one. Cor. 2. Representing the first members of (1) and (2) by S and S\ then S + kS' = is a locus passing through all the points of intersection of (1) and (2) (Art. 37) . When A- = — 1, that is, when the equations are subtracted, the resulting equa- tion, (3), is of the first degree ; hence (3) is the equation of the chord common to the two circles. Examples. Find the intersections of the following loci : 1. / + .^■2 = 25, y' = ^ix. Ans. (3, ±4). 2. y-=\Qx-x\ y' = -2x. Ans. (0,0), (8, ±4). 3. y- + Ax''=2b, 4/ - 25.^-' = - 64. Ans. (2, ±3), (-2, ±3). 4. 2/2 + :^ - 3?/ + 2.x- - 7 = 0, y- + x" - 3y +2x -f- 1 = 0. Ans. Concentric. 5. y2^^i_^y_2x + l = 0, y- + x--\-3y-4.x + 3 = 0. Ans. (1, 0) ; (|, -i). 6. y^-Sy + 8x + lO = 0, y--\-4x + 6 = 0. Ans. (-1, -2); (-|, -1). 7. Sy^ + 2a^-6y + 8x-10 = 0, 3y- + 23^+ 6x-4: = 0. 8. 3?/-+ 2x2- 6?/+ 8a; -10 = 0, 3^/2+ 2ar^- 6?/+ 8a; + 1 = 0. 9. Prove that if two circles intersect, the common chord is perpendicular to the line joining their centres. Let »/2+r2=/J2, and (f + Bxij + Cx-2 + B 'y + E'x + F' = Q, ( 1 ) A}/ + Bxy + Ox' + />"// + E"x + F" = 0, (2) be the equations of the conies, the coefficients of the first three terras l)eino; the same. We have seen that tan2v = . ^ A-C (Art. 81), in which y= the angle made by the axis of the conic with X. Since y depends only upon A, jB, and C, the conies are similaily placed. The axes of the conies being then parallel to each other, we \ may transform to a system of reference whose axes are parallel to those of both curves ; and this transformation does not change the coefficients A and C (Art. 81, Cor. 1), and is pos- sible only when B is the same in both equations. Transforming now each equation separately to an origin at the centre of the corresponding conic and parallel axes, since this transformation does not change the values of A and C (Art. 84, Cor. 3), the equations become Ay- + Cx- + i^i = 0, Ay' + Cx- + i^o = 0, and the squares of the semi-axes are respectively -t^l 1,(2 -^1 --^1 .(19 J^ •> -Lll-y Fr, a'- = t,f2 h'- = --K and a'" = A & b"' = A ,^2 Now e^ = 1 ± "-- = 1 ± fj —-, or the eccentricity is the same for «'- a"'' each conic ; hence they are also similar. / /-e^j' ^ * fi f-^^' / ■^■■^•^ t-w i- ''^ .^.^.J, 3.^1^ £0^^ /}, / c^ c. 132 ANALYTIC GEOMETRY. Cor. 1. All parabolas are similar, since e = l for every parabola. Cor. 2. All circles are similar aud similarly placed. Cor. 3. If two couics are similar and similarly placed, they can intersect each other in but two points and touch each other in but one. For, subtracting the equations (1) and (2), we have {D'-D")y^{E' -E")x+F' -F" = Q, which is a straight line passing through all the points of intersection of (1) and (2). Combining this equation with (1) or (2), there results two sets of coordinates, which may become equal. The above equation is the equation of the common chord, or tangent. Cou. 4. If two conies differ only in their absolute terms, they are concentric. 102. To find the condition that an equation of the second degree may represent two straight lines. We have seen that two intersecting straight lines is a par- ticular case of the hyperbola (Art. 72), and that two parallel straight lines is a particular case of the parabola (Art. 85). It is further evident that if we multiply, member by member, two equations of the form ay + hx -)- c = 0, there will result an equation of the second degree, and that this latter will represent the two straight lines represented by the factors, for it will be satisfied by the values of the coordinates which make either of its factors zero. Conversely', if an equation of the second degree can he resolved into two factors of the first degree, it will rejvesent both the straight lines represented by these factors. Sometimes these factors can be discovered by simple inspection. Thus, xy = can be resolved into the factors a; = 0, y = 0, and, since these are the equations of the axes, xy = is the equation of both axes. Again, x- — y- = can be resolved iuto.r + v = 0, x — y = 0, which are the bisectors of the angles between the axes, and therefore a*^ — y- = is the equation of both bisectors. As these factors are not always readily seen on inspection of the equation, it becomes desirable to determine the general condi- tion for the existence of two factors of the first degree. GENERAL EQUATIONS OF CONIC SECTIONS. 133 Let Af + B.vy + Cx" + Dij + Ex + i^ = be the general equation of a conic. Solving it for 3/, In order that tliis equation may be capable of reduction to the form // = ax ± ft, the quantit}' under the radical must be a perfect square. But the condition that the radical should be a perfect square is {BT- - 4rAC) (I>- - 4 AF) = {BD-2 AE)\ Expanding and reducing, 4.ACF + BDE - AE- - CD'- -FB' = 0, which is the required condition. If the coetiicieuts of the given equation satisfy this relation, the equation represents two straight lines ; to find the lines, solve the equation for y and extract the root indicated by the radical. Examples. Determine which of the following equations represent pairs of straight lines, and find the lines : 1. Ay'-oxi/-hx' + 2y + x-2=0. Applying the test, we find - 32 - 10 - 4 - 4 + 50 = 0. To find the lines, solving for y, we obtain 1/ = ^il^ ^ Vox^- 3(j.r + 36 = ^"^ ~ ^ ± - (.Sx - 6). •^8 8 8 8^^ Hence the lines are ^ = .r — 1 and y — I x + h 2. 'dy--8xy + '3x- 1=0. .3. y- - 2y - .r + 1 = 0. Ans. y — x - 1 =0, y -j- x - I =0. 4. xy — ay — bx -\-(ib = 0. Ans. x=a, y=h. 5. y- + -ixy + -ix- — \ = Q. Ans. y -\-2x=±2. 6. xy + x- — y + ^x-lO = 0. Ans. ic — 1 = 0, ^ + a; + 10 = 0. 134 AN^UiYTIC GEOMETKY. SECTION IX.— TANGENTS AND NORMALS. 103. Defs. Let M3I' be any locus and AB any secant cut- ting the locus in the points P' and P". If AB be turned about P' regarded as fixed, till P". moving in the locus, coincides with P', AB will then have but one point in common with the locus and is called the tangent at that point. The direction of the tangent is that in which the generating X point is moving as it passes through the point of tangency, or the slope of the locus at any point is the slope of its tan- Fig. 69. gent at that point. The per- pendicular to the tangent at the point of tangency, lying iu the plane of the curve, P'A^, is the normal. 104. General equations of the secant and tangent to a conic. Y ^P^ /^ A^ 0/ / \ \ / \ \^ X^/' Let .'/ - ?J = .'/ —y •^ x' — X J,{X-X') (1) be the equation of a straight line passing through any two given points (ic', ?/'), (x",y"). The coefficient of x when the cqua- ?y' - y" y' - y" tion is solved for ?/ being '—. — '—r.^ we have "— ; — '—r, = a = the * x' — x" X — x" slope of the line. Also, let /(.x,?/) = (2) be the equation of any conic. If the two points through which the given line passes are on the conic, we must have TANGENTS AND NORMALS. 135 /(x',2/')=0, aud f{x",y")=0. y' -y" Hence, if we form —, ^, from these two equations aud sub- stitute the vahie thus found in ( 1 ) , we shall introduce the condition that (1) is a secant of (2). ?/' — ?/" Representing this value of ' , '_,, by a,, the equation of the secant will be .'/-?/' = », (;f — a-'). If, now, in the value of a,, we make x" = x' and y" = y', that is, suppose the point {x",y") to coincide with (x',y'), the secant will become a tangent ; hence, representing what a^ becomes under this supposition by (',, the equation of the tangent will be y -y' = a, (x - a;') , in which (x', y') is the point of tangency. Examples. 1. Equation of the tangent to the circle y- + X- = R-. Let (.(', ij'), (j", I/") be any two points of the circle. Then y'- + x>^ = R^, and ij"'^ + x"^ = R^. Subtracting, we liave y'--^"-+-c'--x"-!=0, or (//'-//") (. at= — —, and the normal is ^ — ;/'= ■— (r— .f ') ; or, clearing of fractions, x'y — y'x = 0. Since this equation has no absolute term, the normal passes through the origin, which is the centre ; hence the normal to a circle is the radiux to the point of tangency. 2. Find the normal to the ellipse ahf- + b'x^ = a^b^. b^x' 3. Find the normal to the hyperbola a-y^ — b^a? = — a?b-. Ans. y - y' = -^,(x- x') . Ans. y-y' = -%^,(^-^-" b^x •1 • Find the normal to the parabola y^ = 2px. v' Ans. y- y' = — ^- {x — x'). 5. Find the normal to the circle y^ = 2 Ex — .r A71S. y-y' = ^^^{x-x'). TANGENTS AND NOUMALS. 143 6. Write the equation of :i uormal in tlie following cases : (a) to a circle whose radius is o at the point (o, — 4). (b) to an ellipse whose axes are 6 and 4 at the point x' = 1. (c) to a parabola whose parameter is 9 at the point x' = 4. (d) to an hyperbola whose axes are 6 and 4 at the point .t''=8. '3?/ + 4.b-=0; 3?/ = ± 9 V2a; ip 5 V2 ; 3 // = q: 4 .« ± 34 ; 48^/ = q: 9 Vo5 iK ± 104 V55. Ans. 108. Defs. That portion of the axis of X intercepted between the ordinate from the point of tangency and the tan- gent is called the subtangent. In like manner that portion of the axis of X intercepted between the ordinate and the normal is called the subnormal. Thus (Fig. 70) , T3f and MN are the subtangent and subnormal to the point P'. 4'' 109. To Jind the subtangent and, subnormal at any point of a conic. Let Xi= OT represent the intercept of the tangent on the axis of X ; that is, the value of x when y is made zero in the equation of the tangent. Then, x' being the abscissa OM of P', the point of tan- gency, TM= subtangent = OM -0T= x' - x,. Similarly, lfiV= subnormal = ON— OM=x^—x', Xn being the X-intercept of the nor- mal, or the value of x when y is made zero in the equation of the normal. Examples. 1. To find the subtangent of the ellipse and the hyperbola. The equations of the tangents are ar yij' ± h-.rx' = ± d-b'^. When y — 0, X — Tt= — for both curves. Hence, also, for both curves, .r' '■■-a.2 ^ subt. = x' — .Tj — : .' 144 ANALYTIC GEOMETRY. Cor. Since a;, = — , , x, : a : : a : x' , or OT : OA' : : OA' : 03L x' Hence, the semi-transverse axis is a mean proportional between the intercejits of the tangent and the ordinate of the point of tan- gency. (Figs. 71 and 72.) This principle alTorcls a method of constructing a tangent at any point. First. The ellipse. Let P be tlie point. Describe the circle AP'"A' on the transverse axis, and produce the ordinate Fig. 71. through P' to meet the circle at P'". At P'" draw the tangent to the circle, P"T. Then PT is the required tangent. For, from the right similar triangles, OP'" 31, OP"'T, OT: OP"'{ = OA') : : OP'": 031; or Since both OT=x,= % and 31T = suht. = ^—^^ , areinde- x' x' pendent of b, tangents to all ellipses having the same transverse axis, at points having the same abscissa x' = 031, will evidentlv pass through T. Second. The hyperbola. Let P' be the point. Draw the ordinate P'J/, and on AA', 03r, as diameters, describe circles intersecting at Q. Draw QT perpendicular to X. Then TP' TANGENTS AND NORMALS. 145 is the required tangent. For, joiuiug Q with and M, from the similar triangles 0Q3f, OQT, we have OM:OQ( = OA') ::0Q: OT, or x^: a : : a : X Fig. 72. /^ Cor. Since 0T= ^, Twill be zero only when x' = cc ; that x' is, when the tangent coincides with the asymptote (Art. 91). 2. To find the subtangent of the parabola. Ans. Xf = — x'; subt. = 2a;'. It appears from this result B^ that the subtangent of the pa- rabola is bisected at the vertex. Hence to construct a tangent at a given point P', draw the ordinate P'J/and make 0T= OM. Then TP' is the required tan- gent. Fig. 73. 146 ANALYTIC GEOMETKY. 3. To find the subnormal of the parabola. The equation of the normal being y — ^' = — — (x — r'), we have, when ^ = 0, X — Xn = p + x'. Hence subn. = t,, — x' = p, or the subnormal of the parabola is constant and equal to one-half the parameter. Therefore, to construct a tangent at a given pointy as P', draw the ordinate P'M and make MN=p. Then P'M is the normal, and P'2\ perpendicular to it, is the tangent. 4. The tangent to the ])arabola bisects the angle between the focal radius and the produced diameter through the point of contact. Let P' be the point of contact, F the focus, and P'D the diameter. Then FP' = x' + | (Art. 74, Cor. 2). Also TF=TO+ 0F= x' + | (Ex. 2). Hence Fr=FP', the triangle TFP' is isosce\es,iind DP' T=P'TF=FP'T. Therefore, to draw a tangent at any point, as P', draw the focal ifadius P'F, the diameter P'D, and bisect their included angle. CoR. 1. FN^ OM- OF+MN= x'-^ +p (Ex. 3) ; .-. FN= x' +f = FT=^FP', or the circle described from the focus with a radius equal to the focal radius of any point j^asses through the intersections of the normal and tangent to that iioint with the axis. The triangle FP'N is thus isosceles, and FNP' = FP'N. CoR. 2. P'FN= FP'T+ FTP' = 2 FTP'. Hence, to draw a tangent parallel to a given line, as AB, from F draw FP' mak- ing an angle with the axis equal twice that made by the given line. P' will be the required point of contact and P'T, parallel to AB, the required tangent. Cor. 3. To draio a tangent through a given point without the curve. Let A" be the given point. Join /r with the focus, and with A" as a centre and KF as a radius describe a circle cutting the directrix in D and D'. Draw the diameters throuiih D and D' ; their intersections with the curve, P' and P", are TANGENTS AND NORMALS. 147 the points of tangencj'. To prove that KP' is a tangent, we have P'D = P'F by definition of the parabola ; also KF = KD by construction. Hence KP' bisects the angle FP'D. Simi- larly, P"K may also be shown to be a tangent at P". V. 5. The tangent and normal at any point of the ellipse bisect the angles formed by the focal radii drawn to the point of contact. Since the tangent P'K is perpendicular to the normal P'N, we have only to prove that P'X bisects FP'F', or that FN : FP' : : F'N : F'Pf. Now, FP' and F'P' are the focal radii a + ex', a — ex' (Art. 57), respectively. Fig, 74. FN=FO+ON, in which FO=ae, and ON is the ^intercept of the normal. Making y = m the equation of the normal we have Hence Also Therefore !/-y /_ a^ I/' b'^x' 7/ (^ - ^'). x=ON = a2 _ J2 x' = e^x'. a^ FN = ae + e^x' = e (a + ex') . F'N=F'0-ON=e(a-ex') FN _ e (a + ex') F'N _ e (a - ex') FP' a + ex' F'P' a — ex' or FN ^ F'N FP' F'P'' To draiv a tangent at any point, as P', we have, obviously, only to bisect the angle FPF' and draw P'K perpendicular to the bisector. 148 ANALYTIC GEOMETRY. 6. The tangent and normal at any point of the hyperbola bisect the angles formed by the focal radii drawn to the point of contact. LetP'Tbe the tangent at P'. We have to prove that it bisects the angle FP'F', or that FT : FP' -.-.F'T: F'P'. The focal radii FP', F'P>, are ex' — a and ex' + a (Art. 67), respectively. FT=FO — OT, in which Fig. 75. FO = ae and OT is the Xintercept of the tangent. Making ^ = in the equation of the tangent a^yy' — Ifixx' = — a?l^, we have x=zOT = — ; hence FT = ae = — (ex' — a). X' X' n FT F'T Similarly, F'T ^- {ex' -\- a), and, as before, J^ = jrp,- To draw a tangent at any point, as P', bisect the angle between the focal radii drawn to the point. 7. The principles of Exs. 5 and 6 afford a method for con- structing a tangpnf passing through a given point vdthout the curves. Thus let K (Figs. 74, 75) be the given point. Join K TANGENTS AND NORMALS. 149 with the nearer focus F\ and with /iT as a centre and KF' as a radius describe an arc. With the farther focus i'' as a centre and FH equal to the transverse axis as a radius describe a second arc cutting the first in H and Q. Join H and Q with the farther focus ; the intersections P' and P" of FQ and FH with the curve are the points of tangency. To prove that KP' is a tangent, we have KH== KF'., being radii of the same circle ; also P'H = P'F', since each is equal to 2 a^FP', the upper sign applying to the ellipse and the lower to the hyperbola. Hence KP' bisects the angle F'P'H in the ellipse and FP'F' in the hyperbola. Cor. If an ellipse and an hyperbola have the same foci, at the points of intersection they have the same focal radii, and the tangent to the hyperbola is the normal to the ellipse, and con- versely. Hence confocal conies intersect each other at right angles. o 8. Tangents at two points P', P", of a parabola, meet the axis in T' and T". Prove that T'T" = FP' - FP", F being the focus. 9, Two equal parabolas have a common axis but different vertices. Prove that any tangent to the interior, limited by the exterior, parabola, is bisected at the point of contact. I f lO, *C; 150 ANALYTIC GEOMETPwY. SECTION X.— OBLIQUE AXES. CONJUGATE DIAMETERS. 110. Equation of the ellipse referred to conjugate diameters. Let A'A" = 2a', B'B" = 2b', be conjugate diameters, the axes of reference being taken as in the figure. To transform the equation a^y^-]-b^x^ = a-b^ (1) to these axes, we have the formulae (Ai't. 22, Eq. 7) a; = iCi cos 7 + Vi cos yi, y = x^ sin y + ?/i sin y^. Substituting these in (1), and omitting the subscripts of x and y, (a^ sin2yi+ b'' cos-yO y-+2{a^ sin y sin yi+ b^ cosy cos yi)«i/ 1 + (a^ sin^y + b^ cos^ y) ar = a^ b^. j But, since the diameters are conjugate, they must fulfil the condition tan y tan yi = ^ (Art. 90), Cv OBLIQUE AXES. 151 or a^siny sinyi = — 6-cosy cosyi. Hence the coefficient of the second terra of (2) is zero, and the equation becomes (a^ sin- y, + U' cos^ y,) y^ + (a^ sin- y + h- cos- y) x" = a^ b\ (3) Making y=0, we have x^ = a'- = ffi-Q- . a-sin^y + &-cos^y ' 7 Cfb- andwhena; = 0, y- = b'- = ^^r , , ,., a- sm-yi + 0- cos-yi Substituting from these equations the vahies of the coefficients of y- and cc- in (3), we have the equation in terms of the semi- diameters, a'^ f + h'^ x^ = c^' b'% (4) which is of the same form as the equation of the ellipse referred to its axes, the semi-diameters having replaced the semi-axes. CoR. 1. The equation of the hyperbola referred to conjugate diameters is ^,,^, _ ^,o^, ^ _ ^^„^,,^ ^^^ since the onh' change in the above would be that arising from the minus sign of 6'-' in the equation of tbe hyperbola. Cor. 2. The equations of the tangents to the ellipse and hyperbola referred to conjugate diameters are a'-yy'±b"xx'=±a"b'-, (6) since the only change in the process of Art. 104 would be that arising from the substitution of a' and b' for a and b. 111. The squares of ordinates j^aycdlel to any diameter of an ellipse are to each other as the rectangles of the segments into ivhich they divide its conjugate. Let P'3I', P"M", be the ordinates parallel to any diameter BB', and meeting its conjugate AA' in 3/' and M". Then, a'-y^-{-b''X^= a''^b''^ being the equation of the ellipse referred to these diameters, we have for the points P' and P" 152 ANALYTIC GEOMETRY. a'- a'^ Dividing, ^^^— = = -^^ ■ '-^ '-. y'" a"-x"' {a'-\-x"){a'-x") or P'i¥'2 : P"M"' : : AM' . M'A' : AM" . M"A'. 112. The squares of ordinates 2'>cirallel to any diameter of an hyperbola are to each other as the rectangles of the distances from the feet of the ordinates to the vertices of the conj\igate diameter. 113. The parameter of an ellipse is a third p)roportional to the transverse and conjugate axes. The axes being conjugate diameters, Art. Ill applies, and y'^ : y'2 : : (a + x) (a - x') : (a + x") (a - x") . Let P' coincide with the exti'emity of the conjugate axis, and P" with that of the parameter. Then y'=b, y" = py x'=0, x"=ae, and the proportion becomes b- : 2^': : a' : a' {I - e') . But l-e- = -; .-.a-ib-iib-ip-, or 2a: 2b : : 2b : 2p. a- 114. Any ordinate to the transverse axis of an ellijjse is to the corresponding ordinate of the circumscribed circle as the conjugate axis of the ellipse is to its transverse axis. From the equation of the ellipse ?/- = — (a^— a^), and that of a- the circumscribed circle y{' = cr— x^, where y and y^ are the ordinates corresponding to the same abscissa x^ we have 9 O 115. Tlie sum of the squares of conjugate diameters to the ellipse is constant and equal to the sum of the squares of the axes. OBLIQUE AXES. 153 Let x', y\ be the coordinates of A' (Fig. 76), and a;", ?/", those of B'. Since B"B' is parallel to the tangent at A', its equation is ?/ = — x. Combining this with ci?y-+ b-xr = a-b-, a-y' to determine the intersections B' and B", we find ay' bx' x=:x"=±-^, and y = y"=:f—-. b ' ' a But a~ a- and 6'2 = x"'+ y"'- = ^ + ^ = "' ^ b- a- b- / O 19 \ I b' ,9 9 , Co fO 9 9 19 = o-H ^ — X- = cr— e-.i;-. Hence ce^j^h'- = a-+b-. 116. r/ie difference of the sqtiares of conjugate diameters to the hyperbola is constant and equal to the difference of the squares of the axes. Let a;', y, be the coordinates of A\ and x'\ y", those of B'. Fig. 77. 154 ANALYTIC GEOMETEY. The equation of B"B' is ?/ = x. Combining this with the a'y' equation of the y-hyperbola, c^y"-— h-a? = 0.-11^, to determine the intersections B' and B", we find „ , ay' „ , bx' x = X = ± -^, y=zy"=± b a Hence a"= x''+ 2/"= ^"+ - (x"- a') = ^-'-^x"-- b"= e'x"-- b' ; 'b\^,2_ a^ and 6'2 = a;"2+2/"^ = ^" + ^ = ^, 0" a- Ij- Hence a''- - V' = a' - h\ b\,n + —x' a' 117. The rectangle of the focal radii draicn to the extremities of any diameter of an ellipse is equal to the square of the semi- conjugate diameter. Let {x', y') be one extremitj' of the diameter. Then, if r and r' represent the focal radii, rr' — {a — ex') {a-\- ex') (Art. 57). Let (x", y") be the extremity of the conjugate diameter whose length is 2b'. Then „2,/2 f.2^1-2 b'-^=:x"-'^y"-^ = ^ + ^- (Art. 115) b' a'' = ^.^ (a2_ cc'2)+ ^ = «2_ ^^^co" = a-- e'x'\ This property is also true of the hyperbola. 118. The area of the parallelogram formed by tangents at the extremities of conjugate diameters to the ellipse and the hyperbola is constant, and equal to the area of the rectangle on the axes. Draw OD perpendicular to one side of the parallelogram (Figs. 76, 77). Tlien the area of the parallelogram is 4 OB'. OA' . sin B'OA'= 4 OB'. OA' . sin OA'T=A OB'. OD. OBLIQUE AXES. 155 The normal form of the equation of the tangent at A' is a^yy' ± b'-xx' g: a'-' b'- _ j. ■Va*y"-j-]^x" ~ ' ^ Hence the distance from the origin to the tangent is f= = T7 (Ai'ts- 115, 116), \ 6^ "^ ce and \OB.OD = \y-^^' = ^ah. h' 119. To find the equal conjiKjate diameters of the ellipse. Equating the vahies of a'~ and b''^ (Art. 110), a^ sin"y + 6" cos-y a- sin-yi + Z/- cos^y^' whence a? sin^yi + b' cos-yi = or sin-y + W cos^y ; or, transposing, a-(sin-yi — sin-y) = Zy-(cos^y — cos-yi) =6-(sin-yi — sin^y), since cos-^ = 1 — siu-yl. Hence (a^- 6^) (sin- yi - sin-y) = 0, (1) and therefore sin^yi = sin-y. Since, in the ellipse, if y is acute, yi is obtuse, and the sines are equal, yj=180° — y and tanyi = — tany. 7 - Substituting this in tauy tanyi = — — 5 a- the equation of condition for conjugate diameters to the ellipse, tany = ±-: .•. tanyi= — tany = q: -• a a Hence, rt7ien the diameters are equal, the angles they make toith the transverse axis are supplementary and the diameters fall on the diagonals of the rectangle on. the axes. 156 ANALYTIC GEOMETKY. Cor. 1. If o = b, (1) is satisfied iDdependently of y and yi; or, in the circle every diameter equals its conjugate. Cor. 2. For the hyperbola, (1) becomes (a- +b-) (sin-yi — siiry) = 0, which cannot be satisfied for sin^yi = sin-y, since in the hyper- bola both angles are acute and this condition would make them coincide. Hence the hyperbola has no equal conjugate diameters. From a.'- — b'- = a- — b', however, we see that if a = &, then a' = b' ; or, every diameter in the equilateral hyperbola equals its conjugate. SUPPLEMENTAL CHORDS. * 120. Defs. Straiglit Hues draw^u from auy point of an ellipse or an hyperbola to the extremities of a diameter are called sup- plemental chords. Tjuis, /S"Q, QS' (Figs. 76, 77) are supplemental chords. 121. If a chord of an ellipse or hyperbola is parallel to a diam- eter, the suirplementcd chord is p^araUel to the conjugate diameter. Let A" A' (Figs. 76, 77) be a diameter, and S"Q the parallel chord. Draw the supplemental chord QS', and let x\ y\ be the coordinates of !S\ and therefore — x', — y\ those of S". The equation of S"Q will be y + ?/' = a"{x + x') (Art. 31) , and that of S'Q, y — y' — a'(x — x') . Combining these equations by mul- tiplication, y'^— y'~ = a'a"(x^ — x'-) , in which x and y are the co- ordinates of Q (Art. 36). But S' and Q are on the curve ; hence a-y''^± b-x'"^ = ± d'b- and ahf± b-.n? = ± a^b- ; or, by subtraction, y^ — y'^ = ^: — (a;'- — x'') . f^quatiug these two values of y-— y'-, we have a'a" = 0—- Rut this is the condition for conjugate a- J 2 diameters, viz. : tan y tan y, = q: — (Arts. 90, 93). Hence if a- d= tany, a"= tany,, and conversely. OBLIQUE AXES. 157 Cor. 1 . To drato a tangent at a given point of the curve, as -4', draw the diameter A' A" and any parallel chord as S"Q. Draw the chord QS' supplemental to S"Q. A line parallel to QS' through A' is the required tangent. Cor. 2. To draw a tangent X)arallel to a given line, as MN, draw any chord QS' parallel to it, and the supplemental chord QS". Then the diameter A" A', parallel to S"Q, determines the points of tangenc}' A" aud xV. PARABOLA REFERRED TO OBLIQUE AXES. 122. Equation of the parabola referred to any diameter and the tangent at its vertex. The formulffi for transforming from rectangular to oblique axes, the new origin being at 0', are (Art. 22, Eq. 3) x = Xo-{-XiCosyj-yiCOSyi, 2/ = 2/o + ^isiny + ?/isinyi. (1) But y = 0, since the new axis of X is parallel to the primitive one, hence cosy = 1, siuy = 0. Also tany, = —•> since the new ?h axis of Fis tangent to the curve at (x^, y^) (Art. 104, Ex. 4) ; hence from smy, siny, tanyi = ^ = — ' cosyi Vl-sin-yi we have sin v, = ^ sin yi = Vyo'+P' and therefore cos y^ = Vl — sin- y = — •' ^° Substituting these values in (1), they become x- = a-o + x,+ — M^, y = yo+ ^'^ Substituting these values in the equation to be transformed, y^ = 2px, and remembering that, since 0' is on the curve, 158 ANALYTIC GEOMETRY. t/q- = 2pxo, we have, after omitting the subscripts of x and y, f^-'jy^'+p'U, (2) which is the required equation. Cor. 1 . r/o = MO' = MN tan MNO' = p cot yi, O'X being the normal and MN= subnormal =p. Hence ^M±P^=2p{l + cot^yj) =2pcosec2yi.--^^. and (2) ma}' be written sm' „2_ 2p V = -i~^, — X. sm- Yi 71 (3) Fig. 78. CoR. 2. From the polar equation of the parabola, P r = 1 — cos 6 making 6 = yi, we have r = FQ = P 1 — cosyi making ^= 180° + yi, r=FQ' = P __ — .— . ■■■« 1 + cos yi OBLIQUE AXES. 159 Hence QQ' = FQ + FQ' = 4|- ; sin-yi or, representing QQ' by 2j>', (3) may be written / = •2p'x. (4) Thus the equation of the parabola referred to any diameter and the tangent at its vertex is 'if = 2p'x, 2p' being tlie focal chord parallel to the tangent, and becoming 2p when the diam- eter is the axis. CoR. 3. The equation of the tangent referred to a diameter and the tangent at its vertex is yy'=p\x + x'), since the only change in the process of Art. 104 is that arising from the sub- stitution oi p' for p. 123. The squares of orclinates to any dinmeter of a parabola are to each other as their corresponding abscissas. Referred to any diameter and the tangent at its vertex the equation of the parabola is y^ = 2p' x. Hence for the points pi gnri P" ^ ana i- , ^,o ^ ^^,^,^ ^„, ^ ,^^y^„ . or, by division, ^2=^' ASYMPTOTES. r 124. Equation of the hyperbola referred to its asymptotes. The asymptotes being oblique except when the hyperbola is rectangular (Art. 92), we use the formulae for passing to oblique axes with the same origin, X — x'l cos y + ?/i cos 7i, y = x^ siny + y, sin yj ; and, since the asymptotes coincide with the diagonals of the rectangle on the axes, — h . h a smy= — , smyi= — , cosy = cos yi = Va^ + 6^ Va' + 62 ^a' + b^ 160 ANALYTIC GEOMETRY. The formulae therefore become a x = Va-+ h' (a?i+yi), y iVi-Xi). Substituting these values in a-y^—b-xr = — irb^, and omitting the subscripts, we obtain xy = Fig. 79. Hence the general form of the equation of the hyperbola referred to its asymptotes is xy = m, in which m is constant. CoR. 1. The equation of the ^'-hyperbola referred to the a- + 0- same axes is xy = — (Art. 71). Cor. 2. The equation Bxy + Dy ■j-Ex + F=0 is the general equation of the hyjm-bola referred to axes parallel to its asymptotes. For, transforming xy = m to parallel axes by the formuhie x^Xq+x^, y = yo+yi, we have, after dropping the subscripts, 07/ + .To?/ 4-.'/o-^' + -^"(,?/o = 0, which is the above form. The equations of the asymptotes are evidently ar = — 2/0, c) OBLIQUE AXES. 161 125. The intercepts of the secant between the hyperbola and its asymptotes are equal. Let P', P" (Fig. 79), be any two points of the hyperbola, x'—x" the equation of the secant P'P". Making a; = in this equa- tion, y-y'=D'Q'= f''^-y^f . x'— x" But y'x'=y"x"= m, since tlie points are on the curve. Hence- D'Q'= _y"x- V X x'-x" ■■y"=P"M". Hence the triangles P"M"Q", Q'D'P, being equiangular, and having a side in one equal to a side in the other, are equal, and P"Q"=P'Q'. Cor. To construct the Jiyperhola when the axes are given: draw the asymptotes, the diagonals of the rectangle on the given axes, and through the extremities of the trans- verse axis, as A, draw 11', 22', 33', etc., and make IP', 2P", 3P"', etc., equal respectively to ^1', J. 2', ^3', etc. Then P', P", P'", etc., are points of the curve. By a similar method we may construct the curve when the asymptotes and one point of the curve are given. Fig. go. 126. The area, of the triangle formed by any tangent ivith the asymptotes is constant, and the tangent is bisected at the point of contact. 1B2 AJSTALYTIC GEOMETRY. The equation of the secant P'P" (Fig. 79) is From the equation of the curve, x"u" x^y'z= x"y"= m, .-. y'= " — ^• x' y'— v" The fraction — therefore becomes X —x" x"y" x' -y" = y". X'' x'- ■x" or, when P" coincides with P', — -^ . Hence the equation of the tangent TT' is ^' v' y-y'=-~{x-x'), X and its intercepts are y = 0T=2y', x= 0T'= 2x'. Hence P' is the middle point of TT' (Art. 6). Again, the area of the triangle OTT' is 2 2 = 2 a;'?/' 2 sin TO A cos TO A = 4x'y' , — , = «&, since x'y'= — — — Hence the area of the triangle is constant and equal to the rectangle on the semi-axes. Cor. To construct a tangent at any point, as P', when the asymptotes are given, draw the ordinate P'3I' and make M'T' = OM'. P'T' is the tangent. 127. Tangents at the extremities of conjugate diameters meet on the asymptotes. The equation of the straight line P'B' (Fig. 70), the co- OBLIQUE AXES. 163 ordinates of P' being x', y', and those of B' being -^, — • (Art. IIG), is a , bx' y y-y= — w(^-^)' & or y-y'=-~{x-x'). a But the equation of OT" is y = x', hence P'B' is parallel a to the as3-mptote OT'. Again, the middle point of P'B' is i(x'-\-^\ ify+^'M — f^^'+(^y' bx'+ay' °' '~^6~' ^^r~" which satisfy y = - x. Hence the straight line joining the ex- a tremities of conjugate diameters is parallel to one asymptote and bisected by the other. But the diagonals of a parallelogram bisect each other, and P'B' is one diagonal of a parallelogram of which OP' and OB' are adjacent sides ; hence the other diagonal coincides with the asymptote, or the tangents at P' and B' meet on the asymptote. CHAPTER IV. LOCI. -o-Oj^^OO- 128. Classification of loci. Wlieu the relation between x and y can be expressed by the six ordinary operations of algebra, viz., addition, subtraction, multiplication, division, involution, and evolution, the powers and roots in the latter cases being denoted by constant exponents, the function is called an algebraic function ; and loci whose equations contain only algebraic functions are called algebraic loci. Algebraic loci are classified according to the degree of their equations as loci of the first, second, etc., orders. We have seen that there is but one locus of tlie first order ; that is, whose equation is of the first degree, nameh', the straight line ; and that all loci of the second order are conies. All loci whose equations are above the second degree are called higher plane curves. A function whicli involves a logarithm, as a; = logy, is called a logarithmic function ; one in which the variable enters as an exponent, as y = iij', an exponential function. If a is the base of the logarithmic system, the latter function is evidently another way of expressing the former. Functions involving the trigonometrical elements, as y = smx, .«=sin~^?/, etc., are called circular functions. y = smx and .T=sin"'y ^^re different forms of the same relation, the former being called the direct, and the latter the inverse circular function. It may be shown that logarithmic, exponential and circular functions can- not be expressed by a finite number of algebraic functions, and LOCI. 165 for this reason they are called transcendental functions. A transcendental equation is one involving transcendental func- tions, and the locus of such an equation is called a transcen- dental curve. The exercises which follow will afford the student practice in the production of the equation of a locus from its definition. In all cases the object is to find a relation between the given constants, x, and y ; the latter being the coordinates of any point of the required locus. Any such relation, when stated in the form of an equation, will be the equation sought, whatever the axes ; but the simplicity' of both the solution and the result- ing equation will depend upon the choice of the axes. The student will observe two cases : first, when the given conditions furnish directly a relation between x and y ; second, when the conditions involve other variables ; and in tliis case these con- ditions must afford a sufficient number of independent equations to permit the elimination of all the variables except x and y. Thus, if 71 variables ai'c involved exclusive of x and y, the conditions must furnish w + 1 equations. 166 ANALYTIC GEOMETRY. SECTION XI. — LOCI OF THE FIRST AND SECOND ORDER. 129. 1. Given the base of a triangle and the difference of the squares of its sides, to find the locus of the vertex. Let h be the given base and d^ the constant difiference. Take the base for the axis of X, and its left- hand extrerait}' for the origin, x and y being the coordinates of the vertex. Then, by condition, OP- — BP-=d-, or x- + f-[{b-xY + f]=d\ dr + 6- whence X: M B 2b Fig. 81. to Y, at a distance from it equal to b Hence the locus is a straight line parallel d' + b' 2b If the triangle is isosceles, d = 0, and x = " • In this ease the conditions furnish 2 directly the relation between x and y. 2. To find the locus of the middle point of a rectangle inscribed in a given triangle. Let a = altitude of the triangle, b and c the segments of the base, the axes being taken as in the figure. Then the equations of AB and AC are known ; namely, X V X - + •^=1, and - + -' = 1 .V_ a Now the abscissa of P is the half sum of the abscissas of Q and R ; and if y — k be the altitude of the rectangle, and this value be substituted for y in the above equations, we find x^ = a — k a b, a — k a LOCI OF THE FIRST AND SECOND ORDER. 167 Hence x = abscissa of 2 a But the ordinate of P = ?/ = • 2 This condition enables us to eliminate the variable A: from the above value of x ; substituting therefore A; = 2y, we have 2ax = {a - 2>j) {b + c) , a straight line bisecting the base and altitude, since its intercepts are ^ (b -\-c) and 4- a. 3. To find the locus of a point so moving that the square of its distance from a fixed lioint is in a constant ratio to its dis- tance from a fixed line. ^ Let B be the fixed point, ^X the fixed line and axis of X, the axis of F passing through 5, and 0B = a. Then, m being BP- the constant ratio, = m. But BP- = x- -{-(y — a)-, and PM= y. Hence 2/2 + ar'_(2a + m)^ + a2 = 0, which is the equation of a circle whose centre is at f 2 a + m and whose radius is |^ V4a?» + m^ (Art. 50). If the point is on the line, a = 0, the cen- tre is f 0, — 1 and the radius = — '27 2 4. The squares of the distances of a jjoint from two fixed points are as m to n. Find the locus of the point. Let (a, 0), (0, &), be the coordinates of the fixed points, the axes being assumed to pass through them, and P any point of the locus. Then, A and B being the fixed points, 168 ANALYTIC GEOMETRY. PB" _ x- + {y-h)- _ m . PA'~ xj- + {x-ay~n ' or, clearing of fractions and reducing, o , <> 2 nb , 2 am , nh- — ma^ ^ y- + x- y H X + = 0, 11 — 7)1 n — m n — m a circle whose centre is ( -» — — ), and radius is \ n — III 11 — iiij 1 ■■y/mnia^ + b-) (Art. 50). If 6 = 0, or a = 0, that is, if both the points are on the same axis, the centre is on that axis. If a = b = 0, the centre is at the origin and i2 = 0, or the locus is a point ; unless also m = ?i, when B = — 5. Find the locus of the vertex of a triangle having given the base and the sum of the squares of the sides. Ans. A circle whose centre is the middle x>oint of the base. 6. Given the base of a triangle and the ratio of its sides, find the locus of the vertex. OP Let 6 = base of the triangle (Fig. 81), and — = m, the ratio. Then OP' = m2P5^ or ^^ x'^^f=i,r{y\+{b-xy), whence ?/ + ^ + z z^— z ; = 5 a circle whose centre is f -? ), and radius is \ — 111? j 1 — m' 7. From one extremity A' of a diameter AA' to a circle a secant is drawn meeting the circle at P. At P' a tangent to the circle is drawn, and from A a perpendicular to this tangent. T7ie perpendicidar produced meets the secant at P. Find the locus of P. LOCI OF THE FIRST AND SECOND ORDER. 169 Let the diameter be the axis of X and the centre the origin. Let {x', y') be the coordinates of P' ; then (Art. 32) is the eqnation of the secant ; the equation of the tangent at P' is yy' + xx' = jB^, hence the perpendicular on the tangent from A is , y = l,{x-R). (2) Combining (1) and (2), to find P, we have x=2x' + R, y = 'ly\ (3) But (x', ?/') is on the circle, hence x'- + ?/'- = Bj^. Substituting in this equation the values of xJ aud ?/' from (3) , we have a circle whose centre is at (i?, 0), that is, at A^ and whose radius is '2R = AA . 8. A line is draion parallel to the base of a triangle, and the points where it meets the sides are joined transversely to the extremities of the base; find the locus of their intersection. Take the sides as axes. Ans. A straight line through the middle point of the base and the opposite vertex. 9. Given the base and sum of the sides of a triangle, if the perpendictilar be produced beyond the vertex until its whole length is equal to one of the sides, to find the locus of the extremity of the perpendictdar. Ans. A straight line, 10. Given any parallelogram, ^ Q and PP', QQ', lines parallel to adjacent sides. Prove that the locus of the intersection of PQ p/ and P'Q' is a diagonal of the ^ qi ^ ii parallelogram (Fig. 84) . Fig. 84. 170 ANALYTIC GEOMETRY. 11. In Fig. 84, find the locus of the intersection of BL and PA^ A and B being fixed points, and P and L subject to the con- dition that 0L + OB=OP+ OA. 12. A line cuts two fixed intersecting lines so that the area of the intercepted triangle is constant. Find the locus of the middle point of the line (see Art. 126). Let OX, OT, be the fixed hues and axes, AOB the inter- cepted triangle, m the constant area, <^ the constant angle BOA, and P the middle point of AB. Then OM=x, 3IP=y, and, since P is the middle point of AB, 0A = 2yani\ 0B=2x. Hence area BOA or xy. 2x.2y.smcf) in = ^;r J m 2sin^ an hyperbola whose asymptotes are the fixed lines (Art. 124). 13. Given ttco intersecting fixed lines and a fixed x>oint. A line is drawn through the fixed point. Find the locus of the middle point of the segment iyUercejyted by the given lines. Let OX, OF (Fig. 85), be the fixed lines and axes, Q the fixed point, its coordinates OP = m, PQ = n, AB the line, and /-• its middle point. Then, from similar triangles, 0A{=2y) :0B{=2x) ::PQ{=n) :PB(=2x-m), or 2xy — my — nx = ; an hyperbola passing through Q, whose asymptotes are x = —i 2/ = -(Art. 124, Cor. 2). 14. From a fixed point A (Fig. 85), a line AB is draivn to meet a fixed line OX. From the intersection B, a constant dis- tance BR = h is laid off, and from R a line PQ is drawn, mak- ing a constant angle ivith OX, to meet AB in Q. Find the locus of Q. LOCI OF THE FIRST AND SECOND ORDER. 171 Since the angle BRQ is constant, take a parallel to QR through A for the axis of Y, and the fixed line OX for the axis of X, and let OA = a. Then, OA . JiQ : : OB : MB, or a: y : : x-\-b:b ; whence xy -{-by — ab = 0, an hyperbola through A, one of whose asymptotes is the fixed line and the other x^-b (Art. 124, Cor. 2). 15. To find the locus of the intersection of a perpendicidar from the focus of a parabola on the normal. The equation of the normal is and that of the perpendicular is y y=z-'[^-.) Combining these to find the point of intersection, we find it to be p^4- 2x'y'--{-22yy'- '2px'y'-\-2'ry' ^= 2?/'^+ 2 2/ ' ^= 2.^2+ 2p2 In this pi'oblem the conditions introduce the auxiliary vari- ables a;', y', the coordinates of the point of contact from which the normal is drawn. But this point is on the parabola; hence we have the additional condition y''-= 2px'. Eliminating y' by means of this equation, we have Finally, combining and eliminating a;', we have 2 P P" a parabola on the same axis, whose vertex is at (l^, ), and whose parameter = \ that of the given parabola. 16. The locus of the intersection of the perpendicular from the focus of a parabola upon the tangent is the tangent at the vertex. 172 ANALYTIC GEOMETRY. The equation of the tangent is and of the perpendicular upon the tangent through the focus, Combining these to find the intersection, we obtain, on eUmiuatiug y, x{p -j-'2x')=0, which, since x' cannot be nega- tive, is satisfied only for a;=0; that is, the intersection is always on Y, which is the tangent at the vertex. How does this property enable us to find the focus when the curve and axis are given ? 17. Throvgh any fixed point cliords are draivn to a parabola. Find the locus of the intersections of the tangents to the parcdtola at the extremities of each chord. Let .Tj, 2/1, be the coordinates of the point through which the chords are drawn, and suppose the tangents at the extremities of one of these chords to meet at (/*, k). Then the equation of the chord is (Art. 106) yk^p{x + h). But the chord passes through the fixed point (.rj, yi) , hence y^k = p {Xi+ h). Now this is the equation of a straight line, in which h and k are the variables ; therefore the locus of (/i, k) is a straight liue. If the fixed point is the focus, a^i =^, ?/, = 0, and the equation becomes h = — -- Hence the locus of the intersection of pairs of tangents drawn at the extremities of focal chords is the directrix. 18. Through any fixed point chords are drawn to the ellipse (or hyperbola) ; to find the locus of the intersection of the tan- gents at the extremities of each chord. I^et a^i, ?/,, be the coordinates of the point through which the chords are drawn, the tangents at the extremities of one of LOCI OF THE FIRST AND SECOND ORDER. 173 them meetiug at {h, k) . Then the equation of this chord is a?yk ± b-xh = ± (rl/. (Art. 106) V ^ _.!!..7. 1 1,9.. 7 I ,27,2 But this chord passes through the fixed poiut (x-j, ?/i), hence ahj-^k ± b-xji= ± crV^. Now this is the equation of a straight line in which h and 7i' are the variables ; therefore the locus is a straight line. If the fixed point is the focus, x^= ae, yx= 0, and the equa- tion becomes /* = -• Hence the tangents at the extremities of e focal chords to the ellipse and hyperbola intersect on the directrix. 19. The locus of the intersection of the perpendicular from the focus of an ellipse upon the tangent is the circle described on the transverse axis. In this problem the equation of the tangent in terms of the slope (Art. 105, Ex. 12), y = mx + Va-»i-+ 6", is most convenient. The perpendicular upon it from the focus is 1, . y = {x-ae). 'tn From the former, y~mx= ^a-m^-^b'-, and from the latter, my + tc = ae. Squaring and adding, (v'+a;-)(l +m') = b--\-m-a- + a'e- = b-+ma'+cr^-^—^= a-{i+m'), a^ which eliminates m, giving y"-\- x-= a-. This property is also true of the hj'perbola. 20. Find the locus of the intersection of pairs of tangents to a parabola ivhich intercej^t a constant length, on the tangent at the vertex. The equations of the tangents are yy' =p{x + x'), (1) yy" = p{x + x^'). (2) 174 ANALYTIC GEOMETRY. The equation of the locus will be found by combining these and eliminating x', y\ x", and y". To effect this elimination we have the equations of condition, y'' = 2px', (3) y"-' = 22^x", (4) w'-w" — TT^- = a, a constant. (5) Substituting in (1) and (2) the values of x' and x" from (3) and (4), we have yy'=2:)x + -^, (6) ,112 yy"=px + ^- (7) Substituting from (5) y'= 2a-j-y" in (6) and combining the result with (7), we have y"—y — a\ which substituted in (7) gives y^= 2px + a?, an equal parabola with the same axis, and vertex at ( — — -, 21. Parallel chords, as QQ\ ivhose centime is C, are draivn to a circle. AA' is a diameter parallel to the chords. Find the locus of the intersection of AC loith the radii through the extremi- ties of the chords. Ans. A parabola ivhose axis is the diameter and vertex mid- loay between 0, the centre., and A. 22. Find the locus of the intersection of tangents drawn at the extremities of conjugate diameters of an ellipse. Ans. 2 aY-+ 2 b'x^= 4 o'bK 23. Lines RL, R'L' (Fig. 84), are draivn parallel to one side of a parallelogram and equidistant from the centre. Find the locus of the intersection of a line drawn from through the extremity R of one of the parallels ivith the other, or the other produced. Ans. An hyperbola. LOCI OF THE FIRST AND SECOND ORDER. 175 24. A and B are fixed 2wints. Find the locus of P when PD- = o constant^ D being the foot of the perpendicular ADxDB from P on AB. Ans. An ellipse. 25. To find the locus of the centres of all circles which pass through a given point and are tangent to a given straight line. Ans. A parabola. 26. If a variable circle touch a fixed circle and a fixed straight line., the locus of its centre is a parabola. 27. Given the base of a triangle and the product of the tan- gents of the base angles; the locus of the vertex is an ellipse. 28. The base and area of a triangle is constant; the locus of the vertex is a straight line. 176 ANALYTIC GEOMETRY. SECTION XII.— HIGHER PLANE LOCI. 130. The limits of this work permit a reference to a few ouly of the higher plane curves possessing interesting geometric properties. 1. The cardioid. Through any point of a circle a secant is drawn cutting the circle in Q. Reqnirecl the locus of a point P on the secant tvhen QP= R, the radius of the circle. Let C be the centre of the circle, CO = R the radius, the pole, and OX, a tangent at 0, the polar axis. Then OP=OQ+QP. But OP=r, OQ = OD cos QOD = OD sin XOQ = 2 i? sin 0, and QP = r. Hence r=^2Rsme + R. (1) Discussion of the equation. For 6 = 0°, r = R— OA. As 6 increases, r increases, and when ^ = 90°, r='6R= OB. Fig. 86. HIGHEE PLANE LOCI. 177 As increases from 90° to 180°, r diminishes, and when =■ 180°, r=Ji=OE. When 6 passes 180°, sin ^ becomes negative, but r remains positive nntil sin^ = — ^, when ?• = ; at this point ^=210°. From this value of ^, r is negative and the portion OFC is traced, r being — 7^ when ^ = 270°. From ^=270° to ^ = 360° the portion COA is traced, r becoming positive again when = 330°. Rectangular equation of the cardioid. Transferring to the axes YOX hy the formula (Art. 24, Eq. 4), r = Vic^+2/^5 sin^= — -^ , we have (ar' + y- - 2 By) -' =. {x~ + /) E\ a curve of the fourth order. (2) Trisection of the angle. The cardioid affords a metliod of trisecting an angle, as follows. Let NCO be the given angle. With the vertex C as a centre describe any circle, and construct the cardioid to this circle. Only that portion of the curve in the vicinity of NC produced need be constructed. Produce NC to meet the cardioid at P, and draw FO and QC. Then the tri- angles CQP, CQO, are isosceles by construction. Hence NCO = COP + CPO= CQO + CPO = QCP + 2 CPO = 3 CPO ; or CPO = lNCO. 2. The conchoid. Through a fixed point F a line FP is 178 ANALYTIC geo:metry. drawn cutting a fixed line XX' in Q. Required the locus of P when QP is constant. Let QP= a, FO = b, the distance from the point to the line, and OX, OY, be the axes. Draw FS and PS parallel and per- pendicular respectively to X'X. Then PM : MQ : : PS : FS ; or y : V«^ — y- : : y -\-b : x; whence x-y~= {y + h)- ( «- — y-) , ( 1 ) a curve of the fourth order. Discussion of tue equation. Solving the equation for x, we have X = ± -^— ' — Va- — y- y The curve is evidently symmetrical with respect to Y. When y is positive and equal to a, .^• = 0, locating the point A, which is a limit in the positive direction of Y, since x is imaginary if 2/ > a. As y diminishes, x increases numerically, becoming ± 00 when 2/ =0; hence the curve has infinite branches in the first and second angles with X'X for their common asymptote. Since x is real for negative values of y less than a numerically, there is a branch in the third and fourth angles. When y = — a, or — 6, x=0, locating A' and F; and as x has two values numerically equal with opposite signs for values of y between — a and — b, the locus between these values is an oval. When y is negative and numerically less than b, x increases as y diminishes and becomes ± co for y = 0, giving infinite branches with the axis of X for their common asymptote. ?/ = — a evidently limits the curve in the negative direction of Y. In the above case a>&. If a = ^, i^ coincides with A' and the oval disappears. If a < b, all negative values of y numer- ically greater than a render x imaginary, except y = — b. wliich renders x=0; thus the oval disappears, but .r = 0, y = — b, satisfy the equation, and hence must be considered a ]ioint of the curve. Such a point is called an isolated, or conjugate point. HIGHER PLANE LOCI. 179 Polar equation of the curve. The polar equation ma}- be obtained by transformation, or directly from the figure, thus : Let F be the pole and FS the polar axis ; then r=FP =FQ + QP=FO amO + a, or r=h cosee ^ + a. " ' (2) Trisection of the angle. The conchoid also affords a method of trisecting the angle, as follows : Let AFP be the given angle. Draw any line OX perpendicular to one side, and with i*' as a fixed point, OX a fixed line, and PQ = 2FQ, construct the arc of a conchoid. Only that portion of the curve included within the given angle need be drawn. From Q draw QX perpendicular to OX and join its intersection with the conchoid, X, with F. Bisect QX at R, draw RL parallel to OX, and join L with Q. Then the triangles LXQ^ LFQ, are isosceles ; for, since QR =^ RX, KL =LX^LQ^^QP^FQ. Hence the angle AFX=FXq=LQX=^FLQ = ^LFQ, or AFX ='^ AFP. Mechanical construction. The conchoid may be con- structed mechanically as follows : Let AA\ XX', be two fixed rulers, the latter having a groove on its upper surface. Let FP be a third ruler, having a peg Q fixed on its under side, which is also grooved to slide on a peg at F. A pencil at P will trace the curve. 3. The cissoid. Pairs of equal ordmates are drawn to the diameter of a circle, and throngh one extremity of the diameter a line is draimi through the intersection of one of the ordinates ivith the circle. Find the locus of the intersection of this line with the eqnal ordinate or that ordinate jyrochiced. Let OD be the diameter to which the ordinates are drawn, and the axis of X, the tangent to the circle at being the axis of F. Let QM, Q'M', be equal ordinates. Through 180 ANALYTIC GEOMETRY. draw OQ (or OQ') ; then P' (or P) is a point of the locus. From similar triangles, 0M:3fP:.0M':M'Q'; or, R being the radius of the circle. x: y :: 2R — x: ^x {'2 R — x), whence y- {2R — x) = x^. (1) Discussion of the equation. Solving the equation for y, y / — y- : ?/, whence off = 4Ii\2 Ku -f), (1) a loons of the fourth order. Let the student discuss the equation. 6. To find the locus of the intersection of the perpendicular from the vertex of a parabola upon the tangent. y-—2px being the equation of the parabola, tliat of the re- quired locus is — X' ?r = p 2 -|- X P a cissoid, the diameter of whose circle =j 7. Given two fixed points F and F', to find the locus of P such that PFxPF'=(^'y' Let FF' be the axis of X, and the origin in the middle point oiFF'. Then {tf -\- x-y = 2 c- {or — y-) is the required locus, in which c = ^FF'. The locus is the lemniscate (see Ex. 4), the hyperbola being rectangular, and c = a 8. The corner of a rectangular sheet of paper is folded over so that the sum of the folded edges is constant. Find the locus of the corner. By condition. OB = BP, OA = AP, the angle at P is a right angle, and AP-}- PB = a, a constant. But AP''= AG'^ AE- + EP"- = {AO-yf+x\ .'. 3f+y^=2AO.y = 2AP.y. 184 ANALYTIC GEOMETRY. Also PB''= OJBr- = PD'-+BD'= f+ (x - 0B)\ .-. 0^+ y^= 2 OB . X = 2 PB . X. Substituting the values of AP and PB from these equations in AP-\- PB = a, we have {x-+ ?/-) {x + y)= 2 axy, a locus of the third order. TKANSCENDENTAL CUKVES. 185 SECTION XIII. — TRANSCENDENTAL CURVES. 131. 1. The logarithmic curve. The equation of this curve is x = \ogy. Assumiug the form y = a^, in which a is the base of the logarithmic system, we observe that when x=0, ?/ = l, whatever the base; hence all / Fig. 92. logarithmic curves cut the axis of F at a distance unity from the origin. Again, since negative numbers have no logarithms, y cannot be negative, hence these curves lie wholly above X. If X is positive and increasing, y increases, but more rapidly than X, and the more rapidly as the base is greater ; hence the curve departs rapidly from X in the first angle, and the more so as the base is greater. If x is negative, then y = a~-' = — , from which we see that as x increases numerically, y decreases, and the more rapidly as the base is greater, but becomes zero only when x = — od \ hence the curve approaches X in the second angle, and that axis is an asymptote. 2. The cycloid. To find the locus of a point in the circumfer' ence of a circle ivMch rolls without sliding along a fixed straight line. 186 ANALYTIC GEOMETRY. Let OX be the fixed straight line and axis of X, the initial position of the generating point and origin, r the radins of the circle, and P any point of the locus. Then OJr= 0N—3fN. Fig. 93. But 0M= X, 0X= arc PX= versin"^ QX to the radius r = r versin^ -, r and 3IX =PQ= VXQQT = V2T-^^^'. Hence the required equation is -iV X = r versin "~ —^'Zrv r -V2i r (1) Discussion of the equation. Since x is imaginary if y is negative, the curve lies wholly above X. If y = 0, a; = r vers-^0 = 0, ± 27rr, ±47rr, etc., or there are an infinite number of arcs equal to ODA on each side of F, OA being equal to the circumference of the circle. If y—'2 r, X — r vers~ ^ 2 = ± 7rr, ± 3 tt/-, etc. , locating D, and the corresponding points on the other arcs. Defs. DD' is called the axis of the cycloid, OA the base, and 0, A, etc., the vertices. To put the generating circle in position for any point, as P, draw CC parallel to the base through the centre of tiie axis, and with P as a centre and a radius = CD describe an arc cutting the parallel in C. Then C is the required centre. If the angle PC'N= , 0X= arc PX= r. TRANSCENDENTAL CURVES. 187 a; = r(<^-sin<^)l and ^ h (2) which are called the equations of the c^'cloid, and are more useful than Eq. (1) in determining its properties. If any other point than P of the radius of the rolling circle be the generating poiut, the resulting curve is called the jJ^'olate, or curtate cycloid, according as the generating point is within or without the circle. The locus of a point on the circumfer- ence of a circle rolling without sliding on the circumference of another is called an epicycloid., or liypocycloid^ according as the circle rolls on the exterior or interior of the fixed circle ; if the generating point is not on the circumference of the rolling circle, the curve is called an epitroclioid or hypotrochoid. The general term applied to the locns generated by a point of a rolling curve is roulette. The circular functions. A series of transcendental curves is obtained by assunung the ordinate some trigonometrical func- tion of the abscissa, as y = sin.'C, y = coto;, etc. The length of the arc corresponding to any value of x given in degrees may be found as follows : The length of the arc of 180° in the circle whose radius is unity being 3.1416, the length of any other arc, as that of 10°, will be J^ (3.1416) = .1745 ; this distance being laid off on the axis of X, the corresponding value of y may be taken from the table of natural sines, tangents, etc. The curve may be drawn, however, with sufficient accuracy by observing the general change in the function as the arc increases. 3. y=smx. When .t = 0°, ?/=0, hence the curve passes through the origin. As x increases, y increases, reaching its greatest value y = 1 when x = 90°, locating A. From x = 90° to x= 180°, y decreases from 1 to 0, becoming negative when a:> 180°, and reaching its least value y = —l when x = 270°, locating C. From a: =270° to x' = 360°, y is negative and decreasing numerically, becoming zero again for x = 360°. It is evident that as a; varies from 360° to 720°, a like portion will 188 ANALYTIC GEOMETRY. be traced, as also when x is negative ; hence the curve consists of an infinite number of arcs equal to OABCD, and extends Fifi;. 94. without limit along X to ± go. the sinusoid. The curve is sometimes called 4. y=cosx. Let the student trace the curve. 5. y=t2inx. "When x=0°, y = 0. As x increases, y increases, becoming oo when x = 90°. When x passes 90°, y is negative, and remains negative tilhf=180°, decreasing numer- ically from 00 to 0. From X = 180° to aj = 270°, y is posi- tive and increasing, becoming oo when x-=270°, etc. The curve consists of an infinite number of branches equal to AOB, on either side of the origin, having for as3'mptotcs q the lines x = ± —> x = ± -tt, 2 2 Fig. 95. etc. C. ?/=cot.r. Let the student trace tlie curve. 7. ?/ = versinic. The versine being always positive, the curve lies wholly above X, its limits along F being and 2. student trace the curve ; also : 8. y = coversin x. Let the TRANSCENDENTAL CURVES. 189 9. y= seca;. The curve is given in the figure. Let the stu- dent discuss the equation, and also trace the curve : 10. y = cosecx. O Fig. 96. Spirals. The locus of a point receding from a fixed point along a straight line, u-hich revolves about the fixed point in the same plane, is called a plane spiral. The fixed point is called the pole, and that portion of the spiral traced during one revolution of the line is called a spire. The polar equations of many of the spirals may be derived from the general form r = aO'\ by assigning different values to n. 11. Spiral of Archimedes. The equation of this spiral is obtained from the general form by making n—1; whence r = a6. (1) From this equation- = a ; since the ratio of the radius vector to the vectorial angle is constant, the spiral may be defined as traced by a point ivhich recedes uniformly from, while the line revolves uniformly about, the pole. Assuming as a unit radius the value of r when the line has made one revolution, we have l=a.2 TT a = — 1 27r and Eq. (1) becomes ;7r (2) 190 ANALYTIC GEOMETRY. when ^ = 0, r = 0, or the spiral begins at the pole. The dis- tance between any two consecutive spires measured on the same radius is the same and equal to the unit radius, called the radius of the measuring circle, r increases uniformly with ^, but is oo only where ^ = x . To construct this spiral, let be the pole and OA the polar axis. Through the pole draw any number of straight lines making equal angles with each other, sav 30°, or -• Then when 6 TT 6=-, 6 V — 1_ r = ^'^ = OP. Having laid off 0P=^ on 01, make OP = 2 OP, OP"^i}OP, etc. Tlien OPP'P", etc., is the spiral. OQ is the radius of the measuring circle. 12. The reciprocal ok hyperbolic spiral. The equation of this spiral is obtained from the general form by making Ai = — 1 ; whence TRANSCENDENTAL CURVES. 191 In this spiral the radius vector evidently varies inverse!}- as the angle. Assuming as before that r is unity when ^=27r, we have a = 27r, or 27r ^2) r = 6 When ^ = 0, r= X, and as 6 increases, r diminishes, but is r Fig. 98. zero only when ^ = oo ; hence there are an infinite number of spires between the measuring circle and the pole. To construct the spiral, draw the lines making equal angles with each other, as before. If the angle is 30°, then when 6 = -, r= 12= OP. Make OP' = — = 6, OP" = ^ = 4, 6 2 3 etc., and draw PP'P" ••• . 13. The lituus. This spiral corresponds to n = — i in the general equation. Hence its equation is r = a -Vd (1) Fig. 99. 192 ANALYTIC GEOMETRY. or, if r= 1 when 6=2Tr. a V27r, ?- = (2) For every value of 6, r has two values, one positive and one negative, as shown in the figure. If ^=0, r = cc, and r—0 onl}' when $ =;.-'* ; there are thus an infinite number of spires between the measuring circle and the pole. It may be shown that the polar axis is an asymptote to the spiral. 14. The logarithmic spiral. This spiral is defined by the equation log r=a^, (1) or, if b be the base of the system, r = &«*. (2) Whatever the logarithmic system r = l for ^=0. Hence, if 0A= 1, all logarithmic spirals pass through A ; and OA may be taken as the radius of the measuring circle. From (2) we see that as 6 increases, r increases rapidly, and the more so as the base is greater ; and diminishes rapidly if 6 is nega- tive, but is zero only when 9 = —cc. Also, if ^=00, r = cc. - Hence there are an infinite number of spires within and Fig. luo. without the measuring circle. PART II. SOLID Al^ALYTIO GEOMETEY. CHAPTER V. THE POINT, STRAIGHT LINE, AND PLANE. -oo^O^oo- SECTION XIV. — INTRODUCTORY THEOREMS. 132. Defs. 1. By the angle betiveen ttvo straight lines not in the same ^jlane is meant the angle between any tivo intersecting parallels. Hence, if through any point of one of the lines, a Ijarallel is drawn to the other, the angle between the first and the parallel is the angle between the two lines. Thus, let PQ and KL be any two straight lines which neither intersect nor are parallel. Through any point of FQ, as P, draw FH par- allel to KL. Then HFQ is the angle between FQ and KL. 2. The foot of a ^perpendicular from a point upon a plane is called the projection of the point on the plane. Thus, if F be any point, AB any plane, and the perpendicular to the plane through F meets the plane at il/, M is the projection of F on AB. 3 . The foot of a perpen- dicxdar from a pioint on a line is the projection of the point on the line. Thus, if KL be any straight line and the perpendicular from P meets the line at S, S is the projection of F on KL. Fig. 101. 196 ANALYTIC GEOMETRY. 4. The projection of a straight line of limited length upon a plane is the line joining the feet of the j^erj^endiculars from the extremities of the line upon the plane. Thus, if PQ be any limited straight line, AB any plane, PM, QN, perpendiculars to the plane meeting it at M and N^ MN is the projection of the line PQ upon the plane AB. Since the perpendiculars PM and Q^ determine a plane through PQ perpendicular to AB, the projection of a straight line upon a plane ma}^ also be defined as the intersection of a plane through the line, perpen- dicular to the given plane, with the latter. 5. Tlie p>rojection of a limited straight line upon another straight line is that portion of the latter intercepted bj/ the j)i'ojections of the extremities of the former. Thus, PS and QT being the perpendiculars from P and Q on KL, JST is the projection of PQ on KL. 133. The projection of a limited straight line upon another straight line is equal to the length of the line multiplied by the cosine of the included angle. The projections of any limited straight line upon parallels are equal ; for the perpendiculars from its extremities, being per- pendicular to parallel lines, lie in parallel planes ; these planes, therefore, intercept equal distances on the parallels, and these distances are the projections. Thus, in Fig. 101, KL and 3fN being parallel, the planes PS M aud QTN are parallel, and the intercepts *S'Tand 3/iVare equal. Hence, if we find the pro- jection of PQ on any one of a system of parallels, this projec- tion will be the same for all. Draw P^ parallel to MN. The angle between PQ and any parallel to PH is (Art. 132, 1) HPQ, and IIP = PQ cos HPQ = MN= ST= etc. Hence the propo- sition. CoR. Since the angle between PQ and AB = NRQ = HPQ, the projection of a limited straight line upon a p>lane is equal to the length of the line multiplied by the cosine of the angle which the line makes with the plane. Thus MN= PQ cos NRQ. INTRODUCTORY THEOREMS. 197 134. If AD he the straight line joining A tvith i>, and AB, BC, CD, straight lines forming a broken line from A to D, then the algebraic sum of the projections of the latter upon any straight line OX is equal to the projection of the former on the same line. Draw from A, B, C, aud D, the perpendiculars to OX, meeting OX in A', B', C, D', respectively. It is evident that as A moves to D along the bi'okeu line ABCD, the foot A', of the perpendic- ular from A, moves along A'D\ to the right or the left, according as the angle between the direction of motion of A and OX is acute or obtuse. At -4, B, and C, draw parallels to OX, and denote the angles made by AB, BC, CD, and AD with these parallels by a, (3, y, and 8. As the points are chosen in the figure, in passing from B to C, B' moves to the left along OX, the angle B being obtuse. Now the projection on OX of AB is (Art. 133) A'B'^AB cos a; that of BC is B'C = BC cos (3 ',~ that of CD is CD' = CD cosy ; and the algebraic sum of these projections is A'B' - B'C + CD', since cos^S is negative. But this sum is A'D', which is also equal to AD cos 8, or the projection of AD on OX. Hence AB cos a -{- BC cos (3 + CD cos y = AD cos 8. The same will evidently be true if we take any number of lines between A and D. Hence, if two given pioiMs are joined by a broken line, the algebraic stan of the projections of its parts upon any given straight line is equal to the projection on the same line of the straight line joining the two given points. 198 ANALYTIC GEOMETRY. SECTION XV. — THE POINT. 135. Position of a point in space. The position of any point in space may be determined by referring it to three fixed planes meeting in a point. Thus, if XOY, YOZ, ZOX^ be three planes meeting at 0. and intersecting each other ni Fig. 103. the lines OX, F, OZ, the position of P, relativeh^ to tliese planes, will be known when its distances PQ, PR, PS, from each, measured parallel to the other two, are known. The three l)laues are called the Coor- dinate Planes, their three lines of intersection the Coordinate Axes, their common point the Origin, and the dis- tances PQ, PR, PS, the Coordinates of P. If the planes, and consequently the axes, are at right angles to each other, the coordi- nates are said to be rectan- gular ; otherwise they are oblique. Use will be made only of rectanoular coor- dinates, and they will there- fore be in all cases the perpendicular rh'stances of the point from the coor- dinate planes. It is customary to assume the axes as in the figure, the plane XOF being horizontal and the axis OZ vertical. For brevity, Fig. 104. THE POINT. 199 the coordinate planes will be referred to as the planes XY, YZ, and ZX, and the coordinate axes as the axes of X, Y, and Z. The coordinates PQ, PR, P*S, are represented by the letters X, y, z, corresponding to the axes to which they are parallel. The coordinate i)lanes divide space into eight right triedral angles which are numbered as follows : the Jirst lies above XY, to the right of YZ, and in front of ZX; the second to the left of the first ; the third behind the second ; the fourth behind the first; the Jlfth, sixth, seventh, and eighth, lying under the first, second, third, and fourth, in order. If we extend to Z the convention of signs already adopted for X and F, the positive direction of Z being upward, it is evident that while the absolute values of x, y, 2, may be the same for dilferent points, their signs will determine in which of the eight angles any given point lies, and that a point will thus be completely determined when its coordinates are given in magnitude and sign. 136. Equation of a point. The position of a point may be designated by the equations a^ = a, y=b, z= c ; or b}' the nota- tion (a, b, c), the coordinates being written in the order x, ?/, z. To construct the point {x, y, z), construct the point (x, y), S of Fig. 104, in the plane XY, and at S erect the perpendicular SP=z. Examples. 1. In what angle is the point («, —b, — c) ? 2. Write the coordinates of a point in each of the eight angles. 3. In what plane is the point (o, b, 0) ? 4. "Write the coordinates of a point in each of the three coor- dinate planes. 5. To what plane is the point (x, y, c) restricted? Ans. To a plane parcdlel to XY, at a distance cfrom it. 6. What are the coordinates of points in a plane parallel to YZ at a distance a from it? 200 ANALYTIC GEOMETRY. 7. Where is the point (x, —b, z)? 8. On what axis is the point (a, 0, 0) ? 9. Write the coordinates of a point on each of the three coordinate axes. 10. What are the coordinates of the origin? 137. Distance between two given points. Let x\ y', z', be the coordinates of P' ; x", y", z", those of P". Then But Hence P'P" = VP'IP + KP"^. P'lP = P'H- + HK'. P'P" = ^'P'H' + HIP + KP'"' Now Similarly, Hence, if P'H =2^^' = "^'- ~~ ^P =^" — ^'• HK= y" - y', KP" = z" - z'. P'P" = D, '\2 D = V{x"-x'y + {y" - y'y+ {z" -z'J Cor. 1. If one of the points, as P", is at the origin, x" = y" = z" = 0. Heuce the distance of a point {x\ y', z') from the origin is D = ^x'-' + y'- + z'\ THE POINT. 201 Cor. 2. If z' = 0, that is, if P' is at p in the plane XY, D = Vx'^ + y'- is the distance of j) from the origin = distance of P' from the axis of Z. Hence the distances of {x', y', z') from the axes of X, Y, Z, are y^TqT^^ V.^?M^, V^+F, respectively. 138. Polar coordinates of a point. The position of a point in space, may also be determined by polar coordmates. For this purpose assume any fixed plane, as XY (Fig. 104), and any fixed line in that plane, as OX, being the pole. Then the position of P will be determined when we know OP, its distance from the pole ; the angle SOP, which OP makes with the plane XY; and the angle /SOX which the line SO makes with X. OP is called the radius vector of P and is represented by r, OS is the projection (Art. 134,4) of OP on XY, the angle ^SOP being represented by <^, and ZOS by 6. The point Pmay then be designated as the point (;-, 6, (jy) , $ and cji determining its direction, and r its distance, from the pole 0. 139. Relations between polar and rectangular coordinates. In Fig. 104, we have, X = OL = OS cos 6 = OP cos ^ • cos = r cos ^ cos 6, ( 1 ) y= LS = OSsmO = OP cos <^. sin^ = r cos <^ sin ^, (2) 2 = P>S' = OP sine/) = 'r sin (/>; (3) the rectangular in terms of the polar coordinates. From Art. 137, Cor. 1, we have op=r = V.T- + r + 2'; (4) from the triangle OLS. SL y .rx tan = Trr^ = ■- ; V.^ J OL X 202 ANALYTIC GEOMETRY. from the triangle SOP, tan^: PS ^ z OS Vx-N^^ (6) the polar in terms of the rectangular coordinates. 140. Direction angles and cosines. The angles made by any straight line ivith the axes are called its direction-angles. Since parallel lines make equal angles with the axes, draw OP, parallel to the given line, through the origin. Then LOP = a, 3I0P = (3, NOP= y, are the direction- angles of OP, or of any parallel to OP, and are always esti- mated from the positive directions of the axes. The cosines of the direction-angles are called the direction-cosines. 141. The sum of the squares of the direction-cosines of any straight line is unity. Join P, Fig. lOG, with L. Then, since the plane PSL is perpendicular to OX, the triangle PLO is right-angled at L, and OL = x = OP cos L OP = r cos a ; similarly, drawing PM and PN, OM =y=OP cos MOP = r cos 13, ON = z=OP cos NOP = r cosy. THE POINT. 203 Squaring aud adding, OL^ + OM- + ON^ = r (cos^a + cos-/3 -f- cos^y) . But the first member is r (Art. 137, Cor. 1). Hence COS^a + COS-^ + COS-y = 1. Examples. 1. Two of the direction-angles of a straight line are 60° and 45°. Show that the third is 60°. 2. Find the distances of (4, -7, 4) from the axes, and from the origin. 3. Find the distance of (4, -2, -1) from (6, 3, 2). 142. To find the angle betiveen two straight lines ivhose direc- tion-cosines are given. Let OP, OQ, be parallels to the given lines through the origin, and let a, /3, y, and a', (3\ y', be the angles made by OP and OQ, respectively, with the axes of X, F, and Z. These angles T^h—x Fig. ]07. are, then, the direction-angles of the given lines. Let QOP= 0, and x,y, z, be the coordinates of Q. Then (Art. 134), the projection of OQ upon OP is equal to the algebraic sum of the projections of OL, LS and SQ, upon OP. But the projection of Oi. on OP is (Art. 133) 0Z( COSa= OQ COSa'-COSa. 204 ANALYTIC GEOMETRY. Similarly, the projections of LS and SQ on OP are LS cos (3= OQ cos;8'-cosj8, SQ cosy = OQ cosy' • cosy. The projection of OQ on OP is OQ cos 0. Hence 0Qcos6 = 0Q cosa' cosa+ OQ cos/5' cos/3+OQcosy' cosy, or C0S^= cos a' cos a + cos y8 cos/?' + cosy cosy', or the cosine of the angle inclnded between any two straight lines is the sum of the rectangles of their corresponding direction-cosines. THE PLANE. 205 SECTION XVI.— THE PLANE. 143. General equation of a surface. We have seen that any point (x, y, z) may be constructed by first locating the point (.x, y) in the plane XF", and then laying off, on a perpendicular through this point to XI^, the distance z. Hence, if z be made equal to any constant, as a, X and y remaining variables, the point (.r, y, a) will lie in a plane parallel to XY. If, therefore, f{x,y,z) = (1) be any equation between x-, y, and 2, and in this equation 2 = a, a constant, then /(a;, ?/, a) = 0, being an equation between two variables, will represent a line, all of whose points are in a plane parallel to XY at a distance a from it. Similarly if z = b, f{x, y, h) — will be the equation of a line in a plane parallel to XY^ at a distance b from it. Giving thus, successively, to z, all possible values, i.e., letting z vary continuously between the limits assigned b}' the equation f{x, y, z) = 0, we obtain a series of lines, all of which are plane curves parallel to XY", which, taken together, form a sur- face of which /(a.*, y, z) = is the equation. Hence f{x, y, z) = is the equa- tion of a surface. To illustrate: let P be any point in space subject to the condition that the sum of the squares of its coor- dinates is a constant, or X^+f+z'^R'. (2) - Fig. 108. 200 ANALYTIC GEOMETKY. Since this sum is the square of the distance of P from the origin (Art. 137), it is evident that P is restricted to the sur- face of a sphere whose radius is R and wliose centre is at the origin, and of which (2) is the equation. If we assume 2 = 0, then ar+i/- = J?- is the equation of a circle in the plane Xy, whose radius is that of the sphere ; that is, it is the great circle cut from the sphere by XY. If we make z = a, we have oi?-\-'if= Pr—a^, which is also the equation of a circle, namely, that cut from the sphere by a plane parallel to XY at a dis- tance from it equal to a. As a increases, the radius of the circle, V-B^— or-, diminishes, and when z is made equal to a = P, we have a^+ y^ = 0, or x= 0, ?/ = 0, the plane then touching the sphere at its highest point (0, 0, P). z cannot be made greater than P, for then x- -\- y~ — P- ~ a^ would be impossible, since the sum of two squares cannot be negative, showing that no plane at a greater distance from XY than P can cut the surface of the sphere. The lines cut from an\' surface by a plane are called sections of the surface. If x were made constant in (2), then would be the section cut by a plane parallel to YZ from the surface of the sphere ; and if y were made constant, we should have the section made by a plane parallel to ZX, all of which would in this case be circles. And, in general, if in the equa- tion of any surface, f{^i Vi ^) = 0, one of the variables he made constant, the resulting equation is that of the line cut from the surface by a plane j)araUel to the jilane of the other two axes and at a distance from it equal to the value assigned. 144. Equation of a plane. If, when either .r, y, or z is made constant in the equation f(x,y,z)=0, the resulting equation is of the first degree between the two remaining variables, every section of the surface by planes parallel to the coordinate planes is a straight line, and the surface must be a THE PLANE. 207 plane. But this can be the case only when/(ic, ?/, z) = is of the first degree with respect to all three of the variables. Hence, every equation of the form Ax + By+Cz + F=0 (1) is the equation of a plane. 145. Intercept form of the equation of a plane. If in the equation of a plane, Ax + By+Cz + F=Q, F (1) we make y = z = 0, we obtain x = OQ = — ^, the intercept of F the plane on X. Making z = x — 0, we have y=zOR= F B' the intercept on Y; and for .i- = ?/ = 0. z = OS = — ^, the in- tercept on Z. Representing these intercepts by a, b, and c, respectively, we have F , F F c whence A = -^,B = -^,C = a b F — — • c Substituting these values in (1), we have B a b c (2) -^ Fig. 109. the equation of a plane in terms of its intercepts. This form is not applicable when the plane passes through the origin ; for in this case, since the origin is a point of the plane, (0, 0, 0) must satisfy its equation, and from (1), F=0, and a = b = c = Q. To put the equation of a plane in the intercept form, trans- pose the absolute term to the second member, and, by division, make the second member positive unity. Thus, the intercept 208 ANALYTIC GEOMETRY. 0' z form of 3x — Gy + 2z — 6 = is '— ?/+=!, the intercepts being 2, —1, 3. To write the equation of a plane whose intercepts are given, substitute their values in (2). Thus, the equation of the plane whose intercepts are 4, —3, —6, is --^--=1, or 3.r-4w-2z-12 = 0. 4 3 6' ^ Examples. 1. Write the equation of a plane whose inter- cepts are 2, G, 4; also —2, —3. 1. Alls. 6x + 2y + Sz-12=0; Sx + 2y - Qz-\- 6 = 0. 2. Put the following equations under the intercept form : OX + oy — z + 15 = 0, X — y — z— 1 = 0. 3. Determine the intercepts of the planes of Ex. 2, without putting the equations under the intercept form. 146. Normal equation of the plane. Let QRS, Fig. 109, be any plane ; OD a perpendicular upon it from the origin, its length being p, and a, ^, y, its direction-cosines. Let P be any point of the plane, .t, y, z, being its coordinates. Then the pro- jection on OD of OP is equal to the sum of the projections of ON, NM, and 3fP, on OD (Art. 134). But, whatever the position of P in the plane, the projection of OP on OD is j3, since OD is perpendicular to the plane ; and the projections of ON, NM, MP, are a; cos a, ycosf3, 2 cosy (Art. 133). Hence ajcosa-j- ?/cos/3 + 2: cosy =p (1) is the normal equation of the plane, in which }'> is always posi- tive. Since the sum of the squares of the direction-cosines of any line is unity, to put the equation of a plane under the nor- mal form, we must introduce a factor B fulfilling the condition (BAY + (BB) ' + {BC)-=l, in which A, B, C, are the coefficients of x, y, and z, in the given equation ; hence -J A' + i3= + C THE PLANE. 209 Thus, the normal form of ox + 2y — z+\ =0 is Sx 2y 2 _ 1 Vl4 Vl4 Vl4 Vl4 the second member being made positive ; in which 3 2 1 -? Vl4 Vl4 Vl4 are the direction-cosines, and p = — — = distance of the plane from the origin. V 14 Examples. 1. Put the equations ox-\- 5y — z-\-15 = 0, x — y — z—\— 0, under the normal form. 2. Find the distance of the plane x — y-\-z — l = from the origin. In what angle is the perpendicular from the origin on the plane . Aus. — - ; in the fourth angle. V3 3. Show that 4.t + 7?/ + 4^ — 9 = is at a distance nnity from the origin. 4. Write the equation of a plane whose distance from the origin is 10, the direction-cosines of the perpendicular being 1 1 V23 ,- -' -' — r— Ans. 33; + 2?/ + V232-G0 = 0. 5. Are the direction-cosines of Ex. 4 chosen at random? 6. Write the equation of a plane parallel to YZ at a distance from YZ equal to 6. Since the plane is parallel to YZ, its intercepts on Y and Z are both infinity. Hence, from Eq. 2, Art. 113, b—co, c = aD, and x := a = 6. Or, from Eq. 1, Art. 114, a = 0, ^ = -y = 90° ; hence x = p = 6. 7. Write the equation of a plane parallel to X. 8. What are the equations of the coordinate planes? 147. To write the equation of a plane through three given points. Assuming the general equation Ax-\-By-\-Cz + F=:0, (1) 210 ANALYTIC GEOMETllY. dividiDg by any one of the four constants, as F^ and denoting the resulting coefficients by A', B', C, we obtain A'x + B'y+C'z+\=0. (2) Substituting in this equation the coordinates of the three given points in succession, there results three equations between A', B', and C", from which the values of these latter may be deter- mined. Substituting these values in (2), we have the equation required. Examples. Write the equations of the planes through the following points. (1,0,-2), (3,2,-1), (5,-1,2). .4ns. 9x-4y-10z-20 = 0. (1,2,3), (4,5,6), (-7,8,9). Ans. y-z-{-l = 0. (0,0,0), (1,2,4), {1,-2,6). Ans. Wx-y-2z = 0. (2,0,0), (0,2,0), (0,0,2). Ans. x + y + z-2=0. (0,1,2), (0,2,4), (1,0,2). A71S. 2x + 2y-z=0. 148. To find the angle betiveen tico given planes. The angle between the planes is the same as that between the perpendiculars upon the planes from the origin. Hence, if a, yS, y, and a', /3', y', are the direction-angles of the perpendicu- lars, and 6 the angle between the latter, the required angle is given by the relation (Art. 142) cos 6 = cos a cos a' + COS /? COS P' -f- cos y cos y'. ( 1 ) If the equations of the planes are given in the normal form, we have only to substitute in (1) the coefficients of the variables, they being the direction-cosines (Art. 146). If the equations are in the general form Ax -f By + C^ + F= 0, A'x + B'y + C"z + F' = 0, their normal forms will be A x + By + Cz ^ F VA^B^+C' y/A^+B'-{-C'' A'x + B'v + C'z F' VA"+ B'- -t- C '- VA'' + B''' + C' THE PLANE. 211 in which the radicals have the opposite signs of the absolute terms in order to make the second members positive ; and the direction-cosines are, respectively, ABC — ? ■VA'+B'-\-C' -JA' + B'+G^ ^A' + B' + C A' B' C Substituting these values in ( 1 ) , (2) . AA' + BB' + CC ,., cos6'= — — • (3) VyP + B'-\-C' VA'- + B"+C" Cor. 1. If the given planes are perpendicular to each other, $ = 90°, cos^ = ; hence the condition of perpendicularity is AA' + BB' -\-CC' = 0, (4) or the S}ims of the rectangles of the coefficients of the correspond- ing coordinates in the equations of the planes ninst he zero. Cor. 2. If the planes are parallel, the direction-cosines of their perpendiculars with respect to each axis must be equal. Hence, from (2) A' B' C' ^ ^ smce each ratio is equal to- ^A'-+B''+C" Hence the condition of parallelism is that the ratios of the coef- ficients of the corresponding coordinates in the equations of the planes must be equal. Examples. 1. Find the angle between the planes x -\-2y — 22 + 1 = and 307 -j-Gt/- 62-5 = 0. Ans. 0°. 2. Find the angle between the planes 2x-\-2y + z + 1 = 0, Ax — Ay + 7z-l = 0. Ans. 6 = cos~^^. 3. Show that 3/^a; + ?/- ^^ + 1 = and 12 (.r-j-y) -3^+10 = are parallel. 212 ANALYTIC GEOMETRY. 4. Show that xj-2y — 2z-{-l = Ois perpendicular to2x + 5y -f 62 — 11=0 ; also x-\-2y-\-3z + l=0 to 3x-\-6y— oz —o = 0. 5. Write the equation of a plane parallel to 3.^ + 4?/ — 2 -f- 6 = 0. 6. Write the equation of a plane perpendicular to 3.T + 4y+ z-l = 0. 7. Find the distance between the parallel planes x + 2y — 2z + l = 0, 5x + 6y-6z- 25 = 0. Ans. 2^. 8. Prove that Ax-hBy + Cz-hF-j-k{A'x+B'y+C'z -{-F') = is the equation of a plane through the intersection of the planes Ax + By-\-Cz + F=0 and A'x + B'y + C'z + F'=0. See Art. 37. 9. Write the equation of any plane through the intersection ot 2x + 5y-\-z—l=0 and x — y + z + 2 = 0. 10. Explain how to determine A; in Ex. 8 so that the plane shall pass through a given point. See Art. 37. 11. Write the equation of a plane through the intersection of 2x + y — z-\-l = and oa;4-4?/ + 22; + G = 0, passing also through the point (1, 1, 2). Ans. 28iB + 9^ — 212 + 5 = 0. 12. Write the equation of a plane through the intersection of the planes of Ex, 11, and also passing through the origin. Ans. 9x + 2y-8z = 0. 13. Prove that the distance from the point (x', ?/', z') to the plane a; cos a + ycosf3 + 2:cosy =p, is x' cos a + y' cos;8 + z' cos y — p, Ax'-{-Bx'-{- Cz' ~F c A ^ QQ or ' — See Art. 38. VA' -hB'+C 14. Find the distances from the plane 5x-i-2y — lz-\-d = of the points (1, - 1, 3) and (3, 3, 3). 15. Find tlie equation of a plane through (1, 10, —2), par- allel to the plane 2x-{-y — z-\-6 = 0. Ans.2x-{-y — z — l4:=0. THE PLANE. 213 16. Fiud the equation of a plane through (1,-1,3) per- pendicular to the plane 2a; + ?/ — z + 6 = 0. 17. Find the distance from (8, 14,8) to 4x+7y + 4z — 18 = 0. Ans. 16. 149. Traces of a plane. If in the equation of a plane. Ax + By -{-Cz -{- F= 0^ we make z = 0, the resulting equation Ax + By + F==0 (1) applies to all points of the plane in XY, and is therefore the equation of EQ (Fig. 109), the intersection of the plane with XY. For like reasons B>/ + Cz + F=0, Ax-^Cz + F=0, are the equations of the intersections ES and SQ. These inter- sections are called the traces of the plane. Solving (1) for y, we have A F y = x , \ B B and the corresponding trace for any other plane would be .1' F' y — X . •^ B' B' If the traces are parallel, — = — . or — = — If the corre- ^ B B' A' B' sponding traces on the other coordinate planes are also parallel, in which case the planes themselves are parallel, we obtain in like manner A' B' C' the condition already found. 214 ANALYTIC GEOMETRY, SECTION XVII.— THE STRAIGHT LINE, 150. Equations of the straight line. Assuming the equation of a plane in the intercept form, - + 7 + -=l, a c if we impose the condition that the plane shall be perpendicular to XZ, its ^-intercept b will be infinity, and its equation assumes the form Ax + Cz-i-F=0, (1) whatever the value of y. Hence every equation of the first degree between two variables represents a plane perpendicular to the corresponding coordinate plane, the third variable being indeterminate. Therefore X being indeterminate, represents a plane perpendicular to YZ. Let ABDL be the plane represented by (1), and AHDC that represented by (2). V^alues of x, i/, z, which satisfy both (1) and (2) locate a point in both planes, that is, on AD, their line of intersection. Hence, while taken separately (1) and (2) are equations of planes perpendicular respectively to XZ and YZ, if taken together they represent a straight line in space. Thus, let.T])e the independent variable, and an}' value asx = 03f be substituted in (1). From (1) we may then find z = MS and locate the point S in XZ. Now so long as (1) is considered independent of (2), it represents the plane LABD. and the value of y may be assumed at pleasure. But if (1) and (2) are simultaneous, y must be derived from (2) after the value z = MS has been substituted in it. Since this value of y satis- THE STRAIGHT LINE. 215 fies (2) , y = SP must locate a point P iu the plane AIIDC, and P must lie on the intersection AD. jj Fi2. 110. The Vine AD is evidently completely determined by (1) and (2), since the planes can intersect in but one straight line. Since (1) is true for all values of y, it is true for y= 0, and is then the equation of the trace AB o^ABDL on XZ. Hence (1) is the equation of the plane ABDL, or of its trace AB, accord- ing as ,'/ = ^ or y=0. Similarly (2) is the equation of the plane AHDC, or of its trace AC, according as x= -^ or x = 0. But AB and AC are the projections of AD on XZ and YZ, since the planes ABDL and AHDC are perpendicular res[)ectively to XZ and YZ. Hence the straight line AD is determined when its projections on any two coordinate planes are given. Eliminating z between (1) and (2), we have an equation of the form A"x + B"y + F"^0, (.8) which, in like manner, is the equation of a plane perpendicular to XY, or of its trace on XY, according as we regard z = ^ or z = 0. This plane is evidently the plane AOD, passing through 216 ANALYTIC GEOMETEY. the intersection AD of (1) and (2), since (1), (2), and (3) are simultaneous ; and its trace is OD, the projection of AD on XY. Hence, in general, if we assume any two equations of the first degree between two variables, as f(x,z) = 0, /'(y, 2)=0, and eliminate the common variable, obtaining the third equation, f"{x,y)=0, these three equations may ])e regarded as the projections on the coordinate planes of a straight line in space, any two of them being sufficient to determine the line. 151. Equations of a straight line through a given point haoing a given direction. Let (x', ?/', z') be the given point, P', and a, /?, y, the direction-angles of the given line. Let P be any other point (x, y, z) of the line, and denote P'P by r. Then the projections of P'P on the axes are (Art. 133) x — x'=reosa, ?/ — y'= rcos/S, z — z'—rcosy, from which ^~^' = y~y' = ^^ , ( 1 ) cos a cos/3 cosy which are the required equations, any two being sufficient to determine the line. Eq. (1 ) is called the symmetrical form. 152. To put the equations of a straight line under the sym- metrical form. » The symmetrical form being X — x' _y — y' _z — z' cos a cos/3 cosy' the condition that the equations of a straight line are in this form is that the sum of the squares of the denominators is unity. Let X — x' _ y — ?/' _z — z' L ~~M~~ N THE STRAIGHT LINE. 217 be the given equations. Dividing tlie denominators by Vi/' + M- +N-, we have •r — .>■' II — y' z — z' L " M N in wliich tlie sum of the squares of the denominators = 1. Thus, let 3a; — 22 + 1 = 0, ix — y = 0, be the given line. Then X _2z — I _y 1~ 3~"~4' Dividing the denominators by Vl"+3-+ 4^ = V26, a; 2 z — 1 ?/ the direction-cosines beins; V26 V26 V26 1 3 4 "^ V2q' V26' V26 Examples. 1. Find the equations of the intersection of x — y + z — 2 = 0^ and a; + ^ + 2 2; — 1 = 0, and determine the position of the line. Elimmating >/ and z in succession, we have 2.T+3c-3 = 0, .r-3_y-3 = 0, X 2 — 1 '/ + 1 or - = = iLJ— , 1-2 1 3 3 whence the line passes through (0, —1, 1), and its direction-cosines are 3 2_ 1 Vu Vli' vli 2. Find the intersection of a; -f-^ — 2 + 1=0 and 4a; + y — 22 + 2 = 0. Ans. A line through (0, 0, 1), ichose direction-cosines are 1 2 3 vn' vTi' vii 218 ANALYTIC GEOMETKY. 3. Determine the position of x= 4z + 3, y = 3z — 2. Ans. Aline through (3, —2, 0), irhose direction-cosines are 4 3 1 V26' V2g' V26 4. "Write the equation of a line through (1, 2, —6), having |, 1, |, for du'ection-cosines. Ans. x-2y + 3 = 0, 2y-z-10 = 0. 5. Write the equation of a line through (1, 4, —3) parallel to Z. Ans. a; = 1 , y — 4:. 153. Equations of a straight line through two given points. Let {x', ?/', 2;'), (a;", t/", 2;"), be the given points. The equa- tions of a straight line through {x', y', z') are v' x — x' ^ y — y' ^ z — z .^. cos a cos/? COSy Since the point (.t", ?/", z") is also on the line, its coordinates must satisfj" (1) ; hence x" - x' _ y" - y' _ z"-z ' cos a COS^ COSy Dividing (1) by (2), member by member, (2) X- -x' = V- 2/" -If -y = z- z" -z' 0;" -x' -2' (3) which are the required equations, any two of which determine the line. Examples. 1. Write the equations of the straight line passing through (1, 2, 4), (—3, 6, — 1). . x —\ y — 2 2 — 4 -4 4 -5 2. Find the direction of the line of Ex. 1. 3. Find the points in which the lino of Ex. 1 pierces the co- ordinate planes. Ans. The line pierces XY in {— ^., ^) . THE STRAIGHT LINE. 219 4. Write the equations of lines through the following points, and find their directions : (2,1, -1), (-3, -1, 1); (6, 2, 4),(-G, -3, 1). 5. A line passes through (1, 1, 2) and the origin; find its equations. 6. A line passes through (1, 6, 3,) (1, - 6, 2). Find the equations of its projections on the coordinate planes. 154. To find the angle behceen two given straight lines. Let X — x' _ y — y' __z — z' N' x — x"_ y — y" _z — z" L" M" N" ' be the given lines. The angle between the two lines is given by the relation (Art. 142) COS^ = COS a' cos a" + COS y8' COS /3"+ COS y' COS y", in which a', (3', y', and a", /3", y", are their direction-angles. But (Art. 152), cos a' = COS/3' = cosy' = M' VL" + 3/'- + N^' N' (1) cos a" = cos/3": cosy = L" ' I^ M" N" ^L"- + M"-' + A^"2' J (2) 220 ANALYTIC GEOMETllY. XT ^ L'V'+M'M'^+N'N" ... Hence cos^ = — r===== • (3) Vi'' + M'- + N'-' VX"- + M"'- + N"' Cor. 1. If the lines are parallel, the corresponding direc- tion-cosines are equal, each to each ; hence from (1) and (2), L' ^ M' ^ N' L" M" N" are the conditions of parallelism. CoR. 2. If the lines are perpendicular to each other, ^=90°, cos^ = 0; hence L'L"+ M'M"+ N'N" = is the condition of perpendicularity . Examples. 1. Find the equations of the sides of the tri- angle whose vertices are (1, 2, 3), (3, 2, 1), (2, 3, 1), and the angles of the triangle. '^'''' I ^=2J' .= ir 2x + z = hl C' 2' 3" 2. Find the angle between ?/ = 5.r-f 3, z = ox-\-b^ and y = 2x-\-\, z = x. 3. Show that 4a; -3t/- 10 = 0, ?/ + 4^; + 26 = 0, and 7.c-2?/ + 26 = 0, 34^ + 7^-90=0, are perpendicular to each other. 4. Show that x=-2z + \, ?/ = 32! + 4, and a: = ;3-22;, y = z — 2, are perpendicular to each other. 5. Show that 2.X' - // + 1 = 0, 3 ?/ - 2 2 + = 0, and 2a; — ?/ — 7 = 0, Sr/— 22 + 7 = 0, are parallel. THE STRAIGHT LINE. 221 6. Find the conditions that the straight Hne x — x'_y — y'_z—z' L ~ M ~ N is parallel or perpendicular to the plane Ax + By+Cz + F=0. The line is parallel to the plane when it is perpendicular to the perpen- dicular on the plane. But the direction-cosines of the perpendicular are proportional to A, B, G (Art. 146), and the direction-cosines of the line are proportional to L, M, N (Art. 152). Hence the condition of parallel- ism is (Art. 154) ^^^ ^ ^^^^ ^^ ^ The line is perpendicular to the plane when it is parallel to the perpen- dicular on the plane; hence, the condition of perpendicularity is ABC 7. Find the equation of a line through ( — 2, 3, 5) perpen- dicular to 2x + 8y — z — 4: = 0. Ans. a; + 22-8 = 0, y + Sz-43 = 0. 8. Show that 2x — y = 0, Sy — 2z = is perpendicular to a; + 2?/ + 32-6=0. 9. Show that 2 = 3, x-\-y = o is parallel to .^'-f2/.+ 2-6 = 0, and to the trace of the latter on XY. CHAPTER YI. SURFACES OF REVOLUTION, CONIC SECTIONS, AND HELIX. -OOXKOO- SECTION XVIII. — SURFACES OF REVOLUTION. 155. Defs. A line is said to be revolved about a straight line as an axis when every point of the line describes a circle whose centre is in the axis and ivhose plane is perpendicular to the axis. The moving line is called the generator, and the surface which it generates a surface of revolution. It follows from the definition of revolution that every plane section of a surface of revolution perpendicular to the axis is a circle, and that every plane section through the axis is the generator in some one of its positions. A plane through the axis is called a meridian plane, and the section cut by such a plane is called a meridian. 156. General equation of a surface of revolution. Let the axis of Z be the axis of revolution, the generator a plane curve whose initial posi- tion is in the plane XZ. and P any point of the generator. Since the generator is in the plane XZ, its equation will be X =f{z), but as it revolves about Z, the .T-coordinate of any point FiR. 111. 'V SUEFACES OF REVOLUTION. 223 as P will differ from that of its initial position. Hence, to dis- tinguish the cc-coordinate of the surface from that of the gene- rator in its initial position, represent the latter by r ; then the equation of the generator will he r=f{^). (1) But P remains at the same distance, r, from Z dm'ing the revolution; hence (Art. 137) r^ = .r+r. (2) Substituting in (2) the value of r from (1), ^•^ + ?r = [/(^)T, (3) is the general equation of a surface of revolution. In any par- ticular case substitute in (3) f{z) from the equation of the generator. 157. The sphere. If a circle be revolved about any one of its diameters the surface generated will be a sphere. Let the diameter coincide with Z and the centre with the origin. Then the equation of the generator will be whence 1^ = \_f{'z)f = R- — z-. Substituting this in the general equation x^ -\-y'= [/(^)]'i we have o? + f + z' = R\ which is the required equation. 158. The prolate spheroid, or ellipsoid. This is the surface generated by the revolution of an ellipse about the trans- verse axis. Let the transverse axis coincide with Z and the centre with the origin. Then the equation of the generator is a-r + h-z- = a^h\ whence t''-= -^a'-z') = [f{z)Y. a- 224 ANALYTIC GEOMETRY. Substituting this in ar + ?/- = \^f(z)y, we have cr(^-' + /)+&V = a^Z>% or f^ + t+^[=i, (i) (r b- a- If a^ = b^= Br, the ellipsoid becomes a sphere. B}' definition, plane sections parallel to XY are circles. Let the student prove that plane sections parallel to XZ and YZ are ellipses. 159. The oblate spheroid, or ellipsoid. Tliis is the surface generated by the revolution of the ellipse about the conjugate axis. Let the conjugate axis coincide with Z and the centre with the origin. Then the equation of the generator is whence r' = -' {b' - z') = [/(2)]S and &2(.^- + 2/^)4-aV = a^6^ or t + y'^ + t^l, a- a- b- which is the required equation. If or = br = R-, the ellipsoid becomes a sphere. Let the student determine the plane sections parallel to the coordinate planes. 160. The paraboloid. This is the surface generated by the revolution of a parabola about its axis. Let the vertex of the parabola be at the origin, the axis coinciding with Z. Then the equation of the generator is ,-'=2p2=[/(^)P, and the required equation is x^ + f=2pz. (1) Let the student show that plane sections parallel to YZ and XZ are parabolas. 161. The hyperboloid of two nappes. If an hyperbola be resolved about its transverse axis, the surface generated is SUKFACES OF KEVOLUTION. 225 called the hyperboloid of two nappes. With the centre at the origin and the transverse axis coincident with Z, the equation of the generator is a-r- — b-z- — — a-b^, whence a-(ar + //-) — b'-z- = — a^&^, (1) Q O 9 X- v~ z- or - + !,--,, = - 1, 0- b- a- is the required equation. Let the student determine the plane sections parallel to the coordinate planes. 162. Hyperboloid of one nappe. This is the surface gener- ated by the revolution of an hyperbola about its conjugate axis. Assuming the centre at the origin and conjugate axis coinci- dent with Z, the equation of the generator is a- z- — b- v = — a" O", and that of the surface is b-{x- + f-)-cez'' = a'b\ (1) 099 XT V- Z- or — + ^^,- =1. a- a- b- Let the student determine the sections. 163. Cylinder of revolution. If a straight line revolve about another to which it is parallel, it will generate the surface of a circular cylinder. Let Z be the axis and r = R the equation of the generator parallel to Z in the plane XZ. Then r^R=f{z), and x'-\-y' = Ii' (1) is the required equation, z being indeterminate. Let the student show that sections parallel to z are two par- allel straight lines, or one straight line, elements of the cylinder. 226 ANALYTIC GEOMETRY. 164. Cone of revolution. If a straight line revolves about auotber straight line which it intersects, the surface generated is that of a cone. Any position of the generator is called an element of the cone. Let AB be the generator, and Z the axis of revolution. The cone will be a right cone whose vertex is A, OA = h being the altitude and OB = R the radius of the base in the plane X^. The coordinates of A and B are (0, h). (R, 0), and the equation h of the generator z = R + h, whence R' Fig. 112, h- ,-' = [/(.)]^^ = ^(/i_z)^ (1) and the equation of the surface is (2) or if ^ = angle which the generator makes with X= angle made by the elements of the cone with the plane of the base, (x' +y') tan' e={h-zy. (3) If the vertex A is at an infinite distance, the cone becomes the cylinder. In this case /i = x, and from (1) uw= '^,_2zR- z-R^ h h- = R\ and we obtain the equation of the cyUnder .^" -\-y^ = R\ as before. Let the student prove that every plane section parallel to Z is an hyperbola. THE CONIC SECTIONS. 227 SECTION XIX.— THE CONIC SECTIONS. 165. General equation. Let any plane (Fig. 112) be passed through the axis of 1", cutting the section LPN from the surface of the cone and the line LN from the plaue ZX; and let XON=^ <^, the inclination of the plane to XY. Since the cut- ting plaue is perpendicular to ZX, its equation will be that of its trace LN, or z = tan (f> • X. To refer the curve of intersection LPN to axes in its own plane, let OF be the axis of Y, and OX' = OX produced the new axis of X; then the coordinates of P referred to the primi- tive axes are 0M= x, 3IQ = z, QP = y, and referred to the new axes are x' = OQ, y' = Ql\ Hence y=y\ x=OM=OQ cos(fi=x' cos(j>, z=3IQ=0Q sin<^ = a7'sin<^. If these values, which are true for the point P common to both the plane and the cone, be substituted in Eq. (3) Art. 164, we shall have the equation of the plane section referred to the axes X'OY. Making those substitutions, and omitting the accents, (x- cos- tan- — sin- 0) + 2 hx sin (^ — Jr = 0, or, since siu-<^ = cos-<^ tau-<;i>, 2/2 tan^ e-\-x^ cos' (/> ( tan' ^ - ta n->6, tan-c/) > tan-^, £'-4 AC (Art. 80) is positive, and the section is an h3'perbola. 228 ANALYTIC GEOMETRY. If (f><6, tan^^ ^, the equation takes the form y=± ax\ and represents two straight lines through the origin. If <^ < ^, it is satisfied only for x = 0, ?/ = 0, and represents a point. If <^ = ^, it reduces to y = 0, the equation of X. These are particular cases of the hyperbola, ellipse, and parabola, respectively. If <^ = ; a particular case of 4> <0, the section is a circle ])y definition. If /i=oc, the cone becomes a cylinder. Putting h = cc in 7 2 Eq. (1), after substituting for tan^^ its value -- and dividing through by h^, we have y H- X- cos- (ji = li- ; ie- which is the equation of an ellipse, except when ^ = 0° and (fi = 90°, in which cases the section is a circle, or two paralk'l straight lines, elements of the cylinder. Having tluis given to ^ all possible values from 0° to 90°, and h all possible values from to infinity, we have found every section of the cone except those parallel to the axis of revolu- tion, which latter have been already considered in Arts. 163 and 164. Thus every plane section is seen to be one of the varieties of the conies. THE HELIX. 229 SECTION XX. — THE HELIX. 166. Defs. If a rectangular sheet of paper be rolled up into a right cylinder with a circular base, any straight line drawn on the paper, not parallel to its sides, will become a curve called the helix. Or it may be defined as the curve assumed by the hypothenuse of a right-angled triangle whose base is tangent to the base of the cylinder and whose plane is perpendicular to the radius of the base through the point of contact, when the triangle is wrapped around the cylinder. The helix forms the edge of the common screw. It follows from the definition that the helix makes a constant angle with the elements of the cylin- der ; namely, the acute angle at the base of the triangle. 167. Equations of the helix. Let the axis of the cylinder coincide with Z, OA = li = radius of its base in the plane XY, P being any point of the helix, a = constant angle at the base of the triangle, the vertex of this angle being assumed on the axis of X at A, and <^ = ylOQ = angle made by the projection of the radius vector OP on X5^with X. Then X = 0M= OQ cos cfi = R cos 4>^ y = ^tQ — OQ sin <^ = P sin <^, z = QP = base of triangle x tan a = QA - tana = P<^ tana. Hence, if k = tana, the equations of the helix are a; = Pcos<^, ?/ = Psin(^, z='kR<^. '» '/n^ o