f¥ 4- LIBRARY OF THE University of California. X.cP-ijivo^cLLyv Received , igo I . Accession No, 836^*5 • ^^^^ ^^- .^..fe Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofanalytOOhardrich ELEMENTS OF ANALYTIC GEOMETRY. BY AETHUR SHERBURNE HARDY, Ph.D., pBorBssoB OF Mathematics in Dabtmouth College. BOSTON, U.S.A.: ' PUBLISHED BY GINN & COMPANY. 1895. ^ Copyright, 1888, Bt ARTHUR SHERBURNE HARDY. All Rights Reserted. / Typography bx J. S. Cushing & Co., Boston, U.S.A. Prbsswork by Ginn & Co., Boston, U.S.A. tv PART I. PLANE ANALYTIC GEOMETRY. PEEFAOE. Although writing a text-book for the use of beginners following a short course, the tendenc}* 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 liim 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 : 83645 IV PREFACE. First. A thorough understanding 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 the analytic, as distinguished from the geometric, method of proof. The function of numerical examples — that 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 Geometr}' 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. Hanovtir, N.H., Oct. 6, 1888. OONTEI^TS. Part I. — Plane Ai^alytic Geometry. CHAPTER I. — COORDINATE SYSTEMS. Section I. — The Point. The Rectilinear System. ART. PAGE. m. Position of a point in a plane 1 '%r Definitions 2 tS. Construction of a point 2 t 4. Definitions 2 \^. Equations of a point. Examples 3 6. Division of a line. Examples 4 u?; 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 IT. — The Line. The Rectilinear System. 14. Loci, and their equations 10 15. Distinctions between Analytic Geometry, Geometry, and Algebra 11 16. Quantities of Analytic Geometry 12 17. Construction of loci. Examples 13 VI CONTENTS. The Polar System. 18. Polar equations of loci 21 19. Construction of polar equations 22 20. General notation , , 24 Section III. — Relation between the Rectilinear and Polar Systems. 21. Transformation of coordinates 25 Rectilinear I'ransformations. 22. FormulaB for passing from any 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 line through a given point 45 32. Equation of a straight line through two given points. Examples 45 Plane Angles. 33. Angle between two straight lines 47 34. Equation of a line making a given angle with 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 v./ CONTENTS. VU 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 OP 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'ij^cc:^^Dy + Ex^F=0 73 49. Every equation of the form y"^ ■\- x^ ■\- Dy -\- Ex -{■ F = Q the equa- 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 65. Definitions 80 56. Common form of central equation 80 57. Length of focal radii 81 58. Polar equation 81 69. The ratio 82 \; VIU CONTENTS. PAGE. ABT. 60. Geometrical construction of the ellipse „ 83 61. The circle a particular case 85 62. Varieties of the ellipse. Examples 85 The Hyperbola. 63. Definitions 87 64. Central equation 87 65. Definitions 89 66. Common form of central equation 89 67. Length of focal radii 90 08. Polar equation , 91 69. The ratio - . , 92 70. Geometrical construction of the hyperbola 93 71. The equilateral and conjugate hyperbolas 95 72. Varieties of the hyperbola. Examples 95 The Parabola. 73. 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 Ai/'^ + Cx'^ + i>^ + ^x + F=: 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 the chords bisected by that diameter 114 89. Definitions 114 90. Conjugate diameters of ellipse 114 91. Every straight line through the centre of an hyperbola meets the hyperbola or its conjugate 115 CONTENTS. IX ART. PAGB. 92. Definitions 116 93. Conjugate diameters of the hyperbola 116 Construction of Conies from their Equations. 94. By comparison with the general equation. Examples 117 95. By transformation of axes. Examples 119 96. By conjugate diameters. Examples 122 97. Construction where ^ = 0. Examples 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 Examples 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 third 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. Rectangle 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 - X CONTENTS. Parabola referred to Oblique Axes. ^^'^' PAGE. 122. Equation of the parabola ; 157 li^ Property of ordinates I59 Asymptotes. 124. Equation of the hyperbola I59 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. The cissoid. Duplication of the cube 179 4. The lemniscate 181 5. The witch 182 6-8. Examples 183 Section XIII. — Transcendental Curves. 131. 1. The logarithmic curve 185 2. The cycloid 185 3-10. The circular functions 187 11. Spiral of Archimedes 189 12. Keciprocal spiral 190 13. The lituus 1^1 14. Logarithmic spiral 192 CONTENTS. XI Part II. — Solid Analytic Geometry. CHAPTER v.— THE POINT, STRAIGHT LINE, AND PLANE. Section XIV. — Introductory Theorems. ART. PAGE. 132. Definitions. Projections of lines 195 133. Length of the projection of a line on a line and plane . . o . . . . . 196 134. Projection of a broken line 197 Section XV. — The Point. 135. Position of a point 198 136. 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 206 144. 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 XVIL — The Straight Line, 150. Equations of a straight line 214 151. Symmetrical forms , 216 152. Reduction to symmetrical form. Examples 216 153. Equations of a line through two points 218 154. Angle between straight lines. Examples and problems . , 219 XU COKTENTS. CHAPTER VI. — SURFACES OF REVOLUTION, CONIC SECTIONS, AND THE HELIX. Section XYIII. — Surfaces of Revolution. ART. PAQB. 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. The paraboloid 224 161. The hyperboloid 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 CHAPTER I. COORDINATE SYSTEMS, >>«=:c 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 point in a plane. Thus: if X'X, F'F, 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 distances mP^, nP^, are known, these distances being measured parallel to Y'Y and X'X. If, however, only the numerical lengths of mP\ nP^^ are known, the point may occupy any one of four positions, P% P", P"S P^^. This ambiguity will dis- appear if we distinguish distances laid off above and below X'X, parallel to Y' F, 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 any point in the plane of X'X and Y'Y will be com- pletely determined with reference to these lines when its distances from them, measured parallel to them, are given in magnitude and sign. p ml Fig. 1. - ANALYTIC GEOMETKY. Defs. The fixed lines X'X, Y'Y, are called the axes of reference; their intersection, 0, the origin; the distances mP\ vP\ the coordinates of the point P' ; and to distinguish these co- ordinates, TiP^ is called the abscissa, and mP"^ the ordinate of P\ ^0 Construction of a point. Since the coordinates of a point, when given, fix its position with reference to the axes, and since (Fig. 1) Om = nP\ On = mP' , to determine the position of a point whose coordinates are given we have simply to lav off the given abscissa from O along X'X in the direction indicated by its sign, and at its extremity on a parallel to Y'Y, 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 j)oint. (^ The axes of reference are always lettered as in Fig. 1, and hence are often designated as the axes of X and F, the former being usually taken horizontal. For brevity they will frequently be spoken of as X and Y simply-. The abscissa of a point, being always a distance parallel to the axis of X, is always represented by the letter x ; and for a like reason the ordinate is always represented by the letter y ; hereaflfter, there- fore, these letters will always re})resent 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, FOX', X'OF', F'OX, are known as the first, second, third, and fourth, angles, respectively. It is evident that so long as the angle XO Y 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 \r,i ^ wise, an oblique system. As nothing in- general, is gained b}^ 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 X THE rOlNT. point will thus be (Fig. 2) the perpetidkalar distances of the point from the axes. /S, Equations of a point. It is now evident that we maj: designate the position of a point by giving its coordinates in the form of equations. Thus, a;= 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, tlie abscissa being always written 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 ordinates are zero? Whose ordinates are equal and have the same sign?. AVhose 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 b 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 base? (2) at the left hand extremity of the base ? 6. Construct and name the figures whose vertices are (1). (a,a), {a,-a), (-a, —a), {-a,a), (2).. (a, 6), (a, -6), (-a, -6), (-a, &). (3). (a, 6), (a, -6), (_«, -6), (-a,c), (4). (a, 6), (c, d), (-e, d), (-/,&). ANALYTIC GEOMETRY. 6. To find the coordinates of a point which divides the straight line joining two given points in a given ratio. Let P', P", be the given points, x', y', and x", y", their coordinates, and F sl third point {x, ?/), dividing rP" so that P'F : PF" ::m:n. Then, from similar triangles, V P" ^ - P'R RP Tn. (1) P'F m QF" But FB = x-x\FQ = x"-x,FF=y-y', QP"= y"-y. Substituting these vahies, y - y m /] y"-y Fig. 3. X" — X Solving these equations for x and y, we obtain mx"-\-7ix' my"-{-7iy' m -{-n y m (2) For the middle point of P'P", m = n. Hence the coordinates of the middle point o^ -■. line joining two given points are 2 r— » y y"-i-y' (3) If the line is cut externally, we have from (1) and Fig. 4, ? (' x — x' _y — y m Fig. 4. whence x = x-x'^ y-y" mx"—7ix' m J y — my" — ny' m — n m (4) Oblique Axes. The above formulae hold good for oblique axes, since the triangles remain similar whatever the angle XO Y. Examples. 1 Find the coordinates of the middle point of the line joining (5, 3) and (3, 9); also (5, 8) and ( — 5, —8). Ans, (4, 6); (0, 0). THE POINT. 2. Find the coordinates of the middle points of the sides of the triangle whose vertices are (6, 2), (8, 4), (10, 12). Alls. (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). Ans. I — , \ ^ 3 4. The line whose extremities are (2, 4) and (6, —8) is divided in the ratio 3:2. Find the two points of division fiil- 22 16\ /1 8 4^ filling the condition. Arts. 16\ /18 5 ' 6 J'. \d' 6J^ 5. Find the two points on the line joining (2, 4) and (6, 3) twice as far from (2, 4) as from (6, 3). Ans. (10,2); ~, —V 14 10 3 ' 3 6. 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. ^^^^ /2 3). fy. ' To find the distance betiveen two given x>oints. Let P\ P", be the given points, x', y', and x", if, being their coordinates. P'P" by d, Then P'F" = VP'li' + MF"'; or, representing d = ^{x"-x'y-[-{y"-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 = -y/x'^ + 2/'^. (2) Oblique Axes, In this' case the triangle P'RP" will not be a right triangle. Let /3 = inclination of the axes. Then P'P"= y/P'H^ + JiP"^-2 PR . RP" cos P' RP\ or, since P'RP" = 180^ — /3, Y Fig: ^ -Lv> which, when /3= 90'^, reduces to (1), since cos 90°= 0. y )cos/3, (3) 6 ANALYTIC GEOMETRY. Examples. 1. Find the distance between (2, 4) and (5,8). As the quantities under the radicnl sign are squares, it is immaterial whether we substitute (2, 4) for (a,', //') and (5, 8) for {x", y"), or vice versa. Thus d = V(2 - 5)-^ + (4 - 8)--^ = V(5-2)^+(8-4)--^ = 5. 2. Find the distances between the foUowhig points : (—2,-4) and (-5, -8); (7, -1) and (-6, 1); (7, 2) and (-7, -2). Ans. 5; VTtS ; 2 V53. 3. Find the distance of (6, —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 triangle whose vertices are (4, 5), (4, -5), (-4, 5). Ans. 10; 2V4l; 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 bj' 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 any assumed straight line through a fixed point 0, >P' the position of a point P' in the plane /\ AOP\ with reference to 0, will be ^ known when the angle AOP' and the / distance OP^ are known. The fixed Fig. 6. line OA is called the Polar Axis ; the fixed point, 0, the Pole; the angle AOP and distance OP, the Polar Coordinates, OP being the radius vector and AOP^ the vectorial angle. The radius vector will always be represented by the letter r, and the vectorial angle by the letter 6, THE POINT. 7 9. Signs of the polar coordinates. If the vectorial angle be always laid off above OA (Fig. G) to the left, as in 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 9 often occur in the course of analytic investiga- tions with negative as well as positive signs, it is necessarj^ 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 towards the end of the arc measur- ing the vectorial angle, and negative when laid off in the oppo- site direction. Thus (Fig. 6), for the angle AOP\ OP' is the positive, and OP' the negative, direction of ?\ 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 lay off the given value of 6 above or below OA (Fig. 6) according as 6 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 point. 11. Equations of a point. It is now evident that we may designate the position of a point by giving its coordinates in the form of equations. Thus (Fig. 6), r = 4, ^ = 60°, locate a point P' distant 4 units from on a line inclined at + 60° to OA ; while r = — 4, 6= G0°, locate a point P" on the same line, but on the opposite side of 0. These equations are called the polar equations of a point; but it is more usual to adopt the notation (r, $), writing the radius vector first. Thus, the above points would be (4, G0°) and (—4, 60°), respectively. 8 ANALYTIC GEOMETRY. 12. This system of reference is called the Polar System, and that previously described, whether the axes be oblique or rec- tangular, the Rectilinear System. It will be observed that in each system two 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 two 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 S3'stem a given point can have but one set of coordinates, in the polar system 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 and 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: (5, 90°); (5,*'270°) ; (-3, 120°) ; (-6, -180°). 2. What are the coordinates of the pole? What are all possible values of 6 for points on the polar axis? ' 0^ ^ 3. Construct the points (0, 45°) ; f 0, ^ j ; (4, 0°) ; (-4, 0°). 4. Give three sets of polar coordinates locating (10, 90°) . 5. Construct (8,^); (-8,^); (s,^); (s,-^)- 6. The side of a square is 5 V2, its centre at the pole, and sides parallel and perpendicular to the polar axis. What are the coordinates of its vertices ? pif 13. To find the distance betiveen two /^"^^pi gi'ven points. Let P', P", be the given / / points, r', 0', and r", 0'\ their cooia^inates, /•^ j^ and d the required distance. The*i, from Fig. 7. the triangle P'OP", THE POINT. PP'= ^OP-^Jf OP"'- 2 OP. OP' cos POP", or d=Vr'' + r"^-2r'r" cos {d"-0'). (1) If one of the points, as P", is at the origin, d = 0P= r'. and f 4, — Examples. 1. Find the distance between the points [3, - 3 Since cos^ = cos (— ^), it is immaterial which of the two given points is designated as (?', 6') . Thus d= \/9+l()-24cosOU°= Vl(3 + 9-24 cos (-60°) = Vl3. Observe, also, that if ^''— ^'>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°) ; f-y/2,-) and (1, 0°) ; (10, 30°) and (- 10, - 150°) ; (6, 60°) and (0, 0°) Ans. 5 ; 10 ; 1 ; ; 6. 10 ANALYTIC GEOMETRY. 4y SECTION II.— THE LINE. THE RECTILINEAR SYSTEM. 14. Loci and their equations. Every llne^ straight or curved^ may he regarded as generated by the motion of a poiiit. 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= jR, the radius, then 0P''= Om'-\-mP', :c^ + f = E\ 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 general, the equation of a, line is the algebraic expression of the laio which governs the motion of its generating point. Again, the relation x^ + y^ = Br^ is true for no point within or without the circle, but is true for ever}- point on the circle ; it thus expresses the relation between the coordinates of all points of the circle, and of no other points ; hence, in general, the equation of a line is the algebraic expression of the relation ivhich exists between the coordinates of any and every x>oint 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"^ =^ Br , being the equation of a circle whose radius is 7?, the circle is said to be the locus of the equation; i.e., translat ing the word locus literally. -JJ >4ytTie~ ;place in whinh t,hf> poiril ., nioyi nor- under the law expressed by the equation, is always found ; and, in g^lliirill,^/ie locus of an equation is the path of a x>oint so moving that its coordinates always fulfil the relation expressed 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 adius of the above circle be 5, a;- + 7/^ = 25, and the points '-1^, 4) , (0, — 5) , (—4, — 3) , are al\ points on the circle because . %eir coordinates satisfy the ecmation ibut (2^ -^S^**^' ^^' ^^^ not on the circle. Hence, to "Usf^lWfl^l^ji^therk^iyk^ point lies on a given locus, substitute its coordin^Win the equatio^ of the Jocus and see ivhether they satisfv it. 15. Distinctions between Analytic Geometry' feamgtfj^jyid Algebra. The object of Analytic Geoijififinfi-s^e d^^sklSS^S/tt- deter^nination of the propefties of 4oei. ^ Its "m^hod consists in^ the substitution of the equation of the locus for the in the discussion and determination of its propei oc^ -\-y^ = B^ has been seen to be the equation of Fig. 8. Putting it under the form 2/2=i?2_^2^ (i2 + a!) {B we observe that f- = Fm'-, B-\-x=Ahn. Hence Pm^=:: A'm.mA, or the squa»e7^HIJftalf-cli0i«^l arfy diameter of a circle is a mean proportional betw<^fF t^SKeg- ments iiito which it divides that diametil. This well-tSown property of the circle might be established^ ^eome^rico%, from a figure ; it is here established analytically, from the equation of the circle ; and the object of Analytic Geometry is thus to determine the properties of lines, by discussing their equations, 12 ANALYTIC GEOMETRY, instead of by reasoning upon the lines tliemselves as in Eucli- dean Geometry. Having thus noted the distinction between Analytic Geometry and Geometry, let us note in what way it differs from Algebra. Since the coordinates of every point of the circle must satisfy its equation a? -{- y"^ = Br^ x and y in this equation may have an infinite number of sets of values, corresponding to the infinite number of positions occupied b}' the generating point in tracing the circle. Hence while R is a constant quantit}^ x and y are variable quantities. They differ thus from all the quantities of common Algebra, which, whether known or unknown, are always constants. Observe, also, that while x and y thus admit of an infinite number of values, they do not admit of any val- ues, but only of those which satisfy the relation o? -{-y^z= E^o Again, in Algebra, if only x^-{-y^ = R- were given, x and y being _unknown but constant quantities whose values were required, the solution would be impossible ; for this equation would be satisfied by an infinite number of sets of values of x and y. and without a second independent equation we could not determine the particular values required. Furthermore, if the conditions of the problem were not such as to furnish a second equation, the problem would remain an indeterminate one. It is in virtue of this ver}^ indetermination that we are enabled to represent loci by equations, and, as thus distinguished from Algebra, Analytic Geometry is sometimes called the Indeterminate Analysis. 16. Quantities of Analytic Geometry. g If the centre of the circle were at some y^ I \P point C, whose coordinates are in and 72, in- / I / j \ stead of at the origin, then, from the right- I c}^-- .;i?J angled triangle PCB, CE'+BP'= CP\ \ I \J\ ^^' {x-mY + {y-ny==R\ J- X ! — L__^ This relation between x and v? beinoj 3x Q N ' true for all positions of P on the circle, is the equation of the circle in^ its new THE LINE. "' 13 position with reference to the axes. Now if, in this equation of the circle, we change i2, we change the magnitude of the circle ; and if we change m, or n, or both, we change the j)Ositlon of the circle. Hence the constants in the equation of a locus ] determine the magnitude and position of the locus. The quan- / titles of Analytic Geometry are thus : First. Variable quantities, as x and y of the preceding equa- tion, which, being the coordinates of a moving point, vary cm- tinuously within the limits assigned bN' the equation expressing their mutual relation. Thus x varies continuously between the limits x=:.03f, and x= ON, and y between the limits y= QS, - and y = QT. Since, when y changes, x also changes, and vice versa, x and y are said to be functions of each other. Second. Arbitrary constants, as m, n, B, of the above equa- tion, to which values may be assigned at pleasure, thus locating an}' circle in any position. They do not, however, change whe7i x and y change, that is, they are not functions of x and y, and are thus constants, though arbitrary constants. Third. Absolute coristants, such as m, n, and i?, would become in the above equation, if we placed the centre of the circle at (7, 6) and assumed 5 for its radius; which cannot change under any circumstances. 17. Construction of loci. It is now evident that two general classes of problems will arise. First. Given the law governing the motion of the generating point (usually given in the form of some property of the locus) , to find the equation of the locus. Second. Given the equation of the locus, to determine the locus; i.e., its position, form, and properties. These two fundamental problems form the subject matter of Analytic Geometry and will be fully illustrated in the sequel. Their solution involves on the part of the student a thorough c<5m[)rehension of the relation between a locus and its equation as defined in Art. 14, and to illustrate this relation the following examples of the determination of loci from their equations by points are added. 14 ANALYTIC GEOMETflY. Examples. If any value of either variable, assumed at pleasure, be substituted in the equation of a locus, and the value of the other variable be found from the equation, the set of values thus obtained evidentl3' satisfies the equation ; tliey therefore determine a point of its locus. Hence, to deter- mine points of the locus of an equation, assume in succession any number of values for one variable, and find from the equa- tion the corresponding values of the other. Construct the points thus obtained and draw a line through them. This line will be the locus of the equation. This process is called the construction of the locus. The variable to which values are assigned is called the independent variable; the other, whose values are derived from the equation, is called the dependent valuable. It is evident from the Ature of the process that either variable may be chosen as the independent variable, and it is usual to assign values to x and derive those of y. In such an equation, however, as x = if—2y^-{- 4, it is more convenient to assign values to y and derive those of ic ; i.e., to make the variable which is most involved the independent variable. The illustrations which follow are limited to equations of the first and second degree. 1. y — X — 4: = 0. Solving the equation for ?/, we have y = x-\- 4:, and taking x for the independent variable, we obtain Fig. 10. x = 0, x= 1, x=2, x = S, 2/=7, x = -l, 2/=3, x = -2, y = 2, x = -3, 2/ = ^ ic = -4, ?/ = 0, ic = — 5, ?/ = — 1, etc. y=4:, locating P', 2/ = 5, locating P", 2/= 6, locating P'", etc. Constructing these points, the line MN drawn through them is the locus of y — x — 4: — 0. TPIE LINE. i ft A R y /EHSITT 15 It will subsequently be shown that loc'iirs of every equa- tion of the first degree, between x and y, is a straight line. This being the case, it is necessary to construct but two points for such equations. Assuming this fact, the student may con- struct the straight lines represented by the next four equations, constructing in each case two points ; then verify the construc- tion by locating a third point. 2. y -\-x—l = 0. Solvuig for y, y = — x -\-l\ in which, for fl;=:0, 2^=1, locating P', x= 5, 2/ = — 4, locating P", a; = — 3, y = 4, locating F"'. Constructing P', (0, 1), and P", , (5, —4), F'P" shoukl pass through P'", (-3,4). 3. y — x = 0. 4. y-\-x=0. 5. Sy-2x-l = 0. ^-70-.' 6. 2/^=4 a; — 8. Solving for y, we Assigning values to x, for x = we have ?/ = ± V — 8, which is imaginary ; moreover y will evidently be imaginary for all values of a;<2, algebraically. As the ordinates corresponding to all values of x<2 are imaginary, we conclude that there are no points of the locus having abscissas less than 2 ; and, in gen- eral, tchen either of the coordinates obtained from the equation is imaginary, we conclude there is no corresponding point of the locus. Assuming values for a;>2, we have, for a; = 2, ?/ = 0, - locating P% ic = 3, y = ± 2, loc. P" and P", a; = 4, ?/ = ±2V2, loc. P^^ and P^, ^=5, ?/=±2V3, etc. E-0, y = ±4, etc., ;ry value of a; > 2 locating two points. ^ig- 12. / 16 ANALYTIC GEOMETEY. The above method of constructing a locus bj- 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 pt)sition of the locus. Thus, in the above example, every value of X gives two values of y 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, whenever the equation contains the sqxiare only of either variable^ the other axis is an axis of symmetry. Thus 2/2=9 a; is symmetrical with reference to X; ?/-J-aj^=2 is sym- metrical with respect to Y', while ar-f-?/2=25, af—y^=4:, 9x^+16x^=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 2/ = ; and for a like reason if it has any point on the axis of Y, it will be found b}' making cc = ; and, in general, to find where 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^= 4x — 8 crosses X, make 2/ = 0, whence a;=2=0P^ Making x = 0, y is imaginary, showing that the locus does not meet t' "; axis of Y. The dis- tances from the origin to the points where a locus meets the axes are called the intercepts of the locus. Thev are distinguished as the X-intercept and the F-intercept. ? itrs, ^he X-intercept of 2/2=4x-8 is 0P'=2. o 7. 252/^+9x^=225. We obst ^ that the locus is symmet- rical with respect to both axes. M ,king 2/ = 0, we find x= ±5, or 0^ and OA^ are the X-intercepts ; making x = 0, 2/= ±3, or OB and OB^ are the Y-intercepts, Solving the equation in succession for x and 2/, we have 2/ = ± f V25 — a^, X = ± I V9 — 2/^ 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 tlie right of A or to the left of xi' ; that is, x=±5 gives the limits of the curve in the direction of X, and these values are the roots of the equation obtained by putting the quantity under the radical sign equal to zero. The reason for this is plain : y is real when 25 — ic^ is positive, and imaginary when 25 — a^ is negative ; hence the limiting values of y correspond to 25 — a^ = 0, since in passing through zero 25 — x^ changes sign. For a like reason, placing 9—y^ = 0, y = ±3 are the limits of the curve in the direction of Y, And, in general, whenever the equation of the locus is of the second degree with respect to one of the variables^ if ive solve it for that variable, and place the radical equal to zero, the roots of this equation are Fig. 13. the limits in the direLv, ' ofjhe other axis. (Thus, in Example 6, the equation is of th second degree with respect to y ; solved for 2/, the radical placed e- d to zero gives 4a? —8 =0, or x=2. Beyond this limit the curve' sitends indefinitely in the direction of X.) We have now de irmined the intercepts, symmetry, and limits, of the locus, and so have a general knowledge ot its form and position. Points may now be constructed as before. Thus, for x= 3, or —3, jo; =4, or —4, y ± y , locating P' , P", P"S and P^% I/=±h locating 7^^, P'\ pvii^ pvm^ ^l-g^ 18 ANALYTIC GEOMETRY. 8. 16y^ ~9x^ = —14:4. Making a; = 0, y is imaginary; hence the locus does not meet the axis of Y. Making y = 0, ic = ±4, or OA and OA', the X-intercepts. The curve is symmetrical with respect to both axes. Solving for x, x = ±iVy'' + 9; buty^+9 cannot change sign, or, otherwise, 2/^ + 9 = gives imaginary values for ?/, hence there are no limits in the direc- tion of Y, the curve extending indefinitely in that direction. Solving for y, y=±i-Vx'-U. Placing ic^— 16 = 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 consti'uct a few points. For x=±5, 2/ = ±f, locating P\ P^, P^^, P^, x = ±G, ?/=±fV5, locating P\ P^\ P^'^, P^'^, etc. A curve of this kind, com- posed of two separate branches, is said to be discontinuous. Fig. 14. 9. x^-{-y^ — 8x — 4:y—5=0. Making a; = 0, ?/=5, and — 1, giving the intercepts OB, OB'. For y=0, x=4 ± V2l= 4 ±4.6 nearly, or 8.6 and — .6 for the intercepts OA, OA'. Solving for X, we have ic^ — 8 ic = — 2/^ + 4 ?/ + 5, whence a; = 4± V-2/' + 42/ + 21. 0) f 5 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 Y and 4 units from it. Thus, for 2/ = C, a; = 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 f-iy=-x' + Sx+5, whence 2/ = 2 ± v^a^ + 8a;+9, (2) from which 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 Aoi? -\- Bx -{- etc. with respect to either variable, if the coefficient of the square be made positive unity, then half the coefficient of the first power, with its sign changed, will be the distance from the other axis of a line of symmetry parallel to that axis. Thus, x^-\-y^^lOy + 4: = is symmetri- cal with respect to Y, and also with respect to a line parallel to X and 5 units above it; a^-\-2x-{-y^ — 9y = 0is 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 = d and — 1. Values of y are imaginary for a; > 9 or < — 1 , and the locus lies within these limits. For the limits along Y, (1) gives y = 7 and — 3, or no point of the locus lies above + 7 or below —3. Having now determined the intercepts, limits, and symmetry, we may construct a few points. For r P 5?r— ■^' / 'B T c' f 0, 1 -<- (J v u A ^B^ n' Ypxz — X. Fig. 15. a;=4, y = l, or -3, P'", andP'^ or-1, P% andP^S etc. 20 ANALYTIC GEOMETRY. 10. Show that y^ — Qy -\-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. y^— 10 x-{- oif= 0, Determine the lines of symmetry, intercepts, limits, and construct. 12. .^•^ — 6 ic -{- 9 + 2/^ + 10 ?/ = 0. Lines of symmetry are —5 and 3 from Xand Y, respectively. Limits along X are 8 and —2 ; along F, and —10. Intercepts on Y are —1 and — 9 ; on X, + 3: Construct. 13. 2/2-2a^ + 12ic-22 = 0. Show that the locus has no limits in the direction of X, lies wholly outside the limits ± 2 in the direction of Y, 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 F, has a; = for a limit along X, lying wholly in the first and fourth angles. Construct. Obsei*ve that if x = 0, ?/ = 0, and conversely , or the intercepts are zero on both axes, and hence the locus passes through the origin. Otherwise, the coordinates of the origin satisfy 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, whenever the equation of the locus contains no absolute term, the locus jyasses through the origin. Thus, a^ + y^— lOy = 0, x'^ — y^-\-3x=0 pass through the origin . 10 15. xy = 10. Solving for y, y= — By assigning values X to X, and deriving those of y, we ma}* construct the locus by 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 y 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 ic = 0, ?/ = gc , and as X increases y dimin- ishes, but becomes zero only when a; = oo . In the first angle, then, the locus lies as in the figure, continually ap- proaching the axes as X changes, but touching neitiier within a finite distance from the origin. A line to which a curve thus continually ap- proaches^ hut does not touch within a finite distance is called an asymptote. In the third angle, x being negative and decreasing algebraically, y increases algebraically, becoming zero, however, only when aj = — oo. The axes are thus asymptotes to both branches. Constructing a few points, we have, for x=±l, y= ±10, P',P'\ a;=±2, y=±5, F'", P'^, a; =±5, y=±2, P\ P'\ a;= ±10, y= ±1, P'''\ P^'"S Fig 16. etc. THE POLAR SYSTEM. '•{ 18. Polar equations of loci. We have seen that the equa- 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 oi^ which this law is expressed are the coor- dinates of the moving point and certain constants. Nothing in this statement restricts us to the use of any particular system of 22 ANALYTIC GEOMETRY. coordinates. 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, 0P= r and A0P=6 will be the polar coordinates of P the moving point ; and since OPB will be a right angle for every position of P while tracing the circle. Fig. 17. OP OB = cos BOP, or r=2R cos (9 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 coordinates satisfy the above equation ; hence the circle is the locus of the equation. 19. Construction of polar equations. In a polar equation, the variables 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 the locus as we desire. It is obviously convenient to make B, the vectorial angle, the inde- pendent variable, and derive the values of r. Examples. 1. r=5. This equation is independent of ^, that is, r — b = a constant, for all values of 0. It is then evi- dently the equation of a circle whose radius is 5, the pole being at the centre. We have seen (Art. 14) that the corresponding rectangular equation of the circle is x^ + 2/^ == P'. The student will observe the comparative simplicity of the polar form r = 7?, and will thus see that in many cases it might be preferable to THE LINE. 23 use the polar rather than 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 ^ = 0°, cos has its greatest value, and there- fore also r ; that as 6 increases, cos 0, and therefore also, r, diminishes, becoming zero when = 90°. That as increases from 90° to 180°, r is negative and increasing numerically, becoming — 10 when 6 = 180°, the same numerical value which it had for^ = 0°. Constructing a few points, we Imve, for = p°, 0= 30°, r=10, r = 5V3, -PS |90° pni ^->/Sv\\ \pi 0= G0°, 0= 90°, ^=120°, 7'= 5, r = 0, r = -5. 0, ^=150°, r = -5V3, p^ pTv^S>fr (9=180°, r = -10. Fig. 18. As when ^ = 0° the radius vector coincides with the polar axis, P' is constructed by 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 away from the end of the measuring 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 OP' = 10, and r = 10 cos is true for all- positions of P, OP"P', OP"^PS etc., is always a right angle, and the locus is therefore a circle whose radius is 5. 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 /(a;, 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 ?/, the corresponding general form will bey=/(.T), read ' ?/ a function of ic.' 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 f{x^ y) =0 may be written y =f(x) whenever the former can be solved for y. Similarly /(?', 0) = 0, and r =f(0)^ are general forms for the equation of any locus referred to a polar system. EELATION 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 X^OyYi are Oimj and ?7iiP; referred to the rectangular system XOF, they are Om and mP'^ while if the polar system O-iA is employed they are O2P and AOoP. Again, we have seen that the equation of a circle referred to rectangular axes through the centre (Fig. 20) is a;^ + 2/^ = W^ (Art. 14) , but if it is referred to the system X^O^Yi its equation is (xj — m)^ + (2/1 — ny = R^ Y ^mg. 20. (Art. 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 ?' = 2P cos 6 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 ?-= 72. 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 AKALYTIC GEOMETRY. illustrations, that one form may 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 system of reference (as the equation of the circle (iCi— m)-+ (2/1— w)- = E^ referred to the axes XiOi Yy) , to find its equation when referred to any other system (as the parallel system XOY, to which when the same circle is referred its equation is x^-{-y' = E^). The object of this transformation is to obtain a simpler equation of the same locus ; the metJiod will consist in finding values for the coordinates x^^ y^, 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. 2/, we pass to the polar equation of the same locus by substituting for x and y their values in terms of r and 6 ; th(* 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 : The coordinates of any point P loith respect to one system oj reference being laioivn, to find its coordinates zvith respect 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. (JB). To pass from any rectilinear system to any polar system. (C). To pass from any polar system to an}- rectilinear system. TRANSFORMATION OF AXES. RECTILINEAR TRANSFORMATIONS. 27 22. Formulce for passing from any rectilinear system to another. Let XOY be the primitive system, ^ being the inclination of the axes, and P any point whose primitive coordinates are Om — X, mP=y. Let XiOiTi be the new system, its position being given by the coordi- .^ ,y nates of its origin , OA = Xq, AOi = ?/o, and the angles y, yi, which its axes make with the primitive axis of X, the coor- dinates of P referred to the , -^ a ,„ new system being Oimi = Xi J^^ /\ i and miP= yi. Draw OiB and m^C parallel to OX, and m^D parallel to Y. Then Fig. 21. Om = OA-\- 0,D + miO. But Om = X, OA — Xq, OiD : OiiUi : : sin OimJ) : sin OiDmi, whence ^ Oimi sin OimiD _ x^^ sin (^ — y) sin OiDmi ~~ sin fi and mi (7 :miP:: sin miPC : sin m^CP^ 0,D = 1 whence Substituting these values, sm /? iC — Xll a;i sin (/? - y) -f y^ sin (^ - yQ ^ sin y8 Again, 7?iP = ^ Oj + ^^^^i + CP. But mP= ?/, AO^ — 2/0, />mi : 0{i)\ : : sin Z^OiJJij'sin OiDmi, whence cCi sin y sm /? ANALYTIC GEOMETRY. CP : m^P : : sin CmHiP : sin mi OP, CP^ Vi sm yi y = yo + 28 and whence v.^ — . ^ sm yS Substituting these values, Xi sin y + 2/1 sin yj sin/3 Hence , a;^sin(^-y)+yisin(^-y0 . a?isiny+yisiny i ,.s are the required formulae. The following special cases may arise : First. To pass from any system to a parallel one. In this case (Fig. 22) y = 0, yi = /8, and the general formulae (1) become a; = 0-0 + 0^1, y = yo + yu (2) which are independent of /3 and apply to all parallel axes, oblique or rectangular. Second. To pass from rectangular to oblique axes. In this case (Fig. 23) yS=90° ; and since sin (90°-^) = cos A, sin (90° — y), and sin(90° — yi), become cos y and cos yi, or the general formulae become a; = a^o 4- a^i cos y + 2/1 cos yi, ?/ = 2/0 + ^1 sin y + y^ sin y^. (3) TRANSFORMATION OF AXES. 29 Third. To pass from one rectangular system to another. In this case (Fig. 24) p = 90°, y^ = 90° + y ; and since sin (90° + ^) =cos^, sinyi = sin (90° + y) = cosy, sin(/3 — y) =cosy, sin(^-yi) =sin(90°-90°-y) =sin-y= -siny, and the general formulae become x = Xo-{- 0^1 cosy — ?/isiny, ?/ = 2/o + a^isiny + 2/iCOSy. (4) niA Fig. 25. Fourth. To pass from oblique to rectangular axes. In this case (Fig. 25) yi = 90° + y ; and hence sin (/? - yO = sin [^ - (90° 4- y)] = sin ~ [90° - (/? - y)] = - sin [90° _ (^ - y) ] = _ COS (;8 - y) , and the general formulae become a^isin (^ - y) - vicos (^ - y) sin^ aJi sin y + ?/i cosy sin/3 X = X^)-^ y = yo + (5) The student will observe tliat the special formulae, like the general ones, may be deduced directly from the accompanying figures. 30 ANALYTIC GEOIVIETRY. If the new origin coincides with the primitive origin, Xq and 2/0 in the above formulae become zero. Hence, To pass from one oblique set to another^ ^ ^ a?! sin (^ - y) + Vi sin {fS- yQ ^ a;, sin y + y, sin yi ,gx sinjS ^ siufS ^ ^ To pass from a rectangular to an oblique set, a; = iCj cos y + 2/i cos yi, ?/ = ajj sin y -f- 7/1 sin yi. (7) To pass from one rectangular set to another, x = Xi cos y — 2/1 sin y, 2/ = ^ sin y + 2/i cos y. (8) To j^ciss from an oblique to a rectangular set, ^ ^ Xi sin(/3 - y) - y^ cos(^ - y) ^ a^^ sin y + y, cos y .^. sin/? ^ sin 13 ^ ^ The student will observe that none of the above formulae 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 alwa3's 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 th^ equation. POLAR TRANSFORMATIONS. 23. Formulce for passing 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 = mP. Let Oi be the pole, its coor- dinates being OA = Xq, AOi = y^, and let the polar axis make an angle a with the primitive axis of X. Then OjP=r, and RELATION RECTILINEAR AND POLAR SYSTEMS. 31 6 = -^gOiP, or AiOiP, according as the polar axis lies above or below OiXj, drawn parallel to OX, i.e. according as a is posi- tive or negative. Hence, in general, X,0,P=e±a. Now Om=0^+Oii), in which Om = X, OA = Xq, O^D = OiPcos DOiP= r cos (^ ± a). Hence x = XQ + rcos{0± a) ;^ similarly, 2/ = 2/o + ''sin(<9±a).^ Ki) o^i If the polar axis is parallel to the axis 0/ X, a = 0, and the general formulae become x = Xo-^rcosO, y = yo-{-rs\nO. (2) If the pole coincides with the origin, a^o = 2/0 = 0? and a; = rcos (^ ±a), y = rs\n.{d ±a). (3) If the pole is at the origin and the polar axis coincident with X, a = 0, iCo = 2/0 = 0, and a; = rcos^, y^rsinO. (4) 24. Formidce for passing from any polar system to any rectilinear system. From Equations (1) of Art. 23, x — XQ = r cos (^ ± a) , y -'yQZ=r sin (^ ± a) . Squaring, and adding, and substituting the resulting value of r, we have, since sin^J. + cos^A =1, cos(^±a) = r=V{x-x^y + {y-y,y, X ~~* Xfi '^O ^(^x-x,r-\-(y-y,y i> sin (^ ± a) = y - 2/0 V{x-x,y-\-{y-y,y J (1) / K / 32 Ai^ALYTIC GEOMETRY. Iftlie, polar axis is parallel to X, a = 0, and r = -V{x — Xoy+{y-yoy, cosO ^~^" sin^: ■Vix-x^y+iy-yoY y - ?/o V(a;-a;o)^+(?/-2/o)^ If the new origin is at the pole, XQ = yQ=0, and x (2) r = Va^ + 2/', cos (0 ± a) = sm(0±a)= '^ .(3) 7/* ^7ie neiy origin is at the ])ole and the neio axis of X coincides with the polar axis, a = 0, ic^ = yo = 0, and r = V^H^', cos^ = . sin^ = ^— . (4) Vx'^ + ?/" "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 ?/ — a; — 4 = (Ex. 1, Art. 17) to a new set of parallel axes, the new origin being at (0, 4) . The formulas for passing from any rectiUnear system to any parallel one being x = Xq+ x^, y = y^^ y^, in which Xq= 0, and ?/o= 4, the values of the primitive coordinates in terms of the new are, in this case, x = x^, y = 4:-{. y^. Substituting these values in the given equation, we have y^ — x\ = for the transformed equation. As the subscripts are only used to distinguish the two sets es, they may be omitted after the transformation is effecte' 'o Fig. 10, Art. 17, the student will see that the new origin i. he locus, and that there- fore the transformed equation should have no absolute term. 2. Transform y + x—l = (Ex. 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. 3?/— 2 a; — 9 = 0. 4. Transform ?/2 = 4.t— 8 (Ex. 6, Art. 17) to parallel axes, the new origin being at (2, 0), that is, at F\ Fig. 12. Ans, y^ = Ax. EELATION RECTILINEAR AND 5. Transform y'^=4:X — 8 to anew set of rectangular axes with the same origin, the new axis of X making an angle of — 90° with the primitive axis of X. The formula) are x = x^ cos 7 — ijy sin 7, y = Xj^ sin 7+^1 cos 7, in which 7 = — 90°. They become, then, x = yi, y = — x^. Substituting and omitting subscripts, a:^ = 4 ^ — 8. 6. Transform x" + y^ -8x- Ay - 6 = (Ex. 9, Art. 17) to a new set of parallel axes, the new^ origin being at (4, 2), that is, at 0^ (Fig. 15) . A71S. x^ + y^=2o, 7. Transform xy= 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 a: = Tj cos 7 — i/^ sin 7, y = x^ sin 7 + 3/1 cos 7, in which 7 = 45° and they become x = V^ (x^ — yi'), y='^h (^1 + ^i)- Substituting these in xy = 10, and omitting subscripts, x^ — y"^ — 20. 8. Transform a;^ + ?/- = 25 to a polar system, the pole being at (—5, 0), and the polar axis coincident with X. The formulae are x-=Xq-\-t cos 6, y = y^ + r sin 6, which for Xq = — b, y^ = 0, become x— — ^ -{■ r cos G, y — r sin d. Substituting these in 3.2 _j_ y2 _ 25, we obtain r = 10 cos d. 9. Transform (a^ + y'^y —a^(x^ — y-) to polar coordinates, the pole being at the origin and the polar axis cohicident with X. A)is. ?^ = a^cos2^. 10. Transform the followinfj^ equations, the origin and the pole being coincident, as also tlie' axi?^v -^ X and the polar axis. r = 20 cos 0: xy = a^, . 2 . " * oa a 2 2 a^ ' ^ Ans. x^ 4- ir — 20x- =0 ; r^ = • -^ ' sin2^ 11. Having the distance between two given points in a rec- tangular system, d = V(a;" — ic')^ + (fi^^fy (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 6', y' = r' sin d\ x" — r^' cos 0", y" = r" sin d", d= V(r" cos e" — r' cos B')'^ + (r" sin 6" — /-^ sin d'Y = V?"-(cos"-2*iC SECTION IV. — THE RECTILINEAR SYSTEM. EQUATIONS OP THE STRAIGHT LINE. 25. Every equation of the first degree between two variables is the equation of a straight line. Every such equation may be put under the forru Ax-\-By-{-C^O, (1) in which A and 5 are the collected coefficients of x and y, and is the sura of the absolute terms. Let F, P", P'", be three points on the locus of this equation, whose abscissas in order of magnitude are x\ x'\ a;'". Then, from (1), their ordinates 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=0; whence, by subtraction, A{x"- x') + B(y"-y') = 0, ^ (x'"- x') +B{y"'-y') = 0. Fig. 27. 36 ANALYTIC GEOMETRY. Equating the values of A, y"'-y' y"-y\ x"-x' (2) Let P'Q be drawn parallel to OX. Then, from (2), F'Q P'e' Hence the triangles F"'QP', P"IiP', are similar, and P" is on the straight line P'P"'. In the same manner it may 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 brevity, meaning "the line whose equation is Ax + By + C=0.'' " Oblique Axes. The above demonstration depends only upon the 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 MN with reference to the axes. First, by its intercepts OR, OQ; second, by its F-intercept OQ, and the angle XBQ which the line makes with the axis of X (always measured, as in Trigonom- etry, fi'om OX to the left) ; third, by the length of the perpendicular OD let fall from the origin on the line, and the angle XOD which this perpendicular makes with the 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 Fig. 28. when given in each of these three different wavs. First. Let P be any point of the line, x, y, its coordinates, and 0-K = a. OQ =?>, the given intercepts. Then THE RECTILINEAR SYSTEM. 37 QO: OB:: PL: LB, or b:a::y:a — x, whence hx-\-ay= ab, or, dividing by ab, Second. Draw OS parallel to 3f^, and let tan XBQ = m. Then LP =SP-jSL. But SP=OQ = b, ' SL = tiinSOL.OL = tan OBP . OL = -tan XBQ . 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 JD/r parallel to MN, and PT parallel to OD. Let^QD = a and OD =p. Then 0K+ TP = OD. But 0K=: 0Lco3L0K=xcosa, TP = LP sin TLP = y sina. Hence a; cos a + 2/ sin a = 2^. (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 mcfre 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 + C— 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 ; while 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. Wliencver we have sue- 38 ANALYTIC GEOMETRY. ceeded in establishing this eqiu'tion, we have the equation of the locus, whatever the constants involved. Oblique Axes. Whatever Uie angle XOY {Fig. 23), 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—SP—SL, in which LP^y, SP=h\ but SL : LO : : sin SOL : sin LSO. Let a>= XOF= the inclination of the axes, and A= XRQ, the angle made hy MN with X. Then SL : a; :: sin A : sin (A — to) , whence SL = xsmK ^ ^^^ ^^^^ equation he- fiin (w — A) comes y= ^^- x + b. This may be written sin (w — A) in the form y = inx + 6, understanding that when the axes are oblique. If sin A Fig. 29. sin (w — A) (o = 90°, sin (o) — A) = cos A, and m = tan A, as in (2). For the normal form, EOD being a, as before, let DOQ = ^. Then 0D== 0K+ TP = x cosa + y cos^=p. When XOr= 90"^, DOQ is the complement of XOD, that is of a, cos ^ = sin a, and the equation reduces to (3) y 27. Derivation of the common forms from the general form. Since the equation of every straight line is of the general form Ax-{-By -\- C=^^ it must evidently be possible to derive the common forms from the general form, and to express the par- ticular constants a, 6, m^ p, cosa, sina, in terms of the general constants A, J5, C. First. The intercept form. Assuming ^a;-hi^i!/ + (7=0, transposing C to the second member, and dividing the equation by —C, i.e., making the second member positive unity, we have X y A B which is the required intercept form, the intercepts being C . C Second. The slojje form. Solving the general form for y, we have y B B THE IIKCTILINEAU SYSTEM. 89 A O which is the required slope form, in which ?7i = — - , 6 = as before. Third. The normal form. This form requires that the second member (|)) 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 cos^a + sin^a= 1. Let M be the factor which will transform the general to the normal form. It must fulfil the' condition {RAy -{-{RBy = l. Hence B = — - Introducing this factor and transposing V^P + i^^ (7, we have Ax , By _ -C ■yjA'-\-B' VA' + B' VA' + B' in which A . B -C cos a = — , SHI a — — , p = VA' + B'- ^A' + B' -VA' + B' To make the second member (p) positive, of the two signs of V^^ + B' we must evidently take the opposite one of C. 28. In the preceding article we have found the values of a, b, m, p, cos a, and sin a, in terms of A, B, and C. But it is unnecessary for the student to burden his memory with these relations. Thus, suppose we have giv(Mi the straight line 3a; — iy +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 i_o ' _li) ' and the intercepts are seen to be a = — y , h = i^. A still simpler wa}- of determining the intercepts is to make y and x successively zero (Art. 17, Ex. G) . Thus, for x = 0, y = b — i£- ; 40 ANALYTIC GEOMETIiY. and for 2/=0, x = a = — ^^-. Again, suppose the slope is required. We then put the equation under the slope form by solving it for y, obtaining and the slope is seen to be m = f , the F-intercept being b = -LQ^, as before. Finally, if the distance of the line from the origin (p) is required, we put the equation under the normal form b}' dividing it by ^A'-{-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 w^ould be found from the tables, being the angle whose cosine is — |, or sine is f . 29. Discussion of the common forms. First. The intercept form. This form is X , V ^ . T . , C , C - + ^ = 1 , ni which a = , 6 = a b ' A B If a and b are both positive^ 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 -M"A^", both 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 i}/"'A^"^, and when a is negative and b positive^ as does J^P'^N^^ . We observe, also, that when C and jB, as also C and A, of the general form have like signs, the intercepts are negative, and when they have unlike signs the intercepts are positive. If a = 00, the equation becomes y = b, and since y is b for all values of x, that is, since the equation is independent of x, y = 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 wlien THE HECTILINEAK SYSTEM. 41 a = 00, -4 = 0, and the general form is independent of x. Sim- ilarly, if b=: oo, we have x = a, the equation of a parallel to 1", its position to the right or left of Y depending upon the sign of a ; in this case ^ = 0, and the general form is independent of y. If a = 0, the Hne passes through the origin, therefore b is also zero. In this case the intercept form is inapplicable because there are no intercepts, but we see from the values of a and b that (7=0, and the general form becomes Ax -\-B}/=0, as it should, since when a locus passes through the origin its equation has no absolute term (Art. 17, Ex. 14). Second. TJie slope form. This form is A C y = 7nx + b. in which m = , b= If m is positive, the line makes an acute angle with X and cuts Y sibo\e 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 JfiriJVrni^ y = — mx + 6, 2/ = — wa; — 6, y = mx — 6, y = mx 4- b, If m — 0, the line is parallel to X, and y = bm its equation, as already seen. Ifm = oo, the line must be parallel to F, since the angle whose tangent is 00 is 90°. The equation then becomes y = ccx -\-b. To interpret this form, we observe that, as the line is parallel to ' T, b must also be go , and hence in m = , b = B c the con- ditions m = 00, 6 = 00, will both be fulfilled when B — 0. The general form then becomes Ax+C. 0, or a; = A a, as before. If b = 0, we have y = mx, or Ax -\-By==0, as before, 42 ANALYTIC GEOMETRY. the line passing through the origin. The form y = mx is the most convenient for lines passing through the origin, the value of m fixing the inclination of the line to X. Third. Tlie normal form. This form is iKCOSa -\-ys\i\a =2^1 in which A . B C COSa = — , Sma = — , p VA' + B' -VA' + B' ^/A'+B' the sign of -VA^ + B' being such as to make p positive. IJ both COSa and sina are positive^ a lies between 0° and 90° (IfW). If both are negative, a lies between 180" and 270° (Jf "^^) . If COSa is positive and sin a is negative, a lies between 270° and 3G0° (Jlf"^^"^), and if cos a is negative and sina posi- tive, between 90° and 180° {]\P^N^^), If p = 0, the line passes through the origin, and its inclination is known when sina and COSa are known, its equation taking the form x cosa + y sina = 0, or ^^ 4- ^^ =0, or Ax-{-By=0, as before. ■VA' + B' VA' + B' If a=0° or 180°, sina=0, and the equation becomes COSa V-^' + J^"^ ^i ^ ' as before, the line being parallel to Y. If a — 90° or 270°, cosa=0, and y = -^=b, in like manner, the line being sni a parallel to X. Oblique Axes. The intercept form being the same, the discussion above given applies equally to oblique axes. The slope form is y= . ^/"^ x + b (Art. 26). Since sin A is always positive, the sign of the coefHcient of x depends upon that of sin(w — A), and will be positive or nega- tive as a)> or being a relation between x and y in terms of the given constants x\ ?/', 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 ?/ -h 4 = 1 (a; — 1) , or y — x—5. Oblique Axes. The above equation applies to oblique axes, understanding that sin A m = - . * sin (w — A) 32. Equation of a straight line passiyig 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-\-By+C=0, since the line is a straight line, and must be satisfied for the co- ordinates of the given points; hence (2) Ax' -\- By' + C—0, {3) Ax"+By"+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, ! 2'-2/'=|eJ!(^-»'), ^ (4) ' which is a relation between x and y in terms of the given constants, and hence the equation required, the coefficient of a;, y ~y ^ being the slope (Art. 27). Thus the Hne passing /J.' ^" through (1, 2) and (-3, 4) is 2/- 2 = ?^^ (a;-l), or 1 + 3 2y -^-x — b — O. It is immaterial w^hich point is designated as {x',y'). Thus, y - ^ = _^ ~_^^ (a; + 3) , or 2y + x-b^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 tlie axes are oblique, only the coeflicient y ~y is then the ratio of the sines of the angles which the line makes with A' and Y. x' ~x" Examples. 1. Write the equation of a line through (2, 4) haying the slope 5. Ans. ?/ — 5a;-f-6 = 0. 2. Write the equation of a line through (2, 3) and (1, —2). In place of using Eq. (4), Art. 32, as tliere illustrated, it is quite as expeditious to determine the constants of any one of the three common forms directly. Thus, the form ij = mx + h, satisfied in succession for the two points gives 3 = 2 ?« + ft, and —2 = m^h. Subtracting, we obtain m — 5. Substituting tliis value of m in either of the above, we find i = — 7. Hence »/ = 5 .c — 7. 3. Find the equations of the sides of the triangle whose vertices are (4, 8), (I, 4), (-4, -8). A71S. 3?/-4aj-8 = 0; 5?/ - 12a;— 8 •= , 2r'- 2aj= 0. \.^ THE RECTILINEAJi SYSTEM. 47 4. Find the equations of the meclials of the triangle of Ex. 3. Ans. ny-20x-8 = 0; y-4x = 0; 13?/- 28aj- 8 = 0. 5. Write the equation of a line through (2, 5) and the origin. We may use the equation of a line through two points, making one of the points (0, 0) ; or tlie slope form i/ = mx (since b = 0), which, satisfied for (2, 5), gives m = ^, and therefore y = -| x. 6. Write the equations of the 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 (-6, -3), parallel to X. (5) through (-1, 2), parallel to T. Ans. y — x — 8 = \ y -^x — l = •, y=7x', y = — 3;x = —l, PLANE ANGLES. 33. To find the cu^gle included hetweeyi tioo given straight lines. The slope form is best adapted to this problem. Let y = mx -{-b^ y = m'x + b', be the two given lines ; m and. 7?i' are the tangents of the angles Xi^P=A, and XQP = \'. Then, if c = tan MPQ = tany, from Trig- onometry we have tan \' — tan X tany= 1 + tan A tan V m' — m r> /i\ or c = :; -• y .:- : . x ^ (1) 1 + wm' ^ / Thus the tangent of the angle between Fig. 32. 4 — 2 ^ y—4tx-\-l and ?/=2a; — 1 is c = = -, -and the angle may 1 -f- 8 9 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 2 — 4 2 the latter case c = = , the difference in sis^n being due 1+89 =. » to the fact that the tangents of the supplementary angles MPQ 48 ANALYTIC GEOMETRY. and QPS which the lines 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 making a given angle with a given line. Let y — mx-^h be the given line, y = m'x-{-b' the required line, and c = tangent of the given angle. Then in the relation c = , c and m are known. Solving for m', "we have 1 + mm' m'= — -— • Hence the required equation is 1 — mc 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 liberty 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, . 2/-y=^!L+£(x-a;') (2) 1 — mc is the equation of a straight line passing through a given point and making a given angle ivith a given line. Thus, the line through (2,4), making an angle 45° with ?/= 2 ic— 4, is2/-4=?±i(a;-2), or y = -ox + 10. Constructing, MJSf is the given line /M" y = 2x — 4:; P' the given point (2,4); \Ch^ and P'Q the line y = -?>x-\-\^. The \-J student will observe that P'll makes ir an angle 135° with MN^ the angle be- -f\ X ^"& measured, as always, from the line ' ^ to the left ; and that, therefore, to obtain the equation of P'R we should Fig. 33. 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. Hence (Art. 33) c= ; = 0, or 7n = m'^ that is, two lines ^ 1+mm' a?'e parallel tvhenever, their equations being solved for y, the coefficients of x are equal. This follows obviously from the fact that parallel lines make equal angles with X, and hence the tangents of these angles, m^m\ must be equal. Thus, 2/=2a; + 4,2/ = 2aj — 7,2/ — 2fl;=0, are all parallels. Cor. TJie equation of a line passing through a given point (ic', y') parallel to a given line y — mx + 6, is (Art. 31) y — y' = m {x — x'). (1) Second. If the lines are perpendicular to eadh other, their included angle is 90°, and hence c = — — ■ = oo, 1 + mm' or l+mm' = 0, .-. m'= ; m that is, two lines are perpendicular to each other whenever, their equations being solved for y, the coefficients of x are negative recip- rocals of each other. Thus, ^=-|a; + 4, 2/=-|a;-6, 2/ + |aj = 0, are all perpendicular to y= f ic + 7. Cor. The equation of a line passing through a given point {x\ y') perpendicular to a given lirie y = mx + b is y — y'= (^x — x'). (2) 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. 50 ANALYTIC GEOMETRY. E:^AMPLES. 1. Find the angles between the lines 2/ = —a? + 2, yz=zSx + 7; y = ^x-l, y = ^x + A; y = 2x — S, y=2x+7; yz=^x-3, ?/ = -2a?+9; 3?/ + 4a;+l = 0, 2y + x + 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 y = — Sx-\-4:. Ans. y = b. 3. Write the equations of lines making angles of 45° and 135° with 22/-a; + 3 = 0. Ans. y = 3x-hb', y = -\x-{-h, 4. Write the equation of a line through (—3, 7) making an angle whose tangent is V3 with 2?/ — ic4-l = 0. Ans. (2- V3)2/- (1 + 2V3).T-17 + V3 = 0^ 5. Write the equations of two parallels to 2/ = |^ .^• + 7 ; also to 3 2/ + 7 a; = 0. 6. Write the equation of a parallel to 3?/ — 4a;=2 through (1,2). Ans. 3?/ — 4a;-2 = 0. 7. Write the equations of two perpendiculars to y— — ia? + 4 ; also to ?/ — re + 4 = 0. 8. Write the equation of a line through (7, — 1) perpen- dicular to 2/ = — 4a; 4- 1 ; also through (7, — 1) perpendicular to 32/-2a; = 0. Ans. 42/--a; + 11 = ; 22/ + 3a;- 19 = 0. 9. AYrite the equations of lines through (1, 3) making angles of 0°, 90°, and 45° with X. Ans. 2/ = 3 ; a; = 1 ; y-x~2 = 0. 10. Write the equation of a line through (5, 3) parallel to the line whose intercepts are 3, 2. Ans. 32/ + 2a/' — 19 = 0. 11. Write the equation of a line through (2,3) perpendicular to the line joining (2,1) with (—2,5). Ans. y = x-\-l, 12. The vertices of a triangle are ( — 1 , — 1 ) , ( — 3 , 5) , ( 7, 1 1 ) . Write the equations of its altitudes. Ans. 32/ -a; -26 = 0; 32/4-5a; + 8 = 0; 32/ + 2a;- 9 = 0. THE RECTILINEAR SYSTEM. 51 13. Write the equations of the perpendiculars erected at the middle points of the sides of the triangle of Ex. 12. Ans. Sy — x-8 = 0; 3^/ + 5a; — 34 = ; 3?/ + 2a; - 21= 0. 14. Prove that Ax -\-By -{- C=0 is perpendicular to A'x-\-B'y-{-C' = 0\t AA' + BB' = 0. 15. Prove that Ax -{- By -i- C = is parallel to A'x+B'y+C'±=0 itAB'-A'B = 0, 16 Prove that the angle between Ax -\- By -}- C=0 and A'B — AB' A'x 4- B'v + C = is given by the relation tan y = — .• 17. Write the equation of a straight line perpendicular to Ax -{-By +(7=0 and making an intercept a on the axis of X, 18. Write the equation of a line perpendicular to y = mx -f b and at a distance d from the origin. 19. Write the equation of a line parallel to y = mx-\-b 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. .x"> INTERSECTIONS. 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 tvhich satisfies them both. The above reasoning is obviously entirely general. Whatever 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 y^ 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 ^th and gth degrees, respectively, will be pg, 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 = ^ and x + y=l are parallels. Com- bining, we obtain 3 = 0. Hence non-intersectioyi is shown by the occurrence of impossible or imaginary results. Otherwise, equating the values of £c, 0y = 3, or y = |=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 2?/+a;-10 = 0. Ans, (f,-V-)- 2. a; + 22/-5 = 0, and2a; + 2/-7 = 0. Ans. (3,1). 3. 2/-a; + l = 0, and?/4-aj + l = 0. Ans. (0,-1). 4. e^cc + G?/— 1 = 0, and a;-}-?/ = 4. 5. a; + 2/=0, and ic— ?/ = 0. 6. Find the vertices of the triangle whose sides are 5?/ — 12a;-8 = 0, 3y — 4a; — 8 = 0, y — 2x = 0, Ans. (1,4), (-4, -8), (4,8). THE RECTILINEAK SYSTEM. 53 7. Showthat2/ + 3a; — 1==0, y -\-2x + 7 = 0, y — x+Si = 0, meet in a point. 8. Show that the medials of the triangle of Ex. 4, Art. 32, meet in a point. 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 tlie inter sec- 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 = Q, be the equations of any two straight lines, and h any arbitrary con- stant ; then is (3) Ax -\- By -\- C -\- k {Ax -{-B^y + O) = the equation of a straight line through their intersection. For the values of x and y which satisfy (1) and (2), evidently satisfy (3) also, hence (3) passes through the point 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 — O, $ — 0,be the equations of any two loci, a and )8 representing any functions of x and ij, and k be any arbitrary 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 k may have any value, it may be determiued so that (3) shall fulfil any reasonable condition. Thus : 54 AKALYTIC GEOMETRY. First. To find the equation of a straight line passing through 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') tlie given point. Then (3) is a straight line through their inter- section. Since this line is to pass through {x\y'), we have Ax' + By'+C + k {A'x' + B'y' + C) = 0, in which everything is known but k. Determining k from this equation and substi- tuting its value 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- pendicular) to a given line^ Let (1) and (2) be the given lines and y = mx + h 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 k and sub- 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 a;-|-22/ — 5 = and y — 3a;-f 8 = 0, and the point (6, 4). Substituting a: = 6, .y = 4, in :r -f 2 y — 5 + ^- (3/ — 3 .r + 8) = 0, we find k = f. Hence the required line is 1/ — x + 2 = 0. 2. Through the intersection of 2a; + 2/— 7 = and x + 2y — 5 = 0, parallel to 6 a; — 3 2/ + 5 = 0. We have2a: + y-7 + ^(^ + 22/-5)=rO. Solving for 3/, l-\-2k l + 2k Solving the parallel for y, ?/ = 2 a: + |. Hence — "^ =2. .'.k = — f, and the required Hne is2x — y — 6 = 0. 1 + 2A: 3. Through the intersection of 2a; + 2/ — 7 = and y — x — l = 0, perpendicular to Sx + Sy—1 = 0.. Ans. y — x—l = 0. THE RECTILINEAR SYSTEM. 55 4. Through the intersection of y — x — l = and y — 2x-\-l = 0^ parallel to y = Ax-\-7, Ans. y = 4:X — 5, 5. Through the intersection of y z=Sx -{- 14: and y = x-{-G, making an angle of 45° with y = 2x. Ans. y = --Sx— 10. 6. The line y — mx + h passes through the intersection of y = m'x -}- &' with y = m'^x + 6". Find the value of m. DISTANCES BETWEEN POINTS AND LINES, AND ANGLE-BISECTORS. 38. To find the distance of a given point from ct given straight line. Let a^cos a + ?/ sina=^ 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 perpendiculars. Hence, if OR=p\ the equation oi ST will be X cos a + 2/ sin a = p\ and since P is on ST^ x' cos a + 2/' sin a = p\ Now DP' = QR is the difference be- tween p' and p, hence the required dis- tance is D = x' cos a-\-y' sin a —p. But this is simply what the equation of the given line becomes when p is transposed to the first member and x\ y\ substituted for a;, y. Hence, to find the distance of a given point from a given line, Fig. 34. 56 ANALYTIC GEOMETRY. put the equation of the given line under the normal form, trans* pose the absolute term to the first member, and substitute the co- ordinates of the given point. Since to put Ax -{-By -{- =0 under the normal form we divide by V^^ + JB^, we have -VA' + B' 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 C form (Art. 27), the last term of (1), — , will always he ^A'-{-B' negative, since, when transposed, it must equal -\-p. Now if we make a;' and y' zero, (1) will be the distance of the origin O from the line = — z==l ; hence the orio;in is always con- 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 is2a; + 2/— 2=0. Dividing by V5, the normal form is 2ic v 2 — = + -^ = =0. Substituting the coordinates of P', (|, f), V5 V5 V5 D=hA±i^=.^ = DP<. Vo Vo This being positive, P' is on the opposite side of the line from the origin. But, substituting the coordinates of P", (— f, 2), This being negative, P" is on the same side of the line as the origin. "Were the axes oblique, the equation of the given line being a? cos a + 1/ cos /3 —jy = (Art. 26), as the reasoning above is independent of /3, a;'co8 a + y'coB^—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. 67 given point perpendicular to the given line ; find the intersection of the perpendicular a\id the given line ; then find the distance from the given point to' this intersection by the formula (2 = V(a/-' — x"y-\-(y' — y"y. Thus, to find the distance from (8, 1) to 3ic — 42/4-5 = 0; the perpendicular through (8, 1) to 3a; — 42/ + 5 = is y —1 = — ^ (x —8) ; combining this with 3a; — 42/ + 5 = 0, we have for the point of intersection (5, 5). Hence d = V(8-5)2-f- (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; + 42/-2 = 0, (2, 7). Ans. ^- ;. on the opposite side fjom the origin. 2. 3a;-42/ + 5 = 0, (8, 1). Ans. 5 ; on the side of the origin. 3. 4a;-32/-G = 0, (1, -1). Ans. i ; on the opposite side from the origin. 4. 3x + iy + 2 = 0, (2,4). Ans. ^~ ; on the side of the origin. 5. 2/-2a;+l=0, (-1, -3). Ans. 0. 6. Find the lengths of the altitudes of the triangle whose sides are 4a;- 32/ + 8 = 0, 12a;- 5?/ + 8 = 0, 2x — y = 0. Ans. The vertices are (1, 4), ( — 4, —8), (4, 8), and the altitudes A_, 16 16^ 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. , :, / Vl3 V65 V34 58 ANALYTrC GEOMETRY. 8. Find the area of the triangle whose vertices are (2, 3), (-1,4), (6,5). ^- The line through (—1, 4) and (0, 5) is x — 7 3/ + 29 = ; its normal form is -=:^ +.-^ — = 0. The distance of (2, 3) from this side is =^. V50 \/50 \/60 _ \/50 The length of the line joining (— 1, 4) with (6, 5) is \/50. Hence the area = if-l^X V50'\ = 5. " VV50 / 9. Find the area of the triangte whose sides are 2x-^y—l = 0^ ?/ — a;— 1 = 0, x-\-2y — b = 0. Ans. The vertices are (3, 1), (2, 3), (1, 2), and area f. 10. Find the distance between the parallels 2/= 2 a; — 6, y = 2ic + 8. The line y = 2x — Q crosses Y at (0, — 6) ; the distance of this point from y = 2 .r + 8 is — 14 V5* 11. Find the distance between the parallels yz= 3a;, 2/ = 3a;— 10. Ans. VTO. 40. To find the equation of a line bisecting the angle between two given lines. Let a; cos a +2/ sin a —p = 0, (1) a;cosa' + ?/ sina'— p' = 0, (2) be the two given lines. Then {x cos a' -f 2/ sin a' — i>') + A; (a; cos a + 2/ sin a —p) = (3) is a straight line through their intersection. Now the quantities in the parentheses are the distances of any point (a;, y) from the lines (2) and (1) (Art. 38). ' Thus, if MN, M'JSf'^ be the lines given by (1) and (2), then (3) is the equation of some line VP through their intersection V, and the parentheses are the distances PD', PD, of any of its points from j,.g 35 M'N and M^T. Now if A; = -l, THE BECTILINEAR SYSTEM. 59 PD^ = PD from (3), and (3) will be the equation of the line bisecting the angle MVN\ When 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 'k= — \. For the angle iV'FA^, 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 A;=l. Hence, to bisect the angle between tioo 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 12a; + 5 2/ — 2 = and 3a; — 4?/+ 7 = 0, in which the origin lies. Ans. 99a; — 27?/ + 81 = 0. 2. Find the bisectors of the angles between 2a; + ?/ + 8 = 0, a;-{-22/-3 = 0. Ans. 3a; + 32/ + 5 = 0; a;-^ll=0. 3. Find the bisectors of the angles between 2a;-f^ + 8 = 0, and 2/ = 0. ji^s. 2 aj + (1 ± ■\/l)y +8 = 0. 4. Write the equations of the bisectors of the angles between the axes y = 0, .t = 0. Ans. y ±x — 0. 5. Of what line would Eq. (3), Art. 40, be the equation if fe=2? if ^• = w? 7 60 y 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, any rectangular equation of a straight line may be transformed into the corresponding polar equation (that is, the polar equation expressed in terais of the same constants) by means of the relations x = r cos 0, y = r sin 6 (Art. 23). The simplest and most useful of tlie polar equations is the normal form. 42. Normal polar equation of the straight line. The normal rectangular form being x cos a-\-y sin a=p^ substituting a;=r cos and 2/ = rsin^5 we have r(cos^ cosa-f-sin^sina)=^7, whence »•= ^ r- (1) V COS {0- -a) r — P Discussion. >a and increasing, 6— a is increasing, cos(^ — a) decreases, and hence r increases till = a + 90°, when cos(^ — a) = cos 90° = and r = oo , as it should be, since r is then parallel to MH and must be produced infinitely to meet the line. When 0>a + 90% ^-a>90% and r is negative, showing that it must be produced backwards, or away from the end of the measuring arc, to meet MN, and remains negative THE POLAR SYSTEMgj^ CALIFQ^S^ 61 till ^ = a 4- 270°, or ^ — a = 270°, when r = oo again, and is par- allel to MN, For 6 = 360°, r = ^ — r = -^ = OQ. The cos (—a) COS a entire line is traced for values of 6 between 0° and 180°, for p p which latter value of ^, r = tt^ttb : = = OQ. . COS(180 — a) 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= ^j i^ ^^^ the left of the pole, a = 180°, and the equation becomes P P -P r = cos {0 - 180°) cos - (180° - 0) cos P Hence 7* = ± ^ is the equation of all perpendiculars to the 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 ^ cos ((9 - 90°) ~ cos - (90° - 0) sin ' if below the polar axis, a = 270°, and P P P r = cos (^-270°) cos -(270°-^) sin ^ P 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 pole, i) = 0, and r = 0, except when = 90° -|- a, in which case r = - ; that is, r is zero for all values of 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 inclmed 45° to 62 ANALYTIC GEOMETRY> the polar axis. Find the intercept on the axis, and the values of 6 for which r is infinite. 5 - ^^^^•^= Tn — T^x'' 5V2; 135°; 315°. cos(^ — 4o) ' ' 2. Write the polar equations of lines for which p = 2, a= 60° ; ^ = 10, a = 120° ; and find their intercepts. 2 « 3. Construct r = : — ; r = cos(^-30°) cos (^-60°) 4. Construct r= ± : r = ± cos sin 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 6. Find the rectangular equations of r = ; r= cos 9 sin 9 7. Find the rectangular equation of r = ^ ^ cos (19 -45°) Ans. x + y — dV2 = 0. 8. Find the polar equation ot3x — 4:y-{-l = 0. If the normal form is required, — — ^^-i — = 0, whence /> = |, and o= cos"' (— f), which may be found from the tables. Then substitute p and a in r = -2 -. If the normal form is not specified, substituting directly cos (e — a) - "^ the values of a: = r cos 6 and y = r sin 6, we have r — 4 sin — 3 cos 9. Find the polar equation of 7j = 3 x-\-2. 2 Ans. r = sin ^ — 3 cos V APPLICATIONS. 63 SECTION VI.— APPLICATIONS. 43. Recapitulation. The foregoing formulae 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 formulae, 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 meaning of all the quantities involved. Thus, x = ^ "^^ , y = ^ , 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\ y\ a;", y'\ the coordinates of the given points. aJ = ^^^', y = ^^±^'. Equation (3), Art. 6. d = V (a;"- a;')'+ {y"-y'y. " d = Vr'2+ r"'^- 2 rV" cos (6"- 6') . " x = Xq-^x,, y = yo + yi. " a;=r cos^, 2/ = r sin^, " Ax + By-\-C=0. * " ^ . y a b y = mx-\-b, '* (1), 7. (1), 13. (2), 22. (4), •23. (I)' 25. (1), 26. (2), 26. 64 ANALYTIC GEOMETRY. X COS a-\-y sma=p. Equation (3) , Art. 26. 31. x'—x' c=^^- " (1), y-y' = ^i±^(x-x'). " (2), 1 — mc / 1 m = m , m = -• y-2/' = m(a;-a;'). " (1), y-y' = --{x-x'). " (2), m j^^ Ax'+By'+C ^ ,, .^. xcosa'-\-y sina'— p'± {xcosa-\-y sma—p)= 0. p COS {0 — a) (1), 32. 33. 34. 35. 35. 35. 38. 40. 42. PROPERTIES OP RECTILINEAR PIGURES. ' 44. 1. The diagonals of a square are perpendicular 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. Then (0, 0), (a, 0), (&, c) are the ver- tices, (-, -), [ "^ , I), the middle points, and y = -\B, the \^^ ZJ \ Z ZJ z line joining them. APPLICATIONS. 65 3. The diagonals of a parallelogram bisect each other, * With the axes as in the figure, let the side OB = a, the altitude mD = b, and Oni = c. Then the coordinates of C are (a + c, 5) , and the middle point of OC is ^ The coordinates of B are ^ ^/ Fig. 37. (a, 0), of D, (c, 6), and of the middle point of i)^,(^, I 4. T/ie straight lines joining the middle points of the opposite sides of any quadrilateral bisect each other. Let (0, 0), (a, 0), (b, c), {d, e) be the vertices 0, 5, C7, D, in order. Then the middle point of each line is fa-hb±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 (£ parallelogram form a parallelogram. With the notation of Ex. 3, the sloi^e of the lines joining the middle pointsk. of DC and -BC, DO and OB, is ; hence ^ these lines are parallel. c — a 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, (7, be squares on the sides of a right- angled triangle ORQ, and OT is perpendicular to BQ, then BS, QP, and OT meet in a point. With the axes as in the figure, let c = OQ, d = 07?, the sides. Then the coordinates of S and B are (c, — c), (—c?, 0), and the equation of SB is 66 y = ANALYTIC GEOJVIETRY. cd c + d c + d Similarly the equation of QP is |._1-=1, and y== — — ^ — X — c. The equation of RQ is - d -. —d — c of OT, perpendicular to it, y — -x. Substituting this value of c y in the equations of RS 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 ^4^ = c,* C being given as (x', y') . Then the altitude through G is (1) The equation of BC h y — y'= —^ — (x — x') , and that of the altitude through A is (2) X = X'. x'—c y X' X. y The equation of AC is ?/ = ^a;, and of the altitude through ^ is ^ y = --(x-c). (3) 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 triangle meet in a point. The equation of AC {Y\g. 39) is y = y^^x, and that of the perpendicular to AG through B' is The equation of BC is y — y'=: J — (x — a;') , and that of X — c the perpendicular to BC through A' is The perpendicular to AB at C is aj = ^. (3) /» Combining (1) and (2) , eliminating y, we have a? = — Hence (1) and (2) intersect on (3). 1 1 . The medials of a triangle meet in a point. The middle points A', B', C (Fig. 39), are (^■i>(j.i>(i.«> and the equations of the medials are ^=;^,-'^^^'' W 2' = ^7±^ (*-«). BB; (2) Combining (2) and (3), we find they intersect in /'^-i^, ^\ and these values satisfy (1) ; hence (1), (2), and (3) meet in a point. 68 ANALYTIC GEOMETEY. 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-\-b", (3) y = m"'x-{'b'" be the three lines. The intersection of (1) and (2) is _ j)"-b' _ m'b"-b'm" m'—m" m' — m" But these must satisfy (3) ; hence m'b"-m"b'-^m"'b'- m'6"'+ m"6"'- m'"6"= 0. 13. Shoiv that ny-20x-8 = 0, y-4x = 0, 13?/- 28a;- 8 = 0, m^eet 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"), {x"\y"'),he the vertices. The equation of a line through the first two is y—y'=^ — ^(x — x'), or x'—x" (y"~ y')^-\- (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>iy^(^x'-X"y But the denominator of this expression is the distance between (ic', y') and (ic", y"). Hence the area = -J base X altitude = -J- lx'{y"'-y") +x"(y'-7j"') + a;'" (?/"-?/') J. 15. Find the area of the triangle whose vertices are (2, 3), (-1,4), (6, 5). 16. Find the equation of a straight line passing through a given point and dividing the line joining two given points in a given ratio. y'—y" Substitute in y — y'= •^. — % {x — a;') for a;", ?/", the values of X —X APPLICATIONS. 69 X and y in Equation (2), Art. 6, and for aj', y', the coordinates h, A;, of the given point, and we have ^ m(a;"-/i) + n(a;'-7i) ^ ^' 17. T/ie bisectors of the interior cmgles of a triangle meet in a point. Let the equations of the sides of the triangle be ajcosa' +?/sina' — p' =0, (1) acosa" +2/ sina'"' — jp" = 0, (2) X cosa"'+2/ sina'"— p"'= 0, (3) and let the origin be within the trianojle. Then the oriofin lies within each of the three angles to be bisected, and the equations of the bisectors (Art. 40) are ajcos'a' 4~^-<*Mi^' —p^ — {x cosa^+>-y sina^^ — p^^) =0, (4) ajcostt" 4-y sina" — p^' — (a? cosa'^+tr^Wi^i?'") =0, (5) a;cosa"'+?/sTna"'— ^:»"'— (iccosa' +y^ma^ —!>') =0. (6) But values of x and y which satisfy an}- two of these equations also satisfy the third ; hence these three lines meet in a point. ^ 18. The bisectors of any tivo exterior angles of a triangle and of the third interior angle meet iii 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. 19. Tlie bisectors of the angles between the bisectors of perpen- diculars are the perpendiculars themselves. Let y = mx-\-b, y = x + b', be the perpendiculars. Their normal forms are ^^^ y — mx — b _^ my + x — mb' _ ^ Vl+m^ ' Vl+w? 70 ANALYTIC GEOMETRY. and their bisectors are y — mx — b± (my -f- ^ —mh') = 0. The normal forms of these latter are (1 + m) y + (1 - m) a; - (mb'+ &) ^ q V(r4-m)2 + (l -my (1 — m) y — ( 1 4- m) a; + {mb'— &) _ n V(l+m)=^ + (l -m)2 and their bisectors are {l-\-m)y-\-(l — m)x — (mb'-{- b) ±[_{l-m)y-(l+m)x + (mb'- 5)] = 0, or y — mx — 6 = 0, and my + a; — m6'= 0, which are the given perpendiculars. 20. To find the condition that the three points- {x\ y'), («", y), (a;'", 2/'")? shall be collinear, i.e., lie on the same straight line. 21. Prove that the line which divides two sides of a triangle proportionally is parallel to the third side. CHAPTER III. EQUATION OF TEE SECOND DEGREE. THE CONIC SECTIONS. SECTION VII. — COMMON EQUATIONS OF THE CONIC SECTIONS. 45. The Conic Sections. It has been shown in the previous chapter tliat every complete equation of the j^rs^ degree between X and y, Ax + By -\- C=0, and the various forms wliich such an equation may assume owing to a change in the vahies 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, Af'-^Bxy+Cx^ + Dy + Ex + F^O, and the various forms it may assume when different values and signs are given to the arbitrary constants A^ B^ O, Z>, E^ F, represents some one of a fai..ily of loci called the Conic Sections. These loci, which for brevity may be designated Conies, are so named because every section of tlie 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 famil}^ may thus be generated by a point moving under a single law, it will evidently be possible to express this law in a single equation, and to derive the particular cases froq^ 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 (7, B 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 write the equation of any circle tvhose position and radius are Jcnoivn, substitute the given values of m, n, and R^ in the above equation. Thus, the equation of the circle whose radius is 6 and centre is (6, -2), is (2/ + 2)2+(a;-6)2=36, ov f-^x" + Ay-l'2x+i=0. By assigning different values to m and ?i, 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 oriHn, in which case Fig. 40 m = 0, n = 0, and (1) becomes y' + x' = R\ (2) Fig. 41. COMI^IOK EQUATIONS OF COKIC SECTIONS. 78 called the central equation of the circle. Second : when the origin is at the left-hand extremity of any diameter assumed as the axis of X, in which case m = B, n = 0, and (1) becomes y'=2Mx-a^. (3) Thus, the central equation of the circle whose radius is 6 is f^x'=SQ, \ J and when referred as in Fig. 42, ^"^ ^ The a,bove 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. . 48. Tlie equation of every circle is some form of the equation fj^x" +py + Ex + i^= 0. (1) ' Expanding the general equation of the circle {y-ny^-{x-my=R\ (2) we have y- -\- oi? — 2 ny — 2mx + m^ + n^ — J?^ = 0, (3) which is of the same form as (1). The first two terms of (3) are independent of m, w, and R^ so that no change in the^ position or magnitude of the circle can affect these terms ^ Every equation of a circle, therefore, will contain the squares of X and y with 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 + i^= 0, is the equation of alt circles whose centres are on F; 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 E = 0, D=0, and we have the central form y^ -\-x- = 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 F = 0, and (1) becomes ?/^ + a^ + Dy -f Ex =0 ; if /), E^ and F are all zero, we have y- -{-3? = 0, which is true only foric = 0, y=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 fj^x'+Dy + Ex + F=0, (1) which is not impossible, is the equation pf a circle, >,( C^^^^t/AjL^UiH/^^^L Adding to both members of (1) the squares of half the co- efficients of aj and ?/, we obtain .-/■'.>'" ♦' 4 4 4 4 or {y + ^^+(x+^=\{D-'-+E'-lF), (2) which is of the same form as {y-ny+{x-my=^R\ (3) and in which, therefore, -| = «, -~ = m, \{r^+ir—iF) = R'. (4) If D'+E^>4:F, then E- is positive, R is real, and the equa- / F Ty\ tion represents a circle whose centre is ( , — — ), and whose radius is ^ ^1)'+E'-4.F. If D^^ E'=4.F, then B'- is zero, and the equation becomes (y-\ — ]+(x-\ — )=0, which is satisfied only for x = — —, 2/= — —, or the circle becomes a z z point, namely, the centre, which may be regarded as- a circle whose radius is zero. If Z>2-|-^<4 F, then R- is negative, R is imaginary, and the equation is impossible since the sum of two squares cannot be negative. We have thus three cases, in which R is real, zero, or Imaginary, and for brevity and COMMON EQUATIONS OF CONIC SECTIONS. 75 uniformity of expression we may say that every equation of the form y--\-ir-{-Dy -\-Ex-\-F=0 is the equation of a circle^ real or imaginary. The equation ay'^-\-ax^-\-dy -\-ex -\-f=0 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 {y—ny-\- {x—my=Br^ the centre (m, n) and radius R may, of course, be determined by inspection. If given in the form y'^-\-^+ Dy + Ex + F=0, we may put it under the above form b}^ adding to both members the squares of half the coefficients of x and y, as in Art. 49. Otherwise, by equations (4), Art. 49, the coordinates of the centime are half the coefficients ofx and y with their signs changed, and the radius R = \-\/D^-\-E'-—^F. Thus, given y2+a;^-42/+2a; + l = 0, m = -l, 7i = 2, 7^=1- Vl6+4-4 = ^. 51. The equations of concentric circles differ only in their absolute terms. Since the values of ml = i and nf = ^i are inde- 27 V 2^ pendent of F, and R = ^VD'-{- F^—4tF, 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 whose 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. (2/-8)2-f (a;-0)« = 49, or y^+aP-Uy +15=0. 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 (—5, 5), radius 5. , / 76 ANALYTIC GEOMETKY. 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? af+2y^+Sx—4:y-i-7 = 0;t x--\-y--\-xy -j-x-}-y —1 = } x'-f+x-2y + 4. = 0; x^-Sy'+x-Sy^O; a^+2y-4:x-l = 0; 2x^+2y^-+4:xSy-{-7 = 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 cases) . 6. Determine the position and radius of t he following circles y^-^x^-8y + 4:X-6 = 0, Ans. (-2, 4), 5. y2^a^^l0y-Ax-7 = 0/ Ans. (2, -5), 6. /+ x^+lOy 4-4a; -20 = 0. Ans. (-2, -5), 7. y^+x^—2y+Qx=0. Ans. (-3, 1), VlO. y2^x''+3y-7x-l-=0. Ans. (h -|).4. 2,24. 3^4.42/ _2a; + 5 = 0. Ans. (1, -2),0. 362/-+36aj2-242/-36x-131 = = 0. Ans. (i,i),2. y2^a^^yJ^X-l=0. Ans. (-4. -i),iVG. y'+x'+y + x+l=0. A71S. (-4, -4),iV-2 y^+x'-y-x+i = 0. Ans. (i,i),lV-14. 7. Write the equation of the circle whose centre is at the origin, and which touches the line Sx — iy +25 = 0, Putting the equation of the line under the normal form, 3 a: , 4?/ - which must equal R. Hence y'^-\- x^ = 25. 8. Write the equation of the circle whose centre is (0, 0), and which touches the line Sx + y —C) = 0. 9. Write the equation of the circle whose centre is (2, 3), and which touches Sx-\-4:y -\-12 = 0. COMMON EQUATIONS OF CONIC SECTIONS. 77 10. Prove that the sum of the equations of any 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. The general equation of the circle being {y — nf -]r{'X—mf= W^, let the pole be taken at the origin and the polar ax:is coincident with X. Then, if r\ 6' (Fig. 43), are the polar coordinates of the centre C, and ?', 0, those of any point P, from the formulae for transformation, Eq. (4), Art. 23, we have a; = rcos^, y = r sinO, m = r' cos 0', n = r' sin 6\ which being substituted in the above equation, there results, after reduction, 7-2-2n-' cos {6-6') =B'- r'\ (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 6'. Thus, if the centre is at the pole, r'= 0, and (1) becomes r = R, (2) which is true for all values of 6, (See Ex. 1, Art. 19.) If the pole is on the curve, and the polar axis a diameter, ^'=0, r'=E, and (1) becomes r=21{cos6, (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 GEOMETKY. Discussion of Equation (1). Solving the equation for r, we have r=r' coB(9-0')± ^m-r'^»in^d-d'), which gives two values of r for every value of 6, locating two points P and P^, so long as ^2>r'2 8in2(e — «'), or Jl>r' Bm{0 — e'). If Ji FF'=2c^ or A and A' are without the Fig. 45. foci. Making x = 0, the F-intercepts are ± Vtr— c\ which are real, since a > c, and locate the points B, B\ Solving the equation for y/, y=± -V(a^-c2)(a2-a;2), which is imaginary if a; > a numerically, and therefore the curve lies wholly within the limits A and A' along X. Solving for x, which is imaginary if ±a f >FS' > 1, or y> Va"— c^ numerically, or the curve lies wholly ivithin the limits B and B' along Y. 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 eccentricity by e, we have e — = -, .-. c = ae^ which siibsti AO a II 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'= 2a. 55. Defs. A A' is called the transverse axis of the ellipse, BB' the conjugate axis, A and A' the vertices, FA andi^^' (or FA and FA') the focal distances, the double ordinate through either focus, as GG', the parameter, and the distance from the focus to the centre divided by the semi-transverse axis f — — | \A0J 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 F centricity by e, we have e = — tuted in Eq. (1), Art. 54, gives I f+{l-e'')o?=a\l^e'), (1) another form of the central equation, in terms of the eccen- tricity. Representing the conjugate axis BB by 26, 2h—' 2Va^— c^ .-. a^— 0^=62^ which substituted in Eq. (1), Art. 54, gives ay4-&V=a^6S (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 always less than unity. CoR. 2. Since c^=a^-b\ e = -= ^^'~ ^\ the eccentricity in terms of the semi-axes. ^ ^ c CoR. 3. Since e = -, c = ae^ the distance of either focus from the centre. Note. The student will observe that the form of the ellipse will vary with a, b, c, and e, and therefore that the constants in the equation of a locus may serve to determine its form as well as its magnitude and position (Art. 16). COMMON EQUATIONS OF CONIC SECTIONS. 81 57. Length of the focal radii. P being any point of the ellipse (Fig. 45) , FF'-=FM'+MP^= (ae-^xy+y' (Art. 56, Cor. 3) = (ae + xy+ (a^- af) (1 - e") (Art. 56, Eq. 1) = a^ + 2 aex -\-e^x^= (a + exy ; or FP =a-\-€x. But FP-{-F'P=2a, .-, F'P=2a-(a + ex) = a--' ex. Hence, the focal radii to any poiiit whose 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 r = a(l-e^) (1) 1 — e cos B Discussion of the Equation. When (9=0°, r = a(l+e) = i such that FS = -^ or p = eFS, and draw DD' perpendicular to FS. Then FF= e {FS + FM) = eSM= ePQ, 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 Cor. 1. The ratio is equal to the eccentricity and is always less than unity. A W Cor. 2. Since J. is a point of the curve, —— = e, ...^^ = ^==£(lzL^(Art. 58). e e T? ^u ^'^ AiQ! ^'^ a(l-\-e) TT For the same reason — — - = e, .-. A'/b = = — ^^ — ' — ^. Hence, A'/S e e the distances from the vertices to the directrix are — ^ — ^—1 , CoR. 3. FS = FA + AS=a{l-e)+ ^^^~^^ == ^'^'^~^'^ , the distance from the focus to the directrix. Cor. 4. OS = 0F+ FS=^ae-^ ^(^"^'^ = -, the distance from the centre to the directrix. 60. Geometrical construction of the ellipse when the ratio is given. k Lete = -, in which A;^2_^^2^ 4^2.^4^ V(^^=^^^)M^+(a;-c)^+2/^ or cx — a?=a^{x — c)^+ 2/^. Squaring again, aY+{o?-<^)^ = a\a}-(^), (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 ± a. Take OA=OA'=a., then AA'=:2a, the constant difference, and as the difference between two sides of a triangle is less than the third side, PF- PF' =:2a or, since €"> a, and therefore e = - >1, a I 2/=-(e'-l)ar'=-a»(e=-l), (1) the central equation in terms of the eccentricity. Substituting a^— c^ = — b^, in the same equation, we obtain a',f-b'x' = -a^b^, (2) the common form of the central equation of the hyperbola. Note. The student will observe that Equations (1) and (2) differ from the corre- sponding equations of the ellipse (Art. 56) only iii the value of e and the sign ofb-; also, that while a?=0, in Eq. (2), gives ?/=± V— ^> ^^ imaginary quantity (as it shoul3, be, since the curve does not cross Y), its numerical value is the semi-conjugate axis, as in the case of the ellipse. CoK. 1. Since e = -, and c^ a<, the eccentricity of the hyper- a bola is always greater than unity. \ Cor. 2. Since a^— (? = — W^ e = - = i — , the eccentricity ' in terms of the semi-axes, c CoR. 3. Since e = -, c = ae = Va'^+6'=^5 (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 when the axes are given^ ivith as a centre and AB as a radius, describe an arc; it will cut the transverse^ axis in the foci, 67. Length of the focal radii. P being any point of the hyperbola (Fig. 48) , FP' = FM' + MP' = (ae + xy+ f- (Art. ^^, Cor. 3) ^(^ae-\'xy+{x'-a^){e''-l) (Art. 66, Eq. (1)) =e^a^+2aea; + a^ = (ex + a)^ ; or FP=^ex-\-a. COMMON EQUATIONS OF CONIC SECTIONS. 91 But F'P=FP — 2a = ex + a — 2a = ex — a. Hence tJie focal radii to any point whose 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 the triangle FPF', P being any point of the curve, ^fp2 =i<^p2_|. FF^'^-2FP, FF' cosF'FP. But FP=r, F'FP = 0, FF' = 2ae, Siud F'P=FP— 2 a = r-2 a. Making tliese substitutions, we obtain e cos 6^—1 (1) Discussion of the equation. When ^=0°, r—a(e-\-l)=FA'; when ^ = 180°, r = —a(e — l) = FA', or the focal distances are a{e± 1), numerically » When 6>=90°, r = - a (e^-l) = - a/"-^- A = -~ ; or the parameter^ G0\ is 2a (e^— 1), or — , numerically, \ a I As 6 increases from 0°, cos^ diminishes and r increases, tracing the branch ^'P, r becoming infinity when ecos^ = l, Figo 49. 92 ANALYTIC GEOMETRY. or ^ = cos~^— When e cosO< 1, r is negative, and the branch € G'A is traced, in the direction G'A, r being FA when ^ = 180°. When passes 180°, cos^ is negative and r 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 traces the branch LA\ Eepresenting the parameter GG' = 2a{e^ — l) by 2p, the polar equation (1) may be written e cos — 1 (2) 69. The ratio. TJie hyperbola can he traced by a point so moving that the ratio of its distances from a fixed point and a fixed straight line is constant. From the polar equation of the hyperbola, r = ^ , we e cos 6 — 1 have r = erco^6-p, or (Fig. 49) FP=^eFM-p. Take -F/S p such that FS = -, or ^ = 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. The fixed line DD' is called the directrix. CoR. 1. The ratio is equal to the eccentricity, and is always greater than unity. Cor. 2. Since ^ is a point of the curve, 4f=e, ■•■^^ = ^="('^-1) (Art. 68). AS e e For the same reason -— — = e, . • . AS — = — '^ — ' — ^ • Hence A' S e e the distances from the vertices to the directrix are —^ — — — ^. e COMMON EQUATIONS OF CONIC SECTIONS. 93 Cor. 3. FS = FA -]-AS =a (e - 1) + ^ ^^ ~ "^^ = ^ (^'-^) ^ e e the distance from the focus to the directrix. Cor. 4. OS = OF- FS = ae- ^ (^'-1) ^ ^ ^ the distance e e from the centre to the directrix, y 70. Geometrical construction of the hyperbola when the ratio is given. Let e = -, in which /c > 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 any parallel to GG', as N'L'. With jP as a centre and M']^' as a radius, describe an arc, cutting the parallel in P' and F" ; 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, FP' ^ GF^ ^^ P'Q' FS ' Since J^iMi > MiS by construction, the arc described with FPi—M\Ni, as a radius will determine Pi, Pg? 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 Pas', taken to represent s, 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 AljTALYTlC GEOMETRY. Cor. 1. Since ^ is a point of the curve, AK AjS AF AS' But = e, .'.AF=AK. Similarly, A'K'=A'F. Hence, to find the focal tangents when the axes are given ^ first determine Fig. 50. the focus (Art. 66, Cor. 3), then make AK=^AF and A^IV^A^F. KK^ will be the focal tangent, and its intersec- tion with the axis, /S, a point of the directrix. COMMON EQUATIONS OF CONIC SECTIONS. 95 Cor. 2. Since the curve is symmetrical with respect to its axes, and CF' = CF^ there is another directrix on 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 equals it is said, to be equi- lateral. Making a = 6 in the common form of the central equa- tion a^y^ — b^oi^ = — a^b^, we have for the equation of the equilateral hyperbola. Two 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 hyperbola, the transverse axis had been assumed coinci- dent with Y, the equation of the hyperbola in this position would have been a^x^ — b^y^ = — a^6^, as this supposition simply amounts to interchanging x and y. Interchanging now a and 6, this becomes ^,^, _ ^,^^ ^,y, . ^2) or, the centixd equations of conjugate hyperbolas differ only in the sign of the absolute term. Conjugate hyperbolas are distinguished as the.X- and the Y- 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 6^. CoR. 1. The eccentricity of the F-hyperbola is — — J-—. 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 four foci of a pair of conjugate hyperbolas are equidistant from the centre. 72. Varieties of the hyperbola. Every equation of the form \ Ay''-\-Cx^ + 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 Ay^ — Cx^ = —F, which can be reduced to the form a^y^ — 6^^^ _ _ ^^2^2^ ^^ -^^ ^^^ ^^^^ of the ellipse (Art. 62) , by introducing the factor R = — — • Aiy whence 7^2/^ — — a^ = — -77;? in which a = y\--, and h = a|^^^^ G A Aij \ G ^ A numerically. If ^ = (7, the axes are equal and the hyperbola is equilateral. Second. If -Fis negative, (1) becomes Ay"^ — Caf = F, which is the conjugate hyperbola, since it differs from Ay^ — Cx^ = — F only in the sign of the absolute term (Art 71) . Third. If i^=0, (1) becomes Jl?/^— Oa^ = 0, or y = ±yl^x, 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 92/2 _4aj2 = _ 144? Multiplying by the factor R = — = 4, the equation becomes AG 36 ^2 _ 16 x2 = — 576, which is of the form a^ 3/2 _ 52 ^2 ^ _ ^2 62 ; the axes are therefore 12 and 8. Or, directly, making y — and a: = in succession, we have numerically x = a=Q, y = b = 4:,.'.2a=12, 2 6 = 8. 2. Find the axes, eccentricity, and parameter of 3i/2-2aj2 = -18. Ans. 6, 2V6; JVTS; 4. v/ COMMON EQUATIONS OF CONIC SECTIONS, 97 3. Find the axes, focal distances, and position of the direc- trix of ?/' -a;* =-81. _ g Ans, a = 6 =9 ; 9(V2 ± 1) ; ~;= from centre. 4. AYrite the equation of the h^'perbola whose axes are 18 and 10. Ans. 8l2/'-25a;2 = -2025. 5. Write the equation of the hyperbola whose eccentricity is |- and transverse axis 10. Ans. 9y^ — 7 oc^ = — 175. 6. The conjugate axis of an hyperbola is 4 and its lesser focal distance 1. Find its eccentricity and write its equation. Ans. I; dy'^-lQx'^-Se. 7. Construct the following hyperbolas : (a) e = f . Observe the size is undetermined. (b) e = I ; distance from focus to directrix = 8. (c) a =8, 6 = 6. -C-^0~_v^^;' >^ ^_, 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. 3y -— 2a;- = 18 ; ^P ; 2 ^1- from centre. 9 y^ — af—Sl ; V2 ; —7= from centre. V2 . ' '/ 10. The eccentricity of an h^-perbola being V3, what is the angle between the focal tangents ? Ans. 120°. 11. Find the focal distances, conjugate axis, parameter and directrix of the hyperbola r = — ziz Vlo cos (9 -3 Ans. ^-^ ; 2 V6 ; 4 ; 3^ 5 from centre. Vly=F3 ^5 12. Write the polar equation of the hyperbola whose axes are 8 and 6. , 9 Ans. r = 5 cos 0—4: 98 ANALYTIC GEOMETEY. // THE PARABOLA. 73. Defs. The path of a point so moving that its distance from a fixed point is always equal to its distance from a fixed straight line is called a parabola. The fixed point is the focus, the fixed straight line 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 /iSF perpendicu- lar to DD' and let SF = p. By definition, the middle point 0. of SF is a point of the curve. Let be the origin ancj the axis of X coincident with OF. Then, P being any poijit of the curve, and PQ perpendicular to Z>i>', PF— PQ, or 2px. (1) ^ ^r Squaring and reducing, y^ Discussion of the equation. Solving for y, we have y=± V2^a;, or y has two numerically equal and increasing values for positive increasing values of a;, but is imaginary when X is negative ; hence the curve lies wholly to the right of Y, 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 QJ, and a string, FPI equal in length to the ruler. Fix one end of the string at the focus, the other at the extremity I of the ruler, and, keeping the string pressed against the ruler 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. 61. BAR y COMMON EQUATIONS OF CON\C ^SECTIONS. CAUfOJ OX is called the axis of tlie parabola, the vertex, and the double ordinate GG' through the focus the parameter. Cor. 1. Substituting x—OF = ^- in Eq. (1), we have y=FG =p, or the parameter GG' = 2^) = the coefficient of x in the equation of the curve. Also OS = OF = ^jp = \GG^ \ or SF=^FG=FG'=p. CoR. 2. FP=qP=SO^-OM=\p + x\ or the length of the focal radius to ayiy p)oint 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 formulae for transfor- mation from rectangular axes at to the polar system, are P x = XQ-\-r cos = -^ + r cos 6^ y= yQ-\-r sin 0=r sin 0. Substituting these values in the equation y^ = 2px, we have, r" sin2(9 =2pf^-\-r cos A or ?*^(1 — cos^ 0) = jy^ + 2p)r cos 6. Transposing, 7-2 = 7-2 cos^^ -f- 2pr cos -{-p^ =z{r cos + jp)^ Extracting the root of each member, P „„ „ P orr = -^. (1) 1 — cos ^ vers Let the student discuss the equation. Observe that the equation r = I (2) 1-ecos^ ^ ' 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— ecos for the ellipse (Art. 58), that is, when e 1 ; and r = -— ^ for the parabola, when e = 1 . 76. Geometrical construction of the parabola, the focus and directrix, or the parameter, being given. Lay off SF=p = ^ 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 jP as a centre and 3FN' 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 GF N'M' ^ M'S FS P'F Fig. 52. P'Q' since the triangles GFS and N'M'S are similar, and P'F=N'M',P'Q^ = 3PS, In the same way any number of points may 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 parabola make an angle of 90° with each other. -^ = — TTT. 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 the 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, parabolas, like circles, differ only in size. Examples. 1. Construct the parabola whose parameter is 10, and write its equation. jy """ /^/K 2. Construct the parabola the distance of \^iose vertex from its focus is 2. ^y ^ : y/ 3. Write the polar equations of the parabolas of Exs. 1 and 2. 4 4. The polar equation of a parabola is r = -• Write its rectangular equation. ~~ , ^ ^ Ans, y'- = 8x. 102 ANALYTIC GEOMETKY. SECTION VIII. — GENERAL EQUATIONS OF THE CONIC SECTIONS. 77. Defs. A conic is the locus of a point 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 perpendicular 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, (m, 7i) the focus F, DD' the „. y directrix, its equation being Xi cos a + yi 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 DD\ is the axis. Join F with P, draw PQ perpendicu- lar to i>Z>Vand let e = the constant ratio. Then — — = e, or PQ FF' = e^PQ'. But PF=V{y-ny-}-{x-my (Art. 7) ; and PQ = a; cos a -f 2/ sin a —p (Art. 38) ; hence (y — ny '\-{x — my = e^{x cos a + yBina—pY (1) is the required equation, in which e determines the species, and m, n, a, andp, the position of the conic. Examples. 1. Write the equation of an ellipse whose centre is (1, 2), transverse axis is 6, eccentricity-—, and transverse axis parallel to X. GENERAL EQUATIONS OP CONIC SECTIONS. 103 a=180O, e = ^; .-. cosa = -1, sin a = 0, p= --_--l, m = 1 - \/5, n--=2. Substituting in. Eq. (1), or 91/2 + 4^2-30 ?/-8a: + 4 = 0. 2. Write the equation of a parabola whose axis Is parallel to X, vertex is at ( — 3, —2), and parameter is 9. .Ans. 2/2 4- 4?/ -9a; -23 = 0. 3. Write the equation of an ellipse whose eccentricity is — -<, V3 centre is (1,1), transverse axis 2V3, the latter being inclined at an angle 135° with X. m=l-^, n = l + -^, 79 = 3, e = — , a= 135°, V2 V2 V3 and the equation is 4. Write the equation of a circle whose radius is 5, the axes being tangent to the circle. m = n = 6; y^+ x'^—lOy -lOx -{■ 25 = 0. 5. The centre of an ellipse is (— f , 4), its eccentricity f , and its transverse axis = -J^, and is parallel to X; write its equation. A71S. 92/^4- 5a;2 — 722/+12a; + 144 = 0. 79. Every complete equation of the second degree between x and 2/, and all its forms., is the equation of a conic; and, con- versely, 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 cosaxy + (1 — e^cos^a)iC^ > (1) -\- (2e^p sm a— 2 n)y-\- (2 e^p cosa— 2 m) x-\-m^ -{-71^— e^p^=0. The complete equation of the second degree between x and y, Ay' + Bxy + Cx" + Dy + Ex -f- F= 0, (2) 104 ANALYTIC GEOMETRY. is of the same form, but the coefficients of corresponding terms are not necessarily the same, since an}^ equation may be multi- plied or divided by any factor without affecting the quality. Making these coefficients, therefore, equal, by dividing each equation by its absolute term, and designating the resulting coefficients of (2) by A', B', etc., we have 1 — e^ sin^a e^p^ — 2 easing cosg 1 —e^ COS^g 2e^p sing — 2?i m^-j-n^ — e^p^ 2e^p cosg — 2m m^+n^—e^p^ = -B', = 0', = D', = £'. (3) From these five equations the values of the five constants A\ B', C, etc., may always be determined when g, ?/i, n, p, and e are given ; and as the latter are arbitrary, such values may be assigned to them, that is, the locus may be assumed of such species and in such position, as to give ^^, B\ C, etc., any and every possible set of values. Conversely, the values of g, m, 71, 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. Had they been oblique, the distance FP would have been (Art. 7) V(y-w)2+ (a? — m)2 + 2(y-w) {x-m) cos/3, and the distance PQ would have been (Art. 38) xcosa + y coap'—p, in which /3 is the given inclination of the axes, and ^' the angle made by PQ with Y. The equation PF^ = e^PQ^ would, therefore, have Involved the same arbitrary constants, GENERAL EQUATIONS OF CONIC SECTIONS. 105 and no others. Passing now to rectangular axes, since this transformation 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 B'^-4:A'C' from Eq. (3), Art. 79, we have _ 4:e* sinV cos^a — 4 (1 — e^ sln^g) (1 — e^ cos^g) {m^+n^—e^p^y _ 4ce* sin^g cos^a -^4 + 4e^cQs^a + 4e^ sin^g — 4e^siii^a cos^g "~ {m^+ri^—e^p^y _ 4 (e^- 1) (^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, ^ut, 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 it is positive. Hence the conic will be an ellipse, parabola, or hyperbola, according as B'^ — 4:A' C is negative, zero, or positive. To apply this test it is unnecessary to reduce the given equa- tion to the form AY+B'xy + C'x'+D'y -\-E'x + 1 = 0; for if B'^'-4:A'C' be negative, zero, or positive, then will {KB'y- 4 (KA') (IW) =1P{B''- 4.A'C') also be negative, zero, or positive. Hence, ivhatever the co- ' ' efficients, Ay^ + Bxy + Cx^ -{- Dy -\- Ex -\- F — is the equation of an ellipse, parabola, or hyperbola, according as B^—4:AC is negative, zero, or positive. Examples. Determine the species of the following conies : (1) y^— 5 ic?/ 4- 6 a^ — 14a; -h 5?/ --h 4 = 0, an hyperbola; ■ (2) 2/--8a;2/ + 25aj24-62/-2a; + 49 = 0, an ellipse; (3) ^y'^J^Qxy^4:x'^-^y=^0, an ellipse; 106 ANALYTIC GEOMETRY. (4) y^-\-2xy -\-x'—y -}-l =zO, a x>arabola ; (5) 2/^— 1 -f 3 a; = {x — yY, a^ hyperbola; (6) 2/^= 4 (a; — 1) , a parabola; (7) 4ic?/ — 16 = 0, an hyperbola, -» • — ( 81. The equation Ay''--\-Cx'+Dy+Ex-^F=0 represents all species of the conic sections. The general equation of the conies is Ay''-\-Bxy + Co^+Dy +Ex 4-i^= 0. Passing to any rectangular axes with the same origin by the formulae (Art. 22, Eq. (8)), x = Xi cos y — 2/i sin y, y = ^i sin y -f- 2/i cos y, we have, after omitting the subscripts, A (x^ sin^y •}-2xy sin y cosy -f y^ cos^y) +B(a^ cos y siny + cc?/ cos^ y — xy sin^ y — y^ sin y cos y) 4- {x^ cos^y — 2 .T?/ cosy sin y -\-y~ sin^y) + other terms not involving xy. The term containing xy is [ 2 A sin y cos y + ^ (cos^ y — sin^ y) — 2 (7 sin y cos y] a^, or [(J.— (7) sin2y +5cos2y]a^, whioJjKwill be zero if (A-C) sin2y+5cos2y = 0, tan2y = ^^. (1) Now tan2y can have any and every value from +00 to —00, hence a value can always be found for y which will satisfy (1) whatever the values of A, B, and (7; that is, whatever the species of the conic. To find this value of y, we have (Art. 79) GENERAL EQUATIONS OF CONIC SECTIONS. 107 tan 2 A B -G = - B • F A-C- F A' -B' _ 2e' 'sin a cos a 2 ^ tan 2 a, 1 — e^ sin^a — (1 — e^ cos^a) and since tan2a = tan (180°+ 2a), (1) will be satisfied when- ever 2y=2a or 180°+ 2a; that is, when y = 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'^ Dy +Ex + F=Q is the equation of a conic whose axis is parallel to owe of the axes of reference^ and, since, B^—4:AC = — 4,AC when B=0, the conic will be an ellipse, hyperbola, or parabola, 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 Fj Ay^+Cx-+Dy+Ex+F=0 . (2) represents an ellipse whose axes are parallel to the axes of reference ; ^^2_ Cx'+Dy -\-Ex +i^= (3) represents an hyperbola whose axes are parallel to the axes of reference ; Af-[-Dy-\-Ex + F=0,^ or Ca^+Dy-^Ex-{-F=0,\ ^ ^ represents a parabola whose axis is parallel to X, or Y, respectively. CoR. 1. In the circle B = 0, and A=C (Art. 49). Hence tan 2 y = ^, or there is always a pair of axes parallel to the axes of reference. 108 ANALYTIC GEOMETRY. 82. Defs. The centre of a circle is a point equally distant from ever}' point of the circle. The pohit which has been designated the centre of the ellipse, and hyperbola, is not equally distant from every point of these loci, but it possesses a property in common with the centre of the circle, and in virtue of this common property we may define a centre for all three of these loci. A locus is said to have a centre ichen there is a point through which if any chord of the locus he drawn the chord is bisected at that 2^0 hit. Any chord through the centre is called a diameter. 83. Every locus whose equation is of the form Ay^+Bxy+Cx'+F=Qi (1) ^05 a centre. For if (1) be satisfied for any values x\ y\ of the variables, it is also satisfied for the values — x\ — y'. But the equation of the chord through (x', y') and ( — ic', — y') is x'y — y'x = (Art. 32), which passes through the origin since it has no absolute term. Moreover, the segments of the chord on either side of the origin are equal, since the length of each is ^x''^-\-y'^. Hence the locus has a centre, and the centre is the origin. CoR. 1 . Every locus whose equation is of the form Ay^+Cx^-{-F=:0 has a centre, at the origin. CoR. 2. The circle, ellipse, and hyperbola have centres. 84. The equation Ay^+Ca^-{-F=0 represents all ellipses and hyperbolas. Resuming the general equation of the conies, Ay^-\-Bxy+Cx^+Dy + Ex + F=0, (1) pass to parallel axes, the formulae for transformation being x = Xo-\-Xi, 2/ = 2/o + 2/i; GENERAL EQUATIONS OF CONIC SECTIONS. 109 j(2) and, after omitting subscripts, we obtain ■i-D{yo + y) + E{xo + x) + F=0. The terms containing x and y are {2Ay, + Bxo-{-D)y, {2Cx^-^By, + E)x, which will vanish if 2Ayo-hBxQ+D=0, and 2 Cxo+Byo-\-E=Q ; that is, solving these equations for Xq and ?/o, if the new origin is taken at the point _ 2AE-BD , _ 2CD-BE ; '''- B-4.AG' ^'~ B^-^AG' /\ \ which is always possible when B'—A^AC is not zero; that is, when the locus is not a parabola, in which case Xq and y^ would be infinity. Hence the terms containing x and y may always be made to vanish if the locus is an ellipse or an hyperbola, and, when referred to the new axes, the equation will assume the form Ay^-\- Bxy -\- Cx^-\- F= 0, from which we see that the new origin is the centre (Art. 83). By a second transformation (Art. 81), the equation will finally take the form ^2/2+ (73^4- F=0, the central equation of the ellipse or hyperbola according as A and C have like or unlike signs. CoR. 1. Since, when B^— 4,AC= 0, Xq and y^ are infinity, the parabola has no centre. CoR. 2. Since the above values of Xq and i/o are independent of i^, central conies tvhose equations differ only in their absolute terms are concentric. CoR. 3. By examining Eq. (2) we see that the first three terms of the equation are not altered by the transformation. 85. Varieties of the parabola. ^ We have seen that when B'^—4:AC=0 the centre is at in- finity, and that therefore the terms Dy and Ex cannot be made 110 ANALYTIC GEOMETKY. to vanish from the general equation when it represents a parab- ola ; also (Art. 81) that the term Bxy will vanish if either axis of reference is assumed parallel to the axis of the parabola, in which case B'— iAC becomes — 4: AC, and eithe)- A or C must be zero. Making then ^ = and (7 = in the general equa- *^^"' Ay^+Dy+Ex-{-F=0 (1) represents all parabolas. To see if this form can be still fur- ther simplified, transform to new parallel axes by the formulae x=:Xq-\-Xi, y = yf)-\- 2/i, and we have, omitting subscripts, Ay'-\-{D -\-2Ay,)y + Ex-^Ay,'-\- Dy, + Ex, + F=0. As the terms containing x and y cannot both be made to vanish, let us see if one of them, as ?/, and the absolute term can be made to vanish. This requires that D + 2Ayo = 2ind Ay,' -i- Dyo-{- Exo + F=0, or that 2/o = and Xq = 2A 4.AE The equation then assumes the form Ay'-{-Ex = 0, or. 2 E which is the equation of the parabola referred to its vertex and axis (Art. 74), the curve lying to the right or the left of the origin according as E and A have unlike or like signs. Hence the disappearance of the absolute- term and that containing y involves a system of reference wJiose origin is the vertex. This transformation sis always possible, except in two cases : First, when E = 0, in which case x, = oo. Equation (1) then becomes Ay'-]-Dy-^F=0,ov -D±VD'-4AF ^ 2A which represents two straight lines parallel to X, real and dif- ferent, real and coincident, or both imaginary, according as W^ is greater than, equal to, or less than ^AF. These are the particular cases of the parabola, the vertex receding to infinity. GENERAL EQUATIONS OE CONIC SECTIONS. Ill Second^ when ^ = 0, in which case, however, the equation ceases to be one of the second degree. 86. Defs. A diameter of a conic has been defined as a chord through the centre. As the parabola has no centre it would appear that it has no diameters. A set of lines may, however, be drawn to the parabola which possess a property in common with the diameters of the ellipse and hyperbola ; and in virtue of this common property we may define a diameter for all three species of the conies. A diameter of a conic is the locus of the middle points of par- allel chords. 87. To find the locus of the middle points of parallel chords. First. For the ellipse and hyperbola. Let Ay^+Cx' + F^^), (1) in which A=: a^, C= ± 6^ F= qp a^b^, as the conic is an ellipse or an hyperbola, be the equation of the locus, and y = a'x-\-b' (2) that of any chord PQ. Combining (1) and (2) to find the inter- sections P and Q, we have, after substituting y^ from (2) in (1), . 2a'b'A b"A + F ^a"A + C a'^A + C Fig. 54. 112 ANALYTIC GEOMETRY. or, representing the coefficient of ic by g and the absolute term by r, whence a;=-|±^r+|^ Fig. 55. which are the abscissas of P and Q. Substituting these values of X in (2), we find the ordinates of P and Q are y = a'(^-^±^f^ + b^ Now the coordinates of the middle point M of PQ are given by the formulae x = — - — , y = ^ ^ » Taking, therefore, the half-sum of the above values of x and y, we have for the coordinates of If , or, replacing the value of q, a'b'A ^^ aH'A . ^, GENERAL EQUATIONS OF CONIC SECTIONS. 113 For all other chords parallel to PQ, a' remains the same, but b' differs. Eliminating then b' by substituting its value from the first in the second of the above equations, we obtain G b' a'A a'a^ (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 through the centre. Cor. It A = G, or the locus is a circle, (3) becomes y 1 a' which is perpendicular to y = a'x + 6'. Second. The parabola. Let y^ = 2px be the parabola, and y=a'x-\- 6' any chord PQ. Combining as before, ^+^-'^'-^^^ or, x^-^qx = r, whence, in the same manner the coordinates of the middle point M are ^=-|' y=-~^ + b', or, replacing q by its value, a'b'-v from which we see thaj;/the abscissa x of the middle point varres with ^' the ordinate ?/ is constant(^f a' is constanl that is, if the chords are parallel. Hence, the locus of the middle jioints of parallel chords to the parabola is a straight line parallel to X, 114 ANALYTIC GEOMETRY. The student will observe that if a diameter be defined as a chord through the centre, the diameters of the parabola are necessarily parallel as the centre is infinitel}' 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 3/ approaches T (or T'), P and Q approach each other, and MP, MQ, 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 when it is parallel to the tangents at the vertices of the latter. 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 be ?/ = a'ic. The equation of TT' is y= a"x= --j-^x (Art. 87, Eq. 3). ,2 a a Hence a'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 parallelogram. a'a''=-^-^ (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 hyperbola^ 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_6V=~a262 (2) the equation of the X-hyperbola, and (Art. 71) that of the F-hyperbola. Combining (1) in succession with (2) and (3) , we have a'b'' x^ = — —-'> (4) aj2 = a^a'2 - W (5) Now if a' < -5 a; is real in (4) and imaginary in (5), and the Fig. 57. 116 ANALYTIC GEOMETRY. line intersects the X-hyperbola, as TJ". If a' > -t x is imag* inary in (4) and real in (5) , and the line intersects the F-hyper- bola, as KK'. If a' = ± -? both values of x are infinity. In ct this case (1) becomes y=± -x, 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=b their included angle is 90°, a and the hyperbola is said to be rectangular ; or, when an hyper- bola is rectangular it is also equilateral (Art. 71). 93. Conjugate diameters of the hyperbola. Of two conjugate diameter's, one meets the X-, the other the Y-hyperbola. Let 2T' be any diameter bisecting a system of parallel GENERAL EQUATIONS OF CONIC SECTIONS. 117 chords of which PQ is one. 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 y = a^x^ and that of TT^ is y = a'^x = -^x (p:q. 3, Art. 87). Hence a^a^' = -5 is the relation which must exist between the slopes of h a diameter and the chords which it bisects. If a' < -, a" must evidently be > -, since their product = — , and conversely ; or, .a a- " since - is the slope of the asymptote, if one diameter intersects the X-hyperbola, its conjugate will intersect the Y-hyperbola, and conversely. Again ; since the equation of the F-hyperbola is derived from that of the X-hyperbola by changing the signs of a^ and b^ (Art. 71), a'a" = — is also the relation which must exist a^ between the slopes of any diameter of the y-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 diameters form a ijarallelogram. The equation a'a^' = — a^ is called the equation of condition for conjugate diameters 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 acuto, or both obtuse, or the diameters lie on the same side of the conjugate axis. CONSTRUCTION OF CONICS FROM THEIR EQUATIONS. 94. First Method. By comparison with the general equa- tion . Make the coefficients of like terms in the oriven and the 118 ANALYTIC GEOMETRY. general equation equal by dividing each equation by the co- efficient of the same terra. Equating the resulting coefficients of corresponding teVms, 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?/ + 4a; + 4 = 0. Since B'-AAC=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 y^, 1 — e^ sin^a, and we have, after making e=l, \ r D Q ^ > \ 2L S 1 / — 2 sinacosa :0, (1) 1 - sin^a 1 — COS^a 1 — sin^a 0, (2) 2psina — 271 _ = 4, (3) 1 — sin^a 2;9COSa— 2m _ = 4, (4) 1 - sin^a 1 — sin^a 4. (5) Fig. 59. 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 180°. In either case (3) gives n = — 2. Substituting cosa = ±l in (4), we have ±2p — 2m = 4, or m = ±^ — 2, according as a is 0° or 180°. From (5), since 71 = — 2, m^ = p^\ or, substituting the above values of m, p = ±l. But p is alvva3^s positive; taking, therefore, the upper sign, a = 0°. Finally, from (4), making cosa=l and p=l, we have m^—\, Tiie values of the constants are thus : e = 1, a = 0°, m = — 1, ti = — 2, p = 1. To construct these results, lay off OQ=p = 1 to the right, since a = 0°, and draw the directrix 2)i)'^perpendicular to X. Construct F ^ 125 Solving for y, GENERAL EQUATIONS OF CO B^ — AAC= 0, and the conic is a parabola yz=:-(x+l)± V9a;+9. Construct the diameter QB, y = -{x+l). The radical gives but one value of x = — 1, for which y = 0, lo- cating the vertex T. The X- intercepts are 8, —1, and the y-intercepts 2, —4. Interme- diate points may also be found ; thus for a; = 3, 2/= 2 and —10 (P' and P"). Trace the curve through these points and tangent at T to a parallel to Y. 3. y^-{-2xy — 2a:^ — 4:y — x-j-10=0. .'. the conic is an hyperbola. Fig. 66. y = _ (a; _ 2) ± V3a;2-3a;-6. The diameter is y = — (ic — 2) , its vertices are (2, 0) and ( — 1, 3). The X-intercepts are 2, — f, the F-intercepts being imaginary. . (\ 4. y^-\-2xy-{-S^ — 4:X = 0. ^OtA^-^vW- '- 5. y^-2xy + x^-y-\-2x-l = 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 GEOMETKY. signs, an equilateral hyperbola. Solving the equation for either variable, as ?/, 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, their 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 b}' solving the equation for the variable which enters as a square and placing tlie radical part equal to zero ; this equation determines the limit, ?'.e., the vertex. Examples. 1. 9y^ + 4:x''' - 36y - 8x-{- 4: = 0. A and have like signs, ,*. the conic is an ellipse. Solving for y, 2/ = 2 ± i V-4a;2 + 8aj + 32. The limits along X are found from —4:xF-\-8x-{-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 Yis 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. 2/^ 4- 4 2/ — 6a; — 14 = 0. The locus is a parabola, its axis being parallel to X. Solving for y, y=—2± V6 a; -f- 18 ; hence its vertex is (—3,-2). 3. 42/2 + a^+162/-4a;-i-16 = 0. 4. 92/2 - 4a^ - 362/ + 24aj- 36 = 0. \/ QENERAL THEOREMS. 98. Through any jive points in a plane, one conic may be made to pass. GENERAL EQUATIONS OF CONIC SECTIONS. 127 Let (a?i, 2/1), (a?2, 2/2) » (^3, 2/3)5 (^4, 2/4), (^55 2/5)5 be the five given points. Dividing the general equation of a conic by tb.e coefficient of any of its terms, and distinguishing the new coefficients by accents, we have ^'2/' + B^^y + C"aj2 + D^y -\-Wx-\-l = 0. (1) Substituting in succession the coordinates of the given points, since the conic is to pass through them, we have ^'2/1' + ^'^i2/i + C 'X,' + Z>'2/i + J^'a^i + 1 = 0, A V + ^'^22/2 + G^xi + i>'2/2 + E'x, + 1=0, ^V + ^'^32/3 + C'xi + D'2/3 + E^x, + 1 = 0, [> (2) .4 V + ^'^'42/4 + C V + i>'2/4 + E% +1 = 0, ^'2/5' + B^x.y, + C V + i>'2/5 + E'x, + 1 = 0, in which A\ B', C, i)', 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 absohite term. This results from the fact that the equation sought can have no absolute term. Examples. 1 . Find the equation of the conic passing through (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 have 3/2 4. ^ixy + C'x'^ + D'y + E'x = 0. Substituting the coordinates of the remaining points, 16+16J5'+16C" + 4i>' + 4^' = 0. (1) 16 - 16 ^' + 16 C" - 4 Z)' + 4 £■' = 0. (2) 36 + 54jB'+81C'+6Z>'+9^' = 0. (3) - . 36-54J5' + 81C"-6Z>'+9^' = 0. (4) 128 ANALYTIC GEOMETRY. From (1) and (2), and (3) and (4), by addition, 32+ 32C'+ 8E'=0. (5) 72 + 162C"+18^' = 0. (6) Eliminating E' between these we find C = 0, which in (5) gives E' = — 4. Substituting these values in (1) and (3), we have 16B'-\-4D' = 0, 6iB' + 6D'=0, whence B' = 0, D'= 0. The required equation is therefore y^ = 4iX. 2. Find the equation of the conic passing through (—1, 2), (-f>i). (-1,1). (-f,i), (-4,8). Ans. y^-^x^-^2xy-{-Sx — y-\-4: = 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 Bxy will disappear for every substitution, and B will be undetermined. Ans. The Axes. 4. Through how many points may the conic be made to pass ? 5. Find the equation of the circle circumscribed about the triangle whose vertices are (3, 1), (2, 3), (1, 2). Ans. Sy^ + 3x^-ny-13x-\-20 = 0. 6. Find the equation of a circle through the origin and mak- ing intercepts a and b 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 combining 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 fourth 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 imaginar}', 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 iu 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. ]f the conies are circles, the simplest way of combining their equations is by subtraction. Thus, let (Art. 48) f- 4- x' + Dj/ -\-Ex + F= 0, (1) f + x^-^D'y-\-E'x + F' = 0, (2) be the given circles. Subtracting, 130 ANALYTIC GEOMETRY. {D -D')y + {E- E') x-{-{F-F') = 0, ^ '(3) from which we may find the value of either variable in terms of the other, and substituting it in either (1) or (2) find that of 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-\-liS' = 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) ^s tJie equation of the chord common to the two circles. Examples. Find the intersections of the following loci : 1. 2/' + a;'=25, y^ = ^i-x. Ans. (3, ±4). 2. 2/?=10a;-a^, y' = 2x. Ans. (0,0), (8, ±4). 3. y^ + 4.3^=2o, 42/2-25a^ = -64. Ans. (2, ±3), (-2, ±3). 4. ^^^a^^Sy^2x-7 = 0y y'' + x^-Sy+2x + l = 0. Ans. Concentric, 6. y^ + x'^ + y-2x-\-l = 0, y^ -{-x^ -^Sy - 4.x-\-3 = 0. Ans. (1,0); (h -*)• . a. y^-Sy + Sx-{-10 = 0, 2/^ + 4a;4-6 = 0. Ans. (-1, -2); (-J, -1). 7. 32/2 + 2a^ - 62/ + 8aj- 10 = 0, 3?/' + 2a^+ 6aj - 4 = 0. 8. 32/M-2a;2-62/+8a;-10 = 0, Sy^+ 2x''- Gy+ 8x + 1=0. 9. Prove that if two circles intersect, the common chord is perpendicular to the line joining their centres. Let ?/2 + 3:2 = /j2^ and (// - n)2 + (.r - m)2 = R^^ be the circles. Sub- tracting these equations, the equation of the common chord is '■2ny-{-2mx-n'^-m^ = R^^R^, or y=---r+C n GENERAL EQUATIONS OF CONIC SECTIONS. 131 in which C represents tlie absolute term. The line through the centres is n Ill 10. Prove that the perpendicular from the centre of a circle on a chord bisects the chord. 100. Defs. Conies having the same eccentricity are said to be simila-r. If the corresponding axes are also parallel, each to each, they are said to be similar, and similarly placed. 101. All conies in ibJiose equations the terms of the second degree are the same are both similar and similarly placed. Let Ay'^+Bxy+Cx' + Dhj + E'x + F'^O, (1) Ay-^^ Bxy + Cx" + D^hj + E^^x + i^" = 0, (2) be the equations of the conies, the coefficients of the first three terms being the same. We have seen that tan2y = — -4 — O (Art. 81), in which y= the angle made by the axis of the conic with X. Since y depends only upon A^ B, and (7, the conies are similarly placed. Again, from Equation 3, Art. 79, we have, by division, 1 — e^ sin^ g _A^_A 1 - e' cos'a~ C'~C' 2 e^ sin a cos a B' B l-e^cos^a C C' from which the value of e may be found in terms of ^, J5, and (7, by eliminating a. But A, B^ and C are the same for each curve, by hypothesis. Hence the eccentricities are equal. 132 ANALYTIC GEOMETRY. Cor. 1. All parabolas are similar, since e = l for every parabola. Cor. 2. All circles are similar and similarly placed. Cor. 3. If two conies 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 haze (i)'-i)")2/ + (^'-^")i» + -^'-^" = 0, 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. Cor. 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 represent 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 cc = 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 into x-{-y = 0, x — y = 0, which are the bisectors of the angles between the axes, and therefore x^^ — y^ =.0 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 Aif + Bxy + Ca? + Dij-^Ex-{- F=0 be the general equation of a conic. Solving it for y, 2. jfL 2,xx In order that this equation may be capable of reduction to the form y = ax ± />, the quantity under the radical must be a perfect square. But the condition, that the radical should be a perfect square is (^2 -iAC) (Z>2 - A:AF) = {BD ~2AEy. Expanding and reducing, AACF+BDE - AE" - CLf - FB" = 0, which is the required condition. If the coefficients 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. A.y'^ — bxy +x^-\-2y + x-2^0. Applyhig the test, we find — 32 - 10 — 4 — 4 + 50 = 0. To find the lines, solving for y, we obtain o O O O Hence the Unes are y = x — 1 and y = \x -{■ \. 2. Zf — ^xy + ^x- 1=0. 3. ?/2-27/-aj2 + l = 0. Ans. y-x-l=0, 2/ + a? -1=0. 4. xy — ay — bx-\-ab = 0. Ans. x= a, y= h. 5. 2/- + 4aj?/ + 4a.'2-4 = 0. Ans.y + 2x=±2. G. xy-{-x^ — y-{-dx — 10 = 0. Ans. ic — 1 = 0, y + x-\-10 = 0. 134 ANALYTIC GEOMETKY. SECTION IX.— TANGENTS AND NORMALS. 103. Defs. Let MM ' be any locns and AB any secant cut- ting the locus in the points F' and P". If AB be turned about P' regarded as fixed, till F", 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 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- gent at that point. The per- pendicular to the tangent at the point of tangency, lying in the plane of the curve, P'N, is the normaL 104. General equations of the secant and tangent to a conic. Let 2/-y=!^!(^-^') (1) be the equation of a straight line passing through any two given points (x%y'), {x",y"). The coefficient of x when the equa- tion is solved for y being -. — ^,, we have —. — ^ = a = the ^ x' — x" x' — x" slope of the line. Also, let /(^,2/) = (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 ^OKMALS. l35 f{x\y')=0, and /(x",y') = 0. w' — w" Hence, if we form '^, — ^, from these two equations and sub- stitute the vahie thus found in (1), we shall introduce the condition that (1) is a secant of (2). 7/' — 'y" Representing this value of -, — j, hy a„ the equation of the secant will be y-y' = O's (x-x'). ^ 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 o„ the equation of the tangent will be y-y'==a,{x-x'), in which (a?', y') is the point of tangency. Examples. 1. Equation of the tarigent to the circle y' + ^=R\ Let (.r^ 3/'), (x", //") be any two points of the circle. Then y'-2 j^ x'-2 =r R^, and y"^-^ x"-^ = R^. Subtracting, we have yi2^ytf2^^i2__xif2^0^ or (y'-y")(>j' + yl') = -i^'--^")(^'+^")> whence ■■^, — ^ = — ^ — -• x' — x" y' + y" Substituting this value in the iequation of a line through two given points, x' — x" . x' 4- r" it becomes y — y' =z Ji^ (^x — x'), y' + jj' the general equation of the secant line to the circle. Making now x" = x* x' and y" = y', we have y — y' = (x — z'), y' for the equation of the tangent. This equation may be simpHfied by clear- ing of fractions and replacing 2/'" + x'- by its equal R"^; whence, finally, yy' -\- xx' = R~, 136 ANALYTIC GEOMETRY. the general equation of the tangent to the circle y^-^ x^ = R^, (x', if) being the point of tangency. Note. The process is the same whatever the equation of the conic ; that is, whatever its species or the axes of reference; and the student should thoroughly master the above illustration as exemplifying a method for producing the equation of a tangent to any conic wlien referred to a rectilinear system. Thus, if the circle be referred to a diameter and the tangent at its left-hand vertex, its equation is y'^=2Rx-x'^. Hence y'2=2Rxf - x'^ and y"-=2Bxf' - x"^; and, by subtraction, y'^ - y"^ = 2R {x' - x") - (x'^ - x"^) ; , y'-y" 2R -(x'+x") whence ^ — 2_ = J^ — — ^, x' — x" y' + y'' J> -v,/ which becomes • when the points coincide. The equation of the y' R-x' tangent is therefore y — y' = — (x — ar'). y • 2. Find the equation of the tangent to the ellipse a^'if -f y^x^ = a^h"^. Ans. a^yy' + b^xx' = o?h^. If a = 6, this becomes yy' ■\- xx'= a^, or R"^, the tangent to the circle, as above. Since the equation of the hyperbola differs from that of the ellipse only in the sign of b'^, we have also tlie equation of the tangent to the hyperbola, a^yy'— b^ xx' =—a^ b^. 3. Deduce the equation of the tangent to the hyperbola by the general process. 4. Find the equation of the tangent to the parabola y^ = 2|)aj. Ans. yy' =p(^x-^ x') . When the central equations of the ellipse and hyperbola in terms of the semi-axes are used, and the equation of the parabola referred to its axis and vertex, the corresponding equations of the tangents are easily remembered from the fact that, by dropping the accents which distinguish the coordinates of the point of tangency, they become the equations of the cui'ves themselves. TANGENTS AND NORMALS. 137 105. Problems. Under the head of tangency the following simple prcblems occur. First. To write the equation of a tangent at a given point of a conic, and to find the slope of the conic at that poiyit. Find the general equation of the tangent to the conic by the preceding method, and substitute in this equation for x\ y\ the coordinates of the given point. The coefficient of x in the result- ing equation, when it is put under the slope form, will be the required slope. Thus, the tangent to the circle y^ -\-x^= 100 at the point (—6, — 8) being required, make x' = —Q, y' = — S in the equation of the tangent yy' + xx' = R^, R being 10, and we have — 8?/ — 6aj= 100, or 4?/ + 3ic-f- 50 = 0, which is the tangent. Solving for y-^y=- —fa; — -^^, or a^= — j, and the angle which the tangent makes with X= tan^^f. Second. To find the point on a conic at which the conic has a given slope. In this case the coordinates of the point of tangency are unknown. To find them we have the two equations f{x', y') =0 (since the point is on the conic), and the given condition (X^ = a', where a' is the givep slope. Combining these equations we find x\ y\ the required point of tangency. If more than one set of values for x\ y', are found, there is more than one solu- tion. If the value of either x' or y' proves to be imaginary, there is no point fulfilling the condition. Thus, at what point of the ellipse 9y^ -\- 4:X^ = 36 does the tangent make an angle of 45° with X? By condition, a, = - — = - - - = 1 . Also a-y' 9 y' Substituting fl;' = — f ?/' from the first in the second, we obtain 9 2/" + 4H2/'^ = 36, or 2/' = ± —L- Hence a;' = ip -4=- There* Vl3 Vl3 are therefore two points at which the slope is 1 ; one in the second angle, the other in the fourth. 138 ANALYTIC GEOMETRY. Third. To find the equation of a tangent to a conic which passes through a given point ivithout the curve. Let f{x, ?/) = be the equation of the conic, and {x,y,x',y') = that of the tangent. In this case also the coordinates of the point of taugency are unknown. To find tliem we have f{x', y') = (since the point of tangency is on the conic), and <^ (A, k, x', y') = 0, in which h, k, are the coordinates of the given point (since the tangent passes through it) . Combining these equations, we find x' and y', and there will be as many solutions as there are found sets of values for a;', y'. If either x' or y' should prove imaginary, the problem is impossible. Thus, to find the equation of the tangent to the circle y^-{-x^= 25, passing through (7, 1). Since the point of tangency is on the circle, y^^ -\-x'^=25. Since the point (7, 1) is on the tangent yy' + xx' = B^, we have also y'-\-7x' = 25. Combining, we find a;' = 3, ?/' = 4, and ic' = 4, ?/' = — 3. There are therefore two tangents to the circle through (7, 1), namely, 4y+Sx = 25, and 4aj — 3?/ = 25. Cor. Since the equation of a conic is of the second degree, and that of the tangent of the first degree, no more than two tangents can be drawn having a given slope, or through a given I)oint without the conic. Examples. 1. Find the equation of the tangent to the circle 2/2 + a;2 = 25 at (- 3, 4) . Aris. 4:y-3x=25. 2. Find the slope of the circle y^-{-a^ = E^ at the points whose abscissas and ordinates are numericallv equal. Arts. 45°; 135°. 3. Find the equations of the tangents to the circle y'^-\- xr= 100 passing through the point (10, 5). Ans. 4y-{-Sx = 60y a;=10. 4. Find the slope of the ellipse Sy^-]-x^ = 3 at the points a;' = ; y' = 0; a;' = f • Aus. 0° ; 90° ; 135° and 225°. 1 A TANGENTS AND NORMALS. 139 5. Find the points on the ellipse Sy^ -\- 4x^ = 32 at which the tangent makes an angle of 135° with X. AnS. (-^, —=]'•> ( :::' " — = V VV3 V3y V V3 V3y 6. Find the tangents to the ellipse 20?/^ -f- 93?^ = 324 passing through (-1, 6). Ajis. ^y-\-Sx = 27; 35?/ - 33a;= 243. 7. Write the equations of tangents to the parabola y^ — 8x at the points ic' = 8 ; x' = 2. 8. Find the point on 9?/^ — 4a;^ = — 36 where the tangent makes an angle with X whose tangent is |^- A71S. iVb sucJi point. 9. Show that the focal tangent to the parabola makes an angle of 45° with X. (Find the slope Sit y' =p,) 10. Find the eccentricity of the ellipse 25?/^4-9a;^= 225, by finding the slope at the extremity of the parameter. Ans. |^- 11. Show in the same manner that the eccentricity of the hyperbola 16y^ — dx'^ = — 144 is |- 12. To ivrite the equation of the tangent to the ellipse in terms of the slope. The equation of the tangent is a^yy^ 4- b'^xx' = (filP-, or w = x H Let '— = ni = slope of the a^y' y' ,2 a!^y' tangent, whose equation then becomes yz=mx+ ^- To eliminate y', we have b^x' = - a^y'm, and a'^y'^ + b'^x''^ = a%^ ; whence a^^ + ^}^Jl!^ = a%'^, or v'^ (^ahn'^ + b'^) = b\ 62 from which we obtain — = ^ah)i^ + 6^. Thus the equation of the tangent is y = mx-\- y/dhn^ + b^. Changing the sign of b"^, and making a=: 6, we have the corresponding equations for the hyperbola and circle, y = mx-\- y/a^m^ — 6^, and y = mx + a Vm^ + 1. 13. To write the equation of the tangent to the parabola in terms of the slope. p Ans. y = mx + Tf— • ^ 2m 140 ANALYTIC GEOMETRY. 14. The rectangle of the perpendiculars from the foci of an ellipse upon the tangent is constant and equal to the square of the semi-conjugate axis. Putting the equation of the tangent under the normal form, we have Substituting in succession the coordinates of the foci, ae, 0, and —ae, 0, for X and y, and taking the product of the results, we have, putting the radical = D for brevity, - (b^aex' - a%^) (b^aex' + a^h^) _. a^M - b^a^e'^x'-^ _ 62 [2 ^b^a^M!l±^^^ = b^ Z)2 This property is true also of the hyperbola. 15. The perpendicidar from the focus of an hyperbola upon the asymptote is equal to the semi-conjugate axis. This may be regarded as a particular case of the foregoing, the asymp- tote being a tangent whose point of contact is at an infinite distance. Or, directly, the equation of the asymptote y — -x under the normal form is "• ^~ = ; substituting the coordinates of the focus \/a2 + 62 a: = ae = v a2 +62^ ^ = 0, this expression reduces to — b. 16. To find the length of a tangent from a given point with- out a circle. Let (tj, ^j) be the given point Pp and P' the point of tangency,, (a: — »i)2+ (y — n)2 — i22_o being the equation of the circle and C its centre. Then, since the radius to the point of contact is perpendicular to the tangent, P^P'^ =P^C^- CP'^. But P^^C^ = {x^ - my + {y^ - n)'^ (Art. 7), CP'^ = R^. Hence P,P'-^={_x,-my^{y^-ny-I^. Now this is what the equation of the circle becomes when the coordinates of the given point are substituted for x and y ; hence, put the equation of the given circle under the form f{x, y) — 0, and substitute for x and // the coordinates of the given point. The result will be the square of the required distance. Thus tlie length of the tangent to the circle y2^a:2-6i/ + 8a:-ll = from (5, 1) is Vl + 25-6 + 40- 11 = 7. TANGENTS AND NORMALS. 141 17. If two circles are tangent internally and the radius of the larger is the diameter of the S7naller, all chords of the larger through the j^oint of coritact are bisected by the smaller. Take the origin at the point of contact and the diameter as the axis of X. Then the equation of any chord is y = ax, and the equations of the larger and smaller circles are i/^= 2Rx — x^ and y^ = Bx — x^, respectively. The chord intersects the former at ( , ^1 and the latter at / R Ra \ V«^ + l «^+V U'^i' «-+iJ 106. Chord of contact. Tangents are drawn to a conic from a given external point ; to find the equation of the chord of con- tact. First. The ellipse and hyperbola. Let (/i, k) be the external point, and (a;',?/'), {x",y"), the points of tangency. Then, since both tangents pass through (/i, k)^ these coordinates must satisfy theh' equations ; or a'ky' ±b'hx' =±a-b\ (1) ; a^ky" ±b^hx" = ±a^b\ (2) Then a'ky ±bVix =±a^b^ is the equation of the chord ; for it is satisfied by (ic', y') , (.t", 2/")? as shown by (1) and (2), and is of the first degree with respect to x and y, and therefore represents a straight line through {x\y'), {x",y"). CoR. The chord of contact to the circle is ky -j- hx = B^. Second. The p)arabola. The equations of both tangents must be satisfied for (7i, k) ; hence ky' =p{h-^x'), ky"=p (h-\-x"). Then ky —p {h-]-x) e chord of contact, the reasoning being identical with that 142 ANALYTIC GEOMETRY. It will be observed that the equations of the chord of con- tact are derived from those of the tangent b}* changing the coordinates of the point of tangency, x\ y\ into those of the external point ; these equations are therefore easily memorized. 107. General equation of the normal to a conic. The equa- tion of any line through the point of tangency (.t', ?/') is y — y' = a {x — x') . But the normal is perpendicular to the tangent, hence a must equal , a< being the coefficient of x in the equation of the at tangent when under the slope form ; or the equation of the normal is y — y' = {x-x'). a, Examples. 1. Find the equation of the normal to the circle 2/2 + a;2 = B', The tangent to the circle is yi/' + xx' = R^, ov ^ — ^x -\ ; hence rw*r fir vf \7 at=z -, and the normal is y — y'=^{x — x'); or, clearing of fractions, ij' x' x'y — y'x ■= 0. Since this equation has no absolute terra, the normal passes through the origin, which is the centre ; hence the normal to a circle is the radius to the point of tangency. 2. Find the normal to the ellipse a^y^ -\-h^x^ = a^h^. Arts. y-y' = ^,(x-x'), 3. Find the normal to the hyperbola a^y- — h^x^ = — a^h'^. Ans. y-y' = -j^,(x-x'). b^x^ 4. Find the normal to the parabola y^ = 2px. Ans. y — y' = — ^{x — x'). 5. Find the normal to the circle y^=2Rx — X!- v' Ans. y-y':=-^^-^{x-x'). TANGENTS AND NORMALS. 143 6. Write the equation of a normal in the following cases : (a) to a circle whose radius is 5 at the point (3, — 4). (5) 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 x'=S, 3y + 4x=0; 3?/= ± 9 V2a; if 5 V2 ; 3 2/ = qF4aj±34; 48?/ = q: 9 V55a; ± 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) , TM and MJ^ are tbe subtangent and subnormal to the point P'. 109. To find the subtangent and subriormal at any point of a conic. Let Xt= 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, ri)^= subtangent = OM - OT^x' -x,. Similarly, MN= subnormal = ON— 0M= x^ — x', x„ being the X-intercept of the nor- mal, or the value of x when y is made Fig. 7o. zero in the equation of the normal. Examples. 1. To find the subtangent of the eUipse and the hyperbola. The equations of tlie tangents are a^i/i/' ± b^xx' = ± a^b'^. When y = 0, x = xt=: — for both curves. Hence, also, for both curves, subt = x' — xt. 144 ANALYTIC GEOMETKY. Cor. Since x,=:-. x,: a:i a: x\ or OT : OA' : : OA' : 03L x' Hence, the semi-transverse axis is a mean proportional between the intercepts of the tangent and the ordinate of the point of tan- gency. (Figs. 71 and 72.) This principle affords a method of coyistructing a tangent at any point. First. The ellipse. Let P' be the 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 P'T is the required tangent. For, from the right similar triangles, OP"'M, OP"T, OT: OP"\ = OA') : : OP" : OM; or xr. a: : a: X' Since both OT = x,= ^, and MT= subt. = x' x' , are inde- pendent of b, tangents to all ellipses having the same transverse iixis, at points having the same abscissa x' = OM, will evidently pass through T. Second. TJie hyperbola. Let P be the point. Draw the ordinate P3f^ and on AA\ OM, as diameters, describe circles intersecting at Q. Draw QT perpendicular to X. Then TP' TANGENTS AND NORMALS. 14; is the required tangent. For, joining Q with and Jf, from the similar triangles OQM, OQT, we have OM:OQ(=OA') :: OQiOT, or Xfi 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. iCf = — oj' ; subt. = 2x\ It appears from this result 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 FM and make 0T= OM. Then TF' is the required tan- ent. Fig. 73, 146 ANALYTIC GEOMETEY. 3. To find the subnormal of the parabola. v' The equation of the normal being y — y' = — ~ (a: — a:'), we have, when y = 0, x = Xn — p-{-x'. Hence subn. = x,, — 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'iltf and make MN^p. Then P^M is the normal, and P'2', perpendicular to it, is the tangent. 4. The tangent to the parabola 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'J) the diameter. Then FP' = x' + ^ (Art. 74, Cor. 2). Also TF= TO + 0F= x' + | (Ex. 2). Hence FT= FP', the triangle TFP' is isosceles, and DP'T=P'irF= FP'T. Therefore, to draw a tangent at any point, as P', draw the focal ifadius P'P, the diameter P'D, and bisect their included angle. Cor. 1. FN= OM- 0F+ MN=x^-'^-^p (Ex. 3); .-. FN= x' 4-f = FT=FP, or the circle described from the focus ivith a radius equal to the focal radius of any point passes through the intersections of the normal and tangent to that point with the axis. The triangle FF'N is, thus isosceles, and FNF = FP'N. CoR. 2. PFN= FPT+ 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 draw a tangent through a given point idthout the curve. Let /I'be the given point. Join K with the focus, and with JiTas a centre and KF as a radius describe a circle cutting the directrix in D and D'. Draw the diameters throuoh D and D' ; their intersections with the curve, P' and P", are TANGENTS AND NORMALS. 147 the points of tangenc}*. To prove that KP^ is a tangent, we have P'D=P'Fhy definition of the parabola; also KF = KD by construction. Hence KP bisects the angle FP'D, Simi- larly, P"Kmsiy also be shown to be a tangent at P". y 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'N bisects FP'F', or that FN : FP' :. F'N : F'P. 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 X-intercept of the normal. Making ^/t = in the equation of the normal ^2 A2 1 we have x=ON=- -xf=e'^xf. . Hence FN = ae -{• e"^ x' = e (a + ex') . Also F'N= F'0-ON= e (a^ex'). Therefore ^ = l(li±^, FN^eja-ex') ^^ FN^FN^ FP' a-]- ex' F'Pi a -ex' FP' F'pi To draw 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 tayigent and normal at any point of the hyperbola bisect the angles formed by the focal radii drawn to the point of contact. Let P'T be 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 X-intercept of the tangent. Making t/ = in the equation of the tangent a^yy' — h^xx^^ — a^b^ we have x= 0T = — ; 2 ^' hence FT = ae = — (ex' — a). x' X' Similarly, F'T = —^ (ex' -\- a), and, as before. FT ^F'T X' FP'~ F'P'' 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 tangent passing through a given point without 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 /r 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 t\\Q farther focus ; the intersections P' and P" of FQ and FII with the curve are the points of tangency. To prove that KP is a tangent, WG 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'FH in the elUpse and PP'P' 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. 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 eq lal 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. 150 ANALYTIC GEOMETBY. SECTION X.— OBLIQUE AXES. CONJUGATE DIAMETERS. 110. Equation of the ellipse referred to conjugate diameters. Let A'A" = 2a\ B'B" = 2h'j be conjugate diameters, the axes of reference being taken as in the figure. To transform the equation ay + 62^ = a262 (1) to these axes, we have the formulae (Art. 22, Eq. 7) ic = a;iCOS7' + 2/iCosyi, y = x^ sin y]-y i sin y^. Substituting these in (1), and omitting the subscripts of x and y, ¥ (2) (a^ sin^yi+ b^ cos^yi) y^-\-2 (a^ sin y sin yi+ b^ cosy cos yi)a^ + (a^ sin^y + 6^ cos^ y)oc^=z a^ b^. But, since the diameters are conjugate, they must fulfil the 52 condition tany tanyi = ^ (Art. 90) , OBLIQUE AXES. 151 or a^sinysinyi = — &^ cosy cos yi. Hence the coefficient of the second term of (2) is zero, and the equation becomes (a' sin^yi + h' cos^yO y' + (a' sin^y -f- h' cos^y) x' = a'b\ (3) Making y = 0, we have a^sin^y + 6^cos^y ' and when aj = 0, 2/^=6'2^-^ a^sm^yi-l-o^cos^yi Substituting from these equations the values of the coefficients of if- and X- in (3), we have the equation in terms of the semi- diameters, a'2 2/2 + 6'2a^ = a'-6^ (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 ^,2^2 _ yir^a. ^ _ ^^12^,2^ ^5) since the onl}' change in the above would be that arising from the minus sign of b^ in the equation of the 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 parallel to any diameter of an ellipse are to each other as the rectangles of the segments into ichich they divide its conjugate. Let P'M\ P''M", be the ordinates parallel to any diameter BB\ and meeting its conjugate AA' in M' 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. uiviamg, ^„^ a"-x"' {a'-{-x"){a'-x"y 112. The squares of ordmates parallel to any diameter of an hyperbola are to each other as the rectangles of the distayices from the feet of the ordinates to the vertices of the conjugate diameter. 113. The parameter of an ellipse is a third proportional to the transverse and conjugate axes. The axes being conjugate diameters, Art. Ill applies, and 2/'2 : 2/"2 : : (a 4- x) (a - x') : (a + x") (a - x") . Let P' coincide with the extremity of the conjugate axis, and P" with that of the parameter. Then y'=b, y"=p, aj'= 0, x"=ae, and the proportion becomes But l-e^ = -', .\a^:b^::b^:p^ or 2a: 2b : : 2b :2p. (V- ' 114. Any ordinate to the transverse axis of an ellipse is to the corresponding oi'dinate of the circumscribed circle as the conjugate axis of the ellipse is to its transverse axis. 52 From the equation of the ellipse y'^ = — {a^ — x^) ^ and that of a the circumscribed circle y^z=a^—a?^ where y and 2/1 are the ordinates corresponding to the same abscissa cu, we have 115. The 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. 158 Let x\ y\ be the coordinates of A^ (Fig. 76), and a;", 2/"? those of B\ Since jB"^' is parallel to the tangent at A\ its equation is ?/ = Tl^' Combining this with ari/-\- h^af = a^6^, to determine the intersections B' and B", we find x = x"=±^, and 2/ = 2/"=T— • b a But a'2 = aj'2+ 2/'2 = x"+ -' (a2_ aj'^) = b'-^^^^x'' = 52+ e2^'2 . and 6'2 = a;"=+2,"2^^ + ^' = ^' ^Ja-'-x'^) Hence ^2 ^2 = a^H — x'^ = a^— e^x'^. a'2+6'2 = a2+&2. 116. T^e differeiice of the squares of conjugate diameters to the hyperbola is constant and equal to the difference of the squares of the axes. Let «', y\ be the coordinates of A', and a;", y", those of B', Fig. 77. 154 ANALYTIC GEOMETRY. 52 ^f The equation of B"B' is y = -^ — x. Combining this with the d'y' equation of the F-hyperbola, a^y^— b^x^ = a^b^, to determine the intersections B' and B", we find Hence X=:x"=±-^, yz=y"=± a a"= x''+y''= aj'2+ ~ (x"- a') = ^^!+l'a;'2_ b"-= e^x''- b' ; and &'2^aj"2+2/"2 = ^4 b' a^ b-\_a^ J a^ ' /^i2 x-,2 __ ^2/y ^12 _ ^2_ p2^f2 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). Then the area of the parallelogram is 4 OB'. OA' . sin B'OA'= 4 OB'. OA' . sin 0A'T=4: OB'. OD. OBLIQUE AXES. 155 The normal form of the equation of the tangent at A} is g-yy^ ± b^ xx' g: orlr _ ^ Va'y'^+b^x''' ~ Hence the distance from the origin to the tangent is ^'^' ^^ =4 (Arts. 115,116), V a^y'^+b'x'^ la'y'^ ,^x^ \ 6^ "^ a' and 40J5'.Oi> = 46'.^ = 4a&. b' 119. To find the equal conjugate diameters of the ellipse. Equating the values of a'^ and 6'^ (Art. 110), a^ sin^y + b^ cos^y a^ sin^yi -f- b^ cos^yj whence a^ sin^yi + 6^ cos^yj = a^ sin^y + W cos^y ; or, transposing, a^(sin^yi — sin^y) = 5^(cos^y — cos^yi) =?>^(sin^yi — sin^y), since cos^^l = 1 — sin^^. Hence (a^- 6^) (sin^yi - sin2y) = 0, (1) and therefore sin^yi = sin^y. Since, in the ellipse, if y is acute, yi is obtuse, and the sines are equal, yi = 180° — y and tanyi = — tany. 52 Substituting this in tany tan yi = ? a^ the equation of condition for conjugate diameters to the ellipse, tany=±-? .'. tanyi= — tany = q; -• a a Hence, when the diameters are equals the aiigles they make ivith the transverse axis are supplementary and the diameters fait on the diagonals of the rectangle on the axes. 156 ANALYTIC GEOMETEY. Cor. 1. If a = 6, (1) is satisfied independent!}^ of y and yi; or, in the circle every diameter equals its conjugate. Cor. 2. For the hyperbola, (1) becomes (a' + b') (sin^yi - sin^y) = 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'^ — 6'^ = a^ — 6^, however, we see that if a = b, then a' = 6' ; or, every diameter in the equilateral hyperbola equals its conjugate. SUPPLEMENTAL CHORDS. 120. Defs. Straight lines drawn from any point of an ellipse or an hyperbola to the extremities of a diameter are called sup- plemental chords. Thus, /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 supplemental chord is parallel 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 jS". The equation of S"Q will he y-\-y' = a"(x + x') (Art. 31) , and that of S'Q^y —y' = a'(x — x') . Combining these equations by mul- tiplication, 2/^— 2/'^ = a'a"{a? — x'~) , in which x and y are the co- ordinates of Q (Art. 36) . But /S" and Q are on the curve ; hence aY^± bV = ± a^b^ and aY± b^x" = ± a^'' ; or, by subtraction, 2/2 — 2/'2 = q: _ (x" — ic'2) . Equating these two values of y^— y'\ we have a^a" = q: — • But this is the condition for conjugate a ,2 diameters, viz. : tan y tan yi= ip — (Arts. 90, 93). Hence if a' = tan y, a" = tan yj, and conversely. OBLIQUE AXES. 157 Cor. 1 . To draw a tangent at a given point of the curve,, as A\ 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 tangerit parallel to a given line, as JOT, draw an}^ chord QS' parallel to it, and the supplemental chord QS". Then the diameter A" A', parallel to S"Qj determines the points of tangenc}^ A" and A', PARABOLA REFERRED TO OBLIQUE AXES. 122. Equation of the parabola referred to any diameter and the tangent at its vertex. The formulae for transforming from rectangular to oblique axes, the new origin being at 0', are (Art. 22, Eq. 3) a; = iCo + a?iCosy + 2/iCOSyi, 2/ = 2/o + a^i sin y -f t/i sin yj. (1) But y = 0, since the new axis of X is parallel to the primitive 7) one, hence cosy = 1, siny = 0. 'Also tanyi = — ? since the new axis of Fis tangent to the curve at (xq, 2/0) (Art. 104, Ex. 4) ; hence from sin yi sin yj tanyi cosyi Vl — sin^ 71 we have sin yj = — -^ Vyl+p^ and therefore cos yj = Vl-sin^y, = , ^" - Substituting these values in (1), they become x = xo + x,+ 2/i?/o .,_., , y.P Substituting these values in the equation to be transformed, ^2 = 2px, and remembering that, since 0' is on the curve, 158 ANALYTIC GEOMETRY. y^ = 2j9a;o, we have, after omitting the subscripts of x and ?/, (2) y.^^_M±fl^^ p which is the required equation. Cor. 1 . ?/o = MO' = MN tan MNO' = p cot yi, O'N being the normal and MN= subnormal =p. Hence ^M±^ = 2^ ( 1+ cot^yO = 2l) cosecVi = 4|-, P Sin yi and (2) ma}^ be written y = -^- ^' siir yi (3) Fig. 78. Cor. 2. From the polar equation of the parabola, P . 1 — cos ^ ' making ^ = yi, we have r = FQ = P making ^= 180° + yi, 1 — cos yi 1 + cos yi OBLIQUE AXES. 159 Hence QQ' = FQ + FQ^ = -^ ; sin-'yi or, representing QQ' by 2p', (3) may be written y^ = 2p'x, (4) Thus the equation of the parabola referred to any diameter and the tangent at its vertex is y^ = 2p'x^ 2p' being the 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 otp' tor p. 123. The squares of ordinates to any diameter 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 P'andP", y'2^2p'x', y"'=2p'x"; or, by division, |! = ^,. ASYMPTOTES. 124. Equation of the hyperbola referred to its asymptotes. The asymptotes being oblique except when the hyperbola is rectangular (Art. 92)^ we use the formuliB for passing to oblique axes with the same origin, x — Xi cos y + 2/i cos yj, y = Xi sin y + 2/1 sinyi ; and, since the asymptotes coincide with the diagonals of the rectangle on the axes, sin y = - ~ sin yi = - , cos y = cos yi = ^ J »-'■.■»■• r I — — , \j\jf3 Y — v/vro /I — 160 ANALYTIC GEOMETEY. The formulae therefore become a X (a^i+^i), y (2/1 -aji). Substituting these vahies in a^y^— b^x^ = — a-b^, and omitting the subscripts, we obtain i»2/ = -— 7 — 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 F-hyperbola referred to the same axes is xy = (Art. 71) Cor. 2. The equation Bxy -\-Dy + Ex-{-F=0 is the general equation of the hyperbola referred to axes parallel to its asymptotes. For, transforming xy = m to parallel axes by the formulae a; = a!o+a7i, 2/ = 2/0+2/1? we have, after dropping the subscripts, 3^2/ +^o2/ + 2/o^ + ^o2/o = 0, which is the above form. The equations of the asymptotes are evidently 2/ = — 2/o> X= — Xo. OBLIQUE AXES. 161 125. The intercepts of the secant between the hyperbola and its asymptotes are equal. Let P'j P" (Fig. 79) , be any two points of the hyperbola, the equation of the secant P^P^\ Making a; = in this equa- tion, y _ y = jr>'(2'= ^ — L. a;'— a;" But y'x'=y"x"=m, since the points are on the curve. Hence D'Q ,_y"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 hyperbola 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', ^2', J[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. p-ig go. 126. The area of the triangle formed by any tangent with the asymptotes is constant^ and the tangent is bisected at the point of contact. 162 ANALYTIC GEOMETRY. The equation of the secant P^P'^ (Fig. 79) is x'—x" From the equation of the curve, icy= x"y"= m, .'. y'= — ^. The fraction ^- — ^, therefore becomes x"y" ^n x'-x" ~ x'' or, when P" coincides with P', — ^^. Hence the equation of the tangent TT' is ^ 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 ^^'^^' sin TOT'= ^^''^y' sin 2 TO^ 2 2 = 2xy 2 sin TO J. cos TOA = 4 x'y'—;===, , ^ ^ = «&. Va +^ Va +& 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'M' 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. 79), the co- OBLIQUE AXES. 163 Ctll t)Or ordinates of P' being x\ y\ and those of B^ being -^, — (Art. 116), is ^ ^ h or y-y'= — -{x-x'). But the equation of OT^' is y = x\ hence P'-B' is parallel to the asymptote OT'. Again, the middle point of P'B^ is which satisfy y = -x. Hence the straight line joining the ex- tremities of conjugate diayneters 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. (R. CHAPTER IV. LOCI, 3>©=:c 128. Classification of loci. When the relation between x and y can be expressed by the six ordinar}' 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 t^e degree of their' equations as loci of the first, second, etc., orders. We have seen that there is but one locus of the first order ; that is, whose equation is of the first degree, namely, 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 which involves a logarithm, as ic= log?/, is called a logarithmic function; one in which the variable enters as an exponent, as ?/ = a"", 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 — ^mx^ a;=sin~^?/, etc., are called circular functions, y — ^inx and x=^\ir'^y are 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 ; ^nd in this 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 n variables are involved exclusive of x and ?/, the conditions must furnish n + 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 sqxiares of its sides^ to find the locus of the vertex. Let h be the given base and d? the constant diflference. Take the base for the axis of X, and its left- hand extremity for the origin, x and y being the coordinates of the vertex. Then, by condition, OP^-BP^=d\ or d^ + b^ whence MB 26 Fig. 81. Hence the locus is a straight line parallel (P _L. ^2 to F, at a distance from it equal to — — — If the triangle is h 2b * isosceles, d=0, and x = -- In this case 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 tlie base, the axes being taken as in the figure. Then the equations of AB and AC are known ; namely. b a 1, and - + ^ c a Now the abscissa of P is the half sum of the abscissas of Q and R ; and if ^ = ^ be the altitude of the rectangle, and this value be substituted for y in the above equations, we find h 6, Xj c. LOCI OF THE FIKST AND SECOND ORDER. 167 Hence x = abscissa of ^-i'l^{b+c). 2 a But the ordinate of P = 2/ This condition enables us to eliminate the variable Tc from the above value of x ; substituting therefore k = 2y, we have 2ax = {a-2y)(b + c), a straight line bisecting the base and altitude, since its intercepts are ^ (& + c) and ^ a. 3. To find the locus of a point so moving that the square of. its distance from a fixed point is in a constant ratio to its dis- tance from a fixed line. Let B be the fixed point, OX the fixed line and axis of X, the axis of F passing through JB, and 0B= a. Then, m being the constant ratio. PM = m. But BP' PM= y. Hence 2/- + a;^ — (2 a H- m) 2/ + tt^ = 0, ' which is the equation of a circle whose x^ -^-iy — aY^ and t centre is at ( 0, 2a-}-m and whose radius is ^ V4r a?/i + m^ (Art. 50). If the point is on the line, a = 0, the cen- m tre Ko,f) and the radius = 4. The squares of the distances of a point from two fixed points are as m to n. Find the locus of the point. Let (a, 0), (0, 5), be the coordinates of the fixed points, the axes being assumed to pass througli them, and P ^ny point of the locus. Then, A and B being the fixed points. 168 ANALYTIC GEOMETRY. PB x' + jy-hy _m, PA^~y'' + {x-aY~ n ' or, clearing of fractions and reducing, 2/^ + ar y -\ x ■\ = 0, n — m n — nn n — m a circle whos6 centre is f TIL-^ _!^ — )^ and radius is \^ n — m n — m 1 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 = 6 = 0, the centre is at the origin and i2 = 0, or the locus is a point ; unless also m = n, when ^ = -• x: / 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 point 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"" = m^PB\ or ^^ x^ + f = m\f + {b-xy), V. 2 . ^ . 2m^6 m^b^ ^ whence 2/ + a^ + :; o^- -. = ; f mJ^b \ mb a circle whose centre is f ^^ ), and radius is 7. From one extremity A' of a diameter A A' 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. The perpendicular produced meets the secant at P. Find the locus ofP. LOCI OF THE FIRST AND SECO Let the diameter be the axis of X and the centre the origin. Let (ic', y') be the coordinates of P' ; then (Art. 32) \ is the equation of the secant ; the equation of the tangent at P' is yy' + xx' = i?% hence the perpendicular on the tangent from -4 is , 3, = |(x-i?). (2) Combining (1) and (2), to find P, we have x=2x^-i-E, y = 2y'. (3) But {x', y') is on the circle, hence x'^ + y'^ = P^. Substituting in this equation the values of x' and y' from (3), we have (^x-Py + y^-4:P^=0, - a circle whose centre is at {P, 0), that is, at A, and whose radius is 2P = AA'. ^ 8. A line is drawn 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. ^V^^ 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 perpendicular 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 perpendicular. Ayis. A straight line. 1 . Given any parallelogram , and PP\ QQ', lines parallel to adjacent sides. Prove that the locus of the intersection of PQ and P'Q' is a diagonal of the ^^~Qf l parallelogram (Fig. 84). 170 ANALYTIC GEOMETKY. 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 OL-\-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, OF, be the fixed lines and axes, AOB the inter- cepted triangle, m the constant area, <^ the constant angle BOA, and Pthe middle point of AB. Then OM=x, MP=y^ and, since P is the middle point of AB, 0^1 = 2 2/ and 0B = 2x. Hence area BOA = m = 2x.2y.sm or xy = m 2 sin (f> an hyperbola whose asymptotes are the fixed lines (Art. 124). 13. Given two intersecting fixed lines and a fixed point, A line is drawn through the fixed point. Find the locus of the middle point of the segment intercepted by the given lines. Let OX, OF (Fig. 85), be the fixed lines and axes, Q the fixed point, its coordinates OR = m, BQ= n, AB the line, and Pits middle point. Then, from similar triangles, 0A{=2y) :0B(=2x) ::RQ(=n) :RB(=2x-m), or 2xy — my — nx = ()\ an hyperbola passing through Q, whose m asymptotes are a; = — » <6 (Art. 124, Cor. 2) 14. From a fixed point A (Fig. 85), a line AB is drawn to meet a fixed line OX. From the intersection jS, a constant dis- tance BR = b is laid off, and from R a line RQ is drawn, mak- ing a constant angle with 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, OAiRQ:: OB: MB, or a:y::x-j-b:b', whence xy -i-by — ab = 0, an hj-perbola 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 perpendicular from the focus of a parabola on the normal. The equation of the normal is and that of the perpendicular is Combining these to find the point of intersection, we find it to be _pM^^fcyM-2py2 _2px^y'+p^y^ ^"" 22/'H2p2 ' y~~¥y^^+2^' , In this problem the conditions introduce the auxiliary vari- ables a;', y\ the coordinates of the point of contact from which the normal is drawn. But tliis point is on the parabola; hence we have the additional condition 1/'^= 2px\ Eliminating 2/' by means of this equation, we have Finally, combining and eliminating x\ we have 2 P P^ 2^=2^-4' a parabola on the same axis, whose vertex is at [^, ), 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 tangeut through the focus, Combining these to find the intersection, we obtain, on eliminating y^ a;(p + 2cc') =0, which, since a;' cannot be nega- tive, is satisfied only for a;=0; that is, tlie 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. Through any fixed point chords are drawn to a parabola. Find the locus of the intersections of the tangents to the parabola at the extremities of each chord. Let a^i, 2/i? 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) ylv=p{x-\-h). But the chord passes through the fixed point (a^i, y^) , hence yxlc = p{x^-\- h). Now this is the equation of a straight line, in which h and k are the variables ; therefore the locus of (A, k) is a straight line. If the fixed point is the focus, x^ =^, ?/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. Let iCi, i/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 meeting at {li, k) . Then the equation of this chord is (Art. 106) But this chord passes through the fixed point (oji, 2/1) , hence a^yik ± b'xji= ± a?W. Now this is the equation of a straight hne in which li and Tc are the variables ; therefore the locus is a straight line. If the fixed point is the focus, Xi= ae, 2/1= 0, and the equa- tion becomes U =*— " 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 + ■y/ahn^-\- 6^, is most convenient. The perpendicular upon it from the focus y = (x-ae). m From the former, y — mx = \^a^m^-\- b^, and from the latter, my ■i-x= ae. Squaring and adding, (y'+x")(\ +m'') = b'-{-m'a'+a'e'=b'-i-ma'+a'^^^^= a\l-\-m'\ a? which eliminates m, giving y'^-\- x^= a^. This property is also true of the hyperbola. 20. Find the locus of the intersection of pairs of tangents to a parabola tvhich intercept 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 GEOMETKY. 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 conditipn, y" = 2px\ (3) y"' = 2px'\ (4) v'-v" = a, a constant. (5) Substituting in (1) and (2) the values of x' and x" from (3) and (4), we have yy'=px-i-^, (6) yH2 yy"=px-^^' (7) Substituting from (5) y'= 2a-{-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', whose centre is C, are drawn to a circle. A A' is a diameter parallel to the chords. Find the locus of the intersection of AG with the radii through the extremi- ties of the chords. Ans. A parabola whose axis is the diameter and vertex mid- way 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. 2ay+26V=4a262. 23. Lines BL, R'L' (Fig. 84), are drawn 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 X 24. A and B are fixed points. Find the locus of P when pryi — = a constant^ D being the foot of the perpendicular \ 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 ari 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 only 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. Required the locus of a point P on the secant ivhen 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, 0Q= OD cosQOD = OD s'mXOQ = 2 Rsind, and QP=R, Hence r=2R sinO -{- R. (1) Discussion of the equation. For ^ = 0°, r = R= OA, As d increases, r increases, and when 6 = 90°, r= 3 i? = OB. HIGHER PLANE LOCI. 177 As 6 increases from 90° to 180°, r diminishes, and when 0= 180°, r= R= OE. When passes 180°, sin ^ becomes negative, but ?• remains positive until sin^ = — -J-, when r = 0; at this point ^ = 210°. From this value of ^, r is negative and the portion OFO is traced, r being — E when d = 270°. From ^=270° to ^ = 360° the portion COA is traced, r becoming positive again when = 330°. Rectangdlar equation of the cardioid. Transferring to the axes YOXhj the formulae (Art. 24, Eq. 4), r = Vi»^+2/^, sin^= — ===r» we have (o^ + y'-2 Eyy={^+ y') B\ a curve of the fourth order. (2) Trisection of the angle. The cardioid affords a method of trisecting an angle, as follows. Let NCO be the given angle. With the vertex (7 as a centre describe any circle, and construct the cardioid to tljis circle. Only that portion of the curve in the vicinity of NC produced need be constructed. Produce iV(7 to meet the cardioid at P, and draw FO and QO. Then the tri- angles CQP^ CQO, are isosceles by construction. Hence NCO = COP + CPO= CqO + CFO =QCP+2 CPO = 3 CPO ; or CPO = \NCO. 2. The conchoid. Through a fixed point F a line FP is 178 ♦ ANALYTIC GEOMETRY. drawn cutting a fixed line XX' in Q. Required the locus of P when QF is constant. Let QP= a, FO = 6, the distance from the point to the line, and OX, OF, be the axes. Draw FS and FJS parallel and per- pendicular respectively to X'X, Then FM: MQi.FSiFS; or y : V a^ — y^::y-^b:x; whence a^f = (y + by {a' - f), (1) a curve of the fourth order. Discussion of the equation. Solving the equation for Xy we have x= ± ^ Va^ — y^ y The curve is evidently symmetrical with respect to Y. When y is positive and equal to a, a;=0, locating the point A, which is a limit in the positive direction of Y, since x is imaginary if y > a. As y diminishes, x increases numerically, becoming ± CO 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 ?/ is negative and numerically less than 5, x increases as y diminishes and becomes ± oo for y = 0, giving infinite branches with the axis of X for their common asymptote, y = — a evidently limits the curve in the negative direction of Y. In the above case a>b. If a = b, F 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 z= — b, which renders x=0; thus the oval disappears, but a^=0, y = — b, satisfy the equation, and hence must be considered a yto'mt 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 = FOco^Qce + a, or r = h Q0s>Qc6 -\-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 ii^ 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 QiV perpendicular to OX and join its intersection with the conchoid, iV, with F. Bisect QN at i?, draw BL parallel to OX, and join L with Q. Then the triangles LNQ, LFQ, are isosceles ; for, since QR=BN, KL =LN=LQ = i QP=FQ, Hence the angle uiFN=FJSrQ=LQN=iFLQ = iLFQ, or AFN=^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 ordinates are drawn to the diameter of a circle^ and through one extremity of the diameter a line IS drawn through the intersection of one of the ordinates with the circle. Find the locus of the intei'section of this line with the equal ordinate or that ordinate produced. 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 y. 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, OM'.MPi'.OM^.M^Q'', or, R being the radius of the circle, x:y : : 2M — X: -y/x {2M — x), whence y^ (2R — x) = q?, (1) Discussion op the equation. Solving the equation for 2/, r / I 1 p %' i y 'TT I' CI D ^ ^=*>&E' Fig. 88. which shows that the curve is sym- metrical with respect to X, If x is negative or greater than 2 i?, y is imaginary ; hence the limits along X are zero and 2i?. As x in- creases, y increases, and when x=2 E,y=± CO ; hence the curve has two infinite branches which have the tangent at D for a common asymptote.. Polar equation of the curve. Let be the pole and OX the polar axis. Substituting in (1) the values X = r cos ^, y = r sin (Art. 23, Eq. 4) , we have r^ sin^ 0{2M — r cos 0) = r^ cos^ 0. But sin2(9=l-cos2^; hence 2B — 2E cos^O — r cos6 = 0. 1 Substituting sec^ for cos^, and remembering that sec2^-l = tan20, we obtain finally r = 2Iis[nO tan 0. (2) HIGHER PLANE LOCI. 181 Duplication of the cube. This curve affords a method for finding the edge of a cube whose volume shall be n times that of a given cube, as follows : C being the centre of the circle, take CS=nCD^ and draw SD intersecting the cissoiJ at P. Then the ordinate P3f= n . 3ID, and the cube whose edge is P3f is n times the cube whose edge is OM. For, P being a point of the cissoid, we have from its equation, 03P OM' PM'f MD 1 ' -PM or PM'= n . 0M\ Let c be the edge of the given cube. Find c', so that d:c::PM'.OM, or c^' : c' i : PM^^ : OMK Then c'^= nc^ for PM'^ n . OM'. To duplicate the cube, make ?i = 2 ; that is, take CS = 2 CD. 4. The lemniscate. To find the locus of the intersection of the perpendicular from the centre of an hyperbola upon the tangent. Assuming the form y = mx + -Va^m^—b^ of the equation of the tangent (Art. 105, Ex. 12), that of the perpendicular is - y = X. m Substituting the value of m from the latter in the former, we have {f+xy = a'x'-¥y% a curve of the fourth order. Polar equation of the curve. The formulae for transformation being x^-\-y-=r\ x=rcosO, y=r8\nO, we have 7^ = a^co8\$-b^sm'0, or ')^=a'-{a'-\-b^)sm'0. (2) (1) Fig. 89. 182 ANALYTIC GEOMETRY. Discussion of the polar equation. If ^=0, r=±.a^ locating A and A^. As increases, r diminishes numerically :, when ?• = ± 0, and the portions ABO^ till sin^ = — ^ A'B'O, are traced, r then becomes imaginary, and remains so till sin Va^ -f b' again, 6 being in the second quadrant, to ^=180°, sin^^ is diminishing and From 6' = sin-^ Va^ + b' r increasing numerically, the portions OD'A', ODA, being traced, r being ±a when ^=180°. If the hyperbola is equilateral, a = b, and (1) becomes and (2) in like manner becomes 9-^ = a2 (cos^ - siu2 6) = a^ cos 2 0. In this case r is imaginary for all values of between 45° and 135°. 5. The witch. The ordinate to the diameter of a circle is produced till its entire length is to the diameter as the ordinate is to one of the segments with which it divides the diameter, these segments being taken on the same side. Find the locus of the extremity of the produced ordinate. Let OAB be the circle, R its radius, OD the diameter and axis of y, the origin, and the tangent at the axis of X. HIGHER PLANE LOCI. 183 Then, if P is a point of the curve, pqiDOi'.AQ'.qo, or XI 2 It:: ^2Ry-y^ : y, whence ' x" y^ = 4. R\2 By - f-) , (1) a locus of the fourth order. Let the student discuss the equation. G. 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, that of the re- quired locus is o -x" p ^x a cissoid, the diameter of whose circle = P 7. Given two fixed points F and F', to find the locus of P such thatPFxPF'=f—y. Let FF' be the axis of X, and the origin in the middle point of FF'. Then (?/'+ xy= 2c\x^- y') is" the required locus, in which c — ^FF', The locus is the lemniscate (see Ex. 4), the hyperbola being rectangular, and c = -^• V2 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^=AO^=Am-\-EP''={AO-yf-\-x\ r.x'+y'=2AO.y = 2AP.y. 184 ANALYTIC GEOlvrETRY. Also PB'= OB''=PD''+BD''= y^-i- (x - 0B)\ .-. x'+y^=20B.x = 2PB.x. Substituting the values oi AP and PB from these equations in ^P+P5 = a, wehave {x^+y'^{x-{-y)z=2axy, a locus of the third order. TKANSCENDEKTAL OUKVES. X 185 SECTION XIII. — TRANSCENDENTAL CURVES. 131. 1. The logarithmic curve. The equation of this curve is x = \ogy. Assuming the form y = a', in which a is the base of the logarithmic system, we observe that when x=0, 2/ = l? whatever the base; hence all / T 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 cc is negative, then ?/ = 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 a; = — oo ; 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 which 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 radius of the circle, and P any point of the locus. Then 0M= ON—MN, But 0M= X, 0N= arc PN= versin~^ ON to the radius r = r versin" -» r and MN= PQ = VJSrQ-QT= -\/2ry-y\ Hence the required equation is ni x = r versin"-^ V2 ry — ?/^. (1) Discussion of the equation. Since x is imaginary if y is negative, the curve lies wholly above X. If 2/ = 0, aj = 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=2r^ a; = r vers" ^ 2 = ± 7rr, ±37rr, etc., locating Z), and the corresponding points on the other arcs. Defs. Z)Z>' 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 the 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'W=cf>, ON=iiYcPN=rct>, TRANSCENDENTAL CURVES. 187 and \^ (2) which are called the equations of the cycloid, and are more useful than Eq. (1) in determining its properties. If any other point than F of the radius of the rolling circle be the generating point, the resulting curve is called the p^^olatp, 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 hypocycloid, 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 epitrochoid or hypotrocJioid. The general term applied to the locus generated by a point of a rolling curve is roulette. The circular functions. A series of transcendental curves is obtained by assuming the ordinate some trigonometrical func- tion of the abscissa, as 2/ = sina;, y = cot'ic, etc. The length of the arc corresponding to any value of x given in degi-ees 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/o (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=^mx. When a; = 0°, 2/ = 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 x> 180°, and reaching its least value y = — \ when x = 270°, locating C. From x = 270° to a? = 360°, y is negative and decreasing numerically, becoming zero again for x = .360°. It is evident that as x 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 Fig. 94. without limit along X to ± oo . The curve is sometimes called the sinusoid. 4. y=cosx. Let the student trace the curve. 5. 2/= tana;. 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 till ic= 180°, decreasing numer- ically from 00 to 0. From X = 180° to X == 270°, y is posi- tive and increasing, becoming 00 when a; =270°, etc. The curve consists of an infinite number of branches • equal to AOB, on either side of the origin, having for asymptotes o the lines x = ± -i x = ± -tt. Fig. 95. etc. 6. y=cotx. Let the student'trace the curve. 7. 2/ = versina;. The versine being always positive, the curve lies wholly above X, its limits along F being and 2. Let the student trace the curve ; also r 8. ?/ = coversina;. TRANSCENDENTAL CURVES. 189 9.2/= seca;. The curve is given in the figure. Let ihe stu- dent discuss the equation, and also trace the curve : 10. y = cosecx. Fig. 96. Spirals. The locus of a point receding from a fixed point along a straight line, which 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 = a6'', 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 = ae^ (1) r 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 which 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 1 l=a.27r a = -—9 27r and Eq. (1) becomes r = B (2) 190 ANALYTIC GEOMETKY. 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 Fig. 97. of the measuring circle, r increases uniformly with ^, but is oo only where 6= oo. 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, say 30°, or -• Then when 6 ^ = |, r = ^=OP. Having laid off 0P=^ on 01, make OF' = 2 OP, OP" = 3 OP, etc. Then OPP'P", etc., is the spiral. OQ is the radius of the measuring circle. 12. The reciprocal or hyperbolic spiral. The equation of this spiral is obtained from the general form by making '/I = — 1 ; whence r = l- (1) TRANSCENDENTAL CURVES. 191 In this spiral the radius vector evidently varies inversely as the angle. Assuming as before that r is unity when ^=27r, we have a = 27r, or ' »- = ^- (2) When ^ = 0, r = 00, and as 6 increases, r diminishes, but is P Fig. 98. zero only when ^=oo; hence there are an infinite nuiaber 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, 0F" = ^= 4, 3 = !?:, r= 12= OF. Make 0P'=~. etc., and draw PP'P".... 13. The lituus. This spiral corresponds to n = — i in the general equation. Hence its equation is a r = ^. (i; Fig. 99. 192 ANALYTIC GEOMETRY. or, if r= 1 when B= 27r, .-. a== V2w-, (2) For every value of 6^ r has two values, one positive and one negative, as shown in the figure. If ^ = 0, r=oo, and r=0 onl}' when ^ = oo ; 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 logr=a^, (1) or, if h be the base of the system, r = 6«^ (2) Whatever the logarithmic system r = 1 for ^ = 0. Hence, if OA = 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 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 ^ = — 00. Also, if ^ = 00, r= 00. Hence there are an infinite number of spires within and Fig. 100. without the measuring circle. A^ '^ PART II. SOLID ANALYTIC GEOMETEY. CHAPTER V. THE POINT, STRAIGHT LINE, AND PLANE, SECTION XIV. — INTRODUCTORY THEOREMS. 132. Defs. 1. By the angle betiveen tico straight lines not in the same plane is meant the angle between any two intersecting parallels. Hence, if through any point of one of the lines, a parallel 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 PQ, as P, draw PH par- allel to KL. Then HPQ is the angle between PQ and KL. 2. The foot of a perpendicular from a point upon a plane is called the projection of the point on the pkme. Thus, if P be any point, AB any plane, and the perpendicular to the plane through P meets the plane at 3f, M is the projection of P on AB. 3 . The foot of a perpe n- dicular from a 2)oint 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 P 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 perpendiculars from the exlreynities of the line upon the plane. Thus, if PQ be any limited straight line, AB any plane, PM^ QN, perpendiculars tt) the plane meeting it at M and JV", MN is the projection of the line PQ upon the plane AB. Since the perpendiculars PM and QJV determine a plane through PQ perpendicular to AB, the projection of a straight line upon a plane may also be defined as the intersection of a plane through the line, perpen- dicular to the given plane, with the latter. 5. The projection of a limited straight line upon another straight line is that portion of the latter intercepted by the projections of the extremities of the former. Thus, PS and QT being the perpendiculars from P and Q on KL, ST is the projection of PQ on KL. 133. The projection of a limited straight line upon another straight line is equcd to the length of the line midtiplied 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 MN being parallel, the planes P/S'Jf and QTN are parallel, and the intercepts /ST and 30/" are 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 PH parallel to MN. The angle between PQ and any parallel to PH is (Art. 132, 1) HPQ, and HP = 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 plane is equal to the length of the line multiplied by the cosine of the ayigle which the line makes with the pkwe. Thus MN= PQ cos NRQ. INTRODUCTORY THEOREMS. 197 134. If AD he the straight line joining A with Z), 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, and jy D, tlie perpendiculars to ^ — —- ^^ ; OX, meeting OX in A', \\^ ^^ \ B\ C\ D', respectively. j y^ \ It is evident that as A ^|^^^___J-^B i moves to D along the ^ ' cos = r cos cf) cos 6, (1) y= LS= OSsinO^ OP cos(f> - smO = r cos sin 0, (2) z = PS =OPsin<^ = rsin 1 ABC ... A' = B=G'' ('^ since each ratio is equal to Hence the condition of parallelism is that the ratios of the coef- ficients of the corresponding coordinates in the equations of the planes must he equal. Examples. 1. Find the angle between the planes x-\-2y ^ 22;-f 1 = and 3a; + 62/- 6^-5 = 0. ' Ans.O°. 2. Find the angle between the planes 2x-{-2y-{-z-{-l = ()^ 4:X — ^y-\-lz—\ = Q. Ans. 6 = cos ~^-^Y, 3. Show that 3fx-\-y-'-'\+l = and 12 (a;-f ?/) -32;-f-10=0 are parallel. 212 ANALYTIC GEOMETRY. 4. Show that .'^4-2^ — 22;-hl = 0is perpendicular to 2 a; -f- 5 2^ 4-6^-11 = 0; also a; 4- 22/+ 32; +1 = to Sx-\-6y- 5z -3 = 0. 5. Write the equation of a plane parallel to 3a; + 4?/ — 2 + 6 = 0. 6.. Write the equation of a plane perpendicular to 3a; + 4?/+ 2-1 = 0. 7. Find the distance between the parallel planes x-{-2y — 2z + l = 0,3x+6y-6z-25 = 0. Ans. ^. 8. Prove that Ax-^By-\-Cz+F-{-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 o( 2x + 5y + z—l=0 and x — y-\-z + 2 = 0, 10. Explain how to determine /c 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 2. T + 2/ — 2 + 1 = amd 3a; + 42/ + 22; + 6 = 0, passing also through the point (1, 1, 2) . Ans. 28x + 9y — 21z -\- 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 (a;', y\ z') to the plane a;cosa + 2/cos^ + 2cosy=p, is a^'cosa +2/'cos^ + 2'cosy — p, Ax' + Bx'+ Cz'-F o A ^ QQ or — See Art. 38. ■Va' -\-b'-{-g^ 14. Find the distances from the plane 5x-\-2y — 7z -\- 9 = of the points (1, — 1, 3) and (3, 3, 3). 15. Find the equation of a plane through (1, 10, —2), par- allel to the plane 2x-\-y — z-\-Q = 0. Ans.2x + y — z — 14:=0. ^'4 THE PLANE. 213 16. Find the equation of a plane through (1, —1, 3) per- pendicular to the plane 2x -{-y — z + 6 = 0. 17. Find the distance from (8,14,8) to 4:X-\-7y -{- iz — 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 BQ (Fig. 109), the intersection of the plane with XY. For like reasons By + Cz-{-F=0, Ax-^Cz-{-F=0, are the equations of the intersections RS 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 ^ B'^ B'' A A' A 7? If the traces are parallel, — = — . or — = — If the corre- B B' A' B' sponding traces on the other coordinate planes are also parallel, in whiclr case tlie planes themselves are parallel, we obtain in like manner A = ^ = ^ A' B' C' the condition already found. 214 ANALYTIC GEOMETRY. SECTION XVII.— THE STRAIGHT LINE. i 150. Equations of the straight line. Assuming the equation of a plane in the intercept form, if we impose the condition that the plane shall be perpendicular to XZ, its F-intercept b will be infinity, and its equation assumes the form Ax + Cz-[-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 B'y + C'z + F'=0, (2) X being indeterminate, represents a plane perpendicular to YZ. Let ABDL be the plane represented by (1), and AHDC that represented by (2). Values of x, y, 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 a; be the independent variable, and any value as ic = OM be substituted in (1). From (1) we may tlien 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), 2/ = SP must locate a point P in the plane AHDC, and P must lie on the intersection AD. Fig. 110. The line 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 ofABDL on XZ. Hence (1) is the equation of the plane ABDL^ or of its trace AB^ accord- ing as y = ^ or 2/=0. Similarly (2) is the equation of the plane AIIDC, or of its trace AO, according as cc = ^ or ic = 0. But AB and AC are the projections of AD on XZ and YZ, since the planes ABDL and AHDC are perpendicular respectively to XZ and YZ. Hence the straiglit line AD is determined when its projections on any two coordinate planes are given. Eliminating 2: between (1) and (2), we have an equation of I he form A"x-hB"y + F" = 0, (3) 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 GEOMETRY. the intersection AD of (1) and (2), since (1), (2), and (3) are simultaneous ; and its trace is OZ), 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, f'(y,z)=0, and eliminate the common variable, obtaining the third equation, /"(^,2/)=0, these three equations may be 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 having a given direction. Let {x', y', z') be the given point, P', and a, ^, y, the direction-angles of the given line. Let P be any other point (ic, ?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'=rcosa, y —y'=rcos/3, z — z'=r cosy; £ 1-1 x — x' V— ^' z—z' ..V from which = - — ^ = , (1) cos a cos f3 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 beinor 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 — y^ __z — z' L ~ M ~ N THE STRAIGHT LINE. 217 be the given equations. Dividing the denominators by •sjL- + M^ -f- ^S we have M N in wliich the sum of the squares of the denominators = 1. Thus, let 3 aj — 22; 4- 1 = 0, 4cc — 2/ = 0, be the given line. Then 1 f 4* a/77 Dividing the denominators by Vl^ + (f )^ -|- 4^ = — - — X z—i y ■ V77 V77 V77 2 S 8 the direction-cosines being — iz^, — =, — =• V77 V77 V77 Examples. 1. Find the equations of the intersection of x — y + z — 2 — 0^ and x-^y + 2z—l = ^^ and determine the position of the line. Eliminating y and z in succession, we have 2a:+3z-3 = 0, a: -3^ -3 = 0, or f = izJ::=.y±l, 1-2 1 ' 3 3 whence the line passes through (0, —1, 1), and its direction-cosines are 3 2_ 1 Vii' Vli' Vli 2. Find the intersection of x-\-y — z-\-\ =0 and 4a;-}-2/-- 22 + 2 = 0. Arts. A line through (0, 0, 1), whose direction-cosines are 1 2 3 Vli' vn' vii 218 ANALYTIC GEOMETEY. 3. Determine the position of aj= 4:^ + 3, ?/ = 32; — 2. Ans. Aline through (3, —2, 0), whose direction-cosines are ■ _4_ _3_ 1 ■\/Yq'' V2"6' V26 4. Write the equation of a line through (1, 2, —6), having I , i, |, for direction-cosines. ^?js. a;-22/ + 3 = 0, 2?/ -2; -10 = 0. 5. Write the equation of a line through (1, 4, —3) parallel to Z. Ans. a; = 1 , 2/ = 4. 153. Equations of a straight line through two given points. Let {x, y\ z'), {x'\ y", z"), be the given points. The equa- tions of a straight line through {x', y\ z') are x — x' ^ y — y' ^ z—z\ .^. cos a cos/3 cosy Since the point (x'\ y'\ z'') is also on the line, its coordinates must satisfy (1) ; hence x" - x' ^ y" - y' ^ z''—z \ ^2) cos a cos p cos y Dividing (1) by (2), member by member, (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). Ans. ^Zli=.Vi^ = Zzii. _4 4 -5 2. Find the direction of the line of Ex. 1. 3. Find the points in which tlie line of Ex. 1 pierces the co- ordinate planes. Ans. The line pierces XY in (— ^, ^). x-x' _ y-y' z-z' x" - x' y"-y' z" - z'' 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), (-6, -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 between two given straight lines. Let x — x^ _y — y _ z — z^ V i/' w ' x — x^ _ y — y _ g — g be the given lines. The angle between the two lines is given by the relation (Art. 142) cos 6 = cos a' cos a" + COS /?' COS y8" + COS y' COS y ", in which a', jS', y\ and a", /3", y", are their direction-angles. But (Art. 152), cos a' = COSyS' = cosy' COS a' cos;8" = COSy' VX'2 + 3/'2 _j_ jy/2' A^^ (1) A^" V^^'" + ^"' + iV^'" J (2) 220 ANALYTIC GEOMETEY. Hence cos 6 = — =^ — » (3) Cor. 1. If the lines are parallel, the corresponding direc- tion-cosines are equal, each to each; hence from (1) and (2), L' ^ 3r ^ N' L" J/" N" are the conditions of parallelism. CoR. 2. If the lines are perpendicular to each other, (9 = 90°, cosl9 = 0; hence i>'i."+ Jf' Jf"-f ^'iV^" = 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. (x-\-z='^^ a; + !/ = 5] x — y = —\^ w tt ^ ^'''' \ 2/=2j' . = lj' 2.T + . = 5i' 6' 2' 3* 2. Find the angle between 2/ = 5ic + 3, 2; = 3a; + 5, and y=2x-\-\^ z = x. 3. Show that 4a; -3?/ -10 = 0, ?/ + 4;2 4-26 = 0, and 7a; -2^ + 26 = 0, 34?/ + 7;^ - 90=0, are perpendicular to each other. 4. Showthat x = 2z + l, y = Sz + 4:, and a; = 3 — 22;, y = z — 2, are perpendicular to each other. 5. Show that 2a; — ?/ + l = 0, 3y — 2z-\-5 = 0, and 2aj - y — 7 = 0, 3 ?/ — 2;? + 7 = 0, are paralle\ THE STRAIGHT LINE. 221 6. Find the conditions that the straight line L M N is parallel or perpendicular to the plane Ax^Bij+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 A B g' 7. Find the equation of a line through ( — 2, 3, 5) perpen- dicular to 2a;-|-82/ — 2; — 4 = 0. Ans. aj-f22J-8 = 0, 2/4-8:^-43 = 0. 8. Show that 2x — y=0, 3y — 2z = is perpendicular to x-{-2y-{-Sz — 6=0. 9. Show that z = S, x-\-y = S is. parallel to x + y-}-z-6 = 0, and to the trace of the latter on XY, CHAPTER VI. SURFACES OF REVOLUTION, CONIQ SECTIONS, AND HELIX, 3i«=:c 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 ic-coordinate of any point Fig. 111. / SURFACES OF REVOLUTION. 223 as P will differ from that of its initial position. Hence, to dis- tinguish the a:-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 be r=/(2). (1) But P remains at the same distance, r, from Z during the revolution ; hence (Art. 137) r'^x'-^-yK (2) Substituting in (2) the value of r from (1), ^ + f=\_f{z)J, (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 r' + z'' = R\ whence ?'^ = \_f{zjf = Jf — z^. Substituting this in the general equation x^ -\-y^z= \^f{z)Y, we have x' + y^ + z' = B^ 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 whence r^ = ^, (a'- z') = [f(z)Y. (V- 224 ANALYTIC GEOMETRY. Substituting this in x^ -\-y'^ = \_fiz)f^ we have a2(aj2-|-2/2)+6V = a262, or .^^ + ^' +?! = !. (1) ^ ^^ ' IP h^ o? ^ ^ If a^=zh^z= ]^^ the ellipsoid becomes a sphere. By definition, plane sections parallel to XF are circles. Let the student prove that plane sections parallel to XZ and YZ are ellipses. 159. The oblate spheroid, or ellipsoid. This 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^ = ^^{h'^ -z^) =if{z)Y, and h\x^ + f)^ah' = a'h\ or ^ + ^' + ^ = 1, which is the required equation. If a^ = 6^ = 7^2^ 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 t)arabola be at the origin, the axis coinciding with Z. Then the equation of the generator is and the required equation is o? + f=-^pz. (1) Let the student show that plane sections parallel to TZ and XZ are parabolas. 161. The hyperboloid of two nappes. If an hyperbola be revolved about its transverse axis, the surface generated is \ SURFACIiIS OF BEVOLUTION. ^ 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 whence a^x^ + jr)-b'z' = - a'b\ (1) 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-i'^=~ a^b^, and that of the surface is b'(x' + 7f)-a'z'=a^b\ (1) a^ , y^ z^ or -^ + -.-72 = 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 ;i, 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' + f- = R' (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 GEOMETHY. 164. Cone of revolution. If a straight line revolves about anotiier straight line which it intersepts, 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 = 7i being the altitude and OB = H the radius of the base in the plane XY. The coordinates of A and B are (0, h), (jR, 0), and the equation of the generator z = ^ ?• -f ^, JLi whence '■^ = [/W?=f (A-2)S (1) Fig. 112. and the equation of the surface is (2) (a^+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'')tixn-e= {h-z)\ (3) If the vertex A is at an infinite distance, the cone becomes the cylinder. In this case 7i = oo, and from (1) Lf{^)l ^[ R' 2zR^ . z'R'' h /r = n\ and we obtain the equation of the cylinder x^ -{-if = 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 Y, cutting the section LPJ^ from the surface of the cone and the line LN fr6m the plane ZX] and let XON= <^, the inclination of the plane to XY, Since the cut- ting plane is perpendicular to ZX, its equation will be that of its trace LN^ or z = tan <^ • X, To refer the curve of intersection LPN to axes in its own plane, let OF be the axis of F, and OX' = ON produced the new axis of X; then the coordinates of Preferred to the primi- tive axes are 0M= o:^ MQ = z, QP = ?/, and referred to the new axes are x' = OQ, y' = QP. Hence y=y\ x=OM=OQ coscf>=x' cos, z=MQ=OQ sin<^=a;'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^ ^ _^ ^ ;^^ gj^ cf>-h^ = 0, or, since sin^<^ = cos^<^ tan^<^, 2/2 tan^^ + x^ cos2<^(tan'^ - tan^t^) + 2hx sin ^, tan2<^ > tan^6>, B'-AAC (Art. 80) is positive, and the section is an hyperbola. 228 ANALYTIC GEOMETRY. If (}><0, tan2<^= 0, tsm^cf) = taii^O, B^ — 4^(7 is zero, and the section is a parabola. Hence the section is an hyperbola, ellipse, or parabola, according as the cutting plane makes an angle with the plane of the base greater than, less than, or equal to, that which the elements do. If /i = 0, the equation becomes 2/Han2^ 4-a^ cos2<^(tan2^ -tan^c^) = 0, and the plane passes through the vertex which is at the origin. In this case if <^ > ^, 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 2/ = 0, the equation of AL. These are particular cases of the hyperbola, ellipse, and parabola, respectively. If <^ = ; a particular case of = 0° and <^ = 90°, in which cases the section is a circle, or two parallel straight lines, elements of the cylinder. Having thus 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. Let the axis of the cylinder 167. Equations of the helix coincide with Z, OA = 22 = 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 <^ = ^0Q = angle made by the projection of the radius vector OP on XFwith X Then X = 0M= OQ cos ^ = i? cos cf>, y = MQ = OQsm