the elements of coordinate geometry - s.l. loney, Manuais, Projetos, Pesquisas de Matemática

Baixe the elements of coordinate geometry - s.l. loney e outras Manuais, Projetos, Pesquisas em PDF para Matemática, somente na Docsity! BIBLIOGRAPHIC RECORD TARGET Graduate Library University of Michigan Preservation Office Storage Number: ABR0018 UL FMTBRTaBLmT/C DT07/18/88R/DT07/18/88CC STATmmE/L1 035/1: : |a (RLIN)MIUG86-B48728 035/2: : [a (CaOTULAS)160121345 040; : [a IC] |cMiU 100:1 : |a Loney, Sidney Luxton. 245:04: |a The elements of coordinate geometry, |c by S. L. Loney, M.A. 250: : la 2d edition, revised. 260: : |a London, |a New York, |b Macmillan and Co., |c 1896. 300/1: : ja ix, 416, xiii p. |b diagrs. |c19 cm. 650/1: 0: [a Geometry, Analytic 998: : |c WFA |s 9124 Scanned by Imagenes Digitales Nogales, AZ On behalf of Preservation Division The University of Michigan Libraries Date work Began: Camera Operator:  THE ELEMENTS OF COORDINATE GEOMETRY Ss. 1. LONEY, MA, LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE, PROFESSOR AT THE ROYAL HOLLOWAYX COLLEGE. SECOND EDITION, REVISED. Zondon MACMILLAN AND CO, Lrp, NEW YORK: THE MACMILLAN CO. 1896 Lad! righis reserved.) First Edition (Globe Bvo.), 1895. Reprinted with corrections (Crown 8Bvo.), 1896, CHAP, II. NI, IV. VI. VIL CONTENTS, PAGE INIRODUCFION. ÁLGEBRAIC RESULTS , . 1 CoorDINATES. LENGTHS OF STRAIGHT LINES AND AREAS OF TRIANGLES . . . . . 8 Polar Coordinates . . . . . . 19 Locus. Equarion tro A Locus . . . . 24 The Straight Line. RecrancuLAR COORDINATES, 31 Straight line through two points-. . . . 39 Angle between two given straight lines . . 42 Conditions that they may be parallel and per- pendicular . . . . . . . 44 Length of a perpendicular . . . . 51 Bisectors of angles . . . . . . 58 Tam Seraigar Lise. Porar EQUATIONS AND OBLIQUE COORDINATES . . . . 66 Equations involving an arbitrary constant . . B Examples of loci . . . . , . 8o EQUATIONS REPRESENTINS TWO OR MORE STRAIGHT Lints : . : . . . . .. 88 Angle between two lines given by one equation 90 General equation of the second degree . . 9 TRANSFORMATION OF COORDINATES . . . 109 Tnvaxiants . . . . . . . . 115 viil CHAP, VI IX. XI. NIE XII. CONTENTS, The Circle . Equation to a tangent . Pole and polar . Equation to a circle in polar coordinates Equation referred to oblique axes Equations in terms of one variable SystEMS OF CIRCLES . Orthogonal circles . Radical axis Coaxal circles Conic Sections. Tam PARABOLA Equation to a tangent . Some properties of the parabola Pole and polar Diameters . Equations in terms of one variable Tur PARABOLA (continued) Loci connected with the parabola Three normals passing through a given point. Parabola referred to two tangents as axes Tue ELLIPSE Ausiliary cirele and eccentric angle Equation to a tangent . . Some properties of the ellipse Pole and polar . Conjugate diameters . Four normals through any point Examples of loci THE HyPoRBOLA Asymptotes Equation referred to the asymptotes as axes . One variable. Examples PAGE 118 126 137 145 148 150 160 160 161 166 174 180 187 190 195 198 206 206 211 217 225 281 237 24% 249 54 265 266 s71 284. 296 299 GHAP, XIV. XV. XVI XVII! CONTENTS. Porar Equarion ro A Cont, . . , Polar equation to a tangent, polar, and normal . GmnerAaL Equation. TRAcING OF CURVES Particular casos of conic sections . . Transformation of equation to centre as origin Equation to asymptotes Tracing a parabola, Tracing a central conic. . Eccentricity and foci of general conic GungRAL EQUATION Tangent . Conjugate diameters . Conies through the intersections of iso conies . The equation S=Auy . General equation to the pair of tangente drawm from any point The director circle. The foci The axes . . . Lengths of straight lines drawn in given | divections to meet the conic, Conices passing through four points Conies touching four lines. The conic LM=R2. MiscELLANEOUS PROPOSITIONS . On the four normals from any point to a central conic Confocal conies Circles of curvature and contact of the third order . Envelopes 1x PAGE 306 813 322 322 326 329 332 338 342 349 349 352 356 358 364 365 367 369 370 318 380 382 385 385 392 398 407 ANSWERS . . . . . . . . iii 2 COORDINATE GEOMETRY. Ex. 1. Ifa and 8 be the roots of the equation b e at brrc=0, i.e. Pra +5=0, h =-L and ag=l we have atB=-a ani aB=. Ex. 2. Ifa, 8, and y be the roots of the cubic equation apr ba por +d=0, 6 e a i.e. of gpa? > 0 +-=0, & & & b we have atbty=- e Brrgarag=o, a and aBy=—— a 3. It can easily be shewn that the solution of the equations av+by+es=0, and am + dy +eg=0, Po z ad Da dy is —— = 4 050, Crlty Determinant Notation. Gy Da bi, b, second order and stands for the quantity ed, — ash, so that is called a determinant of the & The quantity Gs dy do dy = hd — God =2x5-4x8=10-12=-2; 2,3 4, 5 -8,-4 -7 Cel=-8M(-6)-(-1x(-4)=18-28= 10, DETERMINANTS, 3 [1 Ca Ca] dis Das Dgleccescceremererertarma (1) Cis Cgy Os is called a determinant of the third order and stands for the quantity 5. The quantity & bas by] ass a Dal ais da + Os Cos Ey Cs Cs! [ax Ca ve. by Art. 4, for the quantity O (Das; — bato) -— Eg (Dies — ba) + Oy (Dt, — bat)» ve. ey (Doc; — ba6a) + dio (Bot — Des) + ds (Dito — Dos). ria (2), 6. A determinant of the third order is therefore reduced to thrce determinants of the second order by the folowing rule: Take in order the quantities which occur in the first row of the determinant; multiply each of these in turn by the determinant which is obtained by erasing the row and column to which it belongs; prefix the sign + and — al. ternately to the products thus obtained and add the results. Thus, if in (1) we omit the row and column to which «, a bi and this is the Cu, Cai belongs, we have left the determinant coefficient of a in (2). Similarly, if in (1) we omit the row and column to which P ds] and this &, Cs] with the — sign prefixed is the coeficient of às in (2). «» belongs, we have left the determinant 1 1,-2,-8 7. Ex. The determinant |-4, 5,-6 -71, 8-9 5 | 4, -6 =1x 8, o (-2 a 7 e 3) xy [0-9 8x ( 02x Ate -(- mn 6) IL 4)x8-(-7)x5) =[-45+48] +2/86-42)-34- 82485) =8-12-9=-18, 1.2 4 COORDINATE GEOMETRY, Grs Ggy Ugy Gy db, ba, dg, da Cs Coy Cs, Og da, dy do, dy is called a determinant of the fourth order and stands for the quantity 8. The quantity das dos b, dy, das bs| Gy X [Ca Og» Ca Ta X |O C3, “| das da, dy di, dg, di | dis boy Ds [is Das Dyl +agX|C 6, Cal —C4X|C Cy Cy] ho ds, da id, das dz and its value may be obtained by finding the value of each of these four determinants by the rule of Art. 6. The rule for finding the value of a determinant of the fourth order in terms of determinants of the third order is clearly the same as that for one of the third order given in Art. 6. Similarly for determinants of higher orders. 9. A determinant of the second order has two terms. One of the third order has 3 x 2, 2.e. 6, terms. One of the fourth order has 4x3 x 2, %e. 24, terms, and so on. 10. Exs. Prove that e 5-8 7 (1) Im “9-2. (2) [4 fes e) |-2, 4- gf=-ss 9% 8, -10) 9,8,7 -a db é (4) 16,5, 4/=0, (5) q -d, el=tade, 8,2,1 La b-e 5h, 9 (6) “hs d, f|=abe+2fgh — af? - dg? - chê, me ELIMINATION. g 14, Tfagain we have the four equations Gê + Coy + dg + ago =0, dyoc+ dby+ bx+bu=0, CGX+ ey +c+cu=0, and dy + day + dg + du =0, it could be shewn that the result of eliminating the four quantities «x, y, &, and w is the determinant Gr, Gas Og; Mg bis Das bas Da. q Cy, Ca Cas Og dh, da, ds, da | A similar theorem could be shewn to be true for w equations of the first degree, such as the above, between m unknown quantities, Tt will be noted that the right-hand member of each of the above equations is zero. 168 GOORDINATE GEOMETRY. [For the radius ZR of this circle is at right angles to the radius 0,R, and so for jts intersection with any other circle of the system.] Fig. IL Hencc the limiting points (being point-circles of the system) are on this orthogonal circle. The limiting points are therefore the intersections with the line of centres of amy circle whose centre is on the common radical axis and whose radius is the tangent from it to any of the circles of the system. Since, in Fig. I., the limiting points are imaginary these orthogonal circles do not meet the line of centres in real points. Tn Fig. II. they pass through the limiting points Z, and Lo. These orthogonal circles (since they all pass through two points, real or imaginary) are therefore a coaxal system, Also if the original circles, as in Fig. 1I., intersect in real points, the orthogonal circles intersect in imaginary points; in Fig. II. the original circles intersect in imaginary points, and the orthogonal eireles in real points. We therefore have the following theorem : A set of comval cireles can be cut orthogonally by another set of conxal circles, the contres of coach set lyimg on the radical axis of the other set; also one set às of the limiting- point species and the other set of the other species. ORTIOGONAL CIRCLES. 169 191. Without reference to thc limiting points of the original system, it may be easily found whether or not the orthogonal cireles meet the original line of centres. For the circle, whose centre is 7 and whose radius is TR, meets or does not mcet the line 0,0, according as TR is > or< TO?, i.e. aecording as TOP-ORºis 2 TO, i.e. according as TO +002-0R2is Z TO, i.e. according as 00, is E OR, à.e. according as the radical axis is without, or within, each of the cireles of the original system. 192, In the next article the above results will be proved analytically. To find the equation to any circle which cuts two given circles orthogonadiy. Take the radical axis of the two circles as the axis of q, so that their equations may be written in the form 2 +9)-Qgare=0 seres (1), and + Igu+c=O.c (2), the quantity c being the same for each. Let the equation to any circle which cuts them or- thogonally be (nu APr(y- BL= Ri (3). The equation (1) can be written in the form (eo grp [NBP (4). The circles (3) and (4) cut orthogonally if the square of the distance between their centres is equal to the sum of the squares of their radi, i.. 1Ê (Agr B=R+VP-cP, i.e, if AA DAg= Re (5). Similarly, (3) will cut (2) orthogonally if AB -BAg= RC. (8). Subtracting (6) from (5), we have A(g -g)=0. Hence 4=0, and Rº=Bº+e. 172 COORDINATE GEOMETRY, Let (1, k) be a point such that the length of the tangent from it to (1) is always À times the length of tho tangent from it to (2). Then RBritcoghro=M[hê4+ kk -2gah+o0]. Hence (h, k) always lies on the circle Ni ato ef Ap o=0 RREO (3). This circle is clearly a cirele of the coaxal system to which (1) and (2) belong. Again, the centre of (1) is the point (g,, 0), the centre of (2) is 2. (9a, 0), whilgt the centre of (3) is E É ,0 Nx Hence, if these three centres be called O,, O,, and O,;, we have Rg 2 00-52 = A=m ló =) GM = 1 and O0,= o ta pa (e a), so that 0,0, : 0,0, :: MN: 2, The required locus.is therefore a circle coaxal with the two given cireles and whose centre divides externally, in the ratio 2:1, the line joining the centres of the two given circles. EXAMPLES. XXIV. 1, Prove that a common tangent to two circles of a coaxal system subtends a right angle at either limiting point of the system. 2. Prove that the polar of a limiting point of a coaxal system with respect to any cirolo of the system is the same for all circles of the system. 8. Prove that the polars of any point with respect to a system of coaxal cireles all pass through a fixed point, and that the two points are equidistant from the radical axis and subtend a right angle at & limiting point of the system. If the first point be one limiting point of the system prove that the second point is the other limiting point. &, A fixed circle is cut by a series of ciroles all of which pass through two given points; prove that the straight line joining the intersections of .the fixed circle with any circle of the system always passes through a fixed point, 5. Prove that tangents drawn from any point of a fixed cirele ot a coaxal system to two other fixed cireles of the system are in a constant ratio, [Exs. XXXIV] COAXAL CIRCLES. EXAMPLES. 178 6. Prove that a system of coaxal circles inverts with respect to either limiting point into a system of concentric circles and find the position of the common centro, 7, A straight line is drawn touching one of a system of coaxal cireles in P and cutting another in Q and R. Shew that PQ and PR subtend equal or supplementary angles at one of the limiting points of the system. 8. Find the locus of the point of contact of parallel tangents which are drawn to each of a series of coaxal circles, 9. Prove that the circle of similitude of the two circles a2.-y3- 2ho +9=0 and a2+y?-2h'2+0=0 (i.e. the locus of the points at which the two cireles subtend the same angle) is the coaxal circle 241 a glk+ô 3=0. + hr dk! t+ô=0. 10. From the preceding question shew that tho centres of simili- tude (i.e, the points in which the common tangents to two circles meet the line of centres) divide the line joining the centres internally and externally in the ratio of the radii. mM, UH sty N/-i=tan (u+o Nf =1), where x, y, “ and v are all real, prove that the curves u=constant give a family of coaxal cireles passing turough the points (0, 1), and that the curves v= constant give à system of cireles cutting the first system orthogonaily. 12. Find the equation to the circle which cuts orthogonally each of the circles Wry+2gare=0, Pryiragie+re=0, and 2+9+2he+2ky+a=0. 13. Find the equation to the circle cutting orthogonally the three cireles e ryi=a, (u-c)ryi=e, and r(y-bP=al, 14. Find the equation to the cirele cutting orthogonally the three circles 2 +y-2r+3y-T=0, 2 +y2+60-54+9=0, and ar yirTao-9y+29=0. 15. Shew that the equation to the circle cutting orthogonally the cireles femoral UP=o (e-bt(y-ap=e, and (u-a-b-cP+yi=ab+e, is atry?-2ar(ard)-y(a+rd) +a?+ Bad d2=0, CONIGC SECTIONS. CHAPTER X. THE PARABOLA. 196. Conic Section. Def. The locus of a point P, which moves so that its distance from a fixed point is always in » constant ratio to its perpendicular distance from à fixed straight line, is called a Conic Section, The fixed point is called the Focus and is usually denoted by 8. The constant ratio is called the Eiccentricity and is denoted by e. The fixed straight line is called the Directrix. The straight line passing through the Focus and per- pendicular to the Directrix is called the Axis. When the eccentricity e is equal to unity, the Conie Section is called a Parabola. When e is less than unity, it is called an Ellipse. When e is greater than unity, it is called a Hyper- bola. [The name Conic Section is derived from the fact that these curves were first obtained by cutting a cone in various ways. | THE PARABOLA, 177 equal in magnitude. The line PP?" is called a double ordinate, As x increases in magnitude, so do the corresponding values of y; finally, when x becomes infinitely great, y becomes infinitely great also. By taking a large number of values of w and the corresponding values of 4 it will be found that the curve is as in the figure o Art. 197. The two branches never meet but are of infinito length. 201, The quantity ay? — Lar às negative, zero, or positive according as the point (a”, y) às within, upom, or arithouê the parabola, Let Q be the point (a, y) and let it be within the curve, 2.º. be between the curve and the axis 4X. Draw the ordinate QN and let it meet the curve in P. Then (by Art. 197), PNº=: dao”, Hence 9º, ic. QN? is <PNº, and hence is < dua. “+ yº- Sax is negative. Similarly, if Q be without the curve, then 7º, i.e. QN”, is> PN?, and hence is > dar”. Hence the proposition, 202. Latus Rectum, Def. The lutus rectum of any conie is the double ordinate ZLSZ/ drawn through the focus 8. Tn the case of the parabola we have SL = distance of L from the directrix = 5Z = 2a, Hence the latus rectum = da, When the latus rectum is given it follows that the equation to the parabola is completely known in its standard form, and the size amd shape of the curve determined, The quantity do is also often culled the principal parameter of the curve. Focal Distance of any point. The focal distance of any point P is the distance SP. This focal distance = PM =ZN=ZA+AN=a +. L. 12 178 COORDINATE GEOMETRY. Ex. Pind the vertez, axis, focus, and latus rectum of the parabola 4yº + 12x — 20y +67=0. The equation can be written W-by=—Be-sr, ie. (y-5P= Se -SEnai= 8 (041). Transform this equation to the point (—%, 3) and it becomes w=— 82, which represents a parabola, whose axis is the axis of « and whose concavity is turned towards the negative end of this axis. Also its latus rectum is 3, Referred to the original axes the vertex is the poini ( —%, $), the axis is y=$, and the focus is the point (-4—&, 8), i.e. (— 3, 8). EXAMPLES, XXV. Find the equation to the parabola with 1. focus (3, —4) and directrix 6x — 7y+5=0. 2. focus (a, b) and directrix + =. Find the vertex, axis, latus rectum, and focus of the parabolas 3 yio-dor dy. 4 r=80-T. 5. 22º-2av+20y=0. 6 gy=4y-da, 7. Draw the curves (1) gê= —daz, (2) a2=4ay, and (3) a?= — day. 8. Find the value of p when the parabola y?=4pa goes through the point (i) (3, — 2), and (ii) (9, — 12). 9. For what point of the parabola yº=18% is the ordinate equal to three times the abscissa ? 10. Prove that the equation to the parabola, whose vertex and focus are on the axis of q at distances « and a! from the origin respectively, is yl=4(a' cata). 11. In the parabola 92=6x, find (1) the equation to the chord through the vertex and the negative end of the latus rectum, and (2) the equation to any chord through the point on the curve whose abscissa is 24, 12. Prove that the equation y2+2427+2By+C=0 represents a parabola, whose axis is parallel to the axis of x, and find its vertex and the equation to its latus rectum. 13. Prove that the locus of the middle points of all chords of the parabola y?=4ax which are drawn throngh the vertex is the parabola y?= 20x. fêxs, XXV.] THE PARABOLA. EXAMPLES. 179 14, Prove that the locus of the centre of a cirele, which intercepts a chord of given length 2a on the axis of x and passos through a given, point on the axis of y distant d from the origin, is the curve 22 — Qyb + bê= 2, Trace this parabola, 15. PQ is a double ordinate of a parabola. Find the locus of its points of trisection, 16. Prove that the locus of a point, which moves so that its distance from a fixed line is equal to the length of the tangent drawn from it to a given cirele, is a parabola. Find the position of the focus and directrix. 17. If a circle be drawn so as always to touch a given straight line and also a given circle, prove that the locus of its centre is a parabola, 18. The vertex 4 of à parabola is joined to any point P on the curve and PQ is drawn at right angles to 4P to meet the axis in Q. Prove that the projection of PQ on the axis is always equal to the latus rectum. 19. Ifon a given base triangles be described such that the sum of the tangents of the base angles is constant, prove that the locus of the vertices is a parabola. 20. A double oxdinate of the curve y?=4pz is of length 8p; prove that the lines from the vertex to its two ends are at right angles. 21. Two parabolas have a common axis ahd concavities in oppo- site directions; if any line parallel to the common axis meet the parabolas in P and P”, prove that the locus of the middle point of PP' ig another parabola, provided that the latera recta of the given para- bolas are unequal, 292. A parabola is dawn to pass through 4 and B, the ends of a diameter of à given circle cf radius «, and to have as directrix a tangent to a concentrice circle of radius b; the axes being AB and a perpendicular diameter, prove that the locus of the focus of the aê y2 parabola is m+ post 203. To find the points of intersection of any straight line with, the parabota Rm Aga ,irrreerrerererrertaeasa (3). The equation to any straight line às YE MU Corrirrerrerreearerams (2). The coordinates of the points common to the straight line and the parabola satisfy both equations (1) and (2), and are therefore found by solving them. 12-—2 182 COORDINATE GEOMETRY, Tbe line (1) will touch (2) if it meet it in two points which are indefinitely close to one another, 2%. in two points which ultimately coincide. The roots of equation (3) must therefore be equal. The condition for this is 4 (me 202 = Ame, te, &-ame=0, so that col, m Substituting this value of c in (1), we have as the equation to a tangent, =mx+ 2 y= = In this equation »m is the tangent of the angle which the tangent makes with the axis of a, The foregoing proposition may also be obtained from the equation of Art. 205. For equation (4) of that article may be written 2a 2a" , Sp RS cereais (1). In this equation put = , £ 2 and hence ret = Ea , and dua = da mf m The oquation (1) then becomes y=ma + a Also it is the tangent at the point (a”, y'), i.e. (E. o) . 207. Equation to the normal at (x, y). The required normal is the straight linc which passes through the point (1º, y) and is perpendicular to the tangent, àe, to the straight line Za , == (mta) 4557 (ua) Tis equation is therefore y—y=m'(e- a), Da, where m xD =-1, de m=- 2 (Art. 69.) y 2a” NORMAL TO A PARABOLA. 183 and the equation to the normal is py é y-y = (8-x) comeram (1). 208. To express the equation of the normal in the form y = ma — Zum — aê, Tn equation (1) of the last article put 4 — : O) om te y =- am, Ze . Y ta 2 Hence =" am da The normal is therefore y + 2am =m (x — am), te. y=mx- 2am - am? and it is a normal at the point (am?, — 2am) of the curve, In this equation m is the tangent of the angle which the normal makes with the axis. It must be carefully distinguished from the m of Art. 206 which is the tangent of the angle which the tangent makes with the axis. The “m” of this article is — 1 divided by the “m” of Art, 206. 209. Subtangent and Subnormal, Def. If the tangent and normal at any point ? of a conic section meet the axis in 7 and'G respectively and PN be the ordinate at ?, then NT is called the Subtangent and NG the Subnormal of P. To find the length of the subtangent amd subnormal. If P be the point (w', 4) the equation to YP is, by Art. 205, gy = 20 (0 +08)... (1). To obtain the length of 47, we have to find the pomt-where this straight line meets the axis of x, i.e. we put y=0in (1) and we have Hence AT=AN, 184 COORDINATE GEOMETRY, [The negative sign in equation (2) shews that 7 and N always lie on opposite sides of the vertex 4.] Hence the subtangent N7-=24N = twice the abscissa of the point ?. Since TPG isa right-angled triangle, we have (Euc. vt. 8) PNº= TN. NG. Hence the subnormal NG PNº PN? CNC aanNO The subnormal is therefore constant for all points on the parabola and is equal to the semi-latus rectum. 2. 210. Ex. 1. Ifa chord which is normal to the parabola at one end subtend a vight angle ai the vertex, prove that it is inclined at an angle tanTi,/2 to the axis, The equation to any chord which is normal is y=ma — 2am — am, i.0. me-y=2am+ am, The parabola is yê= dam, The straight lines joining the origin to the intersections of these two are therefore given by the equation yº (Bam + am?) — das (ma — 9) =0, Tf these be at right angles, then Zam + am — 4am=0, v.e. m= =&/2, Ex. 2. From the point where any normal to the parabdola 9º =4aax meets the axis is drawn a line perpendicular to this normal; prove that this line always touches an equal parabota. The equation of any normal to the parabola is y=ma — 2am — amê, This meets the axis in the point (24 + am?, 0). The equation to the straight line through this point perpendicular to the normal is y=m (2 - 20 — am), where mm=-1. The equation is therefore a =m,(v-24— —— 1 2)» ny a a i.e. y=my (x - 24) = 1 XXVII] TANGENT AND NORMAL. EXAMPLES, 187 98. PNP' is a double ordinate of the parabola ; prove that the locus of the point of intersection of the normal at P and the diameter through P' is the equal parabola 3? = da (x — 40). 29. 'The normal at any point P meets the axis in G and the tangent at the vertex in G'; if 4 be the vertex and the rectangle AGQG be completed, prove that the equation to the locus of Q is sº=2gaº + ay, 30. Two equal parabolas have the same focus and their axes are at right angles; a normal to one is perpendicular to a normal to the other; prove that the locus of the point of intersection of these normals is another parabola. 31. If a normal to a parabola make an angle q with the axis, shew that it will cut the curve again at an angle tan"! (L tan q). 32. Prove that the two parabolas 32 =4aa and 9? =4e (x --b) cannot Pp y . b have à common normal, other than the axis, unless — >2, 93, Ifa?>8b, prove that a point can be found such that the two tangents from it to the parabola 9? = tas are normals to the parabola n2= dby, 34, Prove that thrce tangents to a parabola, which are such that the tangents of their inclinations to the axis arc in a given harmonical progression, form a triangle whose area is constant. 35. Prove that the parabolas 32 = £ax and «2=4by cut one another Jaz bs 2 (aê 453) 86. Prove that two parabolas, having the same focus and their axes in opposite directions, cut at right anglos, 97. Shew that the two parabolas e -4a(y-20-0)=0 and y2=4D (x- 2a 4D) intersect at right angles at a common end of the latus rectum of each. at an angle tan"? 38. A parabola is drawn touching the axis of « at the origin and having its vertex at a given distance k from this axis. Prove that the axis ot the parabola is a tangent to the parabola «2= — Sk (y — 2h). 211. Some properties of the Parabola. (0) If the tamgent and normal at amy poirt P of the parabolu meet the axis in T and G respectively, then ST=8G=5SP, 188 COORDINATE GEOMETRY. and the tangent at P às equally inclined to the axis and the focal distance of P. M p Y Et T Z) NAS N G X = Let ? be the point (a, y). Draw PM perpendicular to the dircetrix. By Art. 209, we have 47 = AM. PS=TArAS=AN+ ZA =ZN=HP=SP, and hence 28TP=: SPT, By the same article, NG=248=45. “ SG=SNANG=Z84 SN=MP=-SP. (8) If the tangent at P meet lhe divectria im K, then KSP às a right amgle. For 2 SPT=1PTS=1 KPM. Hence the two triangles XPS and KPM have the two sides XP, PS and the angle XPS equal respectively to the two sides XP, PM and the angle APM, Hence 2 KSP=:KMP=a right angle. Also 18KP=2 MKP. (y) Tangents at the extremities of any focal chord inter- sect at right angles in the directris, For, if PS be produced to meet the curve in ?;, then, since 2 P'SK is a right angle, the tangent at P' meets the directrix in A. PROPERTIES OF THE PARÁBOLA, 189 Also, by (8), 2 MKP= +: SKP, and, similarly, 2 MKP = SKP'. Hence «PRP =bi SKM + 2 SKM' = a right angle. (3) 1fSTY be perpendicular to the tamgent at P, them Y lies on the tangent at the vertex and S Pe=AS.SP. For the equation to any tangent is & y = mm + — (1) The equation to the perpendicular to this line passing through the focus is 1 y=— a le—a) crrtereraeaseanera (2). "The lines (1) and (2) meet where mas Lo 1 mat, mom mom t.€. where q = 0. Hence Y lies on the tangént at the vertex. Also, by Eue. vi. 8, Cor., 87 = 8A.ST=AS, SP. 212. To prove that through any given point (x, 3) there pass, in general, two tangents to the parabola, The equation to any tangent is (by Art. 206) & = mam ueerer entrem (1). IÉ this pass through the fixed point (2, 44), we have & dA = Ma + mo ENA mig, my ta=0 a (2). For any given values of «, and 4, this equation is in general a quadratic equation and gives two values of qm (real or imaginary). Corresponding to each value of m we have, by substi- tuting in (1), a different tangent. 192 COORDINATE GEOMETRY. Let 7 be the point (2, 74), so that its polar is win = Ba (ão + My) cirerreraeso (1). 'Fhrough 7 draw a straight line parallel to the axis; its equation is therefore US eceereirees (2). Let this straight line meet the polar q in V and the curve in ?. v The coordinates of V, which is the Q intersection of (1) and (2), ave therefore 2 E — 2 and 3 eee (3). R Also P is the point on the curve whose ordinate is 7, and whose coordi- Tig. 1. nates are therefore 8 . . abscissa of 7' + abscissa of V Since abscissa 0P=""""D"" "DD ..., there- 2 fore, by Art. 22, Cor, P is the middle point of 7'F. Also the tangent at P is Yi = 20 (1 + 4) , which is parallel to (1). Hence the polar of Z is parallel tó the tangent at P. Vig. 0. To draw the polar of 7 we therefore draw a line through 7, parallel to the axis, to meet the curve in ? and produce itto VP so that YP= PV; aline through P parallel to the tangent at 2 is then the polar required. 217. If the polar of a point P passes through the point 'E, then the polar of T goes through P. (Fig, Art, 214), Let P be the point (x, y;) and T the point (A, &). The polar of Pisyy=2a(v+x;). Since it passes through T, we have ihi=20 (2,4) ce reerreresiaeaaa (1. PAIR OF TANGENTS FROM ANY POINT. 198 The polar of T is yh=2 (a + h). Since (1) is true, this equation is satisfied by the coordinates x, and qj- Hencee the proposition. Cor. The point oi intersection, 7, of the polars of two points, P amd Q, is the pole of the line PQ. 218. To jindthepole of a given straight line with respect to the parabola. Let the given straight line be Ax+By+0=0. Tf its pole be the point (z,, y), it must be the same straight line as yy=2a (242), i.e. 24x —yy + 242,=0. Since these straight lines are the same, we have 2a —W 2am 1 Bo . CU “Ba i.e. “=q and 4 n=-2" 219. To find the equation to the pair of tangents that cam be dramwn to the porabola from the poi (21, Y). Let (h, k) be any point on. either of the tangents drawn from (o %). The equation to the line joining (ay, 31) to (A, b A ty Lo) h = As Pino has z bm h — a “hm If this be a tangent it must be of the form Yoh= ms 2. y= =ma+ = y= m so that Pon m and land a . h- a h-m nm Hence, by multiplication, by hp — box se “hm hm te. a(h — an) = (kn) (o — hey) 194 COORDINATE GEOMETRY. The locus of the point (A, k) (i.e. the pair of tangents required) is therefore «(00 — 0)? = (3 — 361) Locypy — gas) eseceeoro (1). Tt will be seen that this equation is the same as (3º — Lao) (3 — Aa) = try — 2a (ae + e) P 220. To prove that the middle points of a system of parallel chords of a parabola all tie on q straight line which 2s parallel to the axis, Since the chords are all parallel, they all make the same angle with the axis of a, Let the tangent of this angle be m. The equation to Q2, any one of these chords, is there- fore YSME+C (D, where c is different for the several chords, but m is the same. This straight line meets the parabola 4º = 4a in points whose ordinates are given by mf = da (y—), . o da Ac te. Pout = Ocera (2). Let the roots of this equation, i.e. the ordinates of Q and R, be y and 3”, and let the coordinates of V, the middle point of QR, be (A, k). Then, by Art. 22, from equation (2). The coordinates of V therefore satisfy the equation 20 Vem? so that the locus of V is a straight line parallel to the axis of the curve. THE PARABOLA. EXAMPLES, 197 Substituting these values in (1), we have (2a cot 8 +y sin 0? = 4a (a cot? 0 + w + y cos 6), e. 1 sin? O = Lam. The required equation is therefore 9 = Apa cesirerrerererereeaeenea (2), where P= oo (1 +ctB)=a+AN=SP (by Art. 202). The equation to the parabola referred to the above axes is therefore of the same form as its equation referred to the rectangular axes of Art. 197. The equation (2) states that QV'=4SP. PP. 225. The quantity 4p is called the parameter of the diameter PV. It is equal in length to the chord which is parallel to PY and passes through the focus. For if Q'V'R' be the chord, parallel to PY and passing through the focus and meeting PV in V”, we have PV'=ST=8P=p, so that QVê=dp.PV'= Ap, and hence QR' =20'V' = dp. 226. Justasin Art. 205 it could now be shown that the tangent at any point (x”, 3) of the above curve is gy = Qp(m+ a), Similarly for the equation to the polar of any point. EXAMPLES. XXVII. 1. Prove that the length of the chord joining the points of contact of tangents drawn from the point (x, 44) is ninfa? ny dam, & 92. Prove that the area of the triangle formed by the tangents from the point (x,, y1) and the chord of contact is (y;? — daxjÊ-+2a. 198 COORDINATE GEOMETRY. [Exs, XXVIL] 8. If a perpendicular be let fall from any point P upon its polar prove that the distance of the foot of this perpendicular from the focus is equal to the distance of the point .P from the directrix. 4, What is the equation to the chord of the parabola yº=8a which is bisected at the point (2, -3)? 5. The general equation to a system of parallel chords in the parabola g2=2y is 4x —-y-+k=0, What is the equation to the corresponding diameter? 6. ?,Q,and R axe three points on a parabola and the chord PQ cuts the diameter through Rin V. Ordinates PM and QN are drawn to this diameter. Prove that RM, RN=RV?, 7. 'Two equal paxabolas with axes in opposite directions touch ab a point O. From a point P on one of them are drawn tangents PQ and PQ' to the other. Prove that QQ' will touch the first parabola in P' where PP“ is parallel to the common tangent at O. Coordinates of any point on the parabola ex- pressed in terms of one variable. 227. Tt is often convenient to express the coordinates of any point on the curve in terms of one variable, Tt is clear that the values a 2a Cm “Em always satisfy the equation to the curve. Hence, for all values of m, the point a 2 (rm) lies on the curve. By Art. 206, this m is equal to the tangent of the angle which the tangent at the point makes with the axis. The equation to the tangent at this point is a y=ma+ mº and the normal is, by Art. 207, found to be a my m=2a+—. COORDINATES IN TERMS OF ONE VARIABLE. 199 228. “The coordinates of the point could also he ex- pressed in terms of the »m of the normal at the point; im this case its coordinates are am? and — Zam, The equation of the tangent at the point (am, — Zum) is, by Art. 205, my+e+ram=0, and the equation to the normal is y=ma — Zum — ant, 229. “The simplest substitution (avoiding both nega- tive signs and fractions) is x=at? and y=2at. These values satisfy the equation 3º = Lag, The equations to the tangent and normal at the point q 8 P (at?, 2ut) are, by Arts, 205 and 207, ty=0 + 08, and y+to= at + af, The equation to the straight line joining (ati, 2ut) and (at, 2at,) is easily found to be y (E, + to) = 200 + Zatbito. The tangents at the points (at?, 20) and (at, 2at,) are ty=e+at?, and bg =0+ at? The point of intersection of these two tangents is clearly futito, alb+io)h The point whosc coordinates are (a?, 2ot) may, for brevity, be called the point “4” Tn the following articles we shall prove some important properties of the parabola making use of the above substi- tution, 202 COORDINATE GEOMETRY. 2 Hence AM. AM' =at?. ati=a. Po t=403, o amd PM.PM'=2at, Paty. 4a? ( - 5) =-4a. dO. Cor. If O be the focus, 40=a, and we have . 1 ttg=-L te to= E 1 -2 The points (at;?, 2at,) and (cão = o are therefore at the ends 1 E! of a focal chord, 234. To prove that the orthocentre of any triangle formed by three tangents to a parabola lies on the directris. Let the equations to the three tangents be «4 Penta o erasa manera rar ard can ianaseaes (1), [ía Mg + ocre rrenan eranansreata 2), ums (2) and y=maro cerrereaenemreneranieeraess (8). Mg The point of intersection of (2) and (8) is found, by solving them, to be ta (mm) pablp— Mag Ma My The equation to the straight line through this point perpendicular to (1) is (Art, 69) (a e) = a y-af> + )=- >| e-- |, Ma Ng, m Mo, i.e y+=a 1 + l p— ] ceemereranrrerasa (4). mM Ma Ny My, Similarly, the equation to the straight line through the intersection of (3) and (1) perpendicular to (2) is +Ê=a (atm tas 5 ” Ma Mg My Manari 66), and the equation to the straight line through the intersection of (1) and (2) perpendicular to (3) às a 1. 1 a ptãco(E t+ ) meerenereenereresa (6). Mg My Mag The point which is common to the straight lines (4), (5), and (6), EXAMPLES, ONE VARIABLE. 203 i.e. the orthocentre of the triangle, is easily seen to be the point whose coordinates are 1 1/1 1 v=-6, g=mat— torto + » My My My Mot and this point lies on the directrix. EXAMPLES. XXVIIL 1, If be the anglo which a focal chord of a parabola makes with the axis, prove that the length of the chord is da cosec? w and that the perpendicular on it from the vertex is a sin w, 9. A pointon a parabola, the foot of the perpendicular from it upon the direetrix, and the focus are the vertices of an equilateral triangle. Prove that the focal distance of the point is equal to the latus rectum. 3. Prove that the semi-latus-rectum is a harmonic mean between the segments of any focal chord. 4, Jf£ T be any point on the tangent at any point P of a parabola, and if TL be perpendicular to the focal radius S7 and TN be perpen- dicular to the directrix, prove that SL=TN. Hence obtain a geometrical construction for the pair of tangents drawn to the parabola from any point T. 5. Prove that on the axis of any parabola there is a certain point K whieh has the property that, if a chord PQ of the parabola be drawn throngh it, then 1 1 PRQ is the same for all positions of the chord. 6. The normal at the point (at,?, 2ut,) meets the parabola again in the point (at;2, 2at,); prove that 2 -z 7. A chord is a normal to a parabola and is inclined at an angle 8 to the axis; prove that the area of the triangle formed by it and the tangents at its extremities is 4a? sec? 8 cosee? 0, 8. I£ PQ be a normal chord of the parabola and if S be the focus, prove that the locus of the centroid of the triangle SPQ is the curve 3649? (3x — da) — 81 = 12804, 9. Prove that the length of the intercept on the normal at the point (at?, 2at) made by the circle which is described on the focal distance of the given point as diameter is a q 1+22, b=—t 204 COORDINATE GEOMETRY. [Exs. 10. Prove that the area of the triangle formed by the normals to the parabola at the points (at;?, 2at;), (at, 2at,) and (at;?, Zats) is a? 3 (lato) (tg ty) (io do) (tar ta)?, 1. Prove that the normal chord at the point whose ordinate is equal to its abscissa subtends a right angle at the focus. 12, A chord of a parabola passes through a point on the axis (outside the parabola) whose distance from the vertex is half the latus rectum; prove that the normais at its extremities meet on the curve. 13. The normal at a point P of a parabola meets the curve again in Q, and T is the pole of PQ; shew that T lies on the diameter passing through the other end of the focal chord passing through P, and that PT is bisected by the directrix. 14. If from the vertex of a parabola a pair of chords be drawn at right angles to one another and with these chords as adjacent sides a rectangle be made, prove that the locus of the further angle of the rectangle is the parabola yê=da (x - Ba). 15. A series of chords is drawn so that their projections on a gtraight line which is inclined at an angle a to the axis are all of constant length c; prove that the locus of their miâdle point is the curve (y2- 4ax) (y cosa + 2a sin a) + a2e=0. 16. Prove that the locus of the poles of chords which subtend a xight angle at a fixed point (A, k) is a2? — y? + (49? + 20h) 0 — 2aky ra (+ 1)=0. 17, Prove that the locus of the middle points of all tangente drawn from points on the directrix to the parabola is y(2u-+ra)=a(Be+a). 18. Prove that the orthocentres of the triangles formed by three tangents and the corresponding three normais to a parabola are equidistant from the axis, 19. Tisthe pole of the chord PQ; prove that the perpendiculars from P, T, amd Q upon any tangent to the parabola are in geometrical progression. 20. If w, and r, be the lengths of radii vectores of the parabola which are drawn ab xight angles to one another from the vertex, prove that nErd=16a? (18 + r98). 21. A parabola touches the sides of a triangle ABC in the points D, E, and F respectively ; if DE and DF cut the diameter through the point 4 in b and c respectively, prove that Bô and Cc are parallel. THE PARABOLA. LOCI, 207 ê-dah=(a+hP2tanta, Hence the coordinates of the point T always satisfy the equation yº— d4ag=(a+ mw)! tanta. We shall find in a later chapter that this curve is a hyperbola. As a particular case let the tangents intersect at right angles, so that mnmç= — 1. From (3) we then have h= — a, so thai in this case the point T lies on the straight line «= — a, which is the direetrix. Hence the locus of the point of intersection of tangents, which cut at right angles, is the directrix. Ex. 2. Prove that the locus of the poles of chords which are normal to the parabola y?=4am is the curve vi (e +20) + 40º=0. Let PQ be a chord which is normal at P. Its equation is then y=ma — 2am — am... cerceamento (1). Let the tangents at P and Q intersect in 7, whose coordinates are h and k, so that we require the locus of 7. Since PQ is the polar of the point (h, k) its equation is gh=2a (44). errecrrarnenios (2). Now the equations (1) and (2) represent the same straight line, so that they must be equivalent. Hence 2a 2ah =—— — - Ba. =. and. — Zam —- am E Eliminating m, i.e. substituting the value of mm from the first of these equations in the second, we have da? Ba! 2ah o i.e. k2 (h4 20) 440º=0, The locus of the point 7 is therefore we (x+ 2a) +42=0, Ex. 8. Find the locus of the middle points of chords of a parabola which subtend a right angle at the vertex, and prove that these chords all pass through a fixed point on the axis of the curve, 208 COORDINATE GEOMETRY. First Method. Let PQ be any such chord, and let its equation be YEMBL Css (1). The lines joining the vertex with the points of intersection of this straight line with the parabota PÊ=dar eres (2), are given by the equation ye=darty-ma). (Art, 122) These straight lines are at right angles if crdam=0, (Art. 111) Substituting this valuco of c in (1), the equation to PQ is y=m(r- dA)... (3). This straight line cuts the axis of x at a constant distance 4a from the vertex, i.e. A4'=da, Tf the middle point of PQ be (A, k) we have, by Art, 220, 2a hs eretas (9. Also the point (A, k) lies on (3), so that we have h=m(h- 40)... rrrrrrrrrenea (5). X between (4) and (5) we eliminate m, we have 2a h=— (h— da), tie, h'=2a (h--40), so that (A, k) always lies on the parabola yê=Pa (x — de). This is a parabola one half the gize of the original, and whose vertex ig at the point 4' through which all the chords pass, Pena Method. Let P bethe point (at, 2at,) and Q be the point ato, Berto The tangents of the inclinations of 4P and 4Q to the axis are 2 and 2, H & Since 4P and AQ ate aí right angles, therefore 22.4, hot i.e. th=-4.. Asin Art. 229 the equation to PQ is (hHi)y=22+atyt, THE PARABOLA. LOCL 209 This meets the axis of « at a distance — atyto, i.e., by (Bb), ta, from the origin, Also, (A, k) being the middle point of PQ, we have 2h=e (t+ 3), and 2h=2(tyHto). Hence R-2ah=a? (ty ty) a? (624 to?) =20tht,= - 80º, so that the locus of (A, k) is, as before, the parabola, y2=2a (x — da). Third Method. The equation to the chord which ig bisected ai the point (A, k) is, by Art, 221, E(y-h)=20(e-), i.e. Ey — 200 =K2- Dal... (8). As in Art, 122 the equation to the straight lines joining its points of intersection with the parabola to the vertex is (Kº — 2a) y? = Saw (hy — 200). These lines are at right angles if (k? — 2h) + 802 =0. Hence the locus as before. Also the equation (8) becomes hy — Quo = — Bu?, This straight line always goes through the point (4a, 0). EXAMPLES. XXIX. From an external point P tangents ave drawn to the parabola; find the equation to the locus of P when these tangents make angles à, and 94 with the axis, such that 1. tanô,+tan 9, is constant (=). b3 . tan 0, tan 0, is constant (=c). + Cot 8;+cot 8, is constant (= d). 8, +64 ig constant (=-2e). tan? 9, +-tan? 9, is constant (=). So e dd w cos 8, cos O is constant (=). L. 14, 212 COORDINATE GEOMETRY. This equation, being of the third degree, has three roots, real or imaginary. Corresponding to each of these roots, we have, on substitution in (1), the equation to a normal which passes through the peito 0. Hence three normals, real or imaginary, pass through any point O. If qm, mo, and mg be the roots of the equation (3), we have ma + Ma + mg = 0, If the ordinates of the feet of these normals be 34, %s, and 3, we then have, by (2), Not ya += 2a (my + ma + mm) =0. Hence the second part of the proposition. We shall find, im a subseguent chapter, that, for certain positions of the point O, all three normals are real; for other positions of O, one normal only will be real, and the other two imaginary, 287. Ex. Find the locus of a point which is such that (a) two of the normais drawn from it to the parabola are at right angles, (8) the three normais through it cut the axis in points whose distances from the vertex are in arithmetical progression, Any normal is y=ma — 2am — am?, and this passes through the point (A, k), if ant (Za-h) my h=0........isines (1). If then m, mo, and ms be the roots, we have, by Art. 2, Mag Md Mg =O,.. crereersmererareraransenee (2), Za -h Nega + Mg May = a ementa (3), and Mmamy= + eeceneererenarereraeireranera (4. (4) If two of the normals, say m, and m,, be at right angles, we k. have mymy=—1, and hence, from (4), Mg=— The quantity E; is therefore a root of (1) and hence, by substitution, we have Ea (a Da E =0, RA Exa (h Ba), THREE NORMALS. EXAMPLES. 213 The locus of the point (A, k) is therefore the parabola yº=a (x —- 3a) whose vertex is the point (34, 0) and whose latus rectum is one-quarter that of the given parabola. The student should draw the figure of both parabolas, (8) The normal y=ma — 2am — am? meets the axis of x at a point whoso distance from the vertex is 2a-+am?, The conditions of the question then give Caram)+(QCaram=2 (Ca +am), i.e. my + m=Imo2.. (DB). Tf we eliminate m,, mp, and m, from the equations (2), (3), (4), and (5) we shall have a relation between À and k. From (2) and (3), we have 24h =Myma dmo (ay de mag) = Uma Mg? Lisias (6). Also, (5) and (2) give mo? = (my + ms)? — Qmymy=mo? -- 2mymy, tee. mg + Qmymg= 0... trrrrrrerreerarea (7. Solving (6) and (7), we have 2a —h 2a — h Mmy= a» and m=-2x “80 Substituting these values in (4), we have tah / o! ga-h o & 3a 3a oa i.e. 270h2=2 (A — 208, so that the required locus is 27ay?=2 (e — 2a)3, 238. Ex, If the normais at three points P, Q, and Rmeet in a point O and S de the focus, prove that SP. SQ.SR=a. 802%, As in the previous question we know that the normals at the points (am;?, 7 20), (amo2, — 2ams) and (amp, — Zum) meet in the point (A, k) if May Mat Ma=0 ce reererrenererteranaecas (1), -h Ma + My + Mma= det ceerntceneroencaneros (2), k and Magos — 7 seem eeerentereenentareneass (3). By Art, 202 we have SP=a(l+m)), SQ=a(l+m?), and SÊ=a(l rm?) 214 COORDINATE GEOMETRY, Hence =(Lrm?) (L+mo3) (14 ms?) SP.SQ.SR a =1+(mf+rmftm) + (mm mm mma) mim. Also, from (1) and (2), we have mms mf = (my + tm ma)? — 2 (imnag + egay + mo) and Mon + mm? + mê = (Mota Mig + My)? — Qoyriotito (Ny + Mo +03) = (12 by (1) ana (2). — eme 2 2 Hence SP .80. SR 49h de (18) + a a a a? (h-aPrk so? a i.e. SP.SQ. SR=802,a. EXAMPLES. XXX. Find the locus of a point O when the three normals drawn from it are such that 1. two of them make complementary angles with the axis, 92. two of them make angles with the axis the product of whose tangents is 2, 3. one bisects the angle between the other two. 4, two of them make equal angles with the given line y=ma + c, 5. the gum of the three angles made by them with the axis is constant. 6. the axea of the triangle formed by their feet is constant. 7, the line joining the feet of two of them is always in a given direction. The normals at three points P, Q, and R of the parabola y?--4ax meet in a point O whose coordinates are k and k; prove that 8, the centroid of the triangle PQR lies on the axis. 9. the point O and the orthocentre of tho triangle formed by the tangents at P, Q, and R axe equidistant from the axis, PARABOLA. TWO TANGENTS AS AXES. 217 By Art, 133 we have for « and y to substitute we cos 6 + 3 cos (w' + 0) + À, and esin 8 +ysin (o +0)4k respectively. The equation of Art. 197 then becomes fesinO+ysin(w +0)+kP=4afrcos0+ycos(w +0)+hl, A fe sin 0 +y sin (o + OP + 2a fk sin O — 2a cos 8) + 2y [k sin (o +09) — 2a. cos (wu! + 0)) + Rº — tah =0 ceserererias (1). This equation is therefore the most general form of the equation to a parabola, We notice that in it the terms of the second degree always form a perfect square. 240. To find the equation to À parabola, any two tomgents to it being the axes of coordinates and the points of contact being distant a and b from the origin By the last article the most general form of the equa- tion to any parabola is (d+ By + 2ga+ 2fy + 0=0..... (1). This meets the axis of x in points whose abscissae are given by AÊ + Ig + 00 cics (2). Tf the parabola touch the axis of « at a distance « from the origin, this equation must be equivalent to AS (mg a)=0 essere (3). Comparing equations (2) and (3), we have 9g=— Aa, and c= AMU. (4). Similarly, since the parabola is to touch the axis of y at a distance d from the origin, we have f=— Bd, and co BB is (5). 218 COORDINATE GEOMETRY. From (4) and (5), equating the values of e, we have Br= Ao, so that B=+Á 5 o (6). Taking the negative sign, we have 2 B=-48 =— Ag, = qe, and c= 42, 5 9 1 5 Substituting these values in (1) we have, as the required equation, This equation can be written in the form CV (ED eb (+35) a(s B)ei= [The radical signs in (8) can clearly have both the positive and negative signs prefixed, The different equations thus obtained corre- spond to different portions of the eurve, In the figure of Art. 248, the abseissa of any point on the portion PAQ is <a, and the ordinate -<b, so that for this portion of the curve we must take both signs positive. For tho part beyond P the abseissa is >a, and 5, so that the signs must be + and -. For the part beyond Q the ordinate is >», and io, so that the signs must be — and +. There is clearly no part of the curve corresponding to two negative sigus.] PARABOLA. TWO TANGENTS AS AXES, 219 241. JTf in the previous article we took the positive sign in (6), the equation would reduce to CV 98 MM (+35) 20- Ho, . zo y 2. 1.6. (Er )- This gives us (Fig. Art. 243) the pair of coincident straight lines PQ. This pair of coincident straight lines is also a conic meeting the axes in two coincident points at P and Q, but is not the parabola required. 242, To find the equation to the tamgent at amy point (2, y) 0f the parabola Vi Let (x, y") be any point on the curve close to (2º, 3). The equation to the line joining these two points is Lua) creo (1). But, since these points lie on the curve, we have do Eco /Es ny vo (2) dr dy nb = Neem (3). The equation (1) is therefore , p= NET op op ant em = dado dado E) or, by (3), Eai Po by" + Yy yoy = da dis dd (gm — 0) ecc (4). so that 229 COORDINATE GUHOMETRY, 245. To find the equation to the directrix. Tf we find the point of intersection of OP and a tangent perpendicular to OP, this point will (Art. 211, y) be on the directrix. Similarly we can obtain the point on OQ which is on the directrix. A straight line through the point (7, 0) perpendicular to OX is y=m(e-f), where (Art. 93) 1 4+m cos q = 0. The equation to this perpendicular straight line is then DAMOS m= fre cceerrarireras (1). This straight line touches the parabola if (Art, 242) f f o. ab cos w dt as ei f=- — albeosw? tb / arbeosw The point (Es e. 0) therefore lies on the directrix. “datbeosa Similarly the point (s, pe) is on it. b+acos w The equation to the directrix is therefore e(a+bcosw) +y(b+acosw)=abcoso ...... (2). The latus rectum being twice the perpendicular distance of the focus from the directrix = twice the distance of the point ab? eb (ax mes or” a “Bb eos 618) from the straight line (2) Lab? sin? w É (a? 4 Zad cos o + DP? by Art. 96, after some reduction. 246. To find the coordinates of the vertex and the equation to the tangent at the vertex, PARABOLA. TWO TANGENTS AS AXES. 228 The vertex is the intersection of the axis and the curve, i.e, its coordinates are given by vom. q —b? 1 Boa EIDab cos o EC (1). voy 2 dae 2y o 49 and by (; 2) E BA=O (Art 240), Ro 2 y 2 do te. by (5 5 1) = eee (2). From (1) and (2), we have ro 2 2º a(bracosw)? Cof a +20b cos w +] (a? 4 20h cos w + by! o. 2b (a + b cos w)? Similard ca Ng mea mary y (aê + 2ab cos w + These are the coordinates of the vertex. The tangent at the vertex being parallel to the direcbrix, its equation is 2 o 2 ab (a +bcosw) (a? + 2ab cos w + 6%) % Y ab bracosa” atbcoso. a 4 Bab cos w +56? [The equation of the tangent at the vertex may also be written down by mcans of the example of Art. '242.] EXAMPLES, XXXI 1. If a parabola, whose latus reetum is 4c, slide between two rectangular axes, prove that the locus of its foeus is aY?=c? (22+y?), and that the curve traced out by its vertex is + (bracos a)[ gy — =0, 2.0. 22,2 Ap ryj= e 9. Paxabolas are drawn to touch two given rectangular axes and their foci ave all at a constant distance c from the origin. Prove that the locus of the vertices of these parabolas is the curve 2 2 2 aê yf=eê, 224 COORDINATE GEOMETRY. [Exs. XXXL] 8. The axes being rectangular, prove that the locus of the focus £ the parabola (244 1) = of the parabola (7 + — => that ab=º€2, is the curve (2º + y)?=cêyy. a and b being variables such 4, Parabolas ave drawn to touch two given straight lines which are inclined at an angle w; if the chords of contact all pass through a fixed point, prove that (1) their directrices all pass through another fixed point, and (2) their foei all lie on a circle which goes through the intersection of the two given straight lines. 5. A parabola touches two given straight lines at given points; prove that the locus of the middle point of the portion of any tangent which is intercopted between tho given straight lines is a straight line. 6, TP and TQ are any two tangents to a parabola and the tangent at a third point R cuts them in P' and Q'; prove that mpr , r Tp / 101, md TE OR TP * TQo» Pd p= pp RP 7. Ifa parabola touch three given straight lines, prove that each of the lines joining the points of contact passes through a fixed point. 8. A parabola touches two given straight lines; if its axis pass through the poiut (A, k), the given lines being the axes of coordinates, prove that the locus of the focus is the curve ey horhy=0, 9. A parabola touches two given straight lines, which meet at O, in given points and a variable tangent meets the given lines in P and Q respectively; prove that the locus of the centre of the circumceircle of the triangle OPQ is a fixed straight line. 10. The sides 4B and AC of a triangle ABC are given in position and the harmenie mean between the lengths AB and AC is also given; prove that the locus of the focus of the parabola touching the sides ab B and C isa cirele whose centre lies on the line biseoting the angle BACO. 11, Parabolas are drawn to touch the axes, which are inclined at an angle w, and their directrices all pass throngh a fixed point (h, A), Prove that all the parabolas touch the straight line x y — h+rh seo * Erhseco THE ELIPSE. 297 248. The equation (6) of the previous article may be written y é Pd (arm(a-x) polpa TO a . PNº ANNA ve. NE = Te "3 te PNº: ANNA! :: BO: AC? Def. “The points 4 and 4' are called the vertices of the curve, 4,4 is called the major axis, and BB' the minor axis. Also C is called the centre. 249. Since S is the point (we, 0), the equation to the ellipse referred to S as origin às (Art, 128), (w— ae) Po 2 +G=1 The equation referred to À as origin, and 4X and a perpendicular line as axes, is ua gy fears, . So op 2a te, BR GS Similarly, the equation referred to ZX and ZK as axes is, - a since 0Z = — % a? Ro The equatión to the ellipse, whose focus and directrix are any given point and line, and whose eccentricity is known, is easily written down. For example, if the focus be the point (--2, 3), the directriz be the line 27+3y+4=:0, and the eccentricity be 4, the required equa- tion is (20 +-3y +42 (0+2)2+(y — 3)2= (8)? CEB i.e, 261x2 4. 1819? - 19273 + 1044x — 2834y + 8969=0. 15-—2 228 COORDINATE GEOMETRY, Generally, the equation to the ellipse, whose focus is the point (f, 9), whose directrix is 47+By+-C=0, and whose eccentricity ise, is , (4o+By+CP (e-r(y-gP=e AP Bº 250, There exist q second focus and « second directria Jor the curve. On the positive side of the origin take a point S”, which is such that SC = CS" = ae, and another point Z”, such that Z0=0Z'=*. Draw Z'K' perpendicular to ZZ', and PM" perpen- dicular to Z'K”, The equation (5) of Art. 247 may be written in the form 22 — Quem + e + 9º = a? — Quem + 0, a 2 à.e. (var gi =e (5-5) , ERA SP=A.PMº, Hence any point ? of the curve is such that its distance from 8" is e times its distance from Z'K”, so that we should have obtained the same curve, if we had started with S” as focus, Z'K' as directrix, and the same eccentricity. 251. The sum of the focal distances of any point on the curve is equal to the major axis. For (Fig. Art. 247) we have SP=e. PM, and SP=e.PM”, Hence SP+ SP=e(PM+ PM) =e. MM" =e.ZZ'=2e. CZ=2a (Art, 247.) =the major axis. Also SP-e.PM-=e.NZ=e.CZ+re. CN=a+ex, and SP=e.PM'=e.NZ'=e.0Z' -e. CN=a-=ex', where a! is the abscissa of P referred to the centre. THE ELLIPSE. LATUS-RECTUM. 229 252. Mechanical construction for an ellipse. By the preceding article we can get a simple mechanical method of constructing an ellipse, Take a piece of thread, whose length is the major axis of the required ellipse, and fasten its ends at the points ,S and 8º which are to be the foci. Let the point of a pencil move on the paper, the point being always in contact with the string and keeping the two portions of the string between it and the fixed ends always tight. If the end of the pencil be moved about on the paper, so as to satisfy these conditions, it will trace out the curve on the paper. For the end of the pencil will be always in such a position that the sum of its distances from S and S' will be constant. In practice, it is easier to fasten two drawing pins at S and S”, and to have an endless piece of string whose total length is equal to the sum of SS and AA. This string must be passed round the two pins at $ and Sand then be kept stretched by the pencil as before. By this second arrangement it will be found that the portions of the curve near 4 and 4 can be more casily described than in the first method. 253. Latus-rectum of the ellipse, Let LSZ' be the double ordinate of the curve which passes through the focus S. By the definition of the curve, the semi-latus-rectum SL =g times the distance of Z from the directrix me, 8SZ=e(0Z-CS)=e. CZ-e. CS =a— «é (by equations (3) and (4) of Art. 247) be => +, 247. — (Art, 247.) 254. To trace the curve oa AE Lee cerne (1. 282 COORDINATE GEOMETRY. Hence Art. 248 gives PNº: QNº uu BO: AC?, PN BCG b so that, ON AG a Y , Q [= PP| TSNR PN. A Nº NJ x T The point Q in which the ordinate NP meets the auxiliary circle is called the corresponding point to P. The ordinates of any point on the ellipse and the corresponding point on the auxiliary circle are therefore to one another in the ratio b : a, we. in the ratio of the semiminor to the semi-major axis of the ellipse, The ellipse might therefore have been defined as follows : Take a circle and from each point of it draw perpen- diculars upon a diameter ; the locus of the points dividing these perpendiculars in a given ratio is au ellipse, of which the given circle is the auxiliary circle. 258. Eccentric Angle. Def. The eccentric angle of any point ? on the elhpse is the angle NCQ made with the major axis by the straight line CQ joining the centre O to the point Q ou the auxiliary circle which corresponds to the point 2. This angle is generally called q. THE ECCENTRIC ANGLE. 233 We have CN=CQ.cosp=acos & and NQ=0Qsingp=asin d. Hence, by the last article, b NP=a. NQ=bsin q. The coordinates of any point-P on the ellipse are there- fore a cos q and bsin d. Since P is known when q is given, it is often called “the point q.” 259, To obtain the equation of the straight line joiming tuo points on the ellapse whose eccentrio angles are given. Let the eccentric angles of the two points, P and P”, be & and 4, so that the points have as coordinates (a cos &, bsin db) and (a cos q, bsin db). The equation of the straight line joining them is ., being -bsingd 1 bind cg = acos & (a— a cos p) E celfitmo A Ca rs nabo ei (1). This is the required equation. Cor. The points on the auxiliary circle, corresponding to P and P", have as coordinates (a cos q, a sin q) and (a cos 4”, esin q). The esmation to the line joining them is therefore (Art, 178) HE Lan ti got, BH aos PD 2 234 COORDINATE GEOMETRY. This straight line and (1) clearly make the same intercept on the major axis, Hence the straight line joining any two points on an ellipse, and the straight line joining the corresponding points on the auxiliary circle, meet the major axis in the same point. EXAMPLES. XXXII. 1. Find the equation to the ellipses, whose centres are the origin, whose axes are the axes of coordinates, and which pass through (a) the points (2, 2), and (3, 1), and (8) the points (1, 4) and (— 6, 1). Find the equation of the ellipse referred to'its centre 2. whose latus rectum is 5 and whose eccentricity is 2, 3. whose minor axis is equal to the distance between the foci and whose latus rectum is 10, 4, whose foci are the points (4, 0) and (—4, 0) and whose eccentricity is 3. 5. Find the latus rectum, the eccentricity, and the coordinates of the foci, of the ellipses (1) a2+3y2=a2, (2) 5u?+4y2=1, and (3) 942459? --30y=0, 6. Find the eccentricity of an ellipso, if its latus rectum be equal to one half its minor axis. 7, Find the equation to the cliipse, whose focus is the point (-1, 1), whose directrix is the straight line v —y+8=0, and whogo eccentricity is 3. 8, Is the point (4, - 8) within or without the ellipse 5a? + 7y2=112? 9, Find the lengths of, and the equations to, the focal radii drawn to the point (4/3, 5) of the ellipse 254? + 16y?= 1600, 10. Prove that the sum of the squares of the reciprocals of two perpendicular diameters of an ellipse is constant, 11. Find the inelination to the major axis of the diameter of the ellipse the square of whose length is (1) the arithmetical mean, (2) the geometrical mean, and (8) the harmonical mean, between the squares on the major and minor axes, 12, Yind the locus of the middle points of chords of an ellipse which are drawn through the positive end of the minor axis. 13, Provc that the locus of the intersection of 4P with the straight line through 4' perpendicular to 4'P is a straight line which is perpendicular to the major axis, EQUATION TO THE TANGENT 237 The length of the required chord therefore = (hn y)= (am — a) Tam Lab NE mê rm nom sb e Pd 262. To find the equation to the tangent ot any point (x, 4) of the ellipse. Let ?P and Q be two points on the ellipse, whose coordi- nates are (2º, y) and (a, 3). The equation to the straight line PQ is Since both P and Q lie on the ellipse, we have 12 9 uy EE Die (2), po ga and — + = Dna (3). Tlence, by abracos, a''2 = m'2 gap oo tow S 0, , ww =) (gr! + y) e (a! =2º) (a! + E) Ba = = CT e , Hood Do! ie, py ms er fg dyry On substituting in (1) the equation to any secant PQ becomes bia! —-y=—5 =) citaram é). 1 == ay ET) (4) To obtain the equation to the tangent we take Q indefinitely close to ?, and hence, in the limit, we put q =0 and y'=9y. 288 COORDINATE GEOMETRY. The equation (4) then becomes ba y- rat” ), . a e. ç + ar = + “o =1,by equation (2). The required atos is therefore yy = + bi = Z. Cor. The equation to the tangent is therefore ob- tained from the equation to the curve by the rule of Art. 152, 263. To find the equation to a tangent in terms of the tomgent of its inclination to the major axis, As in Art. 260, the straight line PEMA A Corrcrererrirecrereia (1) meets the ellipse in points whose abscissae are given by 2º (1 an?) + Qmeato + a (é — B) = 0, and, by the same article, the roots of this equation axe coincident if c= mad In this case the straight line (1) is a tangent, and it becomes y=mx+ vaimipbi............... (2). This is the required equation. Since the radical sign on the right-hand of (2) may have either + or — prefixed to it, we see that there are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction. The above form of the equation to the tangent may be dednced from the equation of Art. 262, as in the case of the parabola (Art. 206). I will be found that the point of contact is the point — cm b2 ( amido” fem à) ' EQUATION TO THE TANGENT, 239 264. By a proof similar to that of the last article, it may be shewn that the straight line Wcosa+gysn a=p touches the ellipse, if pi=a?cos? «+ bêsinia., Similarly, it may be shewn that the straight line le + my=n touches the elipse, if 0 + Dm? =? 265. Lquation to the tamgent at the point whose eccentric angle às qd. The coordinates of the point are (« cos &, d sin q). Substituting «= a cos + and 3 = bsin q in the equation of Art. 262, we have, as the required equation, Ecosp+úsing=1 erteertenes (1). This equation may also be deduced from Art. 259. For the equation of the tangent at the point “4” is obtained by making & = & in the result of that article. Ex. Find the intersection of the tangents at the points p and q! The equations to the tangents are a Ya = q es PA; sind 1=0, and Éeos pri sing'-1=0, The required point is found by solving these equations. We obtain z y a b -1 i sin q — sing” = cos 4 — cos p > sa p'cosp-cosg'sin p > sin (p— i.e. o y nl Qucos “ É sin e 2% sin À e E nÊ ss” 2sin &-é Ea os PP 249 COORDINATE GEOMETRY. The equation to the tangent at P is (Art. 262) BD es (1). cê To find where the straight line meets the axis we put y=0 and have e cr- E a=m de UN i.e. CP CN=SP=CA (2). Hence the subtangent N7 =0F CN=S ma =€ * The equation to the normal is (Art. 266) g-" y-y = 7 = y a b To find where it meets the axis, we pub y =0, and have : , g=0 Wo a =p, Ty a b 2 2... 2 ie 0G= = La Lg ns toe CN..(8) Hence the subnormal NG =0N-C6=(1-)09N, i.e, NG NCul-di:l nba, (Art, 247.) Cor. If the tangent meet the minor axis in t and Pn be perpendicular to it, we may, similarly, prove that Cr. Cn=B, 270. Some properties of the ellipse, (1) SG=e.SP, and the tangent and normal at P bisect the external and internal angles between the focal distances of P. By Art. 269, we have OG =e2y'. SOME PROPERTIES OF THE ELLIPSE. 248 Hence SGA=5S0C+CG=ae+tex'=e. SP, by Art. 251, Algo 5'Q=08'-CO=e(a-es)=e. SP. Hence SG:SGuSP:sP, Therefore, by Eue. vr, 3, PG bisects the angle SPS”, Tt follows that the tangent bisects the exterior angle between SP and SP, (8) If SY and S'Y' de the perpendiculars from the foci upon the tangent at amy point P of the elipse, then Y and X' lie on the auxiliary circle, and SY.S'Y'=b. Also CY and S'P ave parallel, The equation to any tangent is E COS AY SIMA riressanerentreaeeras (1), where p= Jc costa rbºsin?a (Art. 264). The perpendicular SY to (1) passes through the point (— ae, 0) and its equation, by Axt. 70, is therefore (zac) gina-gy cos a=0 .ssssrceia (2). If Y be the point (A, k) then, since Y lies on both (1) and (2), we have heosa+ksina=/alcos ab sinta, and hsina-kcosa= -aegin as —- [02 sinta. Squaring and adding these equations, we have A24-h?=a?, so that Y lies,on the auxiliary circle 2? +9º=a?, Similarly it may be proved that Y' lies on this circle. Again S is the point (— ae, 0) and Sis (ae, 0). Hence, from (1), SY=p-+aecosa, and S'Y'=p-aecosa. (Art 75.) Thus 87. S'Y'=pº-a ei cosa =a? cos? a-+ Vi ginê a — (a? - 4º) cos? a, =b2 ni a2 Also OT= a mo, UA ON) and therefore ST= ENT SENT cr a GY “* gro ace CNT FD Hence CY and S'P ate parallel. Similarly CY and SP are parallel, 16-=2 244, COORDINATE GEOMETRY, (y) If the normal at any point P meet the major and minor aves in G and q, and if CF be the perpendicular upon this normal, then PF.PG= and PP.Pg=e. The tangent at any point P (the point “q”) is ” Vaio a OS PH; sin pI. Hence PW= perpendicular from € upon this tangent 1 ab — 4” + ST a b “MD. The normal at Pis - mm =q-b,, cosg sing qb If we put 4=0, we have CG= cos q. 2. h2 b cos 8) +B sin? + PQê= (a cos q — & a [A air 3 = to p+b sin? ep, ie. po=b N bi cosi gra? sin? q. From this and (1), we have PF. PG =b2, Tf we put «=0 in (2), we see that g is the point ap. (o, =p sin 8) . : 2, 2 KHence Py'=aº cos? q + (o sin dt sin ) , so that Pg af N/U2 cost p + a? sin? q, From this result and (1) we thereiore have PF, Pg=e2, 271. To find the locus of the point 97 intersectton of tungents which meet at right amgles. Any tangent to the ellipse is y=ma + Nenê sb, and a perpendicular tangent is od (Vu y=-Vosn/0 (=) +62,