AN INTRODUCTION TO CONIC SECTIONS

Transcription

1 AN INTRODUCTION TO CONIC SECTIONS TOH EE CHOON Abstrct. A set of introductory notes to conic sections. We begin with some remrks on nottion. As fr s possible points re denoted with upper cse letters nd lines with lower cse letters. If nd Q re points, then Q denotes the line tht cuts these two points, while Q denote the line segment joining these two points. We lso use Q to denote the distnce between points nd Q. 1. rbol Let F 1 nd F be pir of distinct points. We denote the locus of points equidistnt from F 1 nd F by (1.1) { : F 1 = F }. Geometriclly, the locus is the perpendiculr bisector of the line segment F 1 F. Wht bout the nlogous question of the locus of points equidistnt from two lines? Before we nswer the question, we require well-defined mesure of distnce from point to line. Definition 1.1. If l is line nd is point, we use l to denote the shortest distnce from to l. In other words, if we drop perpendiculr from to l nd denote by D the foot of this perpendiculr, l = D, where D l nd D l. In the crtesin plne, either two lines intersect or they re prllel nd distinct. In the ltter cse, the locus of points equidistnt from pir of non-intersecting prllel lines l 1 nd l is line l 3 prllel to nd in-between them. In the former cse, if λ 1 nd λ intersect, the pir of ngle bisectors t the point of intersection forms the required locus. λ 1 λ Figure 1.1. Locus of points equidistnt from λ 1, λ. Dte: October 4,

2 TOH EE CHOON Exercise 1.. Let l be line nd F be point distinct from l. Try to sketch the locus defined by { : F = l }. Definition 1.3. Let d be line nd F be point distinct from d. A prbol is defined geometriclly by { : F = d }. The line d is clled the directrix while F is clled the focus of the prbol. D d F Figure 1.. A prbol defined by directrix d nd focus F. Observe tht the line perpendiculr to d tht psses through F forms n xis of symmetry for the prbol. The point of intersection of prbol nd its xis of symmetry is clled its vertex. In order to recover the usul crtesin eqution for the prbol, we let F = (c, 0) nd directrix d be the line x = k. Let = (x, y) be point on the prbol stisfying F = d = (x c) + y = k x. We cn squre this eqution nd simplify the resulting expression to y = x(c k) + k c. We cn further simplify by setting k = c to obtin y = 4cx. We record our observtion below. Theorem 1.4. The prbol with focus F = (c, 0) nd directrix x = c hs the following crtesin form y = 4cx. This is the crtesin eqution of prbol in stndrd position. This mens tht the xis of symmetry of the prbol (y = 0 in this cse) lies on either the x or y xes. Anlogously, we lso hve x = 4cy, if the focus F = (0, c) nd y = c is the directrix. We observe tht in either sitution, the vertex of the prbol lies t the origin. We cn of course perform trnsltion to obtin the following generl crtesin equtions for prbols, which re occsionlly known s prbols in stndrd form: (x x 0 ) = 4c(y y 0 ); (y y 0 ) = 4c(x x 0 ).

3 AN INTRODUCTION TO CONIC SECTIONS 3 We cn lso work bckwrds to obtin the geometric description of prbol from its crtesin eqution. For exmple, consider the eqution y = cx. Since x = y ( c = x + y 1 ) = y ( 4c c + y 4c) 1. We cn simplify the right hnd side nd then tke squre roots to obtin ( x + y 4c) 1 = y + 1 4c. The bove cn be interpreted s F = d where F = (0, 1 4c ) nd the directrix, d, is y = 1 4c. We cn use GeoGebr, dynmic geometry softwre to help us better understnd the prbol. GeoGebr Exmple 1.5. We cn verify the focus-directrix property for the prbol defined by the eqution y = 4xc. (1) Crete slider nmed c with rnge 5 c 5. () In the Input window, enter y^=4*x*c (3) From Theorem 1.4 the focus is F = (c, 0) nd the directrix is x = c. We cn crete these objects by typing the following in the Input window. F=(c,0) x=-c (4) Vry the vlue of c by drgging the slider nd observe the chnge in the prbol, the focus nd the directrix. (5) Crete point on the prbol. (We cn right-click on the point to renme it.) (6) Use the distnce tool to mesure the length F. (7) To mesure distnce d, we need to construct the foot of the perpendiculr from to the directrix d. An lterntive is to crete the point D with c s its x-coordinte nd the y-coordinte of s its y-coordinte. We do this with the following input. D=(-c,y()) (8) Use the distnce tool to mesure the length D nd verify tht it lwys equls F by vrying the position of. It is slightly more tricky to crete prbol directly from its focus nd directrix. GeoGebr hs conics tool which ccomplishes this but it is probbly more instructive to do so from first principles. Let us first fix the focus s F = (1, 0) nd the directrix, d : x = 1. We wish to construct point such tht F = d. Note tht d depends on point D which is the foot of the perpendiculr from to the directrix. (In this cse, it would be more precise to use D p to indicte the dependency on but we elect to scrifice precision for clrity.) There is ctully one-to-one correspondence, i.e. every point D on the directrix corresponds to exctly one point on the prbol. Since F = D, D F forms n isosceles tringle. This mens tht the perpendiculr line from the vertex to the bse DF of this isosceles tringle is ctully the perpendiculr bisector of the bse. So in order to locte the point, we cn begin with n rbitrry point D on the directrix nd construct the perpendiculr bisector of DF. must lie on the intersection of this bisector nd the line through D tht is perpendiculr to the directrix. GeoGebr Exmple 1.6. We use the focus-directrix property to construct prbol.

4 4 TOH EE CHOON (1) Crete sliders nmed c nd k with rnges 5 c, k 5. () Crete the focus F = (c, 0) nd the directrix x = k nd lbel it s d with the following input. F=(c,0) d:x=k (3) Set c = 1 nd k = 1. We cn drg the slider or type directly into the input box. (4) Crete point D on the directrix. (We cn right-click on the point to renme it.) (5) Use the perpendiculr bisector tool to construct the perpendiculr bisector of DF. (6) Use the perpendiculr line tool to construct the perpendiculr line to d from the point D. (7) Crete s the intersection of these two lines. (8) Drg the point D long the directrix nd observe how trces out prbol. We cn right-click on nd toggle trce on or off for better visuliztion. (We cn use the Move Grphics View tool to move the window slightly to remove the trces.) (9) Alterntively, we cn use the locus tool. Select the locus point first, followed by D. (10) Vry c nd k to observe how the prbol chnges with the focus nd directrix. In ddition to the geometric nd crtesin form, prbol cn lso be represented by its polr form. Suppose we hve prbol with focus F t the origin nd directrix x = k for some k > 0. An rbitrry point on the prbol cn be represented by the ordered pir (r, θ) where r = F nd θ is the ngle mde by F nd the x-xis. In this cse, we cn see tht d = k r cos(θ). Equting d = F nd solving for r gives the following result. Theorem 1.7. The prbol with focus F = (0, 0) nd directrix x = k, k > 0 hs the following polr form k r = 1 + cos(θ). Exercise 1.8. Show tht the crtesin eqution of the prbol with polr form: k r = 1 + cos(θ) is given by y = k(x k). Exercise 1.9. Let F = (0, 0) nd k > 0. For ech of the three possible directrices: x = k, y = k nd y = k, find the corresponding polr form. Exercise Given prbol defined by y = 4cx, show tht the polr form is given by r = 4c cot(θ)cosec(θ). The tngent line is n importnt concept in the study of conic sections. We shll define the tngent line to prbol without reference to derivtives nd subsequently show tht this definition is ligned with the usul definition from clculus.

5 AN INTRODUCTION TO CONIC SECTIONS 5 d θ F k Figure 1.3. A prbol with the polr form r = k 1 + cos(θ). r O θ Figure 1.4. olr form of the prbol y = 4cx. Definition Given prbol nd point on the prbol. The tngent line to the prbol t is the unique line tht intersects the prbol t exctly. Consider the prbol y = 4cx nd the stright line defined by y = mx + k. We wish to study the possible intersections of these two curves with respect to the prmeters m nd k. Eliminting y gives (mx + k) = 4cx = m x + (mk c)x + k = 0. The discriminnt of the qudrtic is given by 4(mk c) 4m k = 16c(c mk). So if c < mk then there re no intersections but if c > mk, there re two points of intersection. The most interesting cse is when there is unique point of intersection. Here c = mk nd we see tht the line y = mx + c m

6 6 TOH EE CHOON touches the prbol t exctly one point for ll vlues of m. Let = (x 0, y 0 ) be point on the prbol. If is the vertex, i.e. x 0 = 0, then the tngent line is x = 0. For ny other point on the prbol, the grdient m cn be found since m x 0 y 0 m = c = ( m y 0 x 0 ) = c + y 0 x 0 4x. 0 Since lies on the prbol, we hve y0 = 4cx 0, this mens tht the right hnd side of the bove eqution is zero. So m = y 0 = c x 0 y 0 is the unique solution nd the tngent line is given by y = c x + cy 0 = yy 0 = cx + cx 0. y 0 c We record this observtion below. Theorem 1.1. Let = (x 0, y 0 ), where x 0 0, be point on the prbol y = 4cx. Then the tngent line to the prbol t hs grdient c y 0 nd is given by the eqution yy 0 = cx + cx 0. We remrk tht the grdient cn be found esily using implicit differentition. y dy = 4c = dy = c. dx dx y 0 =(x0,y 0) We further note tht in our previous clcultions, while both x 0 nd y 0 pper in the eqution, they re not independent nd re relted by y 0 = 4cx 0. This reltion ws used on severl occsions to simplify the computtions. This cn be potentilly confusing nd one possible lterntive is to introduce the prmetriztion x = ct nd y = ct, which holds for the prbol y = 4cx. By the chin rule, we lso hve dy dy dx = dt = c ct = 1 = dy = 1 = c. t dx t 0 y 0 dx dt =(x0,y 0) Exercise Let l be the tngent line to the prbol y = 4cx t. Let N be the intersection of l nd the x-xis. Show tht y-xis bisects N. Exercise Let be point on prbol with focus F. If the tngent line t intersects the directrix t R, show tht RF = 90. We next discuss n importnt property of prbol known s the reflection property. It cn be phrsed s: light coming prllel to the xis of prbolic mirror is reflected t the prbol to pss through the focus. Likewise, the reflection of light ry coming from the focus will be prllel to the xis of symmetry of the prbol. The reflection property lies t the hert of mny pplictions, for exmple, the design of torchlight, reflecting telescopes nd stellite dishes. GeoGebr Exmple We verify the reflection property of prbol. (1) Crete slider 5 c 5 nd point T = (c, 0). () Crete prbol y = 4x. Note tht the focus is t (1, 0). (3) Crete point on the prbol nd the line T (4) Use the tngent tool to construct tngent t to the prbol.

7 AN INTRODUCTION TO CONIC SECTIONS 7 F Figure 1.5. Light prllel to the xis of symmetry is reflected to pss through the focus. (5) The reflection of the line T t is ctully the reflection of the line bout the norml to the prbol t. (6) Construct the norml t with the perpendiculr line tool. (7) Use the reflect bout line tool to reflect T bout the norml t. (8) Vry c nd observe the ngle of the reflected line. The reflection property is equivlent to the following result. Theorem Let be point on the prbol defined by y = 4cx with c > 0, focus F nd directrix d. Let D be point on d such tht D is perpendiculr to d. Then the tngent line to the prbol t mkes equl ngles with F nd D. roof. If = (x 0, y 0 ), then D = ( c, y 0 ). Since F = (c, 0), we cn compute the grdient of DF to be y0 c. Theorem 1.1 tells us tht the grdient of the tngent line is c y 0, so the tngent line is perpendiculr to DF. Let the intersection of the tngent line nd DF be M. Since D F is isosceles, M must coincide with the perpendiculr bisector of DF. Hence D M must be congruent to F M nd so D M = F M. d D M F Figure 1.6. roof of Theorem Exercise Let Q = (q, 0) be fixed point with q > 0. A line through Q intersects prbol defined by y = 4px (p > 0) t two points A nd B. Find n expression for the vlue of QA QB nd hence show tht the minimum vlue is given by 4pq.

8 8 TOH EE CHOON Exercise Let A nd B be two points on prbol. The line segment AB is lso clled chord. If this chord psses through F the focus, prove tht the tngent lines t A nd B intersect t right ngles. Exercise rove tht the mid-points of chords through the focus of the prbol y = 4cx (c > 0) lie on second prbol with focus ( 3c, 0) nd directrix x = c.. Ellipse Exercise.1. Let A, F 1 nd F be three points tht form tringle F F 1 A. Find the locus of ll points such tht the tringle F F 1 hs the sme re s F F 1 A. The locus of is the line through A prllel to F 1 F s well s its reflection in F 1 F. We denote the line F 1 F by l nd recll the nottion A l denotes the shortest distnce from A to l. Then the two prllel lines cn be described s { : l = A l } = { : l = k} since A l equls some fixed vlue k. We remrk tht the locus of circle is given by the nlogous form { : F = k} where k is the rdius nd F the centre. Exercise.. Let A, F 1 nd F be three points tht form tringle F F 1 A. Find ll points such tht the tringle F F 1 hs the sme perimeter s F F 1 A. The locus of is { : F + F 1 = AF + AF 1 = k} for some rel constnt k, which defines n ellipse geometriclly. Definition.3 (Geometric). An ellipse is the locus of points whose sum of distnces to the foci F 1 nd F is constnt, i.e. { : F 1 + F = k}. F F 1 Figure.1. An ellipse. We further define chord s the line segment formed by ny two distinct points on n ellipse. The mjor xis is defined s the chord tht lies on the line F 1 F, while the minor xis is defined s the chord tht lies on the perpendiculr bisector of F 1 F. The four vertices re the points of intersections of the mjor nd minor xes with the ellipse. Theorem.4. The mjor nd minor xes correspond to the lrgest nd respectively smllest distnce between nti-podl points of the ellipse.

9 AN INTRODUCTION TO CONIC SECTIONS 9 roof. Although the result is geometriclly evident, we shll provide rigorous proof using Apollonius Theorem. The theorem sttes tht in ny tringle ABC, if AD is the medin, i.e. D bisects BC, then AB + AC = ( AD + BD ). Now let be n rbitrry point on ellipse defined by { : F 1 + F = k} nd let O be the midpoint of F 1 F. If we let F 1 = x, then we hve O + OF 1 = F 1 + F = x + (k x) ( = x k ) + k ( = O = x k ) + k 4 OF 1. So O is minimum when x = k. In other words, is on the perpendiculr bisector of F 1 F, proving tht the minor xis is the smllest distnce. On the other hnd, O is mximized when x is mximized, i.e. when lies on the ry from F 1 to F. Exercise.5. Use ythgors Theorem (three times) to prove Apollonius Theorem. We now obtin the fmilir crtesin eqution of n ellipse from its geometric definition. For simplicity, let us fix F 1 = (c, 0) nd F = ( c, 0) for some c > 0. This mens the mjor nd minor xes lie on the x nd y xes respectively. By symmetry, the vertices re t (±, 0) nd (0, ±b) for some > b > 0. Now when = (, 0), F 1 + F = c + ( + c) =. On the other hnd, when = (0, b), by ythgors Theorem, F 1 + F = c + b =. Hence c = b. Since F 1 + F =, we hve for ny point = (x, y) on the ellipse, (x c) + y + (x + c) + y = = (x + c) + y = ( ) (x c) + y = 4 4 (x c) + y + (x c) + y = 4xc 4 = 4 (x c) + y = ( cx ) = (x c) + y = 4 + c x = x + c + y = ( c ) = ( c )x + y = b = b x + y = 1 = x + y b. We record the bove s the following theorem.

10 10 TOH EE CHOON Theorem.6. An ellipse defined by { : F 1 + F = } with foci F 1 = (c, 0) nd F = ( c, 0) stisfies the eqution where b = c. x + y b = 1, We remrk tht the focl length of the ellipse defined s F 1 F = c = b. Furthermore, if c 0, then b nd the circle is the limiting form of n ellipse. In the cse of circle, there is unique focus nd every chord (i.e. dimeter) is both mjor nd minor xis. The nlogous results hold if we fix the mjor xis on the y-xis. Exercise.7. An ellipse defined by { : F 1 + F = b} with foci F 1 = (0, c) nd F = (0, c) stisfies the eqution where = b c. x + y b = 1, The equtions in Theorem.6 nd Exercise.7 re the crtesin equtions of ellipses in stndrd position. We cn lso perform trnsltion to obtin wht is commonly known s the stndrd form of n ellipse (x x 0 ) + (y y 0) b = 1. In the bove stndrd form, loction of the mjor xis, nd hence the foci, is implicit nd depend of the reltive sizes of nd b. Exercise.8. Use GeoGebr to grph the curve 3x + 5xy + 4y 8x + 10y = 5. Observe tht the curve is n ellipse not in stndrd position. Exercise.9. Let x + y = 1, > b > 0, b represent n ellipse in stndrd position nd F be focus. Find the length of the chord prllel to the minor xis nd pssing through F. This chord is clled ltus rectum, which mens stright side in Ltin. GeoGebr Exmple.10. We cn verify the geometric property of n ellipse defined by the eqution x + y b = 1. (1) Crete sliders 0, b 5. () In the Input window, enter x^/^+y^/b^=1 (3) Vry the vlues of nd b nd observe how the ellipse chnges. (4) Set to vlue lrger thn b nd observe tht mjor xis is on the x-xis. (5) Crete the foci F 1 = ( b, 0) nd F = ( b, 0) by entering F_1=(sqrt(^-b^),0) F_=(-x(F_1),0) (6) Crete point on the ellipse nd use the segment tool to construct F 1 nd F. (7) Use the distnce tool mesure the length of these two segments, nming them f nd g respectively. (8) In the Input window, enter dev=f+g-*

11 AN INTRODUCTION TO CONIC SECTIONS 11 (9) The vlue of dev ppers in the Algebr window nd should lwys equls 0 s long s > b. Vry nd to check. Exmple.11. If the mjor xis of n ellipse given by the eqution 4 + t + y t = 1 lies on the y-xis, find the rnge of vlues of t. Solution. We require 4 + t > 0 = t > 4 nd x t > 0 = t < or t >. Furthermore if the mjor xis is on the y-xis, we need t 4 + t. In other words, t t 6 0 = t 3 or t. Combining the two gives t 3 or 4 < t. We remrk tht in the specil cse of circle (t = 3 nd ), every possible dimeter cn be considered mjor xis. We cn lso use GeoGebr to verify our clcultions. GeoGebr Exmple.1. We use GeoGebr to explore the properties of n ellipse with foci F 1 = (5, 0) nd F = (3, 0) nd mjor xis of length 4. (1) Since the foci lie on the mjor xis, one of the vertex of this ellipse must be (6, 0). () Crete F 1, F nd the vertex (6, 0). (3) Use the ellipse tool to plot the required ellipse. (4) Crete n rbitrry point on this ellipse nd construct the segment F. (5) Vry the position of nd try to deduce reltion between F nd the coordintes of. In the bove exmple, it ppers tht F is lwys hlf the vlue of the x- coordinte of. In other words, ny point is lwys twice s fr wy from the y-xis s its distnce to F. If we define d s the line x = 0, then d = F. This is reminiscent of the geometric definition of prbol in terms of focus nd directrix, with the exception of the fctor of. To see this clerly, we cn define D = (0, y( )), i.e. D is on the directrix x = 0 nd hs the sme y-coordinte s. We continue to explore this concept with the next exmple. Exmple.13. Let F = (0, 0) nd d be the line x = 4. Find the locus of points whose distnce from d is thrice the distnce from F, i.e. { : d = 3 F }. Solution. From definition, we cn see tht (1, 0) stisfy the eqution nd so does (, 0). We further guess tht there my be points on the y-xis. Let = (0, y) nd d = 4, so we hve y = ± 4 3. These four points re insufficient to determine curve. In generl, just s two points determine line, three (non-colliner) points determine circle, five distinct points re required to determine generl conic section. It is instructive to use plot these four points nd use the conics tool in GeoGebr to see how different types of conic sections ll intersect these four points. Returning to the exmple, we represent = (x, y) nd work out the crtesin eqution explicitly. 3 ( x + y = 4 x = 8 x + ) 1 + 9y = 18.

12 1 TOH EE CHOON Definition.14 (Directrix-Eccentricity-Focus). An ellipse is the locus of points whose distnce from focus F nd to directrix d re in constnt rtio less thn 1. This rtio is termed the eccentricity. The ellipse cn be described s { : F } d = e, 0 < e < 1. d D F Figure.. Directrix-Eccentricity-Focus definition of n ellipse. The Directrix-Eccentricity-Focus properties of conic sections were first recorded in the works of ppus of Alexndri ( AD). The eccentricity, e, is mesure of how uncirculr the ellipse is, i.e. s e 0, the ellipse becomes circle. Furthermore, we recognize tht when e = 1 gives us the definition of prbol. We shll see lter tht the sme locus with e > 1 describes hyperbol, so the Directrix-Eccentricity-Focus definition is uniform definition tht works for ll non-degenerte conic sections. We shll work out the crtesin eqution from the Directrix-Eccentricity-Focus definition. Let F = (c, 0) nd directrix d be given by the line x = k, k c. We substitute = (x, y) into F d = e to get (.1) (x c) + y = e (k x). If e = 0, only the point F stisfies the eqution. To see wht is ment by the eccentricity s mesure of how uncirculr the ellipse is, we set e = 1 k nd let k. In the limit, we obtin (x c) + y = 1. We return to eqution (.1). By expnding nd regrouping terms we obtin x (1 e ) x(c ke ) + y = e k c. We note tht if e = 1, the eqution bove describes prbol. If e 1, we cn divide throughout by the fctor (1 e ) nd complete the squre to obtin the following eqution of n ellipse. (.) (x c ke 1 e ) + y 1 e = e (k c) (1 e ). For stndrd position, we cn set k = c/e. The eqution simplifies to e x c + e y c (1 e ) = 1. We cn now set = c/e nd b = 1 e to obtin the usul crtesin eqution in Theorem.6. Furthermore, ll our computtions re reversible, i.e. we cn begin from the crtesin eqution nd work bckwrds to F d = e to obtin the focus,

13 AN INTRODUCTION TO CONIC SECTIONS 13 directrix nd eccentricity in terms of the prmeters nd b. We record the vrious reltionships below. Theorem.15. If > b > 0, the ellipse stisfying the eqution x + y b = 1, hs focus F 1 = (c, 0) where c = b nd directrix d given by x = e where eccentricity e is given by e = 1 b. An lterntive description in terms of these prmeters is { : F } 1 = e, 0 < e < 1. d Two more importnt observtions re in order. Firstly, by symmetry, we could use the focus F = ( c, 0) nd directrix d given by x = e, with the sme vlues of c nd e. The second observtion is tht the foci defined here re exctly the sme foci in Definition.3 nd Theorem.6. For if F 1 = e d nd F = e d, then the sum ( ( F 1 + F = e ( d + d ) = e e )) =. e We sw previously how the perpendiculr bisector ws used to construct prbol directly from its focus nd directrix. For n ellipse defined by F = e d, we use the fcts tht d = m is the locus of two lines prllel to d nd F = em is the locus of circle with rdius em. Thus cn be locted s the intersection of these lines with the circle. GeoGebr Exmple.16. We use the directrix-eccentricity-focus property, F = e d, to construct n ellipse. (1) Crete sliders 0 e 5 nd 5 k 0. () Crete the directrix x = k nd focus F = (0, 0). (3) Crete slider 0 m 100, where m is the vlue of d. This mens tht lies on one of the lines x = k + m or x = k m. Crete these two uxiliry lines. (4) On the other hnd, lies on circle of rdius em with centre F. Use the circle with center nd rdius tool to crete circle with centre t F nd rdius em. (5) Vry the vlues of e, k nd m so tht four possible intersections of this circle with the previous two lines gives. (6) Use the locus tool to drw the locus of ech of the four points depending the vrible m. (7) Declutter the plot by hiding the circle nd the two uxiliry prllel lines. Right-click on the circle nd uncheck show object. (8) Adjust e to vlue ner 0 nd observe how the ellipse is lmost circle. Increse k to increse the rdius. (9) In fct, our construction works well for vlues e 1. The ellipse morphs into prbol t e = 1 nd hyperbol s e increses. We consider yet nother possible wy to obtin n ellipse. We begin with the circle x + y = nd pply scling of fctor b prllel to the y-xis. If Q is

14 14 TOH EE CHOON point of the circle, it cn be prmetrized s Q = ( cos t, sin t) for π < t π. Let be the corresponding point on the ellipse. The coordintes of is given by = ( cos t, b sin t). It is importnt to note tht t is the ngle between OQ nd the x-xis nd not equl θ, the ngle between O nd the x-xis. The two ngles re relted by tn(θ) = b tn(t). The ngle t is sometimes clled the eccentric ngle for the point on the ellipse. Q b O t Figure.3. An ellipse s scled circle. x Using stndrd techniques in clculus, we cn compute the re of the ellipse + y b = 1 by Are = 4 y dx = 4 b 1 x dx. Using the prmetriztion x = cos t, which mens dx dt Are = = 4b = πb. π π 0 b 1 cos t ( sin t) dt 0 0 π sin t dt = sin t. So Likewise, we cn use the rc length formul to write down n expression for the perimeter of the ellipse. However this integrl is not esily evluted. ( ) dy erimeter = dx dx = 4 = 4 0 π b cos t sin sin t dt t 1 e cos t dt where e is the eccentricity. It turns out tht for n ellipse in stndrd position, the expression F 1 hs n elegnt expression.

15 AN INTRODUCTION TO CONIC SECTIONS 15 Exmple.17. Let x + y b = 1 be n ellipse with > b > 0 nd let F 1 be the right focus. Show tht for n rbitrry point = ( cos t, b sin t), F 1 = (1 e cos t). Solution. F 1 = ( cos t b ) + b sin t = cos t (cos t) b + b + b (1 cos t) ) ( = b cos t ( = 1 b cos t 1 ) = (1 e cos t). It is ctully not necessry to use the prmetric equtions nd we cn show tht if = (x 0, y 0 ) then F 1 = ex 0. A similr clcultion shows tht F = (1 + e cos t) = + ex 0 nd consequentilly we recover F 1 + F =. As in the cse of the prbol, we wish to study the conditions for stright line, y = mx + k, to intersect n ellipse t exctly one point. Combining with the crtesin eqution of n ellipse in stndrd position, we hve which simplifies to x (mx + k) + b = 1 ( m + b )x + mkx + (k b ) = 0. For unique solution, we set the discriminnt which reduces to Theorem.18. The line 4 4 m k 4 ( m + b )(k b ) = 0, k = m + b. y = mx ± m + b intersects the ellipse x + y b = 1 t exctly one point. Exercise.19. Let x + y b = 1 be n ellipse nd = (x 0, y 0 ) be point on the ellipse. Show tht the eqution of the tngent line t is given by xx 0 + yy 0 b = 1. Exercise.0. Show tht if the tngents to two points of n ellipse, x + y b = 1, intersect t right ngles, the intersection lies on the director circle, which is circle with rdius + b nd centred t the origin. Similr to prbol, n ellipse lso stisfies reflection property which cn be phrsed s light emitted from one focus is reflected t n elliptic mirror to pss through the other focus. Ellipsoidl reflector spotlights used in stge lighting re designed bsed on the reflection property.

16 16 TOH EE CHOON Theorem.1 (Reflection roperty). Given n ellipse with foci F 1 nd F. Let be point on the ellipse nd N be on the mjor xis such tht N is norml to the tngent line of the ellipse t. Then F N = N F 1. N F F1 Figure.4. Reflection property for n ellipse in stndrd position. We first prove the following result from geometry. Theorem. (Angle Bisector Theorem). In ABC with D on BC,AD bisects BAC if nd only if AB BD = AC CD. roof. To prove BAD = CAD, we extend BA to point C such tht AC = AC. Since AB BD = AC CD = AC CD = BC BC, we see tht BAD is similr to BC C. Thus AD is prllel to C C nd BAD = BC C = BCC = CAD. The converse cn be proven in similr mnner by extending BA to point C such tht C C is prllel to AD. C A B D C Figure.5. roof of Angle Bisector Theorem F NF = roof of Theorem.1. We will show tht.17 tht F 1 = ex 0 nd F = + ex 0. F1 NF 1. Recll from Exmple

17 AN INTRODUCTION TO CONIC SECTIONS 17 Next we cn use Theorem.18 or differentition to find tht the grdient of the tngent line t = (x 0, y 0 ) is given by x0b y 0. So the grdient of N is given by y 0 b x 0. Let N = (x 1, 0), then y 0 x 0 x 1 = y 0 b x 0 = x 1 = x 0 where e is the eccentricity. This mens tht ) (1 b = e x 0, NF 1 = c x 1 = e e x 0 = e( ex 0 ), nd similrly NF = e( + ex 0 ). Thus the rtio F NF = F 1 NF 1 = 1 e. The Directrix-Eccentricity-Focus property of n ellipse cn be reformulted into polr eqution. Suppose tht focus F 1 is now t the origin nd the directrix, d, is x = k for some k > 0. Then Solving for r gives the following. F 1 = e d = r = e(k r cos(θ)). Theorem.3 (olr form with respect to the focus). Let F 1 = (0, 0) be focus nd x = k, k > 0 be the directrix. The polr form of the resulting ellipse with eccentricity e is ek r = 1 + e cos θ. r F F1 θ k Figure.6. olr form of n ellipse. Exercise.4. Let F 1 = (0, 0) nd k > 0. For ech of the three possible directrices: x = k, y = k nd y = k, find the corresponding polr form of n ellipse with eccentricity e. Exmple.5. Given n ellipse in stndrd position. resulting ellipse with respect to the centre is b r = sin θ + b cos θ. The polr form of the Solution. Let = ( cos t, b sin t) be point on the ellipse with polr coordintes (r, θ). Recll tht tn θ = sin θ cos θ = b sin t cos t. So sin θ cos θ + 1 = b sin t cos t + 1 = cos t cos θ = r. On the other hnd, sin θ b cos θ + 1 = sin t cos t + 1 = sin θ + b cos θ b cos = 1 θ cos t.

18 18 TOH EE CHOON Exercise.6. The eccentricity of n ellipse in its stndrd position is. Its foci on the x-xis re denoted by F 1 nd F. For ny point Q on the ellipse, find the coordintes of the point Q when F QF 1 is mximum. Exercise.7. In the digrm AB is the mjor xis of n ellipse in stndrd position. The tngent to the ellipse t intersects the tngents to vertices A nd B t C nd D respectively. Show tht F 1 F = C D. C D A F F 1 B Figure.7. Exercise.7. Exercise.8. Cut out pper disc nd mrk out n rbitrry interior point X distinct from the centre of the disc. Now pick rndom point on the circumference of the disc nd nd fold the disc (with one crese) such tht touches X. Use pencil to mrk out the crese. Repet for s mny points s possible. Wht do you observe? Explin. GeoGebr Exmple.9. We visulize the previous exercise. (1) Crete point A. () Construct circle of rdius 3 with centre A. (3) Crete point X in the interior of the circle. (4) Crete point on the circle. (5) Construct the perpendiculr bisector of X nd. (6) Right-click the perpendiculr bisector nd toggle trce on. (7) Right-click on nd toggle nimtion on. (8) To stop the nimtion right-click on in the Algebr window nd toggle nimtion off. (9) Vry the position of X. Wht hppens when X is on the circle? Wht if X is outside the circle? Exercise.30. In the digrm the tngents to n ellipse in stndrd position t points nd Q meet t R. The point N is the foot of the perpendculr from R to the x-xis. rove tht if N lies inside the ellipse, NR = QNR. 3. Hyperbol Recll tht n ellipse is defined s the locus of points such tht F 1 + F = k. Wht hppens if k F 1 F? If k = F 1 F, then the locus is the line segment F 1 F. We cn think of this s degenerte ellipse. However if k < F 1 F, there will be no solutions for becuse of the the tringle inequlity, nmely, F 1 F F 1 + F.

19 AN INTRODUCTION TO CONIC SECTIONS 19 R Q α β N Figure.8. Exercise.30. We now consider the locus of defined by { : F 1 F = k, k 0}. By the tringle inequlity, F 1 F F 1 F. Thus k F 1 F. If k = F 1 F, the locus is ry from F in the direction of F 1 F. On the other hnd, if k = 0, the locus is the perpendiculr bisector of F 1 F. GeoGebr Exmple 3.1. We study the remining vlues of k with GeoGebr. (1) Crete F 1, F such tht F 1 F = nd slider 5 k 5. () Crete slider 0 r 0. (3) Construct the circles F = r nd F 1 = r + k. (4) The two intersections of the circles give fesible points for. (5) Use the locus tool to drw the locus of depending on r. (6) Vry k to observe the locus. (7) Set k s vlue between 1 nd. Right-click r nd turn nimtion on. We observe tht the locus is tht of hyperbol. In fct, the nimtion ws inspired by the interference pttern of circulr wves mde by dropping two stones into still pool of wter. As seen in the GeoGebr exmple, we obtin the right brnch of the hyperbol when k < 0. This motivtes the following definition. Definition 3. (Geometric). A hyperbol is the locus of points where the distnces to the foci F 1 nd F stisfies { : F 1 F = k}. If we set F 1 = (c, 0) nd F = ( c, 0) for some positive c, we know from our previous discussion tht 0 < k < c. For symmetry we replce k by, where

20 0 TOH EE CHOON 0 < < c. If = (x, y) is point on the hyperbol, we hve (x c) + y (x + c) + y = ± = (x c) + y = (± + ) (x + c) + y = 4 ± 4 (x + c) + y + (x + c) + y = 4xc 4 = ±4 (x + c) + y = ( cx + ) = (x + c) + y = (c )x y = (c ) = x y c = 1. Theorem 3.3. A hyperbol defined by { : F1 F = } with foci F1 = (c, 0) nd F = ( c, 0) stisfies the eqution where b = c. x y b = 1, We sy tht such hyperbol is in stndrd position with trnsverse xis on x-xis. If we nme the vertices A = (, 0), A = (, 0), then the trnsverse xis is ctully the line segment AA. There re lso two symptotes described by y = ± b x. F A A F1 Figure 3.1. Hyperbol in stndrd position with trnsverse xis on x-xis. Exercise 3.4. Show tht hyperbol defined by { : F1 F = b} with foci F 1 = (0, c) nd F = (0, c) stisfies the eqution where = c b. y b x = 1, In the bove, we sy tht the hyperbol is in stndrd position with trnsverse xis BB on y-xis, where B = (0, b) nd B = (0, b). In ddition, the symptotes re lso described by y = ± b x.

21 AN INTRODUCTION TO CONIC SECTIONS 1 F1 B B F Figure 3.. Hyperbol in stndrd position with trnsverse xis on y-xis. The properties of the hyperbol often prllel those of the ellipse with the exception of the chnge in sign. A useful heuristic is to tke result for n ellipse nd replce b by b for the corresponding hyperbol with trnsverse xis on x-xis. Exercise 3.5. Show tht the line y = mx ± m b, provided it exists, intersects the hyperbol x y b = 1 t exctly one point. Exercise 3.6. Let x y b = 1 be hyperbol nd = (x 0, y 0 ) be point on the hyperbol. Show tht the eqution of the tngent line t is given by xx 0 yy 0 b = 1. Exercise 3.7. Show tht if the tngents to two points of hyperbol, x y b = 1, intersect t right ngles, the intersection lies on the director circle, which is circle with rdius b nd centred t the origin. Just s the ellipse cn be prmeterized by ordinry trigonometric functions x = cos(t) nd y = b sin(t), the hyperbol x y b = 1 cn be prmeterized by pir of hyperbolic functions. Definition 3.8. The hyperbolic cosine nd hyperbolic sine functions re defined for ll rel t in terms of the exponentil function s follows. cosh(t) = et + e t nd sinh(t) = et e t. Exercise 3.9. Let t R. Show tht cosh(t) 1, while sinh(t) tkes on every possible rel vlue, i.e. its rnge is R. Exercise rove the differentition formuls: d cosh(t) dt = sinh(t) nd d sinh(t) dt = cosh(t). Exercise Show tht the following identity holds for every t R. cosh (t) sinh (t) = 1. The results of the previous exercise suggest the use of hyperbolic cosine nd sine functions s prmeteriztions of the hyperbol. However, since cosh(t) 1, the left nd right brnch of the hyperbol x respectively s y b = 1 hve to be prmeterized (x, y) = ( cosh(t), b sinh(t) ) nd (x, y) = ( cosh(t), b sinh(t) ).

22 TOH EE CHOON Exercise 3.1. Let x y b = 1 be hyperbol nd = (x 0, y 0 ) be point on the hyperbol. For ech of the cses, x 0 > 0 nd x 0 < 0, use the prmetriztion for hyperbols to show tht the grdient of the tngent line t is given by b x 0 y 0. Recll tht the prmeter t in the prmetriztion (x, y) = ( cos(t), b sin(t)) for n ellipse represents the eccentric ngle. Likewise, the prmeter t in the prmetriztion (x, y) = ( cosh(t), b sinh(t)) hs n interesting geometric interprettion in terms of the re of wht is known s hyperbolic sector. = ( cosh(t 0), b sinh(t 0)) A Figure 3.3. Hyperbolic sector with re bt0. We compute explicitly the re of the hyperbolic sector. Let = (x 0, y 0 ), where x 0 > 0 nd y 0 > 0, be point on the hyperbol x y b = 1. We denote the re bounded by the hyperbol, the x-xis nd the verticl line x = x 0 s A. We first note tht the re of the tringle with vertices t the origin, nd (x 0, 0) is given by x 0 y 0 = cosh(t 0) b sinh(t 0 ) = b et0 + e t0 et0 e t0 = b sinh(t 0). 4 In the bove, we used the prmetriztion (x 0, y 0 ) = ( cosh(t 0 ), b sinh(t 0 )) nd used definitions to compute double-ngle identity for hyperbolic sine. On the other hnd, we cn compute A directly with the integrl x0 x0 x A = y dx = b 1 dx t0 = b cosh (t) 1 sinh(t) dt The first term bove is exctly in the bove Figure is bt0. = b = b 4 = b 8 0 t0 0 t0 0 sinh (t) dt e t + e t dt ( e t 0 e t0) bt 0. b sinh(t0) 4 nd so we conclude tht the shded re

23 AN INTRODUCTION TO CONIC SECTIONS 3 While the hyperbolic functions re the nturl choice for prmeterizing hyperbols, it is lso possible to mke do with the usul trigonometric functions. The hyperbol x y b = 1 cn be prmeterized by x = sec(θ) nd y = b tn(θ), 0 θ < π, θ π, 3π. GeoGebr Exmple We use GeoGebr to investigte the two possible prmetriztions of the hyperbol. (1) Crete sliders 5, b, 5 nd sliders 0 t 0 nd 0 w π. The ltter cn be chieved by selecting the Angle option. () In the Input window enter x=*cosh(t) y=b*sinh(t) (3) Intersect the two stright lines nd use the locus tool to drw the locus of points depending on the prmeter t. (4) Note tht only the right brnch is given nd the left brnch cn be drwn with nother line x = cosh(t). (5) Right-click on the slider t nd toggle nimtion on to observe how the points on the hyperbol vry with t (6) Do the sme with x = sec(z) nd y = b tn(z). (7) Observe tht only one prmetriztion is necessry nd observe which vlues of 0 w π correspond to which section of the hyperbol, pying ttention to the discontinuities. Now let x y b = 1 be hyperbol nd F 1 be the right focus. If = (x 0, y 0 ) is point on the hyperbol, the similr to the cse for ellipses, F 1 hs simple expression. We shll ssume x 0 > 0 nd use the prmetriztion (x 0, y 0 ) = ( cosh(t 0 ), b sinh(t 0 )). F 1 = ( cosh(t 0 ) + b ) + b sinh (t 0 ) = cosh (t 0 ) cosh(t 0 ) + b + + b + b (cosh (t 0 ) 1) = ( + b cosh(t 0 ) = c cosh(t 0 ). ) Exercise Let x y b = 1 be hyperbol with foci F 1 nd F. If = (x 0, y 0 ) is point on the hyperbol, show tht F 1 = cx 0 nd F = cx 0 +. From the bove exercise, we cn lso recover the geometric definition of hyperbol s the curve with locus given by { : F1 F = }, for some > 0. In ddition to the geometric definition, the hyperbol cn lso be defined in terms of directrix-eccentricity-focus. Definition 3.15 (Directrix-Eccentricity-Focus). A hyperbol is the locus of points whose distnce from{ the focus F (focus) } nd the directrix d re in constnt rtio greter thn 1, i.e. : F d = e, e > 1.

24 4 TOH EE CHOON D F D Figure 3.4. Directrix-Eccentricity-Focus definition of hyperbol. We cn work out the crtesin eqution of hyperbol with eccentricity e > 1, F = (c, 0) nd directrix x = c e. In fct, the working proceeds in exctly the sme wy s the cse for ellipse in the previous section s we hd only ssumed e 1. We rrive t Eqution (.) which we rewrite s e x c e y c (e 1) = 1, since e 1 > 0. Thus = c e nd b = (e 1). The directrix cn then be written s x = e nd e = 1 + b. Finlly we lso hve c = e = + b. By symmetry F = ( c, 0) nd x = e lso qulify s focus nd directrix. Exercise Let F = (0, c), c > 0 nd directrix d be the line y = k. Show tht eqution F d = e for e > 1 defines hyperbol nd determine the vlue of e, c nd k in terms of the prmeters for hyperbol in stndrd position with trnsverse xis on y-xis. Exercise Given hyperbol in stndrd position with trnsverse xis on x-xis, show tht the polr form with respect to the centre is given by r = b. b cos (θ) sin (θ) The corresponding reflection property for hyperbol cn be phrsed s light coming from F 1 is reflected in hyperbolic mirror in such wy tht it ppers to hve come from F. An ppliction of this is in the Cssegrin reflector which utilizes primry prbolic reflector nd secondry hyperbolic mirror to bring the focl point to loction behind the primry reflector. Exercise 3.18 (Apollonius). Let T be point on hyperbol in stndrd position with trnsverse xis on x-xis. The tngent line to the hyperbol t T cuts the symptotes t nd Q. Show tht (1) T is the midpoint of Q. () If O is the origin, show tht OQ O is constnt nd hence the tringle O Q hs constnt re.

25 AN INTRODUCTION TO CONIC SECTIONS 5 F F 1 Figure 3.5. Reflection property of hyperbol. Mthemtics & Mthemtics Eduction, Ntionl Institute of Eduction, Nnyng Technologicl University, 1 Nnyng Wlk, Singpore E-mil ddress: peechoon.toh@nie.edu.sg