10.5. ; 43. The points of intersection of the cardioid r 1 sin and. ; Graph the curve and find its length. CONIC SECTIONS

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1 654 CHAPTER 1 PARAETRIC EQUATIONS AND POLAR COORDINATES ; 43. The points of intersection of the crdioid r 1 sin nd the spirl loop r,, cn t be found ectl. Use grphing device to find the pproimte vlues of t which the intersect. Then use these vlues to estimte the re tht lies inside both curves. 44. When recording live performnces, sound engineers often use microphone with crdioid pickup pttern becuse it suppresses noise from the udience. Suppose the microphone is plced 4 m from the front of the stge (s in the figure) nd the boundr of the optiml pickup region is given b the crdioid r 8 8 sin, where r is mesured in meters nd the microphone is t the pole. The musicins wnt to know the re the will hve on stge within the optiml pickup rnge of the microphone. Answer their question Find the ect length of the polr curve. 45. r 3 sin, 46. r e, 47. r stge udience 1 m 3 4 m microphone, 48. r, 49 5 Use clcultor to find the length of the curve correct to four deciml plces. 49. r 3 sin 5. r 4 sin r sin 5. r 1 cos 3 ; Grph the curve nd find its length. 53. r cos r cos 55. () Use Formul 1..7 to show tht the re of the surfce generted b rotting the polr curve (where f is continuous nd b ) bout the polr is is S b r sin r r f b d dr (b) Use the formul in prt () to find the surfce re generted b rotting the lemniscte r cos bout the polr is. 56. () Find formul for the re of the surfce generted b rotting the polr curve r f, (where f is continuous nd b ), bout the line. (b) Find the surfce re generted b rotting the lemniscte r cos bout the line. d b 1.5 CONIC SECTIONS In this section we give geometric definitions of prbols, ellipses, nd hperbols nd derive their stndrd equtions. The re clled conic sections, or conics, becuse the result from intersecting cone with plne s shown in Figure 1. ellipse prbol hperbol FIGURE 1 Conics

2 SECTION 1.5 CONIC SECTIONS 655 PARABOLAS is focus verte FIGURE F(, p) F prbol directri P(, ) A prbol is the set of points in plne tht re equidistnt from fied point F (clled the focus) nd fied line (clled the directri). This definition is illustrted b Figure. Notice tht the point hlfw between the focus nd the directri lies on the prbol; it is clled the verte. The line through the focus perpendiculr to the directri is clled the is of the prbol. In the 16th centur Glileo showed tht the pth of projectile tht is shot into the ir t n ngle to the ground is prbol. Since then, prbolic shpes hve been used in designing utomobile hedlights, reflecting telescopes, nd suspension bridges. (See Problem 18 on pge 68 for the reflection propert of prbols tht mkes them so useful.) We obtin prticulrl simple eqution for prbol if we plce its verte t the origin O nd its directri prllel to the -is s in Figure 3. If the focus is the point, p, then the directri hs the eqution p. If P, is n point on the prbol, then the distnce from P to the focus is FIGURE 3 O =_p p PF s p p nd the distnce from P to the directri is. (Figure 3 illustrtes the cse where p.) The defining propert of prbol is tht these distnces re equl: s p p We get n equivlent eqution b squring nd simplifing: p p p p p p p 4p 1 An eqution of the prbol with focus, p nd directri p is 4p If we write 1 4p, then the stndrd eqution of prbol (1) becomes. It opens upwrd if p nd downwrd if p [see Figure 4, prts () nd (b)]. The grph is smmetric with respect to the -is becuse (1) is unchnged when is replced b. (, p) =_p (, p) =_p =_p ( p, ) ( p, ) =_p () =4p, p> (b) =4p, p< (c) =4p, p> (d) =4p, p< FIGURE 4

3 656 CHAPTER 1 PARAETRIC EQUATIONS AND POLAR COORDINATES If we interchnge nd in (1), we obtin +1= _, 5 FIGURE 5 = 5 4p which is n eqution of the prbol with focus p, nd directri p. (Interchnging nd mounts to reflecting bout the digonl line.) The prbol opens to the right if p nd to the left if p [see Figure 4, prts (c) nd (d)]. In both cses the grph is smmetric with respect to the -is, which is the is of the prbol. EXAPLE 1 Find the focus nd directri of the prbol 1 nd sketch the grph. SOLUTION If we write the eqution s 1 nd compre it with Eqution, we see tht 4p 1, so p 5. Thus the focus is p, ( 5, ) nd the directri is 5. The sketch is shown in Figure 5. ELLIPSES An ellipse is the set of points in plne the sum of whose distnces from two fied points F 1 nd F is constnt (see Figure 6). These two fied points re clled the foci (plurl of focus). One of Kepler s lws is tht the orbits of the plnets in the solr sstem re ellipses with the sun t one focus. P P(, ) F F F (_c, ) F (c, ) FIGURE 6 FIGURE 7 In order to obtin the simplest eqution for n ellipse, we plce the foci on the -is t the points c, nd c, s in Figure 7 so tht the origin is hlfw between the foci. Let the sum of the distnces from point on the ellipse to the foci be. Then P, is point on the ellipse when PF 1 PF tht is, s c s c or s c s c Squring both sides, we hve c c 4 4s c c c which simplifies to We squre gin: s c c c c 4 c c which becomes c c

4 SECTION 1.5 CONIC SECTIONS 657 From tringle F 1 F P in Figure 7 we see tht c, so c nd therefore c. For convenience, let b c. Then the eqution of the ellipse becomes b b or, if both sides re divided b b, 3 b 1 (_, ) (_c, ) (, b) b (, ) c (c, ) (, _b) Since b c, it follows tht b. The -intercepts re found b setting. Then 1, or, so. The corresponding points, nd, re clled the vertices of the ellipse nd the line segment joining the vertices is clled the mjor is. To find the -intercepts we set nd obtin b, so b. Eqution 3 is unchnged if is replced b or is replced b, so the ellipse is smmetric bout both es. Notice tht if the foci coincide, then c, so b nd the ellipse becomes circle with rdius r b. We summrie this discussion s follows (see lso Figure 8). FIGURE 8 + b@ b (, ) (, c) 4 The ellipse b 1 b hs foci c,, where c b, nd vertices,. If the foci of n ellipse re locted on the -is t, c, then we cn find its eqution b interchnging nd in (4). (See Figure 9.) (_b, ) (, _c) (b, ) 5 The ellipse b 1 b FIGURE 9 + =1, b (_4, ) {_œ 7, } (, _) (, 3) (4, ) {œ 7, } hs foci, c, where c b, nd vertices,. V EXAPLE Sketch the grph of nd locte the foci. SOLUTION Divide both sides of the eqution b 144: The eqution is now in the stndrd form for n ellipse, so we hve 16, b 9, 4, nd b 3. The -intercepts re 4 nd the -intercepts re 3. Also, c b 7, so c s7 nd the foci re ( s7, ). The grph is sketched in Figure 1. FIGURE =144 (, _3) V EXAPLE 3 Find n eqution of the ellipse with foci, nd vertices, 3. SOLUTION Using the nottion of (5), we hve c nd 3. Then we obtin b c 9 4 5, so n eqution of the ellipse is Another w of writing the eqution is

5 658 CHAPTER 1 PARAETRIC EQUATIONS AND POLAR COORDINATES Like prbols, ellipses hve n interesting reflection propert tht hs prcticl consequences. If source of light or sound is plced t one focus of surfce with ellipticl cross-sections, then ll the light or sound is reflected off the surfce to the other focus (see Eercise 63). This principle is used in lithotrips, tretment for kidne stones. A reflector with ellipticl cross-section is plced in such w tht the kidne stone is t one focus. High-intensit sound wves generted t the other focus re reflected to the stone nd destro it without dmging surrounding tissue. The ptient is spred the trum of surger nd recovers within few ds. HYPERBOLAS FIGURE 11 P is on the hperbol when PF - PF =. P(, ) F (_c, ) F (c, ) A hperbol is the set of ll points in plne the difference of whose distnces from two fied points F 1 nd F (the foci) is constnt. This definition is illustrted in Figure 11. Hperbols occur frequentl s grphs of equtions in chemistr, phsics, biolog, nd economics (Bole s Lw, Ohm s Lw, suppl nd demnd curves). A prticulrl significnt ppliction of hperbols is found in the nvigtion sstems developed in World Wrs I nd II (see Eercise 51). Notice tht the definition of hperbol is similr to tht of n ellipse; the onl chnge is tht the sum of distnces hs become difference of distnces. In fct, the derivtion of the eqution of hperbol is lso similr to the one given erlier for n ellipse. It is left s Eercise 5 to show tht when the foci re on the -is t c, nd the difference of distnces is, then the eqution of the hperbol is 6 PF1 PF b 1 (_, ) b =_ (_c, ) FIGURE 1 - b@ = b (, ) (c, ) where c b. Notice tht the -intercepts re gin nd the points, nd, re the vertices of the hperbol. But if we put in Eqution 6 we get b, which is impossible, so there is no -intercept. The hperbol is smmetric with respect to both es. To nle the hperbol further, we look t Eqution 6 nd obtin This shows tht, so. Therefore we hve or. This mens tht the hperbol consists of two prts, clled its brnches. When we drw hperbol it is useful to first drw its smptotes, which re the dshed lines b nd b shown in Figure 1. Both brnches of the hperbol pproch the smptotes; tht is, the come rbitrril close to the smptotes. [See Eercise 69 in Section 4.5, where these lines re shown to be slnt smptotes.] 7 The hperbol 1 b 1 s b 1 hs foci c,, where c b, vertices,, nd smptotes b.

6 SECTION 1.5 CONIC SECTIONS 659 =_ b (, c) = b (, ) If the foci of hperbol re on the -is, then b reversing the roles of nd we obtin the following informtion, which is illustrted in Figure The hperbol (, _) b 1 FIGURE 13 - b@ 3 =_ 4 (, _c) = 3 4 hs foci, c, where c b, vertices,, nd smptotes b. EXAPLE 4 Find the foci nd smptotes of the hperbol nd sketch its grph. SOLUTION If we divide both sides of the eqution b 144, it becomes (_5, ) (_4, ) (4, ) (5, ) FIGURE =144 which is of the form given in (7) with 4 nd b 3. Since c , the foci re 5,. The smptotes re the lines 3 nd The grph is shown in Figure 14. EXAPLE 5 Find the foci nd eqution of the hperbol with vertices, 1 nd smptote. SOLUTION From (8) nd the given informtion, we see tht 1 nd b. Thus b 1 nd c b 5 4. The foci re (, s5 ) nd the eqution of the hperbol is 4 1 SHIFTED CONICS As discussed in Appendi C, we shift conics b tking the stndrd equtions (1), (), (4), (5), (7), nd (8) nd replcing nd b h nd k. EXAPLE 6 Find n eqution of the ellipse with foci,, 4, nd vertices 1,, 5,. SOLUTION The mjor is is the line segment tht joins the vertices 1,, 5, nd hs length 4, so. The distnce between the foci is, so c 1. Thus b c 3. Since the center of the ellipse is 3,, we replce nd in (4) b 3 nd to obtin s the eqution of the ellipse.

7 798 CHAPTER 1 VECTORS AND THE GEOETRY OF SPACE to r r nd so we hve 5 n r r which cn be rewritten s 6 n r n r Either Eqution 5 or Eqution 6 is clled vector eqution of the plne. To obtin sclr eqution for the plne, we write n, b, c, r,,, nd r,,. Then the vector eqution (5) becomes or, b, c,, 7 b c (,, 3) Eqution 7 is the sclr eqution of the plne through P,, with norml vector n, b, c. V EXAPLE 4 Find n eqution of the plne through the point, 4, 1 with norml vector n, 3, 4. Find the intercepts nd sketch the plne. SOLUTION Putting, b 3, c 4,, 4, nd 1 in Eqution 7, we see tht n eqution of the plne is (6,, ) FIGURE 7 (, 4, ) or To find the -intercept we set in this eqution nd obtin 6. Similrl, the -intercept is 4 nd the -intercept is 3. This enbles us to sketch the portion of the plne tht lies in the first octnt (see Figure 7). B collecting terms in Eqution 7 s we did in Emple 4, we cn rewrite the eqution of plne s 8 b c d where d b c. Eqution 8 is clled liner eqution in,, nd. Conversel, it cn be shown tht if, b, nd c re not ll, then the liner eqution (8) represents plne with norml vector, b, c. (See Eercise 77.) EXAPLE 5 Find n eqution of the plne tht psses through the points P 1, 3,, Q 3, 1, 6, nd R 5,,. SOLUTION The vectors nd b corresponding to PQ l nd PR l re, 4, 4 b 4, 1,

8 SECTION 1.6 CYLINDERS AND QUADRIC SURFACES 85 CYLINDERS A clinder is surfce tht consists of ll lines (clled rulings) tht re prllel to given line nd pss through given plne curve. V EXAPLE 1 Sketch the grph of the surfce. SOLUTION Notice tht the eqution of the grph,, doesn t involve. This mens tht n verticl plne with eqution k (prllel to the -plne) intersects the grph in curve with eqution. So these verticl trces re prbols. Figure 1 shows how the grph is formed b tking the prbol in the -plne nd moving it in the direction of the -is. The grph is surfce, clled prbolic clinder, mde up of infinitel mn shifted copies of the sme prbol. Here the rulings of the clinder re prllel to the -is. FIGURE 1 The surfce = is prbolic clinder. We noticed tht the vrible is missing from the eqution of the clinder in Emple 1. This is tpicl of surfce whose rulings re prllel to one of the coordinte es. If one of the vribles,, or is missing from the eqution of surfce, then the surfce is clinder. EXAPLE Identif nd sketch the surfces. () 1 (b) 1 SOLUTION () Since is missing nd the equtions 1, k represent circle with rdius 1 in the plne k, the surfce 1 is circulr clinder whose is is the -is. (See Figure.) Here the rulings re verticl lines. (b) In this cse is missing nd the surfce is circulr clinder whose is is the -is. (See Figure 3.) It is obtined b tking the circle 1, in the -plne nd moving it prllel to the -is. FIGURE + =1 FIGURE 3 +@=1 NOTE When ou re deling with surfces, it is importnt to recognie tht n eqution like 1 represents clinder nd not circle. The trce of the clinder 1 in the -plne is the circle with equtions 1,.

TABLE 1 Grphs of qudric surfces Surfce Eqution Surfce Eqution Ellipsoid b c 1 All trces re ellipses. If b c, the ellipsoid is sphere. Cone c b Horiontl trces re ellipses.

9 TABLE 1 Grphs of qudric surfces Surfce Eqution Surfce Eqution Ellipsoid b c 1 All trces re ellipses. If b c, the ellipsoid is sphere. Cone c b Horiontl trces re ellipses. Verticl trces in the plnes k nd k re hperbols if k but re pirs of lines if k. Elliptic Prboloid c b Horiontl trces re ellipses. Verticl trces re prbols. The vrible rised to the first power indictes the is of the prboloid. Hperboloid of One Sheet b c 1 Horiontl trces re ellipses. Verticl trces re hperbols. The is of smmetr corresponds to the vrible whose coefficient is negtive. Hperbolic Prboloid c b Horiontl trces re hperbols. Verticl trces re prbols. The cse where c is illustrted. Hperboloid of Two Sheets b c 1 Horiontl trces in k re ellipses if k c or k c. Verticl trces re hperbols. The two minus signs indicte two sheets.

P 1 (x 1, y 1 ) is given by,.

P 1 (x 1, y 1 ) is given by,. MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce