THE ELLIPSE AND HYPERBOLA

Transcription

1 600 (11-0) Chpter 11 Nonliner Sstems nd the Conic Sections nd E so tht the grph of this eqution is circle. Wht does the grph of look like? B nd D cn be n rel numbers but A must equl C nd AE B D 0. No ordered pirs stisf. 6. Discussion. Suppose lighthouse A is locted t the origin nd lighthouse B is locted t coordintes (0 6). The cptin of ship hs determined tht the ship s distnce from lighthouse A is nd its distnce from lighthouse B is 5. Wht re the possible coordintes for the loction of the ship? 5 nd 5 GRAPHING CALCULATOR EXERCISES GETTING MORE INVOLVED 6. Coopertive lerning. The eqution of circle is specil cse of the generl eqution A B C D E where A B C D nd E re rel numbers. Working in smll groups find restrictions tht must be plced on A B C D Grph ech reltion on grphing clcultor b solving for nd grphing two functions ( 1) ( ) 1 1 ( 1) ( ) In this section 11. THE ELLIPSE AND HYPERBOLA In this section we stud the remining two conic sections: the ellipse nd the hperbol. The Ellipse The Hperbol The Ellipse An ellipse cn be obtined b intersecting plne nd cone s ws shown in Fig We cn lso give definition of n ellipse in terms of points nd distnce. Ellipse An ellipse is the set of ll points in plne such tht the sum of their distnces from two fied points is constnt. Ech fied point is clled focus (plurl: foci). FIGURE 11.0 An es w to drw n ellipse is illustrted in Fig A string is ttched t two fied points nd pencil is used to tke up the slck. As the pencil is moved round the pper the sum of the distnces of the pencil point from the two fied points remins constnt. Of course the length of the string is tht constnt. You m wish to tr this.

2 11. The Ellipse nd Hperbol (11-1) 601 FIGURE = 1 (0 b) b ( 0) ( 0) ( c 0) (c 0) (0 b) FIGURE 11. Like the prbol the ellipse lso hs interesting reflecting properties. All light or sound wves emitted from one focus re reflected off the ellipse to concentrte t the other focus (see Fig. 11.1). This propert is used in light fitures where concentrtion of light t point is desired or in whispering gller such s Sttur Hll in the U.S. Cpitol Building. The orbits of the plnets round the sun nd stellites round the erth re ellipticl. For the orbit of the erth round the sun the sun is t one focus. For the ellipticl pth of n erth stellite the erth is t one focus nd point in spce is the other focus. Figure 11. shows n ellipse with foci (c 0) nd (c 0). The origin is the center of this ellipse. In generl the center of n ellipse is point midw between the foci. The ellipse in Fig. 11. hs -intercepts t ( 0) nd ( 0) nd -intercepts t (0 b) nd (0 b). The distnce formul cn be used to write the following eqution for this ellipse. (See Eercise 55.) Eqution of n Ellipse Centered t the Origin An ellipse centered t (0 0) with foci t (c 0) nd constnt sum hs eqution 1 b where b nd c re positive rel numbers with c b. To drw nice-looking ellipse we would locte the foci nd use string s shown in Fig We cn get rough sketch of n ellipse centered t the origin b using the - nd -intercepts onl. E X A M P L E 1 clcultor close-up To grph the ellipse in Emple 1 grph 1 nd. Grphing n ellipse Find the - nd -intercepts for the ellipse nd sketch its grph. 1 Solution + = 1 To find the -intercepts let 0 in the eqution: (0 ) ( 0) ( 0) (0 ) FIGURE 11. To find -intercepts let 0. We get. The four intercepts re (0 ) (0 ) ( 0) nd ( 0). Plot the intercepts nd drw n ellipse through them s in Fig. 11..

3 60 (11-) Chpter 11 Nonliner Sstems nd the Conic Sections Ellipses like circles m be centered t n point in the plne. To get the eqution of n ellipse centered t (h k) we replce b h nd b k in the eqution of the ellipse centered t the origin. helpful hint When sketching ellipses or circles b hnd use our hnd like compss nd rotte our pper s ou drw the curve. Eqution of n Ellipse Centered t (h k) An ellipse centered t (h k) hs eqution ( h) ( k) 1 b where nd b re positive rel numbers. E X A M P L E An ellipse with center (h k) Sketch the grph of the ellipse: ( ) 1 1 ( 1) ( + ) + = 1 (1 ) 5 (1 0) 1 ( ) (1 ) FIGURE 11. ( 1) ( ) 1 Solution The grph of this ellipse is ectl the sme size nd shpe s the ellipse 1 which ws grphed in Emple 1. However the center for ( 1) ( ) 1 is (1 ). The denomintor is used to determine tht the ellipse psses through points tht re three units to the right nd three units to the left of the center: ( ) nd ( ). See Fig The denomintor is used to determine tht the ellipse psses through points tht re two units bove nd two units below the center: (1 0) nd (1 ). We drw n ellipse using these four points just s we did for n ellipse centered t the origin. The Hperbol A hperbol is the curve tht occurs t the intersection of cone nd plne s ws shown in Fig. 11. in Section 11.. A hperbol cn lso be defined in terms of points nd distnce. Hperbol A hperbol is the set of ll points in the plne such tht the difference of their distnces from two fied points (foci) is constnt. Like the prbol nd the ellipse the hperbol lso hs reflecting properties. If light r is imed t one focus it is reflected off the hperbol nd goes to the

4 11. The Ellipse nd Hperbol (11-) 60 Focus Hperbol Focus other focus s shown in Fig Hperbolic mirrors re used in conjunction with prbolic mirrors in telescopes. The definitions of hperbol nd n ellipse re similr nd so re their equtions. However their grphs re ver different. Figure 11.6 shows hperbol in which the distnce from point on the hperbol to the closer focus is N nd the distnce to the frther focus is M. The vlue M N is the sme for ever point on the hperbol. FIGURE 11.5 M N Focus Focus M N is constnt FIGURE 11.6 Fundmentl rectngle Hperbol Asmptote Asmptote Hperbol FIGURE 11.7 A hperbol hs two prts clled brnches. These brnches look like prbols but the re not prbols. The brnches of the hperbol shown in Fig get closer nd closer to the dshed lines clled smptotes but the never intersect them. The smptotes re used s guidelines in sketching hperbol. The smptotes re found b etending the digonls of the fundmentl rectngle shown in Fig The ke to drwing hperbol is getting the fundmentl rectngle nd etending its digonls to get the smptotes. You will lern how to find the fundmentl rectngle from the eqution of hperbol. The hperbol in Fig opens to the left nd right. If we strt with foci t (c 0) nd positive number then we cn use the definition of hperbol to derive the following eqution of hperbol in which the constnt difference between the distnces to the foci is. = 1 (0 b) b ( 0) ( 0) (0 b) FIGURE 11.8 Eqution of Hperbol Centered t (0 0) A hperbol centered t (0 0) with foci (c 0) nd (c 0) nd constnt difference hs eqution 1 b where b nd c re positive rel numbers such tht c b. The grph of generl eqution for hperbol is shown in Fig Notice tht the fundmentl rectngle etends to the -intercepts long the -is nd

5 60 (11-) Chpter 11 Nonliner Sstems nd the Conic Sections etends b units bove nd below the origin long the -is. The fcts necessr for grphing hperbol centered t the origin nd opening to the left nd to the right re listed s follows. Grphing Hperbol Centered t the Origin Opening Left nd Right To grph the hperbol b 1: 1. Locte the -intercepts t ( 0) nd ( 0).. Drw the fundmentl rectngle through ( 0) nd (0 b).. Drw the etended digonls of the rectngle to use s smptotes.. Drw the hperbol to the left nd right pproching the smptotes. E X A M P L E A hperbol opening left nd right Sketch the grph of 1 nd find the equtions of its smptotes. 6 To grph the hperbol nd its smptotes from Emple grph nd. 1 clcultor close-up Solution The -intercepts re (6 0) nd (6 0). Drw the fundmentl rectngle through these -intercepts nd the points (0 ) nd (0 ). Etend the digonls of the fundmentl rectngle to get the smptotes. Now drw hperbol pssing through the -intercepts nd pproching the smptotes s shown in Fig From the grph in Fig. 11. we see tht the slopes of the smptotes re 1 nd 1. Becuse the -intercept for both smptotes is the origin their equtions re 1 nd 1. 6 ( 6 0) (6 0) FIGURE 11. A hperbol m open up nd down. In this cse the grph intersects onl the -is. The fcts necessr for grphing hperbol tht opens up nd down re summrized on the net pge.

6 11. The Ellipse nd Hperbol (11-5) 605 helpful hint We could include here generl formuls for the equtions of the smptotes but tht is not necessr. It is esier first to drw the smptotes s suggested nd then to figure out their equtions b looking t the grph. Grphing Hperbol Centered t the Origin Opening Up nd Down To grph the hperbol b 1: 1. Locte the -intercepts t (0 b) nd (0 b).. Drw the fundmentl rectngle through (0 b) nd ( 0).. Drw the etended digonls of the rectngle to use s smptotes.. Drw the hperbol opening up nd down pproching the smptotes. E X A M P L E 5 (0 ) (0 ) = 1 5 FIGURE 11.0 A hperbol opening up nd down Grph the hperbol 1 nd find the equtions of its smptotes. Solution If 0 we get 1. Becuse this eqution hs no rel solution the grph hs no -intercepts. Let 0 to find the -intercepts: 1 The -intercepts re (0 ) nd (0 ) nd the hperbol opens up nd down. From we get. So the fundmentl rectngle etends to the intercepts (0 ) nd (0 ) on the -is nd to the points ( 0) nd ( 0) long the -is. We etend the digonls of the rectngle nd drw the grph of the hperbol s shown in Fig From the grph in Fig we see tht the smptotes hve slopes nd. Becuse the -intercept for both smptotes is the origin their equtions re nd. E X A M P L E 5 A hperbol not in stndrd form Sketch the grph of the hperbol. Solution First write the eqution in stndrd form. Divide ech side b to get 1. 5 = 1 ( 1 0) (1 0) There re no -intercepts. If 0 then 1. The hperbol opens left nd right with -intercepts t (1 0) nd (1 0). The fundmentl rectngle etends to the intercepts long the 5 -is nd to the points (0 ) nd (0 ) long the -is. We etend the digonls of the rectngle for the smptotes nd drw the grph s shown in Fig. FIGURE

7 606 (11-6) Chpter 11 Nonliner Sstems nd the Conic Sections WARM-UPS True or flse? Eplin our nswer. 1. The -intercepts of the ellipse 6 1 re (5 0) nd (5 0). Flse 5. The grph of 1 is n ellipse. Flse. If the foci of n ellipse coincide then the ellipse is circle. True. The grph of is n ellipse centered t the origin. True 5. The -intercepts of 1 re (0 ) nd (0 ). True 6. The grph of 1 is hperbol. Flse 7. The grph of hs -intercepts t (0 ) nd (0 ). Flse 6 8. The hperbol 1 opens up nd down. True. The grph of is hperbol. True 10. The smptotes of hperbol re the etended digonls of rectngle. True 11. EXERCISES Reding nd Writing After reding this section write out the nswers to these questions. Use complete sentences. 1. Wht is the definition of n ellipse? An ellipse is the set of ll points in plne such tht the sum of their distnces from two fied points is constnt.. How cn ou drw n ellipse with pencil nd string? Attch string to two thumbtcks nd use pencil to tke up the slck s shown in the tet.. Where is the center of n ellipse? The center of n ellipse is the point tht is midw between the foci.. Wht is the eqution of n ellipse centered t the origin? The eqution of n ellipse centered t the origin is b Wht is the eqution of n ellipse centered t (h k)? The eqution of n ellipse centered t (h k) is ( h) ( k) 1. b 6. Wht is the definition of hperbol? A hperbol is the set of ll points in plne such tht the difference of their distnces from two fied points is constnt. 7. How do ou find the smptotes of hperbol? The smptotes of hperbol re the etended digonls of the fundmentl rectngle. 8. Wht is the eqution of hperbol centered t the origin nd opening left nd right? The eqution of hperbol centered t the origin nd opening left nd right is 1. b Sketch the grph of ech ellipse. See Emple

8 11. The Ellipse nd Hperbol (11-7) Sketch the grph of ech ellipse. See Emple.. ( ) ( 1) 1. ( 5) ( ) ( 1) ( ) 1 6. ( ) ( ) ( ) ( 1) 1 8. ( ) ( 1) Sketch the grph of ech hperbol nd write the equtions of its smptotes. See Emples

9 608 (11-8) Chpter 11 Nonliner Sstems nd the Conic Sections Grph both equtions of ech sstem on the sme coordinte es. Use elimintion of vribles to find ll points of intersection No points of intersection

10 11. The Ellipse nd Hperbol (11-) (10 6) (10 6) (10 6) (10 6) (0 1) (1 0) (1 0) (0 ) ( 0) (5 0) (5 0) (0 ) Solve ech problem. 5. Mrine nvigtion. The lorn (long-rnge nvigtion) sstem is used b boters to determine their loction t se. The lorn unit on bot mesures the difference in time tht it tkes for rdio signls from pirs of fied points to rech the bot. The unit then finds the equtions of two hperbols tht pss through the loction of the bot. Suppose bot is locted in the first qudrnt t the intersection of 1 nd 1. ) Use the grph on the net pge to pproimte the loction of the bot. b) Algebricll find the ect loction of the bot. ) (.5 1.5) b) (7 )

11 610 (11-0) Chpter 11 Nonliner Sstems nd the Conic Sections FIGURE FOR EXERCISE 5 GETTING MORE INVOLVED 55. Coopertive lerning. Let ( ) be n rbitrr point on n ellipse with foci (c 0) nd (c 0) for c 0. The following eqution epresses the fct tht the distnce from ( ) to (c 0) plus the distnce from ( ) to (c 0) is the constnt vlue (for 0): ( c) ( 0) ( (c) ) ( 0) Working in groups simplif this eqution. First get the rdicls on opposite sides of the eqution then squre both sides twice to eliminte the squre roots. Finll let b c to get the eqution 5. Sonic boom. An ircrft trveling t supersonic speed cretes cone-shped wve tht intersects the ground long hperbol s shown in the ccompning figure. A thunderlike sound is herd t n point on the hperbol. This sonic boom trvels long the ground following the ircrft. The re where the sonic boom is most noticeble is clled the boom crpet. The width of the boom crpet is roughl five times the ltitude of the ircrft. Suppose the eqution of the hperbol in the figure is where the units re miles nd the width of the boom crpet is mesured 0 miles behind the ircrft. Find the ltitude of the ircrft. or 6. miles 1. b 56. Coopertive lerning. Let ( ) be n rbitrr point on hperbol with foci (c 0) nd (c 0) for c 0. The following eqution epresses the fct tht the distnce from ( ) to (c 0) minus the distnce from ( ) to (c 0) is the constnt vlue (for 0): ( c) ( 0) ( (c) ) ( 0) Working in groups simplif the eqution. You will need to squre both sides twice to eliminte the squre roots. Finll let b c to get the eqution 1. b GRAPHING CALCULATOR EXERCISES Width of boom crpet 0 0 Most intense sonic boom is between these lines FIGURE FOR EXERCISE Grph nd to get the grph of the hperbol 1 long with its smptotes. Use the viewing window nd. Notice how the brnches of the hperbol pproch the smptotes. 58. Grph the sme four functions in Eercise 57 but use 0 0 nd 0 0 s the viewing window. Wht hppened to the hperbol? 11.5 SECOND-DEGREE INEQUALITIES In this section Grphing Second-Degree Inequlit Sstems of Inequlities In this section we grph second-degree inequlities nd sstems of inequlities involving second-degree inequlities. Grphing Second-Degree Inequlit A second-degree inequlit is n inequlit involving squres of t lest one of the vribles. Chnging the equl sign to n inequlit smbol for n of the equtions of the conic sections gives us second-degree inequlit. Second-degree inequlities re grphed in the sme mnner s liner inequlities.