Section 10.4 Hperbols 751 10.4 HYPERBOLAS Wht ou should lern Write equtions of hperbols in stndrd form. Find smptotes of nd grph hperbols. Use properties of hperbols to solve rel-life problems. Clssif conics from their generl equtions. Wh ou should lern it Hperbols cn be used to model nd solve mn tpes of rel-life problems. For instnce, in Eercise 54 on pge 759, hperbols re used in long distnce rdio nvigtion for ircrft nd ships. Introduction The third tpe of conic is clled hperbol. The definition of hperbol is similr to tht of n ellipse. The difference is tht for n ellipse the sum of the distnces between the foci nd point on the ellipse is fied, wheres for hperbol the difference of the distnces between the foci nd point on the hperbol is fied. Definition of Hperbol A hperbol is the set of ll points, in plne, the difference of whose distnces from two distinct fied points (foci) is positive constnt. See Figure 10.30. Focus d (, ) d 1 Focus d d1is positive constnt. Brnch FIGURE 10.30 FIGURE 10.31 Verte Center c Trnsverse is Verte Brnch U.S. Nv, Willim Lipski/AP Photo The grph of hperbol hs two disconnected brnches. The line through the two foci intersects the hperbol t its two vertices. The line segment connecting the vertices is the trnsverse is, nd the midpoint of the trnsverse is is the center of the hperbol. See Figure 10.31. The development of the stndrd form of the eqution of hperbol is similr to tht of n ellipse. Note in the definition below tht, b, nd c re relted differentl for hperbols thn for ellipses. Stndrd Eqution of Hperbol The stndrd form of the eqution of hperbol with center h, k is h k Trnsverse is is horizontl. Trnsverse is is verticl. The vertices re units from the center, nd the foci re c units from the center. Moreover, c b. If the center of the hperbol is t the origin 0, 0, the eqution tkes one of the following forms. b 1 k b 1 h b 1. Trnsverse is is horizontl. b 1 Trnsverse is is verticl.
75 Chpter 10 Topics in Anltic Geometr Figure 10.3 shows both the horizontl nd verticl orienttions for hperbol. ( h) ( k) = 1 b ( k) ( h) = 1 b ( hk, + c) ( h c, k) ( hk, ) ( h + c, k) ( hk, ) Trnsverse is is horizontl. FIGURE 10.3 ( hk, c) Trnsverse is is verticl. Emple 1 Finding the Stndrd Eqution of Hperbol When finding the stndrd form of the eqution of n conic, it is helpful to sketch grph of the conic with the given chrcteristics. Find the stndrd form of the eqution of the hperbol with foci 1, nd 5, nd vertices 0, nd 4,. B the Midpoint Formul, the center of the hperbol occurs t the point,. Furthermore, c 5 3 nd 4, nd it follows tht b c 3 9 4 5. So, the hperbol hs horizontl trnsverse is nd the stndrd form of the eqution is 5 1. This eqution simplifies to 4 5 1. See Figure 10.33. ( ) ( ) = 1 ( 5 ( 5 4 1 (0, ) (4, ) ( 1, ) (, ) (5, ) 1 3 4 FIGURE 10.33 Now tr Eercise 35.
Section 10.4 Hperbols 753 Conjugte is (h, k + b) Asmptote Asmptotes of Hperbol Ech hperbol hs two smptotes tht intersect t the center of the hperbol, s shown in Figure 10.34. The smptotes pss through the vertices of rectngle of dimensions b b, with its center t h, k. The line segment of length b joining h, k b nd h, k b or h b, k nd h b, k is the conjugte is of the hperbol. (h, k) (h, k b) (h, k) (h +, k) FIGURE 10.34 Asmptote Asmptotes of Hperbol The equtions of the smptotes of hperbol re k ± b h k ± h. b Trnsverse is is horizontl. Trnsverse is is verticl. Emple Using Asmptotes to Sketch Hperbol Sketch the hperbol whose eqution is 4 1. Algebric Divide ech side of the originl eqution b 1, nd rewrite the eqution in stndrd form. 4 1 Write in stndrd form. From this, ou cn conclude tht, b 4, nd the trnsverse is is horizontl. So, the vertices occur t, 0 nd, 0, nd the endpoints of the conjugte is occur t 0, 4 nd 0, 4. Using these four points, ou re ble to sketch the rectngle shown in Figure 10.35. Now, from c b, ou hve c 4 0 5. So, the foci of the hperbol re 5, 0 nd 5, 0. Finll, b drwing the smptotes through the corners of this rectngle, ou cn complete the sketch shown in Figure 10.3. Note tht the smptotes re nd. Grphicl Solve the eqution of the hperbol for s follows. 4 1 4 1 ±4 1 Then use grphing utilit to grph 1 4 1 nd 4 1 in the sme viewing window. Be sure to use squre setting. From the grph in Figure 10.37, ou cn see tht the trnsverse is is horizontl. You cn use the zoom nd trce fetures to pproimte the vertices to be, 0 nd, 0. 1 = 4 1 8 (0, 4) 8 9 9 = 4 1 (, 0) 4 (, 0) 4 (0, 4) ( 5, 0) 4 ( 5, 0) 4 = 1 4 FIGURE 10.37 FIGURE 10.35 FIGURE 10.3 Now tr Eercise 11.
754 Chpter 10 Topics in Anltic Geometr Emple 3 Finding the Asmptotes of Hperbol Sketch the hperbol given b 4 3 8 1 0 nd find the equtions of its smptotes nd the foci. 4 3 8 1 0 4 8 3 1 4 3 1 4 1 3 1 4 4 1 3 1 Write originl eqution. Group terms. Fctor 4 from -terms. Add 4 to ech side. Write in completed squre form. ( 1, 7) 5 4 3 ( 1, ) ( + 1) = 1 ( 1, 0) 1 ( 3) 4 3 1 3 4 5 ( 1, ) 3 ( 1, 7 ) FIGURE 10.38 Divide ech side b 1. Write in stndrd form. From this eqution ou cn conclude tht the hperbol hs verticl trnsverse is, centered t 1, 0, hs vertices 1, nd 1,, nd hs conjugte is with endpoints 1 3, 0 nd 1 3, 0. To sketch the hperbol, drw rectngle through these four points. The smptotes re the lines pssing through the corners of the rectngle. Using nd b 3, ou cn conclude tht the equtions of the smptotes re 1 3 4 1 1 3 1 1 3 nd Finll, ou cn determine the foci b using the eqution c b. So, ou hve c 3 7, nd the foci re 1,7 nd 1, 7. The hperbol is shown in Figure 10.38. Now tr Eercise 19. 1. 3 TECHNOLOGY You cn use grphing utilit to grph hperbol b grphing the upper nd lower portions in the sme viewing window. For instnce, to grph the hperbol in Emple 3, first solve for to get 1 1 1 3 nd 1 1. 3 Use viewing window in which 9 9 nd. You should obtin the grph shown below. Notice tht the grphing utilit does not drw the smptotes. However, if ou trce long the brnches, ou will see tht the vlues of the hperbol pproch the smptotes. 9 9
Section 10.4 Hperbols 755 = 8 Emple 4 Using Asmptotes to Find the Stndrd Eqution 4 4 (3, 1) (3, 5) Find the stndrd form of the eqution of the hperbol hving vertices 3, 5 nd 3, 1 nd hving smptotes 8 nd 4 s shown in Figure 10.39. B the Midpoint Formul, the center of the hperbol is 3,. Furthermore, the hperbol hs verticl trnsverse is with 3. From the originl equtions, ou cn determine the slopes of the smptotes to be FIGURE 10.39 = + 4 m nd m 1 b b nd, becuse 3, ou cn conclude b 3 b b 3. So, the stndrd form of the eqution is 3 3 3 1. Now tr Eercise 43. As with ellipses, the eccentricit of hperbol is e c Eccentricit nd becuse c >, it follows tht e > 1. If the eccentricit is lrge, the brnches of the hperbol re nerl flt, s shown in Figure 10.40. If the eccentricit is close to 1, the brnches of the hperbol re more nrrow, s shown in Figure 10.41. e is lrge. e is close to 1. Verte Focus Verte Focus e = c c e = c c FIGURE 10.40 FIGURE 10.41
75 Chpter 10 Topics in Anltic Geometr Applictions The following ppliction ws developed during World Wr II. It shows how the properties of hperbols cn be used in rdr nd other detection sstems. Emple 5 An Appliction Involving Hperbols Two microphones, 1 mile prt, record n eplosion. Microphone A receives the sound seconds before microphone B. Where did the eplosion occur? (Assume sound trvels t 1100 feet per second.) 3000 Assuming sound trvels t 1100 feet per second, ou know tht the eplosion took plce 00 feet frther from B thn from A, s shown in Figure 10.4. The locus of ll points tht re 00 feet closer to A thn to B is one brnch of the hperbol 00 B 000 000 A where b 1 c 580 40 00 c c c = 580 00 + ( c ) = 580 FIGURE 10.4 nd 00 1100. So, b c 40 1100 5,759,00, nd ou cn conclude tht the eplosion occurred somewhere on the right brnch of the hperbol 1,10,000 1. 5,759,00 Now tr Eercise 53. Hperbolic orbit Verte Ellipticl orbit Sun p Prbolic orbit FIGURE 10.43 Another interesting ppliction of conic sections involves the orbits of comets in our solr sstem. Of the 10 comets identified prior to 1970, 45 hve ellipticl orbits, 95 hve prbolic orbits, nd 70 hve hperbolic orbits. The center of the sun is focus of ech of these orbits, nd ech orbit hs verte t the point where the comet is closest to the sun, s shown in Figure 10.43. Undoubtedl, there hve been mn comets with prbolic or hperbolic orbits tht were not identified. We onl get to see such comets once. Comets with ellipticl orbits, such s Hlle s comet, re the onl ones tht remin in our solr sstem. If p is the distnce between the verte nd the focus (in meters), nd v is the velocit of the comet t the verte (in meters per second), then the tpe of orbit is determined s follows. 1. Ellipse:. Prbol: v < GMp v GMp 3. Hperbol: v > GMp In ech of these reltions, M 1.989 10 30 kilogrms (the mss of the sun) nd G.7 10 11 cubic meter per kilogrm-second squred (the universl grvittionl constnt).
Section 10.4 Hperbols 757 Generl Equtions of Conics Clssifing Conic from Its Generl Eqution The grph of A C D E F 0 is one of the following. 1. Circle:. Prbol: A C AC 0 A 0 or C 0, but not both. 3. Ellipse: AC > 0 A nd C hve like signs. 4. Hperbol: AC < 0 A nd C hve unlike signs. The test bove is vlid if the grph is conic. The test does not ppl to equtions such s 1, whose grph is not conic. Emple Clssifing Conics from Generl Equtions Clssif the grph of ech eqution.. 4 9 5 0 b. 4 8 4 0 c. 4 4 1 0 d. 8 1 0 HISTORICAL NOTE The Grnger Collection Croline Herschel (1750 1848) ws the first womn to be credited with detecting new comet. During her long life, this English stronomer discovered totl of eight new comets.. For the eqution 4 9 5 0, ou hve AC 40 0. Prbol So, the grph is prbol. b. For the eqution 4 8 4 0, ou hve AC 41 < 0. Hperbol So, the grph is hperbol. c. For the eqution 4 4 1 0, ou hve AC 4 > 0. Ellipse So, the grph is n ellipse. d. For the eqution 8 1 0, ou hve A C. Circle So, the grph is circle. Now tr Eercise 1. CLASSROOM DISCUSSION Sketching Conics Sketch ech of the conics described in Emple. Write prgrph describing the procedures tht llow ou to sketch the conics efficientl.