Section 9. Hperols 597 Section 9. Hperols In the lst section, we lerned tht plnets hve pproimtel ellipticl orits round the sun. When n oject like comet is moving quickl, it is le to escpe the grvittionl pull of the sun nd follows pth with the shpe of hperol. Hperols re curves tht cn help us find the loction of ship, descrie the shpe of cooling towers, or clirte seismologicl equipment. The hperol is nother tpe of conic section creted intersecting plne with doule cone, s shown elow 5. The word hperol derives from Greek word mening ecess. The English word hperole mens eggertion. We cn think of hperol s n ecessive or eggerted ellipse, one turned inside out. We defined n ellipse s the set of ll points where the sum of the distnces from tht point to two fied points is constnt. A hperol is the set of ll points where the solute vlue of the difference of the distnces from the point to two fied points is constnt. 5 Proks3 (https://commons.wikimedi.org/wiki/file:conic_sections_with_plne.svg), Conic sections with plne, cropped to show onl hperol L Michels, CC BY 3.0
598 Chpter 9 Hperol Definition A hperol is the set of ll points ( ) difference of the distnces to two fied points F ( ) nd ( ) (plurl for focus) is constnt k: ( ) ( ) k d(q,f ) Q Q, for which the solute vlue of the, F, clled the foci d Q, F d Q F =. d(q,f ), d(q,f ) Q d(q,f ) F F F F The trnsverse is is the line pssing through the foci. Vertices re the points on the hperol which intersect the trnsverse is. The trnsverse is length is the length of the line segment etween the vertices. The center is the midpoint etween the vertices (or the midpoint etween the foci). The other is of smmetr through the center is the conjugte is. The two disjoint pieces of the curve re clled rnches. A hperol hs two smptotes. Which is is the trnsverse is will depend on the orienttion of the hperol. As helpful tool for grphing hperols, it is common to drw centrl rectngle s guide. This is rectngle drwn round the center with sides prllel to the coordinte es tht pss through ech verte nd co-verte. The smptotes will follow the digonls of this rectngle. Verte Focus Asmptote Center Co-verte Trnsverse is Conjugte is
Section 9. Hperols 599 Hperols Centered t the Origin From the definition ove we cn find n eqution of hperol. We will find it for hperol centered t the origin C ( 0,0) opening horizontll with foci t F ( c,0 ) nd F ( c,0) where c > 0. Suppose ( ) ( Q, F ) = ( c) + ( 0) = ( c) Q, is point on the hperol. The distnces from Q to F nd Q to F re: d + ( Q, F ) ( ( c) ) + ( 0) = ( + c) d = +. From the definition, the solute vlue of the difference should e constnt: ( Q, F ) d( Q, F ) = ( c) + ( + c) + k d = Sustituting in one of the vertices (,0), we cn determine k in terms of : ( c) + 0 ( + c) + 0 = k c + c = k Since c >, c = c ( c ) ( + c) = k k = = Using k = nd removing the solute vlues, ( c) + ( + c) + = ± Move one rdicl ( c) + = ± + ( + c) Squre oth sides + ( c) + = 4 ± 4 ( + c) + + ( + c) + Epnd ( + c) + + + c + c c + c + = 4 ± 4 + Comining like terms leves ( + c) 4c = 4 ± 4 + Divide 4 ( + c) c = ± + Isolte the rdicl ± ( + c) + = c 4 ( c) + = + c ( ) c + Squre oth sides gin + Epnd nd distriute c 4 + c + c + = + c + Comine like terms 4 + c = c Fctor common terms ( c ) = ( c ) +
600 Chpter 9 Let = c. Since c >, > 0. Sustituting for c leves + = Divide oth sides + = Rewrite = We cn see from the grphs of the hperols tht the rnches pper to pproch smptotes s gets lrge in the negtive or positive direction. The equtions of the horizontl hperol smptotes cn e derived from its stndrd eqution. = Solve for = Rewrite s = Fctor out = Tke the squre root = ± As ± the quntit 0 nd, so the smptotes re = ±. Similrl, for verticl hperols the smptotes re = ±. 0 depends on whether it opens horizontll or verticll. The following tle gives the stndrd eqution, vertices, foci, smptotes, construction rectngle vertices, nd grph for ech. The stndrd form of n eqution of hperol centered t the origin C (,0)
Section 9. Hperols 60 Eqution of Hperol Centered t the Origin in Stndrd Form Opens Horizontll Verticll Stndrd Eqution = Vertices (-, 0) nd (, 0) (0, -) nd (0, ) = Foci (-c, 0) nd (c, 0) where = c (0, -c) nd (0, c) Where = c Asmptotes Construction Rectngle Vertices Grph = ± = ± (, ), (-, ), (,-), (-, -) (, ), (-, ), (, -), (-, -) (0,c) (0,) (0,) (-c,0) (-,0) (,0) (c,0) (-,0) (,0) (0,-) (0,-) (0,-c) Emple Put the eqution of the hperol 4 = 4 in stndrd form. Find the vertices, length of the trnsverse is, nd the equtions of the smptotes. Sketch the grph. Check using grphing utilit. The eqution cn e put in stndrd form = 4 dividing 4. Compring to the generl stndrd eqution = we see tht = 4 = nd = =. Since the term is sutrcted, the hperol opens verticll nd the vertices lie on the -is t (0,±) = (0, ±).
60 Chpter 9 The length of the trnsverse is is ( ) = ( ) = 4. Equtions of the smptotes re = ± or = ±. To sketch the grph we plot the vertices of the construction rectngle t (±,±) or (-,-), (-,), (,-), nd (,). The smptotes re drwn through the digonls of the rectngle nd the vertices plotted. Then we sketch in the hperol, rounded t the vertices nd pproching the smptotes. To check on grphing utilit, we must solve the eqution for. Isolting gives us = 4 +. ( ) Tking the squre root of oth sides we find + = ±. Under Y= enter the two hlves of the hperol nd the two smptotes s = +, = +, =, nd =. Set the window to comprle scle to the sketch with min = -4, m = 4, min= -3, nd m = 3. Sometimes we re given the eqution. Sometimes we need to find the eqution from grph or other informtion.
Section 9. Hperols 603 Emple Find the stndrd form of the eqution for hperol with vertices t (-6,0) nd (6,0) 4 nd smptote =. 3 Since the vertices lie on the -is with midpoint t the origin, the hperol is horizontl with n eqution of the form =. The vlue of is the distnce from the center to verte. The distnce from (6,0) to (0,0) is 6, so = 6. 3 The smptotes follow the form = ±. From = we see 4 3 sustituting = 6 give us =. Solving ields = 8. 4 6 3 = 4 nd The eqution of the hperol in stndrd form is = 6 8 or = 36 64. Tr it Now. Find the stndrd form of the eqution for hperol with vertices t (0,-8) nd (0,8) nd smptote = Emple 3 Find the stndrd form of the eqution for hperol with vertices t (0, 9) nd (0,-9) nd pssing through the point (8,5). Since the vertices lie on the -is with midpoint t the origin, the hperol is verticl with n eqution of the form =. The vlue of is the distnce from the center to verte. The distnce from (0,9) to (0,0) is 9, so = 9. 5 8 Sustituting = 9 nd the point (8,5) gives =. Solving for eilds 9 ( 8 ) = 6 9 =. 5 9 The stndrd eqution for the hperol is = 9 6 or = 8 36.
604 Chpter 9 Hperols Not Centered t the Origin Not ll hperols re centered t the origin. The stndrd eqution for one centered t (h, k) is slightl different. Eqution of Hperol Centered t (h, k) in Stndrd Form h, depends on whether it opens horizontll or verticll. The tle elow gives the stndrd eqution, vertices, foci, smptotes, construction rectngle vertices, nd grph for ech. The stndrd form of n eqution of hperol centered t C ( k) Opens Horizontll Verticll Stndrd Eqution ( h) ( k) ( k ) ( h) = = Vertices ( h ±, k ) (h, k ± ) Foci ( h ± c, k ) where = c (h, k ± c) where = c Asmptotes k = ± ( h) k = ± ( h) Construction Rectngle Vertices Grph (h-c,k) (h-,k) ( h ±, k ± ) ( h ±, k ± ) (h,k) (h,k+) (h,k-) (h+c,k) (h+,k) (h-,k) (h,k+c) (h,k+) (h,k) (h,k-) (h,k-c) (h+,k)
Section 9. Hperols 605 Emple 4 Write n eqution for the hperol in the grph shown. The center is t (,3), where the smptotes cross. It opens verticll, so the eqution will look like ( 3) ( ) =. The vertices re t (,) nd (,4). The distnce from the center to verte is = 4 3 =. If we were to drw in the construction rectngle, it would etend from = - to = 5. The distnce from the center to the right side of the rectngle gives = 5 = 3. 3 3 The stndrd eqution of this hperol is ( ) ( ) = ( 3) ( ) =. 9, or Emple 5 Put the eqution of the hperol 9 + 8 4 + 6 = 43 in stndrd form. Find the center, vertices, length of the trnsverse is, nd the equtions of the smptotes. Sketch the grph, then check on grphing utilit. To rewrite the eqution, we complete the squre for oth vriles to get 9 + + 4 4 + 4 = 43+ 9 9 ( ) ( ) 6 ( + ) 4( ) = 36 + 4 9 Dividing 36 gives the stndrd form of the eqution, ( ) ( ) = h Compring to the generl stndrd eqution ( ) ( ) = = 4 = nd = 9 = 3. h k we see tht Since the term is sutrcted, the hperol opens horizontll. The center is t (h, k) = (-, ). The vertices re t (h±, k) or (-3, ) nd (,). The length of the trnsverse is is ( ) = ( ) = 4. 3 Equtions of the smptotes re k = ± ( h) or = ± ( + ).
606 Chpter 9 To sketch the grph we plot the corners of the construction rectngle t (h±, k±) or (, 5), (, -), (-3,5), nd (-3,-). The smptotes re drwn through the digonls of the rectngle nd the vertices plotted. Then we sketch in the hperol rounded t the vertices nd pproching the smptotes. To check on grphing utilit, we must solve the eqution for. ( + ) = ± 9. 4 Under Y= enter the two hlves of the hperol nd the two smptotes s ( + ) ( ) + 3 = + 9, = 9 4, = ( + ) +, nd 4 3 = ( + ) +. Set the window to comprle scle to the sketch, then grph. Note tht the gps ou see on the clcultor re not rell there; the re limittion of the technolog. Emple 6 Find the stndrd form of the eqution for hperol with vertices t (, 5) nd 3 (,7), nd smptote = + 4.
Section 9. Hperols 607 Since the vertices differ in the -coordintes, the hperol opens verticll with n eqution of the form ( ) ( ) k h = nd smptote equtions of the form k = ± ( h). 5 + 7 The center will e hlfw etween the vertices, t, = (,). The vlue of is the distnce from the center to verte. The distnce from (,) to (, 5) is 6, so = 6. While our smptote is not given in the form k = ± ( h), notice this eqution would hve slope. We cn compre tht to the slope of the given smptote eqution to find. Setting 3 = nd sustituting = 6 gives us = 4. + 6 4 The eqution of the hperol in stndrd form is ( ) ( ) = ( ) ( + ) 36 6 =. or Tr it Now. Find the center, vertices, length of the trnsverse is, nd equtions of the smptotes for the hperol ( ) ( ) + 5 =. 9 36 Hperol Foci The loction of the foci cn pl ke role in hperol ppliction prolems. To find them, we need to find the length from the center to the foci, c, using the eqution = c. It looks similr to, ut is not the sme s, the Pthgoren Theorem. Compre this with the eqution to find length c for ellipses, which is = c. If ou rememer tht for the foci to e inside the ellipse the hve to come efore the vertices ( c < ), it s cler wh we would clculte minus c. To e inside hperol, the foci hve to go eond the vertices ( c > ), so we cn see for hperols we need minus, the opposite. c
608 Chpter 9 Emple 7 + 3 4 5 Find the foci of the hperol ( ) ( ) = k The center is t (h, k) = (3, -). The foci re t (h, k ± c). The hperol is verticl with n eqution of the form ( ) ( ) =. h. To find length c we use = c. Sustituting gives 5 4 = c or c = 9 = 3. The hperol hs foci (3, -4) nd (3, ). Emple 8 Find the stndrd form of the eqution for hperol with foci (5, -8) nd (-3, -8) nd vertices (4, -8) nd (-, -8). Since the vertices differ in the -coordintes, the hperol opens horizontll with n eqution of the form ( ) ( ) h k =. The center is t the midpoint of the vertices + + 4 + ( ) 8 + ( 8), =, = (, 8) The vlue of is the horizontl length from the center to verte, or = 4 = 3. The vlue of c is the horizontl length from the center to focus, or = 5 = 4. To find length we use = c. Sustituting gives = 6 9 = 7. The eqution of the hperol in stndrd form is ( ) ( ( 8 )) = ( ) ( + 8) 9 7 =.. 3 7 or Tr it Now 3. Find the stndrd form of the eqution for hperol with focus (,9), verte (,8), center (,4).
Section 9. Hperols 609 LORAN Before GPS, the Long Rnge Nvigtion (LORAN) sstem ws used to determine ship s loction. Two rdio sttions A nd B simultneousl sent out signl to ship. The difference in time it took to receive the signl ws computed s distnce locting the ship on the hperol with the A nd B rdio sttions s the foci. A second pir of rdio sttions C nd D sent simultneous signls to the ship nd computed its loction on the hperol with C nd D s the foci. The point P where the two hperols intersected gve the loction of the ship. A C D B P Emple 9 Sttions A nd B re 50 kilometers prt nd send simultneous rdio signl to the ship. The signl from B rrives 0.0003 seconds efore the signl from A. If the signl trvels 300,000 kilometers per second, find the eqution of the hperol on which the ship is positioned. Sttions A nd B re t the foci, so the distnce from the center to one focus is hlf the distnce etween them, giving c = (50) = 75 km. B letting the center of the hperol e t (0,0) nd plcing the foci t (±75,0), the eqution = for hperol centered t the origin cn e used. The difference of the distnces of the ship from the two sttions is km k = 300,000 (0.0003s) = 90km. From our derivtion of the hperol eqution we s determined k =, so = (90) = 45. Sustituting nd c into = c ields = 75 45 = 3600. The eqution of the hperol in stndrd form is = 45 3600 or = 05 3600. To determine the position of ship using LORAN, we would need n eqution for the second hperol nd would solve for the intersection. We will eplore how to do tht in the net section.
60 Chpter 9 Importnt Topics of This Section Hperol Definition Hperol Equtions in Stndrd Form Hperol Foci Applictions of Hperols Intersections of Hperols nd Other Curves Tr it Now Answers. The vertices re on the is so this is verticl hperol. The center is t the origin. = 8 8 Using the smptote slope, =, so = 4. = 64 6. Center (-5, ). This is horizontl hperol. = 3. = 6. trnsverse is length 6, Vertices will e t (-5±3,) = (-,) nd (-8,), Asmptote slope will e 3 6 =. Asmptotes: = ± ( + 5 ) 3. Focus, verte, nd center hve the sme vlue so this is verticl hperol. Using the verte nd center, = 9 4 = 5 Using the focus nd center, c = 8 4 = 4 = 5 4. = 3. ( 4) ( ) 6 9 =
Section 9. Hperols 6 Section 9. Eercises In prolems 4, mtch ech grph to equtions A D. A. = 4 9 B. = 9 4 C. = 9 D. = 9.. 3. 4. In prolems 5 4, find the vertices, length of the trnsverse is, nd equtions of the smptotes. Sketch the grph. Check using grphing utilit. 5. = 4 5 6. = 6 9 7. = 4 8. = 5 9. 9 = 9 0. 4 = 4. 9 6 = 44. 6 5 = 400 3. 9 = 8 4. 4 = In prolems 5 6, write n eqution for the grph. 5. 6.
6 Chpter 9 In prolems 7, find the stndrd form of the eqution for hperol stisfing the given conditions. 7. Vertices t (0,4) nd (0, -4); smptote = 8. Vertices t (-6,0) nd (6,0); smptote = 3 9. Vertices t (-3,0) nd (3,0); psses through (5,8) 0. Vertices t (0, 4) nd (0, -4); psses through (6, 5). Asmptote = ; psses through (5, 3). Asmptote = ; psses through (, 3) In prolems 3 30, mtch ech grph to equtions A H. 9 4 + + 9 4 + + 9 6 9 6 A. ( ) ( ) = B. ( ) ( ) = C. ( ) ( ) = D. ( ) ( ) = 4 9 + + 4 9 + + 4 6 4 6 E. ( ) ( ) = F. ( ) ( ) = G. ( ) ( ) = H. ( ) ( ) = 3. 4. 5. 6. 7. 8. 9. 30.
Section 9. Hperols 63 In prolems 3 40, find the center, vertices, length of the trnsverse is, nd equtions of the smptotes. Sketch the grph. Check using grphing utilit. 3. ( ) ( ) + = 3. ( ) ( ) 3 + 5 = 5 4 6 36 9 33. ( ) ( + ) = 5 34. ( ) ( 6) = 35. 4 8 = 36. 4 + 6 9 = 0 37. 4 6 = 38. 4 6 + 6 = 9 39. 9 + 36 4 + 8 = 4 40. 9 + 36 6 96 = 36 In prolems 4 4, write n eqution for the grph. 4. 4. In prolems 43 44, find the stndrd form of the eqution for hperol stisfing the given conditions. 43. Vertices (-,-) nd (-,6); smptote = ( + ) 44. Vertices (-3,-3) nd (5,-3); smptote + 3 = ( ) In prolems 45 48, find the center, vertices, length of the trnsverse is, nd equtions of the smptotes. Sketch the grph. Check using grphing utilit. 45. = ± 4 9 46. = ± 9 + 4 47. = ± 9 + 8 + 0 48. = ± 9 8 + 8
64 Chpter 9 In prolems 49 54, find the foci. 49. = 6 9 50. = 35 5. ( ) 5 ( 6) = 5. ( 3) ( + 5) 47 = 4 53. = ± + 8 + 5 54. = 3 ± 4 3 5 In prolems 55 66, find the stndrd form of the eqution for hperol stisfing the given conditions. 55. Foci (5,0) nd (-5,0), vertices (4,0) nd (4,0) 56. Foci (0,6) nd (0, -6), vertices (0,0) nd (0,-0) 57. Focus (0, 3), verte (0,), center (0,0) 58. Focus (5, 0), verte (, 0), center (0,0) 8 8 59. Focus (7, 0) nd (-7,0), smptotes = nd = 5 5 4 4 60. Focus (0, 5) nd (0, 5), smptotes = nd = 7 7 6. Focus (0, 0) nd (-0, 0), trnsverse is length 6 6. Focus (0, 34) nd (0, -34), trnsverse is length 3 63. Foci (, 7) nd (, -3), vertices (, 6) nd (,-) 64. Foci (4, -) nd (-6, -), vertices (, -) nd (-4, -) 65. Focus (, 3), verte (4, 3), center (-, 3) 66. Focus (-3, 5), verte (-3, 3), center (-3, -)
Section 9. Hperols 65 67. LORAN Sttions A nd B re 00 kilometers prt nd send simultneous rdio signl to ship. The signl from A rrives 0.000 seconds efore the signl from B. If the signl trvels 300,000 kilometers per second, find n eqution of the hperol on which the ship is positioned if the foci re locted t A nd B. 68. Thunder nd Lightning Anit nd Smir re stnding 3050 feet prt when the see olt of light strike the ground. Anit hers the thunder 0.5 seconds efore Smir does. Sound trvels t 00 feet per second. Find n eqution of the hperol on which the lighting strike is positioned if Anit nd Smir re locted t the foci. 69. Cooling Tower The cooling tower for power plnt hs sides in the shpe of hperol. The tower stnds 79.6 meters tll. The dimeter t the top is 7 meters. At their closest, the sides of the tower re 60 meters prt. Find n eqution tht models the sides of the cooling tower. 70. Clirtion A seismologist positions two recording devices 340 feet prt t points A nd B. To check the clirtion, n eplosive is detonted etween the devices 90 feet from point A. The time the eplosions register on the devices is noted nd the difference clculted. A second eplosion will e detonted est of point A. How fr est should the second eplosion e positioned so tht the mesured time difference is the sme s for the first eplosion? 7. Trget Prctice A gun t point A nd trget t point B re 00 feet prt. A person t point C hers the gun fire nd hit the trget t ectl the sme time. Find n eqution of the hperol on which the person is stnding if the foci re locted t A nd B. A fired ullet hs velocit of 000 feet per second. The speed of sound is 00 feet per second. 7. Comet Trjectories A comet psses through the solr sstem following hperolic trjector with the sun s focus. The closest it gets to the sun is 3 0 8 miles. The figure shows the trjector of the comet, whose pth of entr is t right ngle to its pth of deprture. Find n eqution for the comet s trjector. Round to two deciml plces. 3 0 8
66 Chpter 9 73. The conjugte of the hperol = is =. Show tht 5 + 5 = 0 is the conjugte of 5 + 5 = 0. 74. The eccentricit e of hperol is the rtio c, where c is the distnce of focus from the center nd is the distnce of verte from the center. Find the eccentricit of = 9 6. 75. An equilterl hperol is one for which =. Find the eccentricit of n equilterl hperol. 76. The ltus rectum of hperol is line segment with endpoints on the hperol tht psses through focus nd is perpendiculr to the trnsverse is. Show tht is the length of the ltus rectum of =. 77. Confocl hperols hve the sme foci. Show tht, for 0 < k < 6, ll hperols of the form = re confocl. k 6 k