Chpter 9: Conics Section 9. Ellipses... 579 Section 9. Hperbols... 597 Section 9.3 Prbols nd Non-Liner Sstems... 67 Section 9.4 Conics in Polr Coordintes... 630 In this chpter, we will eplore set of shpes defined b common chrcteristic: the cn ll be formed b slicing cone with plne. These fmilies of curves hve brod rnge of pplictions in phsics nd stronom, from describing the shpe of our cr hedlight reflectors to describing the orbits of plnets nd comets. Section 9. Ellipses The Ntionl Sttur Hll in Wshington, D.C. is n ovl-shped room clled whispering chmber becuse the shpe mkes it possible for sound to reflect from the wlls in specil w. Two people stnding in specific plces re ble to her ech other whispering even though the re fr prt. To determine where the should stnd, we will need to better understnd ellipses. An ellipse is tpe of conic section, shpe resulting from intersecting plne with cone nd looking t the curve where the intersect. The were discovered b the Greek mthemticin Menechmus over two millenni go. The figure below shows two tpes of conic sections. When plne is perpendiculr to the is of the cone, the shpe of the intersection is circle. A slightl titled plne cretes n ovl-shped conic section clled n ellipse. Photo b Gr Plmer, Flickr, CC-BY, https://www.flickr.com/photos/gregplmer/5757950 Pbroks3 (https://commons.wikimedi.org/wiki/file:conic_sections_with_plne.svg), Conic sections with plne, cropped to show onl ellipse nd circle b L Michels, CC BY 3.0 This chpter is prt of Preclculus: An Investigtion of Functions Lippmn & Rsmussen 07. This mteril is licensed under Cretive Commons CC-BY-SA license. This chpter contins content remied from work b Lr Michels nd work from OpenSt Preclculus (OpenSt.org), CC-BY 3.0.
580 Chpter 9 An ellipse cn be drwn b plcing two thumbtcks in piece of crdbord then cutting piece of string longer thn the distnce between the thumbtcks. Tck ech end of the string to the crdbord, nd trce curve with pencil held tught ginst the string. An ellipse is the set of ll points where the sum of the distnces from two fied points is constnt. The length of the string is the constnt, nd the two thumbtcks re the fied points, clled foci. Ellipse Definition nd Vocbulr An ellipse is the set of ll points ( ) fied points F (, ) nd (, ) d ( Q, F ) d( Q F ) = k. +, Q, for which the sum of the distnce to two F, clled the foci (plurl of focus), is constnt k: Q F d(q,f ) d(q,f ) F The mjor is is the line pssing through the foci. Vertices re the points on the ellipse which intersect the mjor is. The mjor is length is the length of the line segment between the vertices. The center is the midpoint between the vertices (or the midpoint between the foci). The minor is is the line perpendiculr to the minor is pssing through the center. Minor is endpoints re the points on the ellipse which intersect the minor is. The minor is endpoints re lso sometimes clled co-vertices. The minor is length is the length of the line segment between minor is endpoints. Note tht which is is mjor nd which is minor will depend on the orienttion of the ellipse. In the ellipse shown t right, the foci lie on the is, so tht is the mjor is, nd the is is the minor is. Becuse of this, the vertices re the endpoints of the ellipse on the is, nd the minor is endpoints (co-vertices) re the endpoints on the is. Vertices Foci Mjor is Minor is endpoints Minor is
Section 9. Ellipses 58 Ellipses Centered t the Origin From the definition bove we cn find n eqution for n ellipse. We will find it for ellipse centered t the origin C ( 0,0) with foci t F ( c,0 ) nd F ( c,0) where c > 0. Suppose ( ) ( Q, F ) = ( c) + ( 0) = ( c) Q, is some point on the ellipse. The distnce from F to Q is d + Likewise, the distnce from F to Q is ( Q, F ) = ( ( c) ) + ( 0) = ( + c) d + From the definition of the ellipse, the sum of these distnces should be constnt: d Q, F d Q F = so tht ( ) + (, ) k ( c) + + ( + c) + = k If we lbel one of the vertices (,0), it should stisf the eqution bove since it is point on the ellipse. This llows us to write k in terms of. ( c) + 0 + ( + c) + 0 = k c + + c = k Since > c, these will be positive ( c) + ( + c) = k = k Substituting tht into our eqution, we will now tr to rewrite the eqution in friendlier form. ( c) + + ( + c) + = Move one rdicl ( c) + = ( + c) Squre both sides + + ( c) = ( + c) + Epnd ( c) + = 4 4 ( + c) + + ( + c) + Epnd more ( + c) + + + c + c c + c + = 4 4 + Combining like terms nd isolting the rdicl leves ( c) + = 4 + c 4 + 4 Divide b 4 ( + c) + = c 4 ( c) + = + + ( ) c c 4 ( c + c + ) = + c c + + Epnd + + Distribute Squre both sides gin
58 Chpter 9 4 + c + c + = + c + c Combine like terms 4 c + = c Fctor common terms ( c ) + = ( c ) Let b = c. Since > c, we know b > 0. Substituting b for c leves b + = b Divide both sides b b + b = This is the stndrd eqution for n ellipse. We tpicll swp nd b when the mjor is of the ellipse is verticl. Eqution of n Ellipse Centered t the Origin in Stndrd Form C depends on whether the mjor is is horizontl or verticl. The tble below gives the stndrd eqution, vertices, minor is endpoints, foci, nd grph for ech. The stndrd form of n eqution of n ellipse centered t the origin ( 0,0) Mjor Ais Horizontl Verticl Stndrd Eqution + b = b + = Vertices (, 0) nd (, 0) (0, ) nd (0, ) Minor Ais Endpoints (0, b) nd (0, b) ( b, 0) nd (b, 0) Foci ( c, 0) nd (c, 0) where b = c (0, c) nd (0, c) where b = c Grph (-,0) (c,0) (0,b) (-c,0) (,0) (-b,0) (0,) (0,c) (b,0) (0,-c) (0,-b) (0,-)
Section 9. Ellipses 583 Emple Put the eqution of the ellipse 9 + = 9 in stndrd form. Find the vertices, minor is endpoints, length of the mjor is, nd length of the minor is. Sketch the grph, then check using grphing utilit. The stndrd eqution hs on the right side, so this eqution cn be put in stndrd form b dividing b 9: + = 9 Since the -denomintor is greter thn the -denomintor, the ellipse hs verticl mjor is. Compring to the generl stndrd form eqution + =, we see the b vlue of = 9 = 3 nd the vlue of b = =. The vertices lie on the -is t (0,±) = (0, ±3). The minor is endpoints lie on the -is t (±b, 0) = (±, 0). The length of the mjor is is ( ) = ( 3) = 6. The length of the minor is is ( b ) = ( ) =. To sketch the grph we plot the vertices nd the minor is endpoints. Then we sketch the ellipse, rounding t the vertices nd the minor is endpoints. To check on grphing utilit, we must solve the eqution for. Isolting ( ) = 9 Tking the squre root of both sides we get = ± 3 gives us Under Y= on our grphing utilit enter the two hlves of the ellipse s = 3 nd = 3. Set the window to comprble scle to the sketch with min = - 5, m = 5, min= -5, nd m = 5.
584 Chpter 9 Here s n emple output on TI-84 clcultor: Sometimes we re given the eqution. Sometimes we need to find the eqution from grph or other informtion. Emple Find the stndrd form of the eqution for n ellipse centered t (0,0) with horizontl mjor is length 8 nd minor is length 6. Since the center is t (0,0) nd the mjor is is horizontl, the ellipse eqution hs the stndrd form + =. The mjor is hs length = 8 or = 4. The minor b is hs length b = 6 or b = 8. Substituting gives + = 6 8 or + = 56 64.. Tr it Now. Find the stndrd form of the eqution for n ellipse with horizontl mjor is length 0 nd minor is length 6. Emple 3 Find the stndrd form of the eqution for the ellipse grphed here. The center is t (0,0) nd the mjor is is verticl, so the stndrd form of the eqution will be + =. b From the grph we cn see the vertices re (0,4) nd (0,-4), giving = 4. The minor-is endpoints re (,0) nd (-,0), giving b =. The eqution will be + = 4 or + = 4 6.
Section 9. Ellipses 585 Ellipses Not Centered t the Origin Not ll ellipses re centered t the origin. The grph of such n ellipse is shift of the grph centered t the origin, so the stndrd eqution for one centered t (h, k) is slightl different. We cn shift the grph right h units nd up k units b replcing with h nd with k, similr to wht we did when we lerned trnsformtions. Eqution of n Ellipse Centered t (h, k) in Stndrd Form h, depends on whether the mjor is is horizontl or verticl. The tble below gives the stndrd eqution, vertices, minor is endpoints, foci, nd grph for ech. The stndrd form of n eqution of n ellipse centered t the point C ( k) Mjor Ais Horizontl Verticl Stndrd Eqution ( h) ( k) ( h) ( k) + b = b + = Vertices ( h ±, k ) (h, k ± ) Minor Ais Endpoints ( h, k ± b ) ( h ± b, k ) Foci ( h ± c, k ) where b = c (h, k ± c) where b = c Grph (h-,k) (h-c,k) (h,k+b) (h+c,k) (h,k) (h+,k) (h-b,k) (h,k+) (h,k+c) (h,k) (h+b,k) (h, k-c) (h,k-b) (h, k-)
586 Chpter 9 Emple 4 Put the eqution of the ellipse + + 4 4 = 33 in stndrd form. Find the vertices, minor is endpoints, length of the mjor is, nd length of the minor is. Sketch the grph. To rewrite this in stndrd form, we will need to complete the squre, twice. Looking t the terms, +, we like to hve something of the form ( + n). Notice tht if we were to epnd this, we d get + n + n, so in order for the coefficient on to mtch, we ll need ( + ) = + +. However, we don t hve + on the left side of the eqution to llow this fctoring. To ccommodte this, we will dd to both sides of the eqution, which then llows us to fctor the left side s perfect squre: + + + 4 4 = 33 + ( + ) + 4 4 = 3 Repeting the sme pproch with the terms, first we ll fctor out the 4. 4 4 = 4( 6) Now we wnt to be ble to write 4( 6) s 4( n) = 4( + n + n ) +. For the coefficient of to mtch, n will hve to -3, giving 4( 3) = 4 6 + 9 = 4 4 +. ( ) 36 To llow this fctoring, we cn dd 36 to both sides of the eqution. ( + ) + 4 4 + 36 = 3 + 36 ( 6 + 9) = 4 ( + ) + 4 ( + ) + 4 ( 3) = 4 Dividing b 4 gives the stndrd form of the eqution for the ellipse ( + ) ( 3) + = 4 Since the -denomintor is greter thn the -denomintor, the ellipse hs horizontl mjor is. From the generl stndrd eqution ( ) ( ) h h k + = we see the vlue b of = 4 = nd the vlue of b = =. The center is t (h, k) = (-, 3). The vertices re t (h±, k) or (-3, 3) nd (,3). The minor is endpoints re t (h, k±b) or (-, ) nd (-,4).
Section 9. Ellipses 587 The length of the mjor is is ( ) = ( ) = 4 The length of the minor is is ( ) = ( ) =. b. To sketch the grph we plot the vertices nd the minor is endpoints. Then we sketch the ellipse, rounding t the vertices nd the minor is endpoints. Emple 5 Find the stndrd form of the eqution for n ellipse centered t (-,), verte t (-,4) nd pssing through the point (0,). The center t (-,) nd verte t (-,4) mens the mjor is is verticl since the - h k + = b vlues re the sme. The ellipse eqution hs the stndrd form ( ) ( ) The vlue of = 4-=3. Substituting = 3, h = -, nd k = gives ( + ) ( ) + =. Substituting for nd using the point (0,) gives b 3 ( 0 + ) ( ) + =. b 3 Solving for b gives b=. The eqution of the ellipse in stndrd form is ( ) ( ) + + = or 3 ( + ) ( ) + =. 4 9. Tr it Now. Find the center, vertices, minor is endpoints, length of the mjor is, nd length of the minor is for the ellipse ( ) ( + ) 4 + =. 4
588 Chpter 9 Bridges with Semiellipticl Arches Arches hve been used to build bridges for centuries, like in the Skerton Bridge in Englnd which uses five semiellipticl rches for support 3. Semiellipticl rches cn hve engineering benefits such s llowing for longer spns between supports. Emple 6 A bridge over river is supported b single semiellipticl rch. The river is 50 feet wide. At the center, the rch rises 0 feet bove the river. The rodw is 4 feet bove the center of the rch. Wht is the verticl distnce between the rodw nd the rch 5 feet from the center? Put the center of the ellipse t (0,0) nd mke the spn of the river the mjor is. 4ft 0ft 50ft Since the mjor is is horizontl, the eqution hs the form + =. b The vlue of = (50) = 5 nd the vlue of b = 0, giving + = 5 5. 5 Substituting = 5 gives + 5 =. Solving for, = 0 = 6. 5 0 65 The rodw is 0 + 4 = 4 feet bove the river. The verticl distnce between the rodw nd the rch 5 feet from the center is 4 6 = 8 feet. 3 Mine Armstrong (https://commons.wikimedi.org/wiki/file:skerton_bridge,_lncster,_englnd.jpg), Skerton Bridge, Lncster, Englnd, CC BY-SA
Section 9. Ellipses 589 Ellipse Foci The loction of the foci cn pl ke role in ellipse ppliction problems. Stnding on focus in whispering gller llows ou to her someone whispering t the other focus. To find the foci, we need to find the length from the center to the foci, c, using the eqution b = c. It looks similr to, but is not the sme s, the Pthgoren Theorem. Emple 7 The Ntionl Sttur Hll whispering chmber is n ellipticl room 46 feet wide nd 96 feet long. To her ech other whispering, two people need to stnd t the foci of the ellipse. Where should the stnd? We could represent the hll with horizontl ellipse centered t the origin. The mjor is length would be 96 feet, so = (96) = 48, nd the minor is length would be 46 feet, so b = (46) = 3. To find the foci, we cn use the eqution b = c. 3 = 48 c c = 48 3 c = 775 ±4 ft. To her ech other whisper, two people would need to stnd (4) = 84 feet prt long the mjor is, ech bout 48 4 = 6 feet from the wll. Emple 8 + 3 4 9 Find the foci of the ellipse ( ) + ( ) = h b The center is t (h, k) = (, 3). The foci re t (h, k ± c). The ellipse is verticl with n eqution of the form ( ) + ( ) = To find length c we use b = c. Substituting gives 4 = 9 c or c = 5 = 5. The ellipse hs foci (, 3 ± 5), or (, 8) nd (, ).. k.
590 Chpter 9 Emple 9 Find the stndrd form of the eqution for n ellipse with foci (-,4) nd (3,4) nd mjor is length 0. Since the foci differ in the -coordintes, the ellipse is horizontl with n eqution of the form ( ) ( ) h h k + =. b + + ( ) + 3 4 + 4 The center is t the midpoint of the foci, =, = (, 4). The vlue of is hlf the mjor is length: = (0) = 5. The vlue of c is hlf the distnce between the foci: c = (3 ( )) = (4) =. To find length b we use b = c. Substituting nd c gives b = 5 =. 4 5 The eqution of the ellipse in stndrd form is ( ) + ( ) = ( ) ( 4) 5 + =. or Tr it Now 3. Find the stndrd form of the eqution for n ellipse with focus (,4), verte (,6), nd center (,). Plnetr Orbits It ws long thought tht plnetr orbits round the sun were circulr. Around 600, Johnnes Kepler discovered the were ctull ellipticl 4. His first lw of plnetr motion ss tht plnets trvel round the sun in n ellipticl orbit with the sun s one of the foci. The length of the mjor is cn be found b mesuring the plnet s phelion, its gretest distnce from the sun, nd perihelion, its shortest distnce from the sun, nd summing them together. 4 Technicll, the re pproimtel ellipticl. The orbits of the plnets re not ectl ellipticl becuse of interctions with ech other nd other celestil bodies.
Section 9. Ellipses 59 Emple 0 Mercur s phelion is 35.98 million miles nd its perihelion is 8.58 million miles. Write n eqution for Mercur s orbit. Let the center of the ellipse be (0,0) nd its mjor is be horizontl so the eqution will hve form + =. b The length of the mjor is is = 35.98 + 8.58 = 64. 56 giving = 3. 8 nd =04.9984. Since the perihelion is the distnce from the focus to one verte, we cn find the distnce between the foci b subtrcting twice the perihelion from the mjor is length: c = 64.56 ( 8.58) = 7. 4 giving c = 3. 7. Substitution of nd c into b The eqution is + =. 04.9984 08.3084 = c ields = 3.8 3.7 = 08. 3084 b. Importnt Topics of This Section Ellipse Definition Ellipse Equtions in Stndrd Form Ellipse Foci Applictions of Ellipses Tr it Now Answers. = 0, so =0. b = 6, so b = 3. + = 00 9. Center (4, -). Verticl ellipse with =, b =. Vertices t (4, -±) = (4,0) nd (4,-4), minor is endpoints t (4±, -) = (3,-) nd (5,-), mjor is length 4, minor is length 3. Verte, center, nd focus hve the sme -vlue, so it s verticl ellipse. Using the verte nd center, = 6 = 5 Using the center nd focus, c = 4 = 3 b = 5 3. b = 4. ( ) ( ) 6 + 5 =
59 Chpter 9 Section 9. Eercises In problems 4, mtch ech grph with one of the equtions A D. A. + = 4 9 B. + = 9 4 C. + = 9 D. + = 9.. 3. 4. In problems 5 4, find the vertices, the minor is endpoints, length of the mjor is, nd length of the minor is. Sketch the grph. Check using grphing utilit. 5. + = 4 5 6. + = 6 4 7. + = 4 8. + = 5 9. 5 = + 5 0. 6 + = 6. 6 + 9 = 44. 6 + 5 = 400 3. 9 + = 8 4. + 4 = In problems 5 6, write n eqution for the grph. 5. 6. In problems 7 0, find the stndrd form of the eqution for n ellipse stisfing the given conditions. 7. Center (0,0), horizontl mjor is length 64, minor is length 4 8. Center (0,0), verticl mjor is length 36, minor is length 8 9. Center (0,0), verte (0,3), b = 0. Center (0,0), verte (4,0), b = 3
Section 9. Ellipses 593 In problems 8, mtch ech grph to equtions A-H. A. ( ) ( ) + = E. ( ) + ( + ) + = 4 9 4 9 B. ( ) ( ) + = F. ( ) + ( + ) + = 4 6 4 6 C. ( ) ( ) + = G. ( ) + ( + ) + = 6 4 6 4 D. ( ) ( ) ( + ) ( + ) + = H. + = 9 4 9 4.. 3. 4. 5. 6. 7. 8. In problems 9 38, find the vertices, the minor is endpoints, length of the mjor is, nd length of the minor is. Sketch the grph. Check using grphing utilit. ( ) ( + ) ( + 5) ( 3) 9. + = 30. + = 5 4 6 36 ( 3) ( ) 3. ( + ) + = 3. + ( 6) = 5 5 33. 4 + 8 + 4 + = 6 34. + 4 + 6 + 6 = 36 35. + + 4 + 6 = 36. 4 + 6 + 8 = 4 37. 9 36 + 4 + 8 = 04 38. 4 + 8 + 9 + 36 = 4
594 Chpter 9 In problems 39 40, write n eqution for the grph. 39. 40. In problems 4 4, find the stndrd form of the eqution for n ellipse stisfing the given conditions. 4. Center (-4,3), verte(-4,8), point on the grph (0,3) 4. Center (,-), verte(-5,-), point on the grph (,0) 43. Window A window in the shpe of semiellipse is feet wide nd 4 feet high. Wht is the height of the window bove the bse 5 feet from the center? 44. Window A window in the shpe of semiellipse is 6 feet wide nd 7 feet high. Wht is the height of the window bove the bse 4 feet from the center? 45. Bridge A bridge over river is supported b semiellipticl rch. The river is 50 feet wide. At the center, the rch rises 60 feet bove the river. The rodw is 5 feet bove the center of the rch. Wht is the verticl distnce between the rodw nd the rch 45 feet from the center? 46. Bridge A bridge over river is supported b semiellipticl rch. The river is 50 feet wide. At the center, the rch rises 75 feet bove the river. The rodw is 3 feet bove the center of the rch. Wht is the verticl distnce between the rodw nd the rch 600 feet from the center? 47. Rcetrck An ellipticl rcetrck is 00 feet long nd 90 feet wide. Wht is the width of the rcetrck 0 feet from verte on the mjor is? 48. Rcetrck An ellipticl rcetrck is 50 feet long nd 50 feet wide. Wht is the width of the rcetrck 5 feet from verte on the mjor is?
Section 9. Ellipses 595 In problems 49-5, find the foci. 49. + = 9 3 50. + = 38 ( ) ( 3) 5. ( + 6) + + = 5. + ( + 5) = 6 0 In problems 53-7, find the stndrd form of the eqution for n ellipse stisfing the given conditions. 53. Mjor is vertices (±3,0), c= 54. Mjor is vertices (0,±7), c=4 55. Foci (0,±5) nd mjor is length 56. Foci (±3,0) nd mjor is length 8 57. Foci (±5,0), vertices (±7,0) 58. Foci (0,±), vertices (0,±3) 59. Foci (0,±4) nd -intercepts (±,0) 60. Foci (±3,0) nd -intercepts (0,±) 6. Center (0,0), mjor is length 8, foci on -is, psses through point (, 6) 6. Center (0,0), mjor is length, foci on -is, psses through point ( 0,4) 63. Center (-,), verte (-,5), focus (-,3) 64. Center (-,-3), verte (-7,-3), focus (-4,-3) 65. Foci (8,) nd (-,), mjor is length 66. Foci (-,5) nd (-,-3), mjor is length 4 67. Vertices (3,4) nd (3,-6), c= 68. Vertices (,) nd (-4,), c= 69. Center (,3), focus (0,3), psses through point (,5) 70. Center (-,-), focus (,-), psses through point (,-) 7. Focus (-5,-), vertices (-9,-) nd (5,-) 7. Focus (-3,), vertices (-3,4) nd (-3,-8)
596 Chpter 9 73. Whispering Gller If n ellipticl whispering gller is 80 feet long nd 5 feet wide, how fr from the center of room should someone stnd on the mjor is of the ellipse to eperience the whispering effect? Round to two deciml plces. 74. Billirds Some billirds tbles re ellipticl nd hve the foci mrked on the tble. If such one is 8 feet long nd 6 feet wide, how fr re the foci from the center of the ellipse? Round to two deciml plces. 75. Plnetr Orbits The orbits of plnets round the sun re pproimtel ellipticl with the sun s focus. The phelion is plnet s gretest distnce from the sun nd the perihelion is its shortest. The length of the mjor is is the sum of the phelion nd the perihelion. Erth s phelion is 94.5 million miles nd its perihelion is 9.40 million miles. Write n eqution for Erth s orbit. 76. Stellite Orbits The orbit of stellite round Erth is ellipticl with Erth s center s focus. The stellite s mimum height bove the Erth is 70 miles nd its minimum height bove the Erth is 90 miles. Write n eqution for the stellite s orbit. Assume Erth is sphericl nd hs rdius of 3960 miles. 77. Eccentricit e of n ellipse is the rtio c where c is the distnce of focus from the center nd is the distnce of verte from the center. Write n eqution for n ellipse with eccentricit 0.8 nd foci t (-4,0) nd (4,0). 78. Confocl ellipses hve the sme foci. Show tht, for k > 0, ll ellipses of the form + = re confocl. 6 + k k 79. The ltus rectum of n ellipse is line segment with endpoints on the ellipse tht b psses through focus nd is perpendiculr to the mjor is. Show tht is the length of the ltus rectum of + = where > b. b
Section 9. Hperbols 597 Section 9. Hperbols In the lst section, we lerned tht plnets hve pproimtel ellipticl orbits round the sun. When n object like comet is moving quickl, it is ble to escpe the grvittionl pull of the sun nd follows pth with the shpe of hperbol. Hperbols re curves tht cn help us find the loction of ship, describe the shpe of cooling towers, or clibrte seismologicl equipment. The hperbol is nother tpe of conic section creted b intersecting plne with double cone, s shown below 5. The word hperbol derives from Greek word mening ecess. The English word hperbole mens eggertion. We cn think of hperbol 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 hperbol is the set of ll points where the bsolute vlue of the difference of the distnces from the point to two fied points is constnt. 5 Pbroks3 (https://commons.wikimedi.org/wiki/file:conic_sections_with_plne.svg), Conic sections with plne, cropped to show onl hperbol b L Michels, CC BY 3.0
598 Chpter 9 Hperbol Definition A hperbol 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 bsolute 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 hperbol which intersect the trnsverse is. The trnsverse is length is the length of the line segment between the vertices. The center is the midpoint between the vertices (or the midpoint between the foci). The other is of smmetr through the center is the conjugte is. The two disjoint pieces of the curve re clled brnches. A hperbol hs two smptotes. Which is is the trnsverse is will depend on the orienttion of the hperbol. As helpful tool for grphing hperbols, 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. Hperbols 599 Hperbols Centered t the Origin From the definition bove we cn find n eqution of hperbol. We will find it for hperbol 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 hperbol. The distnces from Q to F nd Q to F re: d + ( Q, F ) ( ( c) ) + ( 0) = ( + c) d = +. From the definition, the bsolute vlue of the difference should be constnt: ( Q, F ) d( Q, F ) = ( c) + ( + c) + k d = Substituting 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 bsolute vlues, ( c) + ( + c) + = ± Move one rdicl ( c) + = ± + ( + c) Squre both sides + ( c) + = 4 ± 4 ( + c) + + ( + c) + Epnd ( + c) + + + c + c c + c + = 4 ± 4 + Combining like terms leves ( + c) 4c = 4 ± 4 + Divide b 4 ( + c) c = ± + Isolte the rdicl ± ( + c) + = c 4 ( c) + = + c ( ) c + Squre both sides gin + Epnd nd distribute c 4 + c + c + = + c + Combine like terms 4 + c = c Fctor common terms ( c ) = ( c ) +
600 Chpter 9 Let b = c. Since c >, b > 0. Substituting b for c leves + b = b Divide both sides b b + = Rewrite = b b We cn see from the grphs of the hperbols tht the brnches pper to pproch smptotes s gets lrge in the negtive or positive direction. The equtions of the horizontl hperbol smptotes cn be derived from its stndrd eqution. b = Solve for = b Rewrite s = b Fctor out = b Tke the squre root = ± b As ± the quntit b 0 nd, so the smptotes re = ±. Similrl, for verticl hperbols the smptotes re = ±. b 0 depends on whether it opens horizontll or verticll. The following tble gives the stndrd eqution, vertices, foci, smptotes, construction rectngle vertices, nd grph for ech. The stndrd form of n eqution of hperbol centered t the origin C (,0)
Section 9. Hperbols 60 Eqution of Hperbol Centered t the Origin in Stndrd Form Opens Horizontll Verticll Stndrd Eqution b = Vertices (-, 0) nd (, 0) (0, -) nd (0, ) b = Foci (-c, 0) nd (c, 0) where b = c (0, -c) nd (0, c) Where b = c Asmptotes Construction Rectngle Vertices Grph b = ± = ± b (, b), (-, b), (,-b), (-, -b) (b, ), (-b, ), (b, -), (-b, -) (0,c) (0,b) (0,) (-c,0) (-,0) (,0) (c,0) (-b,0) (b,0) (0,-) (0,-b) (0,-c) Emple Put the eqution of the hperbol 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 be put in stndrd form = 4 b dividing b 4. Compring to the generl stndrd eqution = we see tht = 4 = nd b b = =. Since the term is subtrcted, the hperbol 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 = ±. b To sketch the grph we plot the vertices of the construction rectngle t (±b,±) or (-,-), (-,), (,-), nd (,). The smptotes re drwn through the digonls of the rectngle nd the vertices plotted. Then we sketch in the hperbol, 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 both sides we find + = ±. Under Y= enter the two hlves of the hperbol nd the two smptotes s = +, = +, =, nd =. Set the window to comprble 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. Hperbols 603 Emple Find the stndrd form of the eqution for hperbol 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 hperbol is horizontl with n eqution of the form =. The vlue of is the distnce b from the center to verte. The distnce from (6,0) to (0,0) is 6, so = 6. b 3 The smptotes follow the form = ±. From = we see 4 3 b substituting = 6 give us =. Solving ields b = 8. 4 6 3 b = 4 nd The eqution of the hperbol in stndrd form is = 6 8 or = 36 64. Tr it Now. Find the stndrd form of the eqution for hperbol with vertices t (0,-8) nd (0,8) nd smptote = Emple 3 Find the stndrd form of the eqution for hperbol 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 hperbol is verticl with n eqution of the form =. The vlue of is the distnce from b the center to verte. The distnce from (0,9) to (0,0) is 9, so = 9. 5 8 Substituting = 9 nd the point (8,5) gives =. Solving for b eilds 9 b ( 8 ) = 6 9 b =. 5 9 The stndrd eqution for the hperbol is = 9 6 or = 8 36.
604 Chpter 9 Hperbols Not Centered t the Origin Not ll hperbols re centered t the origin. The stndrd eqution for one centered t (h, k) is slightl different. Eqution of Hperbol Centered t (h, k) in Stndrd Form h, depends on whether it opens horizontll or verticll. The tble below gives the stndrd eqution, vertices, foci, smptotes, construction rectngle vertices, nd grph for ech. The stndrd form of n eqution of hperbol centered t C ( k) Opens Horizontll Verticll Stndrd Eqution ( h) ( k) ( k ) ( h) b = b = Vertices ( h ±, k ) (h, k ± ) Foci ( h ± c, k ) where b = c b (h, k ± c) where b = c Asmptotes k = ± ( h) k = ± ( h) Construction Rectngle Vertices Grph (h-c,k) (h-,k) ( h ±, k ± b ) ( h ± b, k ± ) (h,k) (h,k+b) (h,k-b) (h+c,k) (h+,k) (h-b,k) b (h,k+c) (h,k+) (h,k) (h,k-) (h,k-c) (h+b,k)
Section 9. Hperbols 605 Emple 4 Write n eqution for the hperbol in the grph shown. The center is t (,3), where the smptotes cross. It opens verticll, so the eqution will look like ( 3) ( ) =. b 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 b = 5 = 3. 3 3 The stndrd eqution of this hperbol is ( ) ( ) = ( 3) ( ) =. 9, or Emple 5 Put the eqution of the hperbol 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 both vribles to get 9 + + 4 4 + 4 = 43+ 9 9 ( ) ( ) 6 ( + ) 4( ) = 36 + 4 9 Dividing b 36 gives the stndrd form of the eqution, ( ) ( ) = h Compring to the generl stndrd eqution ( ) ( ) = = 4 = nd b = 9 = 3. h k b we see tht Since the term is subtrcted, the hperbol 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. b 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±b) 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 hperbol 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 hperbol nd the two smptotes s ( + ) ( ) + 3 = + 9, = 9 4, = ( + ) +, nd 4 3 = ( + ) +. Set the window to comprble 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 hperbol with vertices t (, 5) nd 3 (,7), nd smptote = + 4.
Section 9. Hperbols 607 Since the vertices differ in the -coordintes, the hperbol opens verticll with n eqution of the form ( ) ( ) k h = nd smptote equtions of the form b k = ± ( h). b 5 + 7 The center will be hlfw between 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) b, notice this eqution would hve slope b. We cn compre tht to the slope of the given smptote eqution to find b. Setting 3 = b nd substituting = 6 gives us b = 4. + 6 4 The eqution of the hperbol 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 hperbol ( ) ( ) + 5 =. 9 36 Hperbol Foci The loction of the foci cn pl ke role in hperbol ppliction problems. To find them, we need to find the length from the center to the foci, c, using the eqution b = c. It looks similr to, but is not the sme s, the Pthgoren Theorem. Compre this with the eqution to find length c for ellipses, which is b = c. If ou remember tht for the foci to be inside the ellipse the hve to come before the vertices ( c < ), it s cler wh we would clculte minus c. To be inside hperbol, the foci hve to go beond the vertices ( c > ), so we cn see for hperbols we need minus, the opposite. c
608 Chpter 9 Emple 7 + 3 4 5 Find the foci of the hperbol ( ) ( ) = k The center is t (h, k) = (3, -). The foci re t (h, k ± c). The hperbol is verticl with n eqution of the form ( ) ( ) =. h b. To find length c we use b = c. Substituting gives 5 4 = c or c = 9 = 3. The hperbol hs foci (3, -4) nd (3, ). Emple 8 Find the stndrd form of the eqution for hperbol with foci (5, -8) nd (-3, -8) nd vertices (4, -8) nd (-, -8). Since the vertices differ in the -coordintes, the hperbol opens horizontll with n eqution of the form ( ) ( ) h k =. b 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 b we use b = c. Substituting gives b = 6 9 = 7. The eqution of the hperbol in stndrd form is ( ) ( ( 8 )) = ( ) ( + 8) 9 7 =.. 3 7 or Tr it Now 3. Find the stndrd form of the eqution for hperbol with focus (,9), verte (,8), center (,4).
Section 9. Hperbols 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 hperbol 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 hperbol with C nd D s the foci. The point P where the two hperbols 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 before the signl from A. If the signl trvels 300,000 kilometers per second, find the eqution of the hperbol 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 between them, giving c = (50) = 75 km. B letting the center of the hperbol be t (0,0) nd plcing the foci t (±75,0), the eqution = for hperbol centered t the origin cn be used. b 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 hperbol eqution we s determined k =, so = (90) = 45. Substituting nd c into b = c ields = 75 45 = 3600 b. The eqution of the hperbol 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 hperbol 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 Hperbol Definition Hperbol Equtions in Stndrd Form Hperbol Foci Applictions of Hperbols Intersections of Hperbols nd Other Curves Tr it Now Answers. The vertices re on the is so this is verticl hperbol. The center is t the origin. = 8 8 Using the smptote slope, =, so b = 4. b = 64 6. Center (-5, ). This is horizontl hperbol. = 3. b = 6. trnsverse is length 6, Vertices will be t (-5±3,) = (-,) nd (-8,), Asmptote slope will be 3 6 =. Asmptotes: = ± ( + 5 ) 3. Focus, verte, nd center hve the sme vlue so this is verticl hperbol. Using the verte nd center, = 9 4 = 5 Using the focus nd center, c = 8 4 = 4 b = 5 4. b = 3. ( 4) ( ) 6 9 =
Section 9. Hperbols 6 Section 9. Eercises In problems 4, mtch ech grph to equtions A D. A. = 4 9 B. = 9 4 C. = 9 D. = 9.. 3. 4. In problems 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 problems 5 6, write n eqution for the grph. 5. 6.
6 Chpter 9 In problems 7, find the stndrd form of the eqution for hperbol 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 problems 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. Hperbols 63 In problems 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 problems 4 4, write n eqution for the grph. 4. 4. In problems 43 44, find the stndrd form of the eqution for hperbol stisfing the given conditions. 43. Vertices (-,-) nd (-,6); smptote = ( + ) 44. Vertices (-3,-3) nd (5,-3); smptote + 3 = ( ) In problems 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 problems 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 problems 55 66, find the stndrd form of the eqution for hperbol 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. Hperbols 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 before the signl from B. If the signl trvels 300,000 kilometers per second, find n eqution of the hperbol 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 bolt of light strike the ground. Anit hers the thunder 0.5 seconds before Smir does. Sound trvels t 00 feet per second. Find n eqution of the hperbol 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 hperbol. 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. Clibrtion A seismologist positions two recording devices 340 feet prt t points A nd B. To check the clibrtion, n eplosive is detonted between 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 be detonted est of point A. How fr est should the second eplosion be 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 hperbol on which the person is stnding if the foci re locted t A nd B. A fired bullet 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 hperbolic 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 hperbol = is =. Show tht b b 5 + 5 = 0 is the conjugte of 5 + 5 = 0. 74. The eccentricit e of hperbol 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 hperbol is one for which = b. Find the eccentricit of n equilterl hperbol. 76. The ltus rectum of hperbol is line segment with endpoints on the hperbol tht psses through focus nd is perpendiculr to the trnsverse is. Show tht b is the length of the ltus rectum of =. b 77. Confocl hperbols hve the sme foci. Show tht, for 0 < k < 6, ll hperbols of the form = re confocl. k 6 k
Section 9.3 Prbols nd Non-Liner Sstems 67 Section 9.3 Prbols nd Non-Liner Sstems To listen for signls from spce, rdio telescope uses dish in the shpe of prbol to focus nd collect the signls in the receiver. While we studied prbols erlier when we eplored qudrtics, t the time we didn t discuss them s conic section. A prbol is the shpe resulting from when plne prllel to the side of the cone intersects the cone 6. Prbol Definition nd Vocbulr A prbol with verte t the origin cn be defined b plcing fied point t F ( 0, p) clled the focus, nd drwing line t = p, clled the directri. The prbol is the set of ll points Q (, ) tht re n equl distnce between the fied point nd the directri. (0,p) Q Focus Ais of smmetr =-p Verte Directri For generl prbols, The is of smmetr is the line pssing through the foci, perpendiculr to the directri. The verte is the point where the prbol crosses the is of smmetr. The distnce from the verte to the focus, p, is the focl length. 6 Pbroks3 (https://commons.wikimedi.org/wiki/file:conic_sections_with_plne.svg), Conic sections with plne, cropped to show onl prbol, CC BY 3.0
68 Chpter 9 Equtions for Prbols with Verte t the Origin From the definition bove we cn find n eqution of prbol. We will find it for prbol with verte t the origin, C ( 0,0), opening upwrd with focus t F ( 0, p) nd directri t = p. Suppose ( ) d( Q, F ) = ( 0) + ( p) = + ( p) Q, is some point on the prbol. The distnce from Q to the focus is The distnce from the point Q to the directri is the difference of the -vlues: d = ( p) = + p From the definition of the prbol, these distnces should be equl: + ( p) = p ( p) = ( ) + + + p Epnd + Squre both sides + p + p = + p p Combine like terms = 4 p This is the stndrd conic form of prbol tht opens up or down (verticl is of smmetr), centered t the origin. Note tht if we divided b 4p, we would get more fmilir eqution for the prbol, =. We cn recognize this s trnsformtion of 4 p the prbol =, verticll compressed or stretched b. 4 p Using similr process, we could find n eqution of prbol with verte t the origin opening left or right. The focus will be t (p,0) nd the grph will hve horizontl is of smmetr nd verticl directri. The stndrd conic form of its eqution will be = 4 p, which we could lso write s =. 4 p Emple Write the stndrd conic eqution for prbol with verte t the origin nd focus t (0, -). With focus t (0, -), the is of smmetr is verticl, so the stndrd conic eqution is = 4 p. Since the focus is (0, -), p = -. The stndrd conic eqution for the prbol is = 8 = 4( ), or
Section 9.3 Prbols nd Non-Liner Sstems 69 For prbols with verte not t the origin, we cn shift these equtions, leding to the equtions summrized net. Eqution of Prbol with Verte t (h, k) in Stndrd Conic Form The stndrd conic form of n eqution of prbol with verte t the point ( h, k) depends on whether the is of smmetr is horizontl or verticl. The tble below gives the stndrd eqution, verte, is of smmetr, directri, focus, nd grph for ech. Stndrd Eqution Horizontl Verticl ( k) = 4 p( h) ( ) h = 4 p ( k ) Verte (h, k) (h, k) Ais of smmetr = k = h Directri = h - p = k - p Focus (h + p, k) (h, k + p) An emple with p < 0 An emple with p > 0 =h-p =h Grph =k (h,k) (h,k+p) (h+p,k) (h,k) =k-p Since ou lred studied qudrtics in some depth erlier, we will primril eplore the new concepts ssocited with prbols, prticulrl the focus.
60 Chpter 9 Emple Put the eqution of the prbol = 8( ) + in stndrd conic form. Find the verte, focus, nd is of smmetr. From our erlier work with qudrtics, ou m lred be ble to identif the verte s (,), but we ll go hed nd put the prbol in the stndrd conic form. To do so, we need to isolte the squred fctor. = 8( ) + Subtrct from both sides = 8( ) Divide b 8 ( ) 8 = ( ) This mtches the generl form for verticl prbol, ( h) = 4 p( k) 4 p =. Solving this tells us 8 ( ) = 4 ( ). 3 The verte is t (,). The is of smmetr is t =. 63 The directri is t = =. 3 3 65 The focus is t, + =,. 3 3, where p =. The stndrd conic form of the eqution is 3 Emple 3 A prbol hs its verte t (,5) nd focus t (3,5). Find n eqution for the prbol. Since the verte nd focus lie on the line = 5, tht is our is of smmetr. The verte (,5) tells us h = nd k = 5. Looking t the distnce from the verte to the focus, p = 3 =. Substituting these vlues into the stndrd conic form of n eqution for horizontl prbol gives the eqution 5 = 4() ( ) ( ) ( 5) = 8( ) Note this could lso be rewritten b solving for, resulting in = ( 5) 8 +
Section 9.3 Prbols nd Non-Liner Sstems 6 Tr it Now. A prbol hs its verte t (-,3) nd focus t (-,). Find n eqution for this prbol. Applictions of Prbols In n erlier section, we lerned tht ellipses hve specil propert tht r eminting from one focus will be reflected bck to the other focus, the propert tht enbles the whispering chmber to work. Prbols lso hve specil propert, tht n r eminting from the focus will be reflected prllel to the is of smmetr. Reflectors in flshlights tke dvntge of this propert to focus the light from the bulb into collimted bem. The sme propert cn be used in reverse, tking prllel rs of sunlight or rdio signls nd directing them ll to the focus. Emple 4 A solr cooker is prbolic dish tht reflects the sun s rs to centrl point llowing ou to cook food. If solr cooker hs prbolic dish 6 inches in dimeter nd 4 inches tll, where should the food be plced? We need to determine the loction of the focus, since tht s where the food should be plced. Positioning the bse of the dish t the origin, the shpe from the side looks like: The stndrd conic form of n eqution for the prbol would be = 4 p. The prbol psses through (4, 8), so substituting tht into the eqution, we cn solve for p: 8 = 4( p)(4) 8 p = = 6 4 The focus is 4 inches bove the verte. This mkes for ver convenient design, since then grte could be plced on top of the dish to hold the food. 8 4 Tr it Now. A rdio telescope is 00 meters in dimeter nd 0 meters deep. Where should the receiver be plced?
6 Chpter 9 Non-Liner Sstems of Equtions In mn pplictions, it is necessr to solve for the intersection of two curves. Mn of the techniques ou m hve used before to solve sstems of liner equtions will work for non-liner equtions s well, prticulrl substitution. You hve lred solved some emples of non-liner sstems when ou found the intersection of prbol nd line while studing qudrtics, nd when ou found the intersection of circle nd line while studing circles. Emple 4 Find the points where the ellipse + = 4 5 intersects the circle + = 9. To strt, we might multipl the ellipse eqution b 00 on both sides to cler the frctions, giving 5 + 4 = 00. A common pproch for finding intersections is substitution. With these equtions, rther thn solving for or, it might be esier to solve for or. Solving the circle eqution for gives into the ellipse eqution. 5 + 4 = 00 Substitute ( ) + 4 = 00 5 9 = 9. We cn then substitute tht epression for Distribute = 9 5 5 + 4 = 00 Combine like terms = 5 Divide b - = 5 Use the squre root to solve 5 5 5 = ± = ± We cn substitute ech of these vlues bck in to = 9 = ± 64 5 = ± 8 5 89 5 = 9 = = There re four points of intersection: ± 64 8 5 5, ±. = 9 to find
Section 9.3 Prbols nd Non-Liner Sstems 63 It s worth noting there is second technique we could hve used in the previous emple, clled elimintion. If we multiplied the circle eqution b -4 to get 4 4 = 36, we cn then dd it to the ellipse eqution, eliminting the vrible. 5 + 4 = 00 4 4 = 36 Add the left sides, nd dd the right sides = 64 Solve for = ± 64 = ± 8 Emple 5 Find the points where the hperbol = 4 9 intersects the prbol =. We cn solve this sstem of equtions b substituting = into the hperbol eqution. ( ) = Simplif 4 9 4 4 = 4 9 Simplif, nd multipl b 9 4 9 = 9 Move the 9 to the left 4 9 9 = 0 While this looks chllenging to solve, we cn think of it s qudrtic in disguise, 4 since = ( ). Letting u =, the eqution becomes 9u u 9 = 0 Solve using the qudrtic formul ( ) ± ( ) 4(9)( 9) ± 35 u = = Solve for (9) 8 ± 35 = But 35 < 0, so 8 + 35 = ± This leds to two rel solutions 8.08, -.08 Substituting these into =, we cn find the corresponding vlues. The curves intersect t the points (.08,.4) nd (-.08,.4).
64 Chpter 9 Tr it Now 3. Find the points where the line = 4 intersect the ellipse = 4 6 Solving for the intersection of two hperbols llows us to utilize the LORAN nvigtion pproch described in the lst section. In our emple, sttions A nd B re 50 kilometers prt nd send simultneous rdio signl to the ship. The signl from B rrives 0.0003 seconds before the signl from A. We found the eqution of the hperbol in stndrd form would be = 05 3600 A C D B P Emple 0 Continuing the sitution from the lst section, suppose sttions C nd D re locted 00 km due south of sttions A nd B nd 00 km prt. The signl from D rrives 0.000 seconds before the signl from C, leding to the eqution ( 00) + =. Find 5 75 the position of the ship. To solve for the position of the bot, we need to find where the hperbols intersect. This mens solving the sstem of equtions. To do this, we could strt b solving both equtions for. With the first eqution from the previous emple, = Move the term to the right 05 3600 = + Multipl both sides b 05 05 3600 05 = 05 + Simplif 3600 9 = 05 + 6 With the second eqution, we repet the sme process ( 00) + = 5 75 Move the term to the right nd multipl b 5 5( + 00) = 5 + 75 Simplif
Section 9.3 Prbols nd Non-Liner Sstems 65 9( + 00) = 5 + 9 Now set these two epressions for equl to ech other nd solve. 9 9( + 00) 05 + = 5 + 6 9 Subtrct 5 from both sides 9 9( + 00) 800 + = 6 9 Divide b 9 ( + 00) 00 + = 6 9 Multipl both sides b 6 9= 456 900 + 9 = 6( + 00) Epnd nd distribute 900+ 9 = 6 + 6400 + 640000 Combine like terms on one side 75 6400 348800 = 0 Solve using the qudrtic formul ( 6400) ± ( 6400) 4(75)( 348800) = 3. km or -37.78 km (75) We cn find the ssocited vlues b substituting these -vlues into either hperbol eqution. When 3., 9(3.) 05 + 6 ±0.7 When -37.78km, 9( 37.78) 05 + 6 ±53.8 This provides 4 possible loctions for the ship. Two cn be immeditel discrded, s the re on lnd. Nvigtors would use other nvigtionl techniques to decide between the two remining loctions.