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# Chapter 10 Analytic Geometry

Chapter 10 Analytic

Section 10.1 13. (e); the graph has ()()hk,1,1= and opens to the right. Therefore, the of the graph Not applicable has the form ()yax−=142 () − 1. Section 10.2 14. (d); the graph has vertex ()()hk,0,0= and ()()−+−22 1. xx21 yy 21 opens down. Therefore, the equation of the graph has the form xay2 =−4 . The graph passes 2 −4 through the ()−−2, 1 so we have 2. = 4 2 ()−=−−2412 a () 2 = 3. ()x +=49 44a = x +=±43 a 1 2 xx+=43 or +=− 4 3 Thus, the equation of the graph is xy=−4 . xx=−1 or =− 7 15. (h); the graph has vertex ()(hk,1,1=− − ) and The solution set is {7,1}−−. opens down. Therefore, the equation of the graph 2 4. (2,5)−− has the form ()xay+=−+141 ().

5. 3, up 16. (a); the graph has vertex ()()hk,0,0= and 6. parabola opens to the right. Therefore, the equation of the graph has the form yax2 = 4 . The graph passes 7. c through the point ()1, 2 so we have 8. (3, 2) ()2412 = a () 44= a 9. (3,6) 1 = a 2 10. y =−2 Thus, the equation of the graph is yx= 4 .

11. (b); the graph has a vertex ()()hk,0,0= and 17. (c); the graph has vertex ()()hk,0,0= and opens up. Therefore, the equation of the graph opens to the left. Therefore, the equation of the 2 has the form xay2 = 4 . The graph passes graph has the form yax=−4 . The graph passes through the point ()2,1 so we have through the point ()−−1, 2 so we have 2 ()2412 = a () ()−=−−241a () 44= a 44= a 1 = a 1 = a 2 Thus, the equation of the graph is xy2 = 4 . Thus, the equation of the graph is yx=−4 .

12. (g); the graph has vertex ()()hk,1,1= and opens 18. (f); the graph has vertex ()(hk,1,1=− − ) and to the left. Therefore, the equation of the graph opens up; further, a = 1 . Therefore, the equation 2 has the form ()yax−=−−1412 (). of the graph has the form ()xy+=141 () +.

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19. The is (4, 0) and the vertex is (0, 0). Both 21. The focus is (0, –3) and the vertex is (0, 0). Both lie on the horizontal y = 0 . a = 4 and since lie on the vertical line x = 0 . a = 3 and since (4, 0) is to the right of (0, 0), the parabola opens (0, –3) is below (0, 0), the parabola opens down. to the right. The equation of the parabola is: The equation of the parabola is: 2 yax2 = 4 xay=−4 2 yx2 =⋅⋅44 xy=−43 ⋅ ⋅ 2 yx2 = 16 xy=−12 2 Letting xyy===±4, we find 2 64 or 8 . Letting yxx=−3, we find = 36 or =± 6 . The points (4, 8) and (4, –8) define the latus The points ()−−6, 3 and ()6,− 3 define the latus rectum. rectum.

20. The focus is (0, 2) and the vertex is (0, 0). Both 22. The focus is (–4, 0) and the vertex is (0, 0). Both lie on the vertical line x = 0 . a = 2 and since lie on the horizontal line y = 0 . a = 4 and since (0, 2) is above (0, 0), the parabola opens up. The (–4, 0) is to the left of (0, 0), the parabola opens equation of the parabola is: to the left. The equation of the parabola is: 2 xay= 4 yax2 =−4 2 xy=⋅⋅42 yx2 =−44 ⋅ ⋅ 2 xy= 8 yx2 =−16 2 Letting yxx===±2, we find 16 or 4 . Letting xyy=−4, we find 2 = 64 or =± 8 . The The points (–4, 2) and (4, 2) define the latus points (–4, 8) and (–4, –8) define the latus rectum. rectum.

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23. The focus is (–2, 0) and the directrix is x = 2 . equation of the parabola is: The vertex is (0, 0). a = 2 and since (–2, 0) is to xay2 = 4 the left of (0, 0), the parabola opens to the left. 1 The equation of the parabola is: xy2 =⋅⋅4 2 yax2 =−4 xy2 = 2 2 =− ⋅ ⋅ yx42 1 Letting yxx===±, we find 2 1 or 1. yx2 =−8 2 2 1 1 Letting xyy===±– 2, we find 16 or 4 . The The points 1, and −1, define the latus 2 2 points (–2, 4) and (–2, –4) define the latus   rectum. rectum.

24. The focus is (0, –1) and the directrix is y = 1. 1 The vertex is (0, 0). a = 1 and since (0, –1) is 26. The directrix is x =− and the vertex is (0, 0). 2 below (0, 0), the parabola opens down. The 1 1 1 equation of the parabola is: The focus is ,0 . a = and since ,0 is 2 2 2 xay2 =−4   to the right of (0, 0), the parabola opens to the 2 =− ⋅ ⋅ xy41 right. The equation of the parabola is: xy2 =−4 yax2 = 4 ===±2 1 Letting yxx–1, we find 4 or 2 . The yx2 =⋅⋅4 points (–2, –1) and (2, –1) define the latus 2 rectum. yx2 = 2 1 Letting xyy===±, we find 2 1 or 1 . The 2 1 1 points ,1− and ,1 define the latus 2 2 rectum.

1 25. The directrix is y =− and the vertex is (0, 0). 2 1 1 1 The focus is 0, . a = and since 0, is 2 2 2 above (0, 0), the parabola opens up. The

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27. Vertex: (0, 0). Since the axis of symmetry is vertical, the parabola opens up or down. Since (2, 3) is above (0, 0), the parabola opens up. The equation has the form xay2 = 4 . Substitute the coordinates of (2, 3) into the equation to find a : 2432 =⋅a 412= a 1 a = 3 4 The equation of the parabola is: xy2 = . The 3 1 1 focus is 0, . Letting y = , we find 29. The vertex is (2, –3) and the focus is (2, –5). 3 3 Both lie on the vertical line x = 2 . 2 42 21 =− −−() = xx==± or . The points , and a 532 and since (2, –5) is below 93 33 (2, –3), the parabola opens down. The equation 21 − , define the latus rectum. of the parabola is: 33 ()xh−=−−2 4 ayk () ()xy−=−−−24232 ()()() ()xy−=−+2832 () Letting y =−5 , we find ()x −=2162

xxx−=±24 =− 2 or =6 The points (–2, –5) and (6, –5) define the latus rectum.

28. Vertex: (0, 0). Since the axis of symmetry is horizontal, the parabola opens left or right. Since (2, 3) is to the right of (0, 0), the parabola opens to the right. The equation has the form yax2 = 4 . Substitute the coordinates of (2, 3) into the equation to find a : 3422 =⋅a 98= a 9 a = 30. The vertex is (4, –2) and the focus is (6, –2). 8 Both lie on the horizontal line y =−2 . 2 9 The equation of the parabola is: yx= . The a =−= 2 46 2 and since (6, –2) is to the right of 9 9 (4, –2), the parabola opens to the right. The focus is ,0 . Letting x = , we find 8 8 equation of the parabola is: ()yk−=2 4 axh () − 2 81 9 99 yy==± or . The points , and  2 16 4 84 ()yx−−()2424 =()( − ) 99 , − define the latus rectum. +=−2 84 ()()yx284

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Letting x = 6 , we find 32. The vertex is (3, 0) and the focus is (3, –2). Both 2 = =− − = ()y +=216 lie on the horizontal line x 3 . a 20 2

yyy+=±24 =− 6 or =2 and since (3, –2) is below of (3, 0), the parabola opens down. The equation of the parabola is: The points (6, –6) and (6, 2) define the latus 2 rectum. ()xh−=−−4 ayk () ()xy−=−−34202 ()() ()xy−=−382 Letting y =−2 , we find ()x −=3162

xxx−=±34 =− 1 or =7 The points (–1, –2) and (7, –2) define the latus rectum.

31. The vertex is (–1, –2) and the focus is (0, –2). Both lie on the horizontal line y =−2 . a =−10 − = 1 and since (0, –2) is to the right of (–1, –2), the parabola opens to the right. The equation of the parabola is: ()yk−=2 4 axh () − 2 ()yx−−()2411 =()() −− ( )

()()yx+=2412 + 33. The directrix is y = 2 and the focus is (–3, 4). = Letting x 0 , we find This is a vertical case, so the vertex is (–3, 3). ()y +=242 a = 1 and since (–3, 4) is above y = 2 , the

yyy+=±22 =− 4 or =0 parabola opens up. The equation of the parabola −=2 − The points (0, –4) and (0, 0) define the latus is: ()4()xh ayk rectum. (xy−− ( 3))2 = 4 ⋅⋅ 1 ( − 3)

(3)4(3)xy+=2 − Letting y = 4 , we find (3)4x +=2 or x +=±32. So, xx=−1 or =− 5 . The points (–1, 4) and (–5, 4) define the latus rectum.

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34. The directrix is x =−4 and the focus is (2, 4). 36. The directrix is y =−2 and the focus is (–4, 4). This is a horizontal case, so the vertex is (–1, 4). This is a vertical case, so the vertex is (–4, 1). a = 3 and since (2, 4) is to the right of x =−4 , a = 3 and since (–4, 4) is above y =−2 , the the parabola opens to the right. The equation of parabola opens up. The equation of the parabola the parabola is: ()4()yk−=2 axh − is: ()4()xh−=2 ayk − (4)43((1))yx−=⋅⋅−−2 ((4))43(1)xy−−2 = ⋅ ⋅ − (yx−= 4)2 12( + 1) +=2 − Letting x = 2 , we find (4)36y −=2 or (4)12(1)xy 2 y −=±46. So, yy=−2 or = 10 . The points Letting y = 4 , we find (4)36x += or (2, –2) and (2, 10) define the latus rectum. x +=±46. So, xx=−10 or = 2 . The points (–10, 4) and (2, 4) define the latus rectum.

35. The directrix is x = 1 and the focus is (–3, –2). This is a horizontal case, so the vertex is (–1, –2). a = 2 and since (–3, –2) is to the left 37. The equation xy2 = 4 is in the form xay2 = 4 of x = 1 , the parabola opens to the left. The where 44aa== or 1. equation of the parabola is: Thus, we have: ()yk−=−−2 4() axh Vertex: (0, 0) Focus: (0, 1) ((2))42((1))yx−−2 =−⋅ ⋅ −− Directrix: y =−1 (2)8(1)yx+=−+2 Letting x =−3 , we find (2)16y +=2 or y+2=± 4 . So, yy==−2 or 6 . The points (–3, 2) and (–3, –6) define the latus rectum.

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38. The equation yx2 = 8 is in the form yax2 = 4 41. The equation (2)8(1)yx−=+2 is in the form where 48aa== or 2. Thus, we have: ()4()yk−=2 axh −where 48aa== or 2, Vertex: (0, 0) hk=−1, and = 2 . Thus, we have: Focus: (2, 0) Directrix: x =−2 Vertex: (–1, 2); Focus: (1, 2); Directrix: x =−3

39. The equation yx22=−16 is in the form yax =− 4 where 42. The equation (xy+= 4)2 16( + 2) is in the form −=−416aa or =4. Thus, we have: 2 Vertex: (0, 0) ()4()xh−= ayk −where 416aa== or 4, Focus: (–4, 0) hk=−4, and =− 2 . Thus, we have: = Directrix: x 4 Vertex: (–4, –2); Focus: (–4, 2) Directrix: y =−6

2 =− 2 =− 40. The equation xy4 is in the form xay4 where −=−44aa or =1. Thus, we have: 2 Vertex: (0, 0) 43. a. The equation (3)(1)xy−=−+ is in the Focus: (0, –1) form ()xh−=−−2 4() ayk where Directrix: y = 1 1 –4aa=− 1 or = , hk==−3, and 1 . Thus, 4 we have: Vertex: (3, –1); Focus: ()3, − 5 ; 4 Directrix: y =−3 4

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46. The equation (2)4(3)xy−=−2 is in the form ()4()xh−=2 ayk − where 44aa== or 1, hk==2, and 3 . Thus, we have: Vertex: (2, 3); Focus: (2, 4); Directrix: y = 2

44. The equation (1)4(2)yx+=−−2 is in the form ()yk−=−−2 4() axh where –4aa=− 4 or = 1, hk==−2, and 1 . Thus, we have: Vertex: (2, –1);

Focus: ()1,− 1 47. Complete the to put in standard form: Directrix: x = 3 yyx2 −++=4440 yy2 −+=−444 x ()yx−=−242 The equation is in the form ()yk−=−−2 4() axh where −=−44aa or =1, hk==0, and 2 . Thus, we have: Vertex: (0, 2); Focus: (–1, 2); Directrix: x = 1

45. The equation (3)8(2)yx+=−2 is in the form ()4()yk−=2 axh − where 48aa== or 2, hk==−2, and 3 . Thus, we have: Vertex: (2, –3); Focus: (4, –3) Directrix: x = 0

48. Complete the square to put in standard form: xxy2 +−+=6410 xx2 ++=−+69419 y ()()xy+=3422 + The equation is in the form ()4()xh−=2 ayk − where 44aa== or 1, hk=−3, and =− 2 . Thus, we have: Vertex: (–3, –2); Focus: (–3, –1) Directrix: y =−3

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49. Complete the square to put in standard form: 51. Complete the square to put in standard form: xxy2 +=−848 yyx2 +−=20 xx2 ++=8164816 y −+ yy2 ++=+21 x 1 (4)4(2)xy+=+2 ()yx+=+112 The equation is in the form ()4()xh−=2 ayk − The equation is in the form ()4()yk−=2 axh − where 44aa== or 1, hk=−4, and =− 2 . 1 where 41aa== or , hk=−1, and =− 1 . Thus, we have: 4 Vertex: (–4, –2); Thus, we have: Focus: (–4, –1) 3 =− Vertex: (–1, –1); Focus: − ,–1 Directrix: y 3 4 5 Directrix: x =− 4

50. Complete the square to put in standard form:

yyx2 −=−281 yy2 −+=−+21811 x 52. Complete the square to put in standard form: xxy2 −=42 ()yx−=182 xx2 −+=+4424 y The equation is in the form ()4()yk−=2 axh − ()()xy−=2222 + where 48aa== or 2, hk==0, and 1. Thus, we have: The equation is in the form ()4()xh−=2 ayk − Vertex: (0, 1); 1 where 42aa== or , hk==−2, and 2 . Focus: (2, 1) 2 =− Directrix: x 2 Thus, we have: 3 Vertex: (2, –2); Focus: 2, − 2 5 Directrix: y =− 2

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53. Complete the square to put in standard form: 55. (1)(0)ycx−=2 − xxy2 −=+44 (1)ycx−=2 xx2 −+=++44 y 44 (2−= 1)2 cc (1) 1 = (2)xy−=+2 8 (1)yx−=2 The equation is in the form ()4()xh−=2 ayk − 1 −=2 − where 41aa== or , hk==−2, and 8 . Thus, 56. (1)(xcy 2) 4 −=−2 we have: (2 1)c (1 2) =− =− 31 11cc Vertex: (2, –8); Focus: 2, − 4 (1)(xy−=−−2 2) 33 Directrix: y =− 4 57. (1)(2)ycx−=2 − (0−=− 1)2 c (1 2) 11=−cc =− (1)(2)yx−=−−2

58. (0)((1))xcy−=−−2 xcy2 =+(1) 2(01)42 =+cc = xy2 =+4( 1)

−=−2 54. Complete the square to put in standard form: 59. (0)(1)xcy yyx2 +=−+12 1 xcy2 =−()1 yy2 ++=−++12 36 x 1 36 2212 =−c() (6)(37)yx+=−−2 4 = c The equation is in the form xy2 =−41() ()yk−=−−2 4() axh where 2 1 60. (1)(xcy−= −− (1)) −=−=41aa or , hk==−37, and 6 . Thus, 4 ()xcy−=112 () + we have: 1 147 (0−=+ 1)2 ccc (1 1) 1= 2  = Vertex: (37, –6); Focus: ,–6 ; 4 2 149 −=2 1 + Directrix: x = (1)(1)xy 4 2 977 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

61. (0)((2))ycx−=−−2 24(1)442 = aaa =  = 1 ycx2 =+(2) a is the distance from the vertex to the focus. Thus, the bulb (located at the focus) should be 1 1 1(02)122 =+ccc =  = inch from the vertex. 2 1 yx2 =+(2) 66. Set up the problem so that the vertex of the 2 parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: xay2 = 4 . −=−2 62. (0)(1)ycx Since the focus is 1 inch from the vertex and the ycx2 =−()1 depth is 2 inches, a = 1 and the points (xx , 2) and (− , 2) are on the parabola. 1012 =−c() Substitute and solve for x : 1 =−c xxx22= 4(1)(2) = 8 =± 2 2 c =−1 The diameter of the headlight is 42≈ 5.66 yx2 =−() −1 inches.

63. Set up the problem so that the vertex of the 67. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: equation of the parabola has the form: xcy2 = . xay2 = 4 . Since the parabola is 10 feet across The point (300, 80) is a point on the parabola. and 4 feet deep, the points (5, 4) and (–5, 4) are Solve for c and find the equation: on the parabola. Substitute and solve for a : 3002 = cc (80) = 1125 2 25 54(4)2516= aaa =  = 2 = 16 xy1125 a is the distance from the vertex to the focus. y Thus, the receiver (located at the focus) is (300,80) 25 = 1.5625 feet, or 18.75 inches from the base (150,h) 80 16 x (0,0) h of the dish, along the axis of the parabola.

64. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: Since the height of the cable 150 feet from the xay2 = 4 . Since the parabola is 6 feet across center is to be found, the point (150, h) is a point and 2 feet deep, the points (3, 2) and (–3, 2) are on the parabola. Solve for h: on the parabola. Substitute and solve for a : 1502 = 1125h 9 = 34(2)982 = aaa =  = 22,500 1125h 8 20 = h a is the distance from the vertex to the focus. The height of the cable 150 feet from the center Thus, the receiver (located at the focus) is is 20 feet. 9 = 1.125 feet, or 13.5 inches from the base of 8 68. Set up the problem so that the vertex of the the dish, along the axis of the parabola. parabola is at (0, 10) and it opens up. Then the equation of the parabola has the form: 65. Set up the problem so that the vertex of the xcy2 =−( 10) . parabola is at (0, 0) and it opens up. Then the The point (200, 100) is a point on the parabola. 2 equation of the parabola has the form: xay= 4 . Solve for c and find the equation: Since the parabola is 4 inches across and 1 inch deep, the points (2, 1) and (–2, 1) are on the parabola. Substitute and solve for a :

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2002 =−c() 100 10 71. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the 40,000= 90c equation of the parabola has the form: xay2 = 4 . 444.44 ≈ c Since the parabola is 20 feet across and 6 feet 2 xy=−444.44() 10 deep, the points (10, 6) and (–10, 6) are on the parabola. Substitute and solve for a : 102 = 4a (6) (–200,100) (200,100) 100= 24a a ≈ 4.17 feet The heat will be concentrated about 4.17 feet from the base, along the axis of symmetry.

(0,10) h 72. Set up the problem so that the vertex of the 50 parabola is at (0, 0) and it opens up. Then the

Since the height of the cable 50 feet from the equation of the parabola has the form: xay2 = 4 . center is to be found, the point (50, h) is a point Since the parabola is 4 inches across and 3 on the parabola. Solve for h: inches deep, the points (2, 3) and (–2, 3) are on 502 =− 444.44()h 10 the parabola. Substitute and solve for a : 2 2500=− 444.44h 4444.4 24(3)= a 6944.4= 444.44h 412= a ≈ 1 15.625 h a = inch The height of the cable 50 feet from the center is 3 about 15.625 feet. The collected will be concentrated 1/3 inch from the base of the mirror along the axis of 69. Set up the problem so that the vertex of the symmetry. parabola is at (0, 0) and it opens up. Then the 73. Set up the problem so that the vertex of the 2 = equation of the parabola has the form: xay4 . parabola is at (0, 0) and it opens down. Then the a is the distance from the vertex to the focus equation of the parabola has the form: xcy2 = . (where the source is located), so a = 2. Since The point (60, –25) is a point on the parabola. the opening is 5 feet across, there is a point Solve for c and find the equation: (2.5, y) on the parabola. 602 =−cc ( 25) =− 144 Solve for y: xy2 = 8 xy2 =−144 2.52 = 8y 6.25= 8y y = 0.78125 feet y The depth of the searchlight should be 0.78125

feet. x

70. Set up the problem so that the vertex of the 25 parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: xay2 = 4 . (60,–25) a is the distance from the vertex to the focus

(where the source is located), so a = 2. Since To find the height of the bridge 10 feet from the the depth is 4 feet, there is a point center the point (10, y) is a point on the parabola. (x, 4) on the parabola. Solve for x: Solve for y: xyx22= 8843242 =⋅ x 2=  x=± 102 =− 144y The width of the opening of the searchlight 100=− 144y ≈ should be 82 11.31 feet. −≈0.69 y

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The height of the bridge 10 feet from the center 75. a. Imagine placing the Arch along the x-axis is about 25 – 0.69 = 24.31 feet. To find the with the peak along the y-axis. Since the height of the bridge 30 feet from the center the Arch is 625 feet high and is 598 feet wide at point (30, y) is a point on the parabola. its base, we would have the points Solve for y: ()−299,0 , ()0,625 , and ()299,0 . The 2 =− 30 144y equation of the parabola would have the =− 900 144y form yaxc=+2 . Using the point ()0,625 −=6.25 y we have The height of the bridge 30 feet from the center =+()2 is 25 – 6.25 = 18.75 feet. To find the height of 625ac 0 the bridge, 50 feet from the center, the point 625 = c (50, y) is a point on the parabola. Solve for y: The model then becomes yax=+2 625 . 502 =− 144y Next, using the point ()299,0 we get 2500=− 144y 2 y =−17.36 0=+a() 299 625 2 The height of the bridge 50 feet from the center −=625() 299 a is about 25 – 17.36 = 7.64 feet. 625 a =− 2 74. Set up the problem so that the vertex of the ()299 parabola is at (0, 0) and it opens down. Then the Thus, the equation of the parabola with the equation of the parabola has the form: xcy2 = . same given dimensions is The points (50, –h) and (40, –h+10) are points on 625 c h yx=−2 + . the parabola. Substitute and solve for and : 2 625 2 () 50=−ch ( ) 402 =−+ch ( 10) 299 =− ch 2500 1600=−ch + 10 c 625 b. Using yx=−2 + , we get 2 625 ()299 y Width (ft)x Height (ft), model 567 283.5 63.12 x h (40,–h+10) 478 239 225.67 10 308 154 459.2 (50,–h)

c. No; the heights computed by using the 1600=−() − 2500 + 10c model do not fit the actual heights. 1600=+ 2500 0c 76. Ax2 += Ey00,0 A ≠ E ≠ −=900 10c 22E −=90 c AxEyx=−  =− y A −=− 90h 2500 This is the equation of a parabola with vertex at h ≈ 27.78 (0, 0) and axis of symmetry being the y-axis. The height of the bridge at the center is about E The focus is 0, − . The directrix is 27.78 feet. 4A E E y = . The parabola opens up if −>0 and 4A A E down if −<0 . A

980 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.2: The Parabola

77. Cy2 += Dx00,0 C ≠ D ≠ d. If E = 0 , then 2 =− −±DD2 −4 AF Cy Dx Ax2 ++= Dx F0  x = 2A 2 D yx=− 2 −< C If DAF40, there is no real solution. This is the equation of a parabola with vertex at The graph contains no points. (0, 0) and whose axis of symmetry is the x-axis. 2 +++= ≠ D 79. Cy Dx Ey F00 C The focus is − ,0 . The directrix is 4C a. If D ≠ 0 , then: D 2 +=−− x = . The parabola opens to the right if Cy Ey Dx F 4C 22 2 EE E D D Cy++ y =−−+ DxF −>0 and to the left if −<0 . CC2 4 C C 4C 2 EE1 2 2 yDxF+=−−+ 78. Ax+++= Dx Ey F00 A ≠  24CC C a. E ≠ 2 2 If 0 , then: EDFE−  yx+= +− Ax2 +=−− Dx Ey F  24CCDCD 22 2 DD D 2 −−2 Ax++ x =−−+ EyF +=EDECF −4 AA4A2 4 yx  24CC CD 2 DD1 2 This is the equation of a parabola whose xEyF+=−−+ 24AA A ECFE2 − 4 vertex is ,− , and whose 2 2 42CD C DEFD−   xy+= +−  axis of symmetry is to the x-axis. 24A AEAE 2 b. If D = 0 , then DEDAF−−2 4 xy+= − 2  −±EE −4 CF 24AA AE Cy2 ++= Ey F0  y = C This is the equation of a parabola whose 2 E DD2 − 4 AF If ECF2 −=40, then y =− is a single vertex is − , and whose 2C AAE 24 horizontal line. axis of symmetry is parallel to the y-axis. c. If D = 0 , then b. If E = 0 , then −±EE2 −4 CF Cy2 ++= Ey F0  y = −±DD2 −4 AF Ax2 ++= Dx F0  x = 2C 2A If ECF2 −>40, then D 2 −= =− 2 If DAF40, then x is a single −+EE −4 CF 2A y = and vertical line. 2C c. E = −−EE2 −4 CF If 0 , then y = are two horizontal 2C −±DD2 −4 AF Ax2 ++= Dx F0  x = lines. 2A = If DAF2 −>40, then d. If D 0 , then −±2 − −+DD2 −4 AF 2 ++= = EE4 CF x = and Cy Ey F0  y 2A 2C If ECF2 −<40, then there is no real −−DD2 −4 AF x = are two vertical lines. solution. The graph contains no points. 2A

981 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

Section 10.3 16. (a); the major axis is along the x-axis and the vertices are at ()−2,0 and ()2,0 . 2 1. d =−+−−−=+=()422 () 2 () 5 222 3 13 xy22 17. +=1 2 25 4 39 2. −= The center of the is at the origin. 24  ab==5, 2 . The vertices are (5, 0) and (–5, 0). Find the value of c: 3. x-intercepts: 016422=−x cab222=−=−=→=25 4 21 c 21 416x2 = The foci are ()21,0 and ()− 21,0 . x2 = 4 x =±2 →()() − 2,0 , 2,0

y-intercepts: y2 =−16 4() 0 2 y2 = 16 y =±40,4,0,4 →()() − The intercepts are ()−2, 0 , ()2, 0 , ()0,− 4 , and ()0, 4 .

() − −→−−() = xy22 4. 2,5 ; change x to x : 222 18. +=1 94 5. left 1; down 4 The center of the ellipse is at the origin. ab==3, 2 . The vertices are (3, 0) and (–3, 0). 2 6. ()xy−+−−=2312 () () 2 Find the value of c: cab222=−=−=→=945 c 5 ()()xy−++=23122 The foci are ()5,0 and ()− 5,0 . 7. ellipse

8. major

9. (0,− 5);(0,5)

10. 5; 3; x

11. ()−−2, 3 ()6,− 3

12. ()1, 4 xy22 19. +=1 13. (c); the major axis is along the x-axis and the 925 vertices are at ()−4,0 and ()4,0 . The center of the ellipse is at the origin. ab==5, 3 . The vertices are (0, 5) and (0, –5). 14. (d); the major axis is along the y-axis and the Find the value of c: vertices are at ()0,− 4 and ()0, 4 . cab222=−=−=25 9 16

c = 4 15. (b); the major axis is along the y-axis and the The foci are (0, 4) and (0, –4). vertices are at ()0,− 2 and ()0, 2 .

982 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

22. xy22+=918 y2 20. x2 +=1 Divide by 18 to put in standard form: 16 xy22918 += The center of the ellipse is at the origin. 18 18 18 ab==4, 1 . The vertices are (0, 4) and (0, –4). xy22 Find the value of c: +=1 18 2 cab222=−=−=16 1 15 The center of the ellipse is at the origin. c = 15 ab==32, 2. The vertices are ()32,0 The foci are ()0, 15 and () 0,− 15 and ()−32,0. Find the value of c: cab222=−=−=18 2 16

c = 4 The foci are (4, 0) and (–4, 0).

21. 416xy22+= Divide by 16 to put in standard form:

416xy22 += 22 16 16 16 23. 48yx+= xy22 Divide by 8 to put in standard form: +=1 48yx22 416 += The center of the ellipse is at the origin. 888 ab==4, 2 . xy22 +=1 The vertices are (0, 4) and (0, –4). Find the 82 value of c: The center of the ellipse is at the origin. cab222=−=−=16 4 12 == = ab822, 2. == c 12 2 3 The vertices are ()22,0 and ()− 22,0. Find − The foci are ()0, 2 3 and () 0, 2 3 . the value of c: cab222=−=−=82 6

c = 6 The foci are ()6, 0 and ()− 6, 0 .

983 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

xy22 26. xy22+=→4 + = 1 44 This is a whose center is at (0, 0) and radius = 2. The focus of the ellipse is ()0,0 and the vertices are ()−2,0 , ()2,0 , ()0,− 2 , ()0, 2 .

24. 4936yx22+= Divide by 36 to put in standard form: 4936yx22 += 36 36 36 xy22 +=1 49 27. Center: (0, 0); Focus: (3, 0); Vertex: (5, 0); The center of the ellipse is at the origin. Major axis is the x-axis; ac==5; 3 . Find b: == ab3, 2 . The vertices are (0, 3) and (0, –3). bac222=−=−=25 9 16

Find the value of c: b = 4 222 cab=−=−=→94 5 c = 5 22 xy+= − Write the equation: 1 The foci are ()0, 5 and () 0, 5 . 25 16

25. xy22+=16 28. Center: (0, 0); Focus: (–1, 0); Vertex: (3, 0); == xy22 Major axis is the x-axis; ac3; 1 . Find b: +=1 16 16 bac222=−=−=91 8

This is a circle whose center is at (0, 0) and b = 22 radius = 4. The focus of the ellipse is ()0,0 and xy22 Write the equation: +=1 the vertices are ()−4,0 , ()4,0 , ()0,− 4 , ()0, 4 . 98

984 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

29. Center: (0, 0); Focus: (0, –4); Vertex: (0, 5); 32. Foci: (0, ±2); length of the major axis is 8. Major axis is the y-axis; ac==5; 4 . Find b: Center: (0, 0); Major axis is the y-axis; == bac222=−=−=25 16 9 ac4; 2 . Find b : 222 b = 3 bac=−=−=→=16 4 12 b 2 3 22 xy22 xy Write the equation: +=1 Write the equation: +=1 925 12 16

30. Center: (0, 0); Focus: (0, 1); Vertex: (0, –2); 33. Focus: ()−4,0 ; Vertices: ()−5, 0 and ()5, 0 ; Major axis is the y-axis; ac==2; 1. Find b: Center: ()0,0 ; Major axis is the x-axis. 222=−=−=→= bac41 3 b 3 a = 5 ; c = 4 . Find b: xy22 222=−=−=→= Write the equation: +=1 bac25 16 9 b3 34 xy22 Write the equation: +=1 25 9

31. Foci: (±2, 0); Length of major axis is 6. Center: (0, 0); Major axis is the x-axis; ac==3; 2 . Find b: 34. Focus: (0, –4); Vertices: (0, ±8). Center: (0, 0); Major axis is the y-axis; 222 bac=−=−=→=945 b 5 ac==8; 4 . Find b: 22 xy 222 Write the equation: +=1 bac=−=−=64 16 48 →= b 4 3 95 xy22 Write the equation: +=1 48 64

985 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

35. Foci: (0, ±3); x-intercepts are ±2. 38. Vertices: (±5, 0); c = 2 ; Major axis is the x- Center: (0, 0); Major axis is the y-axis; axis; ab= 5; Find : cb==3; 2 . Find a : bac222=−=−=25 4 21 abc222=+=+=→4913 a = 13 b = 21 xy22 Write the equation: +=1 xy22 413 Write the equation: +=1 25 21

36. Vertices: (±4, 0); y-intercepts are ±1. Center: (0, 0); Major axis is the x-axis; ab==4; 1 . 39. Center: ()−1,1 Find c: Major axis: parallel to x-axis cab222=−=−=16 1 15 Length of major axis: 42=→=aa 2 Length of minor axis: 22=→=bb 1 a = 15 2 2 (1)x + x +−2 = Write the equation: +=y2 1 (1)1y 16 4 40. Center: ()−−1, 1 Major axis: parallel to y-axis Length of major axis: 42=→=aa 2 Length of minor axis: 22=→=bb 1 (1)y + 2 (1)x ++2 = 1 4

41. Center: ()1, 0

Major axis: parallel to y-axis 37. Center: (0, 0); Vertex: (0, 4); b = 1; Major Length of major axis: 42=→=aa 2 axis is the y-axis; ab==4; 1 . Length of minor axis: 22=→=bb 1 2 y2 2 y −+2 = Write the equation: x +=1 (1)x 1 16 4 42. Center: ()0,1 Major axis: parallel to x-axis Length of major axis: 42=→=aa 2 Length of minor axis: 22=→=bb 1 x2 +−(1)1y 2 = 4

986 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

(3)(1)xy−+22 45. Divide by 16 to put the equation in standard 43. The equation +=1 is in the form: 49 (5)4(4)16xy++22 −= −−22 ()()xh+= yk form 1 (major axis parallel (5)4(4)16xy+−22 ba22 += to the y-axis) where ab==3, 2, 16 16 16 (5)(4)xy+−22 hk==−3, and 1 . Solving for c: +=1 16 4 cab222=−=−=→=945 c 5 The equation is in the form Thus, we have: ()()xh−−22 yk Center: (3, –1) +=1 (major axis parallel to the 22 Foci: ()3,−+ 1 5 ,() 3, −− 1 5 ab x-axis) where ab==4, 2, hk=−5, and = 4 . Vertices: (3, 2), (3, –4) Solving for c: cab222=−=−=→=16 4 12 c 12 = 2 3 Thus, we have: Center: (–5, 4) Foci: ()−−5 2 3,4 ,() −+ 5 2 3,4 Vertices: (–9, 4), (–1, 4)

(4)(2)xy++22 44. The equation +=1 is in the 94 ()()xh−−22 yk form +=1 ab22 (major axis parallel to the x-axis) where ab==3, 2, hk=−4, and =− 2 . 46. Divide by 18 to put the equation in standard Solving for c: form: 222 cab=−=−=→=945 c 5 22 9(xy−++= 3) ( 2) 18 Thus, we have: −+22 Center: (–4, –2); Vertices: (–7, –2), (–1, –2) 9(xy 3)+= ( 2) 18 Foci: ()−+45,2,45,2 −() −− − 18 18 18 (3)(2)xy−+22 +=1 218 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ba22 y-axis) where ab==32, 2, hk==−3, and 2 . Solving for c: cab222=−=−=→=18 2 16 c 4 Thus, we have: Center: (3, –2) Foci: (3, 2), (3, –6) Vertices: ()3,−+ 2 3 2 ,() 3, −− 2 3 2

987 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ab22 x-axis) where abhk====3, 1, 0, and 2 . Solving for c: cab222=−=−=→=31 2 c 2 Thus, we have: Center: (0, 2) − 47. Complete the square to put the equation in Foci: ()2, 2 ,() 2, 2 standard form: Vertices: ()− 3, 2 ,() 3, 2 xxyy22++44 −+= 840 (xx22+++ 4 4)4( yy −+=−++ 2 1)444 (2)4(1)4xy++−=22 +−22 (2)4(1)4xy+= 444 (2)x + 2 +−(1)1y 2 = 4 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ab22 49. Complete the square to put the equation in x-axis) where abh===−=2, 1, 2, and k 1 . standard form: 222 22+−++= Solving for c: cab=−=−=→=41 3 c 3 238650xyxy Thus, we have: 2(xx22−+ 4 ) 3( yy + 2 ) =− 5 Center: (–2, 1) 2(xx22−++ 4 4) 3( yy ++=−++ 2 1) 5 8 3 Foci: ()−−23,1,23,1() −+ 2(xy−++= 2)22 3( 1) 6 Vertices: (–4, 1), (0, 1) −+22 2(xy 2)+= 3( 1) 6 666 (2)(1)xy−+22 +=1 32 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ab22 x-axis) where

abh===3, 2, 2, and k =− 1. 48. Complete the square to put the equation in 222 standard form: Solving for c: cab=−=−=→=321 c 1 xy22+−+=31290 y Thus, we have: Center: (2, –1) 22+−+=−+ xyy3( 4 4) 9 12 Foci: (1, –1), (3, –1) xy22+−=3( 2) 3 Vertices: ()23,1,23,1−−() +− 22− xy+=3( 2) 3 333 x2 +−(2)1y 2 = 3

988 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

9(xy−+ 1)22 4( 2) 36 += 36 36 36 (1)(xy−+22 2) +=1 49 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ba22 y-axis) where abh===3, 2, 1, and k =− 2 .

Solving for c: 50. Complete the square to put the equation in cab222=−=−=→=945 c 5 standard form: Thus, we have: 22 43865xyxy++−= Center: ()1,− 2 22++ − = 4(xx 2 ) 3( yy 2 ) 5 Foci: ()1,−+ 2 5 , ()1,−− 2 5 4(xx22+++ 2 1)3( yy −+=++ 2 1)543 Vertices: ()1,1 , ()1,− 5 4(xy++ 1)22 3( −= 1) 12 +−22 4(xy 1)+= 3( 1) 12 12 12 12 (1)(1)xy+−22 +=1 34 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ba22 y-axis) where ab==2, 3, h =−= 1, and k 1. Solving for c: cab222=−=−=→=431 c 1 52. Complete the square to put the equation in Thus, we have: standard form: Center: (–1, 1) xyxy22++−+=961890 Foci: (–1, 0), (–1, 2) (6)9(2)9xx22++ yy − =− Vertices: (–1, –1), (–1, 3) (xx22+++ 6 9)9( yy −+=−++ 2 1)999 (3)9(1)9xy++22 −= +−22 (3)9(1)9xy+= 999 (3)x + 2 +−(1)1y 2 = 9 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the ab22 51. Complete the square to put the equation in x-axis) where abh===−=3, 1, 3, and k 1 . standard form: Solving for c: cab222=−=−=→=91 8 c 22 9xy22+−+−= 4 18 xy 16 11 0 Thus, we have: 9(xx22−+ 2 ) 4( yy + 4 ) = 11 Center: (–3, 1) 9(xx22−++ 2 1) 4( yy ++=++ 4 4) 11 9 16 Foci: ()−+3 2 2,1 ,() −− 3 2 2,1 9(xy−+ 1)22 4( + 2) = 36 Vertices: (0, 1), (–6, 1)

989 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

54. Complete the square to put the equation in standard form: 9180xy22+− x = 9(xx22−++= 2 1) y 9 9(xy−+= 1)22 9 − 22 9(xy 1)+= 9 999 y2 (1)x −+2 = 1 53. Complete the square to put the equation in 9 standard form: The equation is in the form xy22++= y ()()xh−−22 yk 440 +=1 (major axis parallel to the 22 4444xy22+++= y ba y-axis) where abh===3, 1, 1, and k = 0 . 4(2)4xy22++ = Solving for c: 22+ 4(2)4xy+= cab222=−=−=→=91 8 c 22 444 Thus, we have: (2)y + 2 Center: (1, 0) x2 +=1 4 Foci: ()1, 2 2 ,() 1,− 2 2 The equation is in the form Vertices: (1, 3), (1, –3) ()()xh−−22 yk +=1 (major axis parallel to the ba22 y-axis) where abh===2, 1, 0, and k =− 2 . Solving for c: cab222=−=−=→=41 3 c 3 Thus, we have: Center: (0,–2) Foci: ()0,−+ 2 3 ,() 0, −− 2 3 Vertices: (0, 0), (0, –4) 55. Center: (2, –2); Vertex: (7, –2); Focus: (4, –2); Major axis parallel to the x-axis; ac==5; 2 . Find b: bac222=−=−=→=25 4 21 b 21 (2)(2)xy−+22 Write the equation: +=1 25 21

990 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

56. Center: (–3, 1); Vertex: (–3, 3); Focus: (–3, 0); 59. Foci: (5, 1), (–1, 1); Major axis parallel to the y-axis; ac==2; 1. Length of the major axis = 8; Center: (2, 1); Find b: Major axis parallel to the x-axis; ac==4; 3 . bac222=−=−=→=41 3 b 3 Find b: 222=−=−=→= (3)(1)xy+−22 bac16 9 7 b 7 Write the equation: +=1 (2)(1)xy−−22 34 Write the equation: +=1 16 7

57. Vertices: (4, 3), (4, 9); Focus: (4, 8); Center: (4, 6); Major axis parallel to the y-axis; 60. Vertices: (2, 5), (2, –1); c = 2 ; ac==3; 2 . Find b: Center: (2, 2); Major axis parallel to the y-axis; == bac222=−=−=→=945 b 5 ac3; 2 . Find b: 222 (4)(6)xy−−22 bac=−=−=→=945 b 5 Write the equation: +=1 59 (2)(2)xy−−22 Write the equation: +=1 59

58. Foci: (1, 2), (–3, 2); Vertex: (–4, 2); Center: (–1, 2); Major axis parallel to the 61. Center: (1, 2); Focus: (4, 2); contains the point == x-axis; ac3; 2 . Find b: (1, 3); Major axis parallel to the x-axis; c = 3 . bac222=−=−=→=945 b 5 The equation has the form: 22 (1)(xy+−22 2) (1)(xy−− 2) Write the equation: +=1 +=1 95 ab22 Since the point (1, 3) is on the : 01 +=1 ab22

1 =→111bb2 =→ = b2 Find a : abc222=+=+=→=1 9 10 a 10 (1)x − 2 Write the equation: +−(2)1y 2 = 10

991 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

64. Center: (1, 2); Vertex: (1, 4); contains the point (1+ 3, 3) ; Major axis parallel to the y-axis; 62. Center: (1, 2); Focus: (1, 4); contains the point = (2, 2); Major axis parallel to the y-axis; c = 2 . a 2 . The equation has the form: (1)(xy−−22 2) The equation has the form: +=1 (1)(xy−−22 2) ba22 +=1 ba22 Since the point (1+ 3, 3) is on the curve: Since the point (2, 2) is on the curve: 31 +=1 10 2 +=1 4 b 22 ba 11 =→bb2 =→=42 1 2 =→111bb2 =→ = b 4 2 b (1)(xy−−22 2) Find a : Write the equation: +=1 44 222=+=+=→= abc14 5 a 5 Solve for c: (2)y − 2 cab222= − =−=44 0. Thus, c = 0 . Write the equation: (1)x −+2 = 1 5

65. Rewrite the equation: =−2 yx16 4 yxy22=−16 4 , ≥ 0 63. Center: (1, 2); Vertex: (4, 2); contains the point (1, 5); Major axis parallel to the x-axis; a = 3 . 416,0xy22+= y ≥ The equation has the form: 22 xy+= ≥ (1)(xy−−22 2) 1,y 0 +=1 416 ab22 Since the point (1, 5) is on the curve: 2 03+= 2 1 9 b 9 =→193bb2 = → = b2 Solve for c: cab222=−=−=99 0. Thus, c = 0 .

(1)(xy−−22 2) Write the equation: +=1 99

992 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

66. Rewrite the equation: 69. The center of the ellipse is (0, 0). The length of 2 a = yx=−99 the major axis is 20, so 10 . The length of half the minor axis is 6, so b = 6 . The ellipse is 22=− ≥ yxy99 , 0 situated with its major axis on the x-axis. The 22+= ≥ 99,0xy y xy22 equation is: +=1. xy22 +=1,y ≥ 0 100 36 19 70. The center of the ellipse is (0, 0). The length of the major axis is 30, so a = 15 . The length of half the minor axis is 10, so b = 10 . The ellipse is situated with its major axis on the x-axis. The xy22 equation is: +=1. 225 100 The roadway is 12 feet above the axis of the ellipse. At the center ( x = 0 ), the roadway is 2 feet above the arch. At a point 5 feet either side of the center, 67. Rewrite the equation: evaluate the equation at x = 5 : yx=−64 − 16 2 522y +=1 yxy22=−64 16 , ≤ 0 225 100 22+= ≤ y2 25 200 16xy 64, y 0 =−1 = xy22 100 225 225 +=1,y ≤ 0 200 464 y =≈10 9.43 225 The vertical distance from the roadway to the arch is 12−≈ 9.43 2.57 feet. At a point 10 feet either side of the center, evaluate the equation at x = 10 : 1022y +=1 225 100 y2 100 125 =−1 = 100 225 225 125 68. Rewrite the equation: y =≈10 7.45 yx=−44 − 2 225 The vertical distance from the roadway to the 22=− ≤ yxy44 , 0 arch is 12−≈ 7.45 4.55 feet. 44,0xy22+= y ≤ At a point 15 feet either side of the center, the xy22 roadway is 12 feet above the arch. +=1,y ≤ 0 14 71. Assume that the half ellipse formed by the gallery is centered at (0, 0). Since the hall is 100 feet long, 2aa== 100 or 50 . The distance from the center to the foci is 25 feet, so c = 25 . Find the height of the gallery which is b : bac222=−=2500 − 625 = 1875 b =≈1875 43.3 The ceiling will be 43.3 feet high in the center.

993 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

72. Assume that the half ellipse formed by the has a span of 100 feet, the length of the major gallery is centered at (0, 0). Since the distance axis is 100, or 2aa== 100 or 50 . Let h be the between the foci is 100 feet and Jim is 6 feet maximum height of the bridge. The equation is: from the nearest wall, the length of the gallery is xy22 == +=1. 112 feet. 2aa 112 or 56 . The distance 2 from the center to the foci is 50 feet, so c = 50 . 2500 h The height of the arch 40 feet from the center is Find the height of the gallery which is b : 10 feet. So (40, 10) is a point on the ellipse. 222=−= − = bac3136 2500 636 Substitute and solve for h : b =≈636 25.2 4022 10 +=1 The ceiling will be 25.2 feet high in the center. 2500 h2 102 1600 9 73. Place the semi-elliptical arch so that the x-axis =−1 = coincides with the water and the y-axis passes h2 2500 25 through the center of the arch. Since the bridge 9h2 = 2500 has a span of 120 feet, the length of the major 50 axis is 120, or 2aa== 120 or 60 . The h =≈16.67 maximum height of the bridge is 25 feet, so 3 The height of the arch at its center is 16.67 feet. xy22 b = 25 . The equation is: +=1 . 3600 625 75. If the x-axis is placed along the 100 foot length The height 10 feet from the center: and the y-axis is placed along the 50 foot length, 1022y xy22 +=1 the equation for the ellipse is: +=1 . 3600 625 5022 25 y2 100 Find y when x = 40: =−1 625 3600 4022y +=1 22 2 3500 50 25 y =⋅625 3600 y2 1600 ≈ =−1 y 24.65 feet 625 2500 The height 30 feet from the center: 9 y2 =⋅625 3022y 25 +=1 3600 625 y = 15 feet y2 900 To get the width of the ellipse at x = 40 , we =−1 625 3600 need to double the y value. Thus, the width 10 2700 feet from a vertex is 30 feet. y2 =⋅625 3600 76. Place the semi-elliptical arch so that the x-axis y ≈ 21.65 feet coincides with the major axis and the y-axis The height 50 feet from the center: passes through the center of the arch. Since the height of the arch at the center is 20 feet, b = 20 . 5022y +=1 The length of the major axis is to be found, so it 3600 625 is necessary to solve for a . The equation is: 2 y =−2500 xy22 1 +=1 . 625 3600 a2 400 1100 y2 =⋅625 The height of the arch 28 feet from the center is 3600 to be 13 feet, so the point (28, 13) is on the y ≈ 13.82 feet ellipse. Substitute and solve for a :

74. Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge

994 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.3: The Ellipse

22 Therefore, c = 1.5 million. Find b: 28+= 13 1 222=− a2 400 bac 784 169 231 6622 =−1 = =×()()93 10 −× 1.5 10 a2 400 400 =×8.64675 1015 231a2 = 313600 =×8646.75 1012 a2 = 1357.576 =×6 a = 36.845 b 92.99 10 The span of the bridge is 73.69 feet. The equation of the orbit is: xy22 +=1 77. Because of the pitch of the roof, the major axis 22 will run parallel to the direction of the pitch and ()93×× 1066() 92.99 10 the minor axis will run to the We can simplify the equation by letting our units direction of the pitch. The length of the major for x and y be millions of miles. The equation then axis can be determined from the pitch by using becomes: the . The length of the xy22 minor axis is 8 inches (the diameter of the pipe). +=1 8649 8646.75

80. Since the mean distance is 142 million miles, a = 142 million. The length of the major axis is 284 million. The aphelion is 2(5) = 10 284 million – 128.5 million = 155.5 million miles. The distance from the center of the ellipse to the (focus) is 142 million – 128.5 million = 13.5 million miles. Therefore, c = 13.5 million. Find b: 2(4) = 8 bac222=− The length of the major axis is 22 22 =×66 − × ()8+== ( 10 ) 164 2 41 inches. ()()142 10 13.5 10 =×1.998175 1016 78. The length of the football gives the length of the b =×141.36 106 major axis so we have 211.125a = or The equation of the orbit is: a = 5.5625 . At its center, the prolate spheroid is xy22 a circle of radius b. This means +=1 22 228.25πb = ()142×× 1066() 141.36 10 28.25 b = We can simplify the equation by letting our units 2π for x and y be millions of miles. The equation then 2 becomes: 44ππ2 =≈() 28.25 Thus, ab 5.5625 471 . xy22 33 2π +=1 The football contains approximately 471 cubic 20,164 19,981.75 inches of air. 81. The mean distance is 79. Since the mean distance is 93 million miles, 507 million – 23.2 million = 483.8 million miles. a = 93 million. The length of the major axis is The perihelion is 186 million. The perihelion is 483.8 million – 23.2 million = 460.6 million 186 million – 94.5 million = 91.5 million miles. miles. The distance from the center of the ellipse to the Since ac=×483.8 1066 and =× 23.2 10 , we can sun (focus) is find b: 93 million – 91.5 million = 1.5 million miles.

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222=− FF bac If A ≠ C , then −≠−. So, this is the A C 6622 =×−×()()483.8 10 23.2 10 equation of an ellipse with center at (0, 0). =×2.335242 1017 b. If A = C , the equation becomes: 6 −F b =×483.2 10 Ax22+=−→+= Ay F x 22 y The equation of the orbit of Jupiter is: A xy22 This is the equation of a circle with center at +=1 22 −F ()483.8×× 1066() 483.2 10 (0, 0) and radius of . A We can simplify the equation by letting our units for x and y be millions of miles. The equation 84. Complete the square on the given equation: then becomes: Ax22++++=≠≠ Cy Dx Ey F0, A 0, C 0 xy22 +=1 Ax22+++=− Cy Dx Ey F 234,062.44 233,524.2 22DE Ax++ x C y +=− y F 82. The mean distance is AC 22 4551 million + 897.5 million = 5448.5 million DEDE22 AxCy+++=+− F miles. 2244ACAC The aphelion is where AC⋅>0 . 5448.5 million + 897.5 million = 6346 million DE22 miles. Let UF=+−. AC Since ac=×5448.5 1066 and =× 897.5 10 , we 44 can find b: a. If U is of the same sign as A (and C ) , then bac222=− DE22 xy++ 22AC =×−×6622 +=1 ()()5448.5 10 897.5 10 UU AC =×2.8880646 1019 This is the equation of an ellipse whose 6 b =×5374.07 10 −−DE center is , . The equation of the orbit of Pluto is: 22AC xy22 +=1 b. If U = 0 , the graph is the single point ××6622 ()5448.5 10() 5374.07 10 −−DE , . We can simplify the equation by letting our units 22AC for x and y be millions of miles. The equation c. If U is of the opposite sign as A (and C ) , this then becomes: graph contains no points since the left side xy22 +=1 always has the opposite sign of the right side. 29,686,152.25 28,880,646 83. a. Put the equation in standard ellipse form: 85. Answers will vary. 22++= Ax Cy F 0 22 AxCyF+=− Ax22 Cy +=1 −−FF xy22 +=1 (/)(/)−−FA FC where AC≠≠≠0, 0, F 0 , and −FA/ and −FC/ are positive.

996 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The

Section 10.4 10. ()2, 4 ; ()2,− 2 =−−+−−()2 () ()2 1. d 23 1 4 11. ()2, 6 ; ()2,− 4 =−()522 + () 5 = 25 + 25 = 50 = 5 2 12. 4

2 525 13. 2; 3; x 2. = 24  4 4 14. yx=− ; yx= 9 9 3. x-intercepts: 09422=+x 49x2 =− 15. (b); the hyperbola opens to the left and right, and ()± 9 has vertices at 1, 0 . Thus, the graph has an x2 =− (no real solution) 4 y2 equation of the form x2 −=1. 2 y-intercepts: y2 =+940()2 b y2 = 9 16. (c); the hyperbola opens up and down, and has y =±3 → ()() 0, − 3 , 0,3 vertices at ()0,± 2 . Thus, the graph has an ()− () y2 x2 The intercepts are 0, 3 and 0,3 . equation of the form −=1. 4 b2 4. True; the graph of yx22=+9 is a hyperbola

with its center at the origin. 17. (a); the hyperbola opens to the left and right, and 5. right 5 units; down 4 units has vertices at ()±2, 0 . Thus, the graph has an x2 − 9 x2 y2 6. ypxxqxx==−=−; ()229, () 4 equation of the form −=1. x2 − 4 4 b2 The vertical are the zeros of q . 18. (d); the hyperbola opens up and down, and has qx()= 0 vertices at ()0,± 1 . Thus, the graph has an 2 −= x 40 2 2 x 2 = equation of the form y −=1. x 4 b2 xx=−2, = 2 The lines xx=−2, and = 2 are the vertical 19. Center: (0, 0); Focus: (3, 0); Vertex: (1, 0); == asymptotes. The degree of the numerator, Transverse axis is the x-axis; ac1; 3 . Find the value of b: px()=− x2 9, is n = 2 . The degree of the bca222=−=−=91 8 denominator, qx()=− x2 4, is m = 2 . b ==822 1 Since nm= , the line y ==1 is a horizontal 1 . Since this is a rational and there is a horizontal asymptote, there are no oblique asymptotes.

7. hyperbola

8. transverse axis

9. b

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y2 22. Center: (0, 0); Focus: (–3, 0); Vertex: (2, 0) Write the equation: x2 −=1. == 8 Transverse axis is the x-axis; ac2; 3 . Find the value of b: bca222=−=−=945

b = 5 x2 y2 Write the equation: −=1. 45

20. Center: (0, 0); Focus: (0, 5); Vertex: (0, 3); Transverse axis is the y-axis; ac==3; 5 . Find the value of b: bca222=−=−=25 9 16

b = 4 y2 x2 23. Foci: (–5, 0), (5, 0); Vertex: (3, 0) Write the equation: −=1. 916 Center: (0, 0); Transverse axis is the x-axis; ac==3; 5 . Find the value of b: bca222=−=−=25 9 16 b= 4 x2 y2 Write the equation: −=1. 916

21. Center: (0, 0); Focus: (0, –6); Vertex: (0, 4) Transverse axis is the y-axis; ac==4; 6 . Find the value of b: bca222=−=−=36 16 20

b ==20 2 5 y2 x2 Write the equation: −=1. 16 20

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24. Focus: (0, 6); Vertices: (0, –2), (0, 2) 26. Vertices: (–4, 0), (4, 0); asymptote: yx= 2 ; Center: (0, 0); Transverse axis is the y-axis; Center: (0, 0); Transverse axis is the x-axis; ac==2; 6 . a = 4 . Find the value of b using the slope of the Find the value of b: bb asymptote: ==28 b = bca222=−=−=36 4 32 b= 4 2 a 4 y2 x2 Find the value of c: Write the equation: −=1. 222=+=+= 432 cab16 64 80 c = 45 x2 y2 Write the equation: −=1. 16 64

25. Vertices: (0, –6), (0, 6); asymptote: yx= 2 ; Center: (0, 0); Transverse axis is the y-axis; a = 6 . Find the value of b using the slope of the a 6 asymptote: ==226 bb=  = 3 27. Foci: (–4, 0), (4, 0); asymptote: yx=− ; bb Center: (0, 0); Transverse axis is the x-axis; Find the value of c: c = 4 . Using the slope of the asymptote: 222=+=+= cab36 9 45 b −=−1 −=−baba = . c = 35 a y2 x2 Find the value of b: Write the equation: −=1. 222 222 36 9 bca=− abc+=() c =4 bb22+=16 2 b 2= 16 b 2= 8

b ==822 aab==822( = ) x2 y2 Write the equation: −=1. 88

999 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

28. Foci: (0, –2), (0, 2); asymptote: yx=− ; y2 x2 30. −=1 Center: (0, 0); Transverse axis is the y-axis; 16 4 c = 2 . Using the slope of the asymptote: The center of the hyperbola is at (0, 0). a == −=−1 −=−baba = ab4, 2 . The vertices are (0, 4) and b (0, –4). Find the value of c: Find the value of b: 222 cab=+=+=16 4 20 bca222=− c ==20 2 5 ab222+= c() c =2 The foci are ()0, 2 5 and () 0,− 2 5 . bb22+=42 b 2= 4 The transverse axis is the y-axis. The asymptotes bb2 = 22 = are yxyx==−2; 2 . aab==2( ) y2 x2 Write the equation: −=1. 22

31. 416xy22−= Divide both sides by 16 to put in standard form: 2222 x2 y2 416xx−=yy−= 29. −=1  1 25 9 16 16 16 4 16 The center of the hyperbola is at (0, 0). The center of the hyperbola is at (0, 0). == ab==5, 3 . The vertices are ()5, 0 and ab2, 4 . ()− The vertices are (2, 0) and (–2, 0). 5, 0 . Find the value of c: Find the value of c: cab222=+=+=25 9 34 c= 34 cab222=+=+=41620

The foci are ()34,0 and ()− 34,0 . c ==20 2 5 The transverse axis is the x-axis. The asymptotes The foci are ()25,0 and ()− 25,0. 33 are yxyx==−; . The transverse axis is the x-axis. The asymptotes 55 are yxyx==−2; 2 .

1000 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

32. 416yx22−= 34. xy22−=4 Divide both sides by 16 to put in standard form: Divide both sides by 4 to put in standard form: 4yy22xx2216 xx22yy224 −= −=1 −= −=1. 16 16 16 4 16 444 44 The center of the hyperbola is at (0, 0). The center of the hyperbola is at (0, 0). ab==2, 4 . The vertices are ()0, 2 and ab==2, 2 . The vertices are ()2,0 and ()0,− 2 . Find the value of c: ()−2,0 . Find the value of c: cab222=+=+=41620 →= c 2025 = cab222=+=+=→=448 c 8 = 22 The foci are ()0, 2 5 and () 0,− 2 5 . The foci are ()22,0 and ()− 22,0. The transverse axis is the y-axis. The asymptotes The transverse axis is the x-axis. 11 The asymptotes are yxy==−; x. are yxy==− and x. 22

35. yx22−=25 22−= 33. yx99 Divide both sides by 25 to put in standard form: Divide both sides by 9 to put in standard form: 2 2 y −=x yy2299x2 1. −= −=x2 1 25 25 9999 The center of the hyperbola is at (0, 0). The center of the hyperbola is at (0, 0). ab==5, 5 . The vertices are ()0,5 and ab==3, 1 . ()0,− 5 . Find the value of c: The vertices are (0, 3) and (0, –3). Find the value of c: cab222=+=+=25 25 50

cab222=+=+=9110 == c 50 5 2 c = 10 The foci are ()0,5 2 and () 0,− 5 2 . − The foci are ()0, 10 and () 0, 10 . The transverse axis is the y-axis. The transverse axis is the y-axis. The asymptotes are yxy==−; x. The asymptotes are yxyx==−3; 3 .

1001 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

36. 24xy22−= 39. The center of the hyperbola is at (0, 0). == Divide both sides by 4 to put in standard form: ab6, 3 . 2 − x2 y The vertices are ()0, 6 and ()0,6 . Find the −=1. 24 value of c: The center of the hyperbola is at (0, 0). cab222=+=+=36 9 45

ab==2, 2 . c ==45 3 5 − The vertices are ()2, 0 and () 2, 0 . The foci are ()0,− 3 5 and () 0,3 5 . Find the value of c: The transverse axis is the y-axis. cab222=+=+=→=246 c 6 The asymptotes are yxyx==−2; 2 . The The foci are ()6, 0 and ()− 6, 0 . y2 x2 equation is: −=1. The transverse axis is the x-axis. 36 9 The asymptotes are yxyx==−2; 2 . 40. The center of the hyperbola is at (0, 0). ab==2, 4 . The vertices are ()()−2, 0 and 2, 0 . Find the value of c: cab222=+=+=41620

c ==20 2 5 The foci are ()−25,0 and ()25,0. The transverse axis is the x-axis. ==− The asymptotes are yxyx2; 2 . x2 y2 37. The center of the hyperbola is at (0, 0). The equation is: −=1. 416 ab==1, 1 . The vertices are ()1, 0 and (− 1, 0 ). Find the value of c: 41. Center: (4, –1); Focus: (7, –1); Vertex: (6, –1); cab222=+=+=11 2 Transverse axis is parallel to the x-axis; ac==2; 3 . c = 2 Find the value of b: ()()− The foci are 2, 0 and 2, 0 . bca222=−=−=945 b= 5 The transverse axis is the x-axis. (4)(1)xy−+22 ==− Write the equation: −=1 . The asymptotes are yxyx; . 45 The equation is: xy22−=1 .

38. The center of the hyperbola is at (0, 0). ab==1, 1 . The vertices are ()()0,− 1 and 0,1 . Find the value of c: cab222=+=+=11 2

c = 2 The foci are ()0,− 2 and () 0, 2 . The transverse axis is the y-axis.

The asymptotes are yxyx==−; . The equation is: yx22−=1 .

1002 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

42. Center: (–3, 1); Focus: (–3, 6); Vertex: (–3, 4); 44. Center: (1, 4); Focus: (–2, 4); Vertex: (0, 4); Transverse axis is parallel to the y-axis; Transverse axis is parallel to the x-axis; ac==3; 5 . Find the value of b: ac==1; 3 . Find the value of b: bca222=−=−=25 9 16 b= 4 bca222=−=−=91 8

(1)(3)yx−+22 == Write the equation: −=1. b 822 916 (4)y − 2 Write the equation: (1)x −−2 = 1. 8

43. Center: (–3, –4); Focus: (–3, –8); Vertex: (–3, –2); 45. Foci: (3, 7), (7, 7); Vertex: (6, 7); Transverse axis is parallel to the y-axis; Center: (5, 7); Transverse axis is parallel to the ac==2; 4 . x-axis; ac==1; 2 . Find the value of b: Find the value of b: bca222=−=−=16 4 12 bca222=−=−=41 3

b ==12 2 3 b = 3 22 (4)(3)yx++ (7)y − 2 Write the equation: −=1. Write the equation: (5)x −−2 = 1. 412 3

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46. Focus: (–4, 0); Vertices: (–4, 4), (–4, 2); 48. Vertices: (1, –3), (1, 1); Center: (1, –1); Center: (–4, 3); Transverse axis is parallel to Transverse axis is parallel to the y-axis; a = 2 . the y-axis; ac==1; 3 . 3 Asymptote: yx+=11() − ind the value of b: 2 bca222=−=−=91 8 Using the slope of the asymptote, find the value of b: == b 822 a 23 4 2 ==  34bb=  = 2 (4)x + bb23 Write the equation: (3)y −− = 1. 8 Find the value of c: 16 52 52 2 13 cab222=+=+=4  c== 99 9 3 (1)9(1)yx+−22 Write the equation: −=1 . 416

47. Vertices: (–1, –1), (3, –1); Center: (1, –1); Transverse axis is parallel to the x-axis; a = 2 . 3 Asymptote: yx+=11() − 2 Using the slope of the asymptote, find the value of b: bb3 ==  b = 3 (2)(3)xy−+22 a 22 49. −=1 49 Find the value of c: The center of the hyperbola is at (2, –3). cab222=+=+=4913 ab==2, 3 . c = 13 The vertices are (0, –3) and (4, –3). (1)(1)xy−+22 Find the value of c: Write the equation: −=1. 49 cab222=+=+=4 9 13 c= 13 Foci: ()2−− 13, 3 and () 2 +− 13, 3 . Transverse axis: y =−3 , parallel to x-axis. 3 Asymptotes: yx+= 3 ( − 2); 2 3 yx+=− 3 ( − 2) 2

1004 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

(3)(2)yx+−22 52. (4)9(3)9xy+−−=22 50. −=1 49 Divide both sides by 9 to put in standard form: The center of the hyperbola is at (2, –3). (4)x + 2 == −−(3)1y 2 =. ab2, 3 . The vertices are (2, –1) and (2, –5). 9 Find the value of c: The center of the hyperbola is (–4, 3). cab222=+=+=4 9 13 c= 13 ab==3, 1 . Foci: ()2,−− 3 13 and ()2,−+ 3 13 The vertices are (–7, 3) and (–1, 3). Find the value of c: Transverse axis: x = 2 , parallel to the y-axis cab222=+=+=9110 c= 10 2 Asymptotes: yx+= 3 ( − 2); −− −+ 3 Foci:() 4 10, 3 and () 4 10, 3

2 = yx+=− 3 ( − 2) Transverse axis: y 3 , parallel to the x-axis. 3 11 Asymptotes: yxyx−=3 ( + 4), −=−3 ( + 4) 33

51. (2)4(2)4yx−−+=22 Divide both sides by 4 to put in standard form: 53. (1)(xy+−+=22 2)4 (2)y − 2 Divide both sides by 4 to put in standard form: −+(2)1x 2 =. 4 (1)(xy++22 2) −=1. The center of the hyperbola is at (–2, 2). 44 ab==2, 1 . The center of the hyperbola is (–1, –2). The vertices are (–2, 4) and (–2, 0). Find the ab==2, 2 . value of c: The vertices are (–3, –2) and (1, –2). cab222=+=+=41 5 c= 5 Find the value of c: Foci: ()−−2, 2 5 and () −+ 2, 2 5 . cab222=+=+=448 c== 8 22 Transverse axis: x =−2 , parallel to the y-axis. Foci:()−− 1 2 2, − 2 and () −+ 1 2 2, − 2 Asymptotes: yxyx−=22(2); + −=−22(2) + . Transverse axis: y =−2 , parallel to the x-axis. Asymptotes: yxy+=+ 2 1; +=−+ 2 ( x 1)

1005 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

54. (3)(2)4yx−−+=22 56. Complete the squares to put in standard form: 22 Divide both sides by 4 to put in standard form: yx−−+−=4410 yx (3)(2)yx−+22 (yy22−+−−+=+− 4 4) ( xx 4 4) 1 4 4 −=1. The center of the 44 (2)(2)1yx−−−=22 hyperbola is at (–2, 3). ab==2, 2 . The center of the hyperbola is (2, 2). The vertices are (–2, 5) and (–2, 1). ab==1, 1 . The vertices are ()2,1 and ()2,3 . Find the value of c: Find the value of c: cab222=+=+=448 c== 8 22 cab222=+=+=11 2 c= 2 Foci:()−− 2, 3 2 2 and () −+ 2, 3 2 2 Foci: ()2, 2−+ 2 and () 2, 2 2 . Transverse axis: x =−2 , parallel to the y-axis. Transverse axis: x = 2 , parallel to the y-axis. Asymptotes: yx−=+ 3 2; Asymptotes: yxy−=−22; −=−−2(2) x. yx−=−+ 3 ( 2)

57. Complete the squares to put in standard form: yxyx22−−−−= 44840 (44)4(21)444yy22−+− xx ++=+− 55. Complete the squares to put in standard form: −−+=22 xy22−−−−=2210 xy (2)4(1)4yx 2 22 (2)y − (xx−+−++=+− 2 1) ( yy 2 1) 1 1 1 −+(1)1x 2 = 4 −−+=22 (1)(1)1xy The center of the hyperbola is (–1, 2). The center of the hyperbola is (1, –1). ab==2, 1 . == ab1, 1 . The vertices are (0, –1) and The vertices are (–1, 4) and (–1, 0). (2, –1). Find the value of c: Find the value of c: cab222=+=+=11 2 c= 2 cab222=+=+=41 5 c= 5 Foci: ()12,1−− and ()12,1 +−. Foci: ()−−1, 2 5 and () −+ 1, 2 5 . Transverse axis: y =−1, parallel to x-axis. Transverse axis: x =−1, parallel to the y-axis. −= + −=− + Asymptotes: yxy+=11; − +=−−1(1) x. Asymptotes: yxyx22(1); 22(1).

1006 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

58. Complete the squares to put in standard form: 24440xy22−++−= xy 2(xx22++−−+=+− 2 1) ( yy 4 4) 4 2 4 2(xy+−−= 1)22 ( 2) 2 (2)y − 2 (1)x +−2 = 1 2 The center of the hyperbola is (–1, 2). ab==1, 2 .

The vertices are (–2, 2) and (0, 2). Find the value of c: 60. Complete the squares to put in standard form: cab222=+=+=12 3 c= 3 22830yx22−+++= xy Foci: ()−−1 3,2 and () −+ 1 3,2 . 2(yy22++−−+=−+− 4 4) ( xx 2 1) 3 8 1 Transverse axis: y = 2 , parallel to the x-axis. 2(yx+−−= 2)22 ( 1) 4 Asymptotes: 22 (2)(1)yx+− −= + −=− + −=1 yxyx22(1); 22(1). 24 The center of the hyperbola is (1,–2). ab==2, 2 . Vertices: ()1,−− 2 2 and ()1,−+ 2 2 Find the value of c: cab222=+=+=246

c = 6 Foci: ()1,−− 2 6 and () 1, −+ 2 6 . Transverse axis: x = 1 , parallel to the y-axis. 2 Asymptotes: yx+=2 ( − 1); 2

2 59. Complete the squares to put in standard form: yx+=− 2 ( − 1) 4244160xy22−− xy −+= 2 4(xx22−+− 6 9) ( yy ++=−+− 4 4) 16 36 4 4(xy−−+= 3)22 ( 2) 16 (3)(2)xy−+22 −=1 416 The center of the hyperbola is (3, –2). ab==2, 4 . The vertices are (1, –2) and (5, –2). Find the value of c: cab222=+=+=41620

c ==20 2 5 Foci: ()3−− 2 5, 2 and () 3 +− 2 5, 2 . Transverse axis: y =−2 , parallel to x-axis. Asymptotes: yx+= 2 2( − 3);

yx+=− 2 2( − 3)

1007 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

61. Complete the squares to put in standard form: yx22−−−−=4162190 xy (yy22−+− 2 1) 4( xx ++=+− 4 4) 19 1 16 (1)4(2)4yx−−22 + = (1)y − 2 −+(2)1x 2 = 4 The center of the hyperbola is (–2, 1). ab==2, 1 . The vertices are (–2, 3) and (–2, –1). Find the 63. Rewrite the equation: value of c: =+2 cab222=+=+=41 5 yx16 4 22=+ ≥ c = 5 yxy16 4 , 0 22−= ≥ Foci: ()−−2, 1 5 and () −+ 2, 1 5 . yx416, y 0 y2 x2 Transverse axis: x =−2 , parallel to the y-axis. −=1,y ≥ 0 16 4 Asymptotes: yxyx−=12(2); + −=−1 2(2) + .

64. Rewrite the equation: yx=−99 + 2 62. Complete the squares to put in standard form: 22=+ ≤ xyxy22−+−+=38640 yxy99 , 0 22−= ≤ (816)3(21)4163xx22++ − yy + +=−+− yx99, y 0 2 22 y x2 (4)3(1)9xy+−+= −=1,y ≤ 0 91 (4)(1)xy++22 −=1 93 The center of the hyperbola is (–4, –1). ab==3, 3 . The vertices are (–7, –1) and (–1, –1). Find the value of c: cab222=+=+=9312

c ==12 2 3 Foci: ()−−4 2 3, − 1 and () −+ 4 2 3, − 1 .

Transverse axis: y =−1 , parallel to x-axis. Asymptotes: 33 yxyx+=1(4); + +=−1 (4) + 33

1008 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

65. Rewrite the equation: yx=− −25 + 2 yxy22=−25 + , ≤ 0

xy22−=25, y ≤ 0 x2 y2 −=1,y ≤ 0 25 25

(2)(2)yx+−22 68. −=1 16 4 The graph will be a hyperbola. The center of the hyperbola is at (2, –2). ab==4, 2 . The vertices are (2, 2) and (2, –6). Find the value of c: 222 cab=+=+=16 4 20 c== 20 2 5 Foci: ()2,−− 2 2 5 and ()2,−+ 2 2 5 66. Rewrite the equation: = yx=−+1 2 Transverse axis: x 2 , parallel to the y-axis Asymptotes: yxyx+=22(2); − +=−−22(2) yxy22=−1, + ≥ 0 xy22−=1, y ≥ 0

69. xy2 =−16( 3)

The graph will be a parabola. The equation is in (3)xy− 22 xh−=2 ayk − a = 67. −=1 the form ()4()where 416 or 425 a = 4 , hk==0, and 3 . Thus, we have: The graph will be a hyperbola. The center of the Vertex: (0, 3); Focus: (0, 7) ; Directrix: y =−1 hyperbola is at (3, 0). ab==2, 5 . The vertices are (5, 0) and (1, 0). Find the value of c: cab222=+=+=42529 c= 29 Foci: ()3−+ 29, 0 and () 3 29, 0 Transverse axis is the x-axis. 55 Asymptotes: yx=− ( 3); y =−− ( x 3) 22

1009 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

70. yx2 =−12( + 1) 72. The graph will be an ellipse. Complete the square to put the equation in standard form: The graph will be a parabola. The equation is in xyxy22+−++=36 2 288 541 0 the form ()yk−=−−2 4() axh where –4a =− 12 22−+ + =− or a = 3 , hk=−1, and = 0 . Thus, we have: (2)(36288)541xx y y Vertex: (1,0)− ; Focus: (4,0)− ; Directrix: x = 2 (2)36(8)541xx22−+ yy +=− (xx22−++ 2 1) 36( yy ++ 8 16) =−++ 541 1 576 (xy−+ 1)22 36( + 4) = 36 −+22 (xy 1)+= 36( 4) 36 36 36 36 (1)x − 2 ++(4)1y 2 = 36 The equation is in the form −−22 ()()xh+= yk 1 (major axis parallel to the ab22 71. The graph will be an ellipse. Complete the x-axis) where abh===6, 1, 1, and k =− 4 . square to put the equation in standard form: Solving for c: 25xy22+− 9 250 x += 400 0 cab222=−=−=→=36 1 35 c 35 (25xxy22−+=− 250 ) 9 400 Thus, we have: Center: (1,− 4) 25(xxy22−+=− 10 ) 9 400 Foci: ()1−− 35, 4 ,() 1 +− 35, 4 25(xx22−++ 10 25) 9 y =−+ 400 625 Vertices: (7,− 4) , (5,4)−− 25(xy−+ 5)22 9 = 225 − 22 25(xy 5)+= 9 225 225 225 225 (5)xy− 22 +=1 925 The equation is in the form ()()xh−−22 yk +=1 (major axis parallel to the 22 ba y-axis) where abh===5, 3, 5, and k = 0 . 222=−=−=→= 73. The graph will be a parabola. Complete the square Solving for c: cab25 9 16 c 4 to put the equation in standard form: Thus, we have: 2 −−−= Center: (5, 0) xxy68310 Foci: (5, 4), (5, –4) xxy2 −=+6831 Vertices: (5, 5), (5, –5) 2 xx−+=++698319 y (3)840xy−=+2 (3)8(5)xy−=2 + The equation is in the form ()4()xh−=2 ayk −where 48a = or a = 2 , hk==−3, and 5 . Thus, we have: Vertex: (3,− 5) ; Focus: (3,− 3) ; =− Directrix: y 7

1010 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

75. First note that all points where a burst could take place, such that the time difference would be the same as that for the first burst, would form a hyperbola with A and B as the foci. Start with a diagram: ()x, y Ν

B A

2 miles 74. The graph will be a hyperbola. Complete the Assume a coordinate system with the x-axis squares to put in standard form: containing BA and the origin at the midpoint of 9188880xy22−− xy −−= BA . 22 ()xy 9(xx−+−++ 2 1) ( yy 8 16) =+− 88 9 16 The ordered pair , represents the location of −−+=22 the fireworks. We know that travels at 1100 9(xy 1) ( 4) 81 feet per second, so the person at point A is 1100 (1)(xy−+22 4) feet closer to the fireworks display than the person −=1 981 at point B. Since the difference of the distance The center of the hyperbola is (1,− 4) . from ()xy, to A and from ()xy, to B is the ab==9, 81 . constant 1100, the point ()xy, lies on a hyperbola The vertices are (2,4)−− and (4,− 4) . whose foci are at A and B. The hyperbola has the Find the value of c: equation 2 2 cab222=+=+=98190 x −=y 1 ab22 c ==90 3 10 where 2a = 1100 , so a = 550 . Because the Foci: ()1−− 3 10, 4 and () 1 +− 3 10, 4 . distance between the two people is 2 miles (10,560 Transverse axis: x =−2 , parallel to the y-axis. feet) and each person is at a focus of the hyperbola, we have += − +=−− Asymptotes: yxyx43(1); 4 3(1). 210,560c = c = 5280 bca222=−=5280 2 − 550 2 = 27,575,900 The equation of the hyperbola that describes the location of the fireworks display is x2 y2 −=1 5502 27,575,900 Since the fireworks display is due north of the individual at A, we let x = 5280 and solve the equation for y. 52802 y2 −=1 5502 27,575,900

y2 −=−91.16 27,575,900 y2 = 2,513,819,044 y = 50,138 Therefore, the fireworks display was 50,138 feet (approximately 9.5 miles) due north of the person at point A.

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76. First note that all points where the strike could 77. To determine the height, we first need to obtain take place, such that the time difference would the equation of the hyperbola used to generate be the same as that for the first strike, would the . Placing the center of the form a hyperbola with A and B as the foci. hyperbola at the origin, the equation of the Start with a diagram: xy22 ()x, y Ν hyperbola will have the form −=1. ab22 The center diameter is 200 feet so we have 200 a ==100 . We also know that the base B A 2 diameter is 400 feet. Since the center of the 1 mile ()− Assume a coordinate system with the x-axis hyperbola is at the origin, the points 200, 360 containing BA and the origin at the midpoint of and ()−−200, 360 must be on the graph of our BA . hyperbola (recall the center is 360 feet above The ordered pair ()xy, represents the location ground). Therefore, 22 of the lightning strike. We know that sound ()200()− 360 −=1 travels at 1100 feet per second, so the person at ()100 22b point A is 2200 feet closer to the lightning strike than the person at point B. Since the difference 3602 41−= of the distance from ()xy, to A and from ()xy, b2 to B is the constant 2200, the point ()xy, lies on 3602 3 = a hyperbola whose foci are at A and B. The b2 hyperbola has the equation b2 = 43,200 x2 y2 −=1 b ==43,200 120 3 22 ab The equation of the hyperbola is where 22200a = , so a = 1100 . Because the xy22 distance between the two people is 1 mile (5,280 −=1 feet) and each person is at a focus of the 10,000 43, 200 hyperbola, we have 300 = At the top of the tower we have x ==150 . 25,280c 2 = c 2,640 15022y 222 2 2 −=1 bca=−=2640 − 1100 = 5,759,600 10,000 43, 200 The equation of the hyperbola that describes the y2 location of the lightning strike is = 1.25 43,200 x2 y2 −=1 2 = 11002 5,759,600 y 54000 Since the lightning strike is due north of the y ≈ 232.4 individual at A, we let x = 2640 and solve for y. The height of the tower is approximately 26402 y2 232.4+= 360 592.4 feet. −=1 11002 5,759,600 y2 −=−4.76 5,759,600 y2 = 27,415,696 y = 5236 The lightning strikes 5236 feet (approximately 0.99 miles) due north of the person at point A.

1012 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.4: The Hyperbola

78. First note that all points where an explosion 80. Assume the origin lies at the center of the could take place, such that the time difference hyperbola. From the equation we know that the would be the same as that for the first detonation, hyperbola has a transverse axis that is parallel to would form a hyperbola with A and B as the the y-axis. The foci of the hyperbola are located foci. Start with a diagram: at ()0, ±c Ν 222=+=+= = D2 cab91625 or c 5 (1200, y ) Therefore, the foci of the hyperbola are at ()0,− 5 and ()0,5 . − ( 1200, 0) D1 (1200, 0) If we assume the parabola opens up, the common (0, 0) (,a 0) A B focus is at ()0,5 . The equation of our parabola 2400 ft will be xayk2 =−4 (). The focal length of the Since A and B are the foci, we have 2cc= 2400  = 1200 parabola is given as a = 6 . We also know that the distance focus of the parabola is located at Since D1 is on the transverse axis and is on the += hyperbola, then it must be a vertex of the ()()0,ka 0,5 . Thus, hyperbola. Since it is 300 feet from B, we have ka+=5 = a 900 . Finally, k +=65 222=−= 2 − 2 = bca1200 900 630,000 k =−1 Thus, the equation of the hyperbola is and the equation of our parabola becomes x2 y2 −=1 xy2 =−−46()() ( 1 ) 810,000 630,000 2 =+() The point ()1200, y needs to lie on the graph of xy24 1 the hyperbola. Thus, we have or 2 1 ()1200 y2 yx=−2 1 . −=1 24 810,000 630,000 y2 7 x2 y2 −=− 81. Assume −=1. 630,000 9 ab22 y2 = 490,000 If the eccentricity is close to 1, then ca≈≈and b0 . When b is close to 0, the y = 700 hyperbola is very narrow, because the slopes of The second explosion should be set off 700 feet the asymptotes are close to 0. due north of point B. If the eccentricity is very large, then 79. a. Since the particles are deflected at a 45° cabis much larger than and is very large. The , the asymptotes will be yx=± . result is a hyperbola that is very wide, because the slopes of the asymptotes are very large. b. Since the vertex is 10 cm from the center of = the hyperbola, we know that a 10 . The yx22 b For −=1, the opposite is true. When the slope of the asymptotes is given by ± . ab22 a eccentricity is close to 1, the hyperbola is very Therefore, we have wide because the slopes of the asymptotes are bb close to 0. When the eccentricity is very large, = 1  = 1  b = 10 a 10 the hyperbola is very narrow because the slopes Using the origin as the center of the of the asymptotes are very large. hyperbola, the equation of the particle path would be xy22 −=1 , x ≥ 0 100 100

1013 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

82. If ab= , then caa222=+=2 a 2. Thus, 85. Put the equation in standard hyperbola form: 22++= ⋅<≠ cc2 Ax Cy F00,0 A C F ==2 or 2 . The eccentricity of an a2 a Ax22+=− Cy F equilateral hyperbola is 2 . Ax2 Cy2 +=1 −−FF 2 x 2 2 83. −=yab2 1( = 2,1) = x y 4 +=1 FF This is a hyperbola with horizontal transverse −− AC axis, centered at (0, 0) and has asymptotes:  1 Since −−FA/ and FC / have opposite signs, yx=± 2 this is a hyperbola with center (0, 0). 2 The transverse axis is the x-axis if −>FA/ 0 . 2 x yab−=1( = 1,2) = y- −

parentheses. Since we are multiplying by C, we 2 2 y x 22 84. −=1 EE 22 add C ⋅= to the right side of the ab 4C 2 4C Solve for y: equation. Therefore, we obtain the following: 2 2 y x 2 2 22 =+1 A()xCy+++=+−DEDE() F ab22 2244ACAC x2 =+−DE22 ya22=+1 Let UF. 2 44AC b a. If U ≠ 0 , then 22 2 ax b 22 y2 =+1 DE 22 xy++ bx 22AC +=1 ax b2 UU y =± +1 AC b x2 U U b2 with and having opposite signs. This As xx→−∞ or as →∞, the term gets A C 2 x is the equation of a hyperbola whose center close to 0, so the expression under the radical DE gets closer to 1. Thus, the graph of the hyperbola is −−, . 22A C aa gets closer to the lines yxyx=− and = . bb

These lines are the asymptotes of the hyperbola.

1014 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: ; General Form of a Conic

b. If U = 0 , then ACAC>≠0 and , the equation defines an 2 2 ellipse. Ax()+++=DE Cy()0 22AC 2 −A 2 14. 28420xy22+−++= xy ()yx+=ED() + 22CAC ACAC====2 and 1; (2)(1) 2. Since −A >≠ yx+=±ED() + ACAC0 and , the equation defines an 22CAC ellipse. which is the graph of two intersecting lines, DE 15. 32640xyx22−++= through the point −−, , with 22A C ACAC==−=−=−3 and 2; (3)( 2) 6 . Since A AC < 0 , the equation defines a hyperbola. slopes ± . C 16. 438610xyxy22−−++= Section 10.5 ACAC==−=−=−4 and 3; (4)( 3) 12. Since < 1. sinA cosBAB+ cos sin AC 0 , the equation defines a hyperbola.

2. 2sinθθ cos 17. 20yxyx22−−+= ACAC=−1 and = 2; = ( − 1)(2) =− 2 . Since 1cos− θ < 3. AC 0 , the equation defines a hyperbola. 2 18. yxxy22−−−=82 0 1cos+ θ =− = = − =− 4. ACAC8 and 1; ( 8)(1) 8. Since 2 AC < 0 , the equation defines a hyperbola.

A − C 22 5. cot() 2θ = 19. xy+−+=84 xy 0 B ACAC====1 and 1; (1)(1) 1. Since 6. parabola ACAC>=0 and , the equation defines a circle.

7. BAC2 −<40 20. 22880xyxy22+−+= ACAC====2 and 2; (2)(2) 4 . Since 8. True ACAC>=0 and , the equation defines a circle. 9. True 21. xxyy22++−=430 A − C == = 10. False; cot() 2θ = AB1, 4, and C 1; B AC−−11 0 cot() 2θ ==== 0 2 +++= B 44 11. xxy430 ππ ACAC====1 and 0; (1)(0) 0 . Since 2θθ=  = 24 AC = 0 , the equation defines a parabola. ππ xx=−′′cos y sin 44 12. 2330yyx2 −+= 22 2 ACAC====0 and 2; (0)(2) 0 . Since =−xy′′ =() xy ′′ − 22 2 AC = 0 , the equation defines a parabola. ππ22 yx=+′′sin y cos =+ x ′′ y 13. 631260xy22+−+= xy 4422 2 ACAC====6 and 3; (6)(3) 18 . Since =+()xy′′ 2 1015 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

22. xxyy22−+−=430 25. 13xxyy22−+−= 6 3 7 16 0 AB==−1, 4, and C = 1; AB==−13, 6 3, and C =7; AC−−11 0 −− cot() 2θ ==== 0 ()θ ====−AC 13 7 6 3 B −−44 cot 2 B −−63 63 3 ππ 2θθ=  = 2ππ 24 2θθ=  = 33 ππ ′′22 ′′ ππ xx=−=−cos y sin x y ′′13 ′′ 4422 xx=−=−cos y sin x y 3322 =−2 ′′ 1 ()xy =−()xy′′3 2 2 ππ =+′′ =+22 ′′ ππ31 yxsin y cos x y yx=+′′sin y cos =+ x ′′ y 4422 3322 2 1 =+()xy′′ =+()3xy′′ 2 2

22 23. 56580xxyy++−= 26. 11xxyy22++−= 10 3 4 0 == = AB5, 6, and C5; AB==11, 10 3, and C =1; AC−−55 0 cot() 2θ ==== 0 AC−−11 1 10 3 B 66 cot() 2θ ==== ππ B 10 3 10 3 3 2θθ=  = ππ 24 2θθ=  = 36 ππ22 xx=−=−′′cos y sin x ′′ y ππ31 4422 xx=−=−′′cos y sin x ′′ y 6622 2 =−()xy′′ 1 2 =−()3xy′′ 2 ππ22 yx=+′′sin y cos =+ x ′′ y ππ13 4422 yx=+′′sin y cos =+ x ′′ y 6622 =+2 ()′′ 1 xy =+()xy′′3 2 2

24. 3103xxyy22−+−= 320 AB==−3, 10, and C = 3; AC−−33 0 cot() 2θ ==== 0  B −−10 10 ππ 2θθ=  = 24 ππ22 xx=−=−′′cos y sin x ′′ y 4422 2 =−()xy′′ 2 ππ22 yx=+′′sin y cos =+ x ′′ y 4422 2 =+()xy′′ 2 1016 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: Rotation of Axes; General Form of a Conic

27. 44xxyy22−+− 851650 x − y = 29. 25xxyy22−+ 36 40 − 12 13 x − 8 13 y = 0 AB==−=4, 4, and C1; AB==−=25, 36, and C40; AC−−41 3 3 AC−−25 40 5 5 cot() 2θθ===−=− ; cos 2 cot() 2θθ== = ; cos 2 = B − 44 5 B −36 12 13

3 − 5 1−− 1 5 4225 13 42213 sinθ ==== ; sinθ = === ; 255 5 2131313 3 + 5 1+− 1  θ ====13 93313 θ =5 ===11 5 cos cos 2131313 255 5 ′′313 ′ 213 ′ ′′525 ′′ xx=−=cosθθ y sin x − y xx=−=−cosθθ y sin x y 13 13 55 13 ′′ 5 ′′ =−()32xy =−()xy2 13 5 ′′213 ′ 313 ′ ′′25 ′′ 5 yx=+sinθθ y cos = x + y yx=+sinθθ y cos =+ x y 13 13 55 13 ′′ 5 ′′ =+()23xy =+()2xy 13 5

28. xxyy22+++445550 y += AB==1, 4, and C =4; AC−−14 3 3 cot() 2θθ===−=− ; cos 2 B 44 5 3 1−− 5 4225 sinθ ==== ; 255 5 3 1+− 5 11 5 cosθ = === 255 5 525 xx=−=−′′cosθθ y sin x ′′ y 55 5 =−()xy′′2 5 25 5 yx=+′′sinθθ y cos =+ x ′′ y 55 5 =+()2xy′′ 5

1017 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

30. 34xxyy22−+−= 24 41 25 0 AB==−=34, 24, and C41; AC−−34 41 7 7 cot() 2θθ== = ; cos() 2 = B − 24 24 25 77 11−+ 93 164 sinθθ======25 ; cos 25 2255 2255 431 xx=−=−=−′′cosθθ y sin x ′′′′ y() 4 x 3 y 555 341 yx=+′′sinθθ y cos =+=+ x ′′′′ y() 3 x 4 y 55 5

31. xxyy22++−=430; θ = 45º

22y 2222 ()xy′′−+() xy ′′ −() xy ′′ ++() xy ′′ +−= 3 x' 430 2222 y' 11 ()()x′222222−++−+++−=22 xyy ′′′ x ′ y ′() x ′ 230 xyy ′′′ (1, 0) 22 x 11 11 –3 xxyy′′′′′222222−+ +22 x − y ′ + xxyy ′′′′ ++ = 3 (–1, 0) 22 22 33xy′′22−= xy′′22 –3 −=1 13 Hyperbola; center at the origin, transverse axis is the x′ -axis, vertices (±1, 0).

32. xxyy22−+−=430; θ = 45º

22y 2222 x' ()xy′′−−() xy ′′ −() xy ′′ ++() xy ′′ +−= y' 3 430 2222 11 ()()xxyyxy′′′′′′222222−+−−+22() xxyy ′′′′ ++−= 230(0, 1) 22 x 11 11–3 3 xxyy′′′′222222−+ −22 x ′ + y ′ + xxyy ′′′′ ++ = 3 (0, –1) 22 22 −+xy′′2233 = yx′′22 –3 −=1 13 Hyperbola; center at the origin, transverse axis is the y′ -axis, vertices (0, ±1).

1018 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: Rotation of Axes; General Form of a Conic

33. 56580xxyy22++−=; θ = 45º

22y 2222 ()xy′′−+() xy ′′ −() xy ′′ ++() xy ′′ +−= 3 x' 56 580 2222 y' 55(0, 2) ()()x′′′′′′222222−++−+++−=23 xyy x y() x ′′′′ 280 xyy 22 x 55222222 55–3 3 xxyyx′′′′′−+53358 +−+++ y ′ xxyy ′′′′ = 22 22 (0, –2) 828xy′′22+= xy′′22 –3 +=1 14 Ellipse; center at the origin, major axis is the y′ -axis, vertices (0, ±2).

34. 3103xxyy22−+−= 320; θ = 45º

22y 2222 ()xy′′−−() xy ′′ −() xy ′′ ++() xy ′′ +−= 5 x' 3103320 y' 2222 33 ()()x′222222−+−−+++−=25 xyy ′′′ x ′ y ′() x ′ 2320 xyy ′′′ (0, 2) 22 x 33 33–5 5 xxyyxy′′′′′′222222−+355332 −++++ xxyy ′′′′ = (0, –2) 22 22 −+2832xy′′22 = yx′′22 –5 −=1 416 Hyperbola; center at the origin, transverse axis is the y′ -axis, vertices (0, ±2).

35. 13xxyy22−+−= 6 3 7 16 0 ; θ = 60º y 2 2 1111 x' 13()xy′′−− 3 6 3() xy ′′ − 3() 3 xy ′′ +++−= 7() 3 xy ′′ 16 0 3 2222 (2, 0) 13 3 3 7 y' ()xxyy′′′′222222−+−23 3() 3 xxyyxxyy ′′′′′′′′ −−+++= 2 3() 3 23 16 42 4 x –3 3 13 13 3 39 9 9 21 7 3 7 xxyyxxyyxxyy′′′′′′′′′′′′222222−+−+++++=33 16 42 42 2424 (–2, 0) ′2 + ′2 = 416x y 16 –3

xy′′22 +=1 41 Ellipse; center at the origin, major axis is the x′ -axis, vertices (±2, 0).

1019 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

36. 11xxyy22++−= 10 3 4 0 ; θ = 30º

2 2 y 1111′′ ′′ ′ ′ ′ ′ y' 11() 3xy−+ 10 3() 3 xy −() x + 3 y +() x + 3 y −= 4 0 1 2222 1, 0 x' 2 11 5 3 1 ()323xxyy′′′′222222−++() 32 xxyyxxyy ′′′′′′′′ +−+++= 3() 233 4 42 4 x –1 1 33 11 3 11 15 15 1 3 3 xxyyxxyyxxyy′′′′′′′′′′′′222222−+++−+++=53 4 –,1 0 42 42 2424 2 16xy′′22− 4 = 4 –1

41xy′′22−= xy''22 −=1 1 1 4 Hyperbola; center at the origin, transverse axis is the x′ -axis, vertices (±0.5, 0).

37. 44xxyy22−+− 851650 x − y =; θ ≈ 63.4º

2 2 y 5555 x' ()xy′′−−() xy ′′ −() xy ′′ +++() xy ′′ 42422 2 4 5555 55y' −−−+=()xy′′() xy ′′ (2, 0) 85 2 165 2 0 55 x 44 1 –4 4 ()xxyy′′′′222222−+44 −() 232 xxyy ′′′′ −− +() 44 xxyy ′′′′ ++ 55 5 − 8x′ +−−=16yxy′′′ 32 16 0 4161681284 4 1 –4 x′222222−+−+++++−= xyy ′′′ x ′ xyyx ′′′′ xyyx ′′′′40 0 55 555 5555 5yx′′2 −= 40 0 yx′′2 = 8 Parabola; vertex at the origin, axis of symmetry is the x ' axis, focus at (2, 0).

1020 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: Rotation of Axes; General Form of a Conic

38. xxyy22+++445550 y +=; θ ≈ 63.4º 2 2 5555 ()xy′−+ ′() xy ′ − ′() xy ′′ ++() xy ′′ + 24 2 2 42 5555 5 +++=()xy′′ 5 5 2 5 0 5 14 4 ()xxyy′′′′222222−+44 +() 232 xxyy ′′′′ −− +() 44 xxyy ′′′′ ++ 55 5 +++= 10xy′′ 5 5 0 14 4812816164 x′222222−++− xyyx ′′′′ xyy ′′′ −++ x ′ xyy ′′′ + 555555555 +++= 10xy′′ 5 5 0 5xxy′′′2 +++= 10 5 5 0 xx′′2 ++=−−+ 2 1 y ′ 1 1 (xy′′+=− 1)2 ′ 1 Parabola; vertex at (–1, 0), axis of symmetry parallel to the y -axis; focus at ()xy', '=−− 1, . 4

39. 25xxyy22−+− 36 40 12 13 x − 8 13 y = 0 ; θ ≈ 33.7º 2 2 13 13 13 13 ()xy′′−−() xy ′′ −() xy ′′ ++() xy ′′ + 25 3 2 36 3 2 2 3 40 2 3 13 13 13 13 13 13 −−−+=()xy′′() xy ′′ 12 13 3 2 8 13 2 3 0 13 13 25 36 40 ()9124x′′′′222222−+− xyy() 65 x ′′′′ +−+ xyy 6() 4129 x ′′′′ ++ xyy 13 13 13 −+ 36xyxy′′′′ 24 −− 16 24 = 0 225 300 100 216 180 216 xxyyxxyy′′′′′′′′2222−+−−+ 13 13 13 13 13 13 160 480 360 ++xxyyx′′′′′22 + −= 52 0 13 13 13

1021 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

13xyx′′′22+−= 52 52 0 xxy′′′22−+44 = 0 ()xy′′−+242 2 = 4 (2)xy′′− 22 +=1 41 Ellipse; center at (2, 0), major axis is the x′ -axis, vertices (4, 0) and (0, 0). y y' 5

x' (2, 1) (4, 0)

x –3 (2, –1) –1

40. 34xxyy22−+−= 24 41 25 0 ; θ ≈ 36.9º 2 2 1111′′ ′′ ′′ ′′ 34() 4xy−− 3 24() 4 xy − 3() 3 xy ++ 4 41() 3 xy +−= 4 25 0 5555 34 24 41 ()16xxyy′′′′222222−+− 24 9() 12 xxyy ′′′′ +−+ 7 12() 9 xxyy ′′′′ ++ 24 16 = 25 25 25 25 544 816 306 288 168 288 369 984 656 xxyyxxyy′′′′′′′′2222−+−−++xxyy′′′′22++=25 25 25 25 25 25 25 25 25 25 25xy′′22+= 50 25 xy′′22+= 2 1 xy''22 +=1 1 1 2 Ellipse; center at the origin, major axis is the x′ -axis, vertices (±1, 0). y 2 y' x'

(1, 0) x –2 2

(–1, 0)

–2

1022 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: Rotation of Axes; General Form of a Conic

41. 16xxyyxy22++−+= 24 9 130 90 0 AC−−16 9 7 7 AB==16, 24, and C =9; cot() 2θθ = = = cos() 2 = B 24 24 25 77 11−+ 93 164 sinθθθ======25 ; cos 25  ≈ 36.9o 2255 2255 431 xx=−=−=−′′cosθθ y sin x ′′′′ y() 4 x 3 y 555 341 yx=+′′sinθθ y cos =+=+ x ′′′′ y() 3 x 4 y 55 5 22 1111′′ ′′ ′′ ′′ 1643()xy−+ 2443() xy −() 34 xy ++ 934() xy + 5555 11′′ ′′ −−++=130() 4xy 3 90() 3 xy 4 0 55 16 24 ()16xxyy′′′′2222−++ 24 9() 12 xxyy ′′′′ +− 7 12 25 25 9 +++−() 9xxyy′′′′22 24 16 1047854720xyxy′′′′+++= 25 256 384 144 288 168 288 xxyyxxyy′′′′′′′′2222−+++− 25 25 25 25 25 25 81 216 144 ++xxyyxy′′′′′′22 + −+=50 150 0 25 25 25 25x′2 −+50xy′′ 150 = 0 xx′′2 −=− 2 6 y ′ (xy′′−=−+ 1)2 6 1

′′2 1 (xy−=−− 1) 6 6 1 4 Parabola; vertex 1, , focus 1, − ; axis of symmetry parallel to the y ' axis. 6 3 y y' 5 1, 1 6 x'

x –5 1, – 4 3

1023 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

42. 16xxyyxy22++−+= 24 9 60 80 0 AC−−16 9 7 7 AB==16, 24, and C =9; cot() 2θθ = = = cos() 2 = B 24 24 25 77 11−+ 93 164 sinθθθ======25 ; cos 25  ≈ 36.9 2255 2255 431 xx=−=−=−′′cosθθ y sin x ′′′′ y() 4 x 3 y 555 341 yx=+′′sinθθ y cos =+=+ x ′′′′ y() 3 x 4 y 55 5 22 1111′′ ′′ ′′ ′′ 1643()xy−+ 2443() xy −() 34 xy ++ 934() xy + 5555 11′′ ′′ −−++=60() 4xy 3 80() 3 xy 4 0 55 16 24 ()16xxyy′′′′2222−++ 24 9() 12 xxyy ′′′′ +− 7 12 25 25 9 +++−+() 9xxyyx′′′′′22 24 16 48 3648640yxy′′′++= 25 256 384 144 288 168 288 xxyyxxyy′′′′′′′′2222−+++− 25 25 25 25 25 25 81 216 144 ++xxyyy′′′′′22 + += 100 0 25 25 25 25xy′2 +=100′ 0 xy′′2 =− 4 Parabola; vertex (0, 0), axis of symmetry is the y ' axis, focus (0, –1). y 2 y' x'

x –2 2 (0, –1)

–2

43. ABC==1, 3, =−−=−−=> 2 BAC22 4 3 4(1)( 2) 17 0 ; hyperbola

44. AB==−=2, 3, C 4 BAC22 −=−−=−< 4 ( 3) 4(2)(4) 23 0 ; ellipse

45. AB==−=1, 7, C 3 BAC22 −=−−=> 4 ( 7) 4(1)(3) 37 0 ; hyperbola

46. AB==−=2, 3, C 2 BAC22 −=−−=−< 4 ( 3) 4(2)(2) 7 0 ; ellipse

47. AB==9, 12, C = 4 BAC22 −=− 4 12 4(9)(4) = 0 ; parabola

1024 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.5: Rotation of Axes; General Form of a Conic

48. ABC===10, 12, 4 BAC22 −=−=−< 4 12 4(10)(4) 16 0 ; ellipse

49. AB==−=10, 12, C 4 BAC22 −=−−=−< 4 ( 12) 4(10)(4) 16 0 ; ellipse

50. AB==4, 12, C = 9 BAC22 −=− 4 12 4(4)(9) = 0 ; parabola

51. AB==−=−=−−=−<3, 2, C 1 BAC22 4 ( 2) 4(3)(1) 8 0 ; ellipse

52. ABC===3, 2, 1 BAC22 −=−=−< 4 2 4(3)(1) 8 0 ; ellipse

53. AA′ =+cos22θθθθ B sin cos + C sin BB′ =−+−(cos22θθ sin ) 2( CA )(sin θθ cos ) CA′ =−sin22θθθθ B sin cos + C cos DD′ =+cosθθ E sin ED′ =−sinθθ + E cos FF′ =

54. AC′′+=()() Acos2222θθθθ + B sin cos + C sin + A sin θθθθ − B sin cos + C cos

=+++=+=+AC()()cos22θθ sin sin 2 θ cos 2 θ ACAC (1) (1)

55. BB'[(cossin)2()sincos]222=−+−θθ CA θθ 2 =−−[cos2(BACθθ )sin2]2 =−−+−B22cos2θθθθ 2( BAC )sin2cos2 ( AC )sin2 22 4AC ' '=+ 4[ A cos2222θθθθ B sin cos + C sin ][ A sin θθθθ − B sin cos + C cos ] 1cos2+−θθ sin21cos2−+θθ  1cos2 θθ sin2 1cos2 =+4 ABCABC+−()  +() 2222  22 =+θθ + +− θ − θ − θθ++ [(1cos2)ABCAB (sin2) (1cos2)][(1cos2) (sin2)(1cos2)]C =++[(AC ) B sin 2θθ +− ( AC )cos 2 ][( AC +− ) ( B sin 2 θθ +− ( AC )cos 2 )] =+()[sin2()cos2]AC22 − Bθθ +− AC =+()[sin22()sin2cos2()cos2]AC222 − Bθθθθ + BAC − +− AC 22 BACB'222−= 4' ' cos2θθθθ −− 2( BAC )sin2cos2 +− ( AC )sin2 22 −+(AC )222 + B sin2θθθθ + 2( BAC − )sin2cos2 +− ( AC )cos2 22 =++−+−+BACAC22(cos 2θθ sin 2 2 ) ( ) 22 (cos 2 θθ sin 2 2 ) ( ) 2 =+−B22222222()() AC −+ AC =+ B (2 A − ACC + )(2 − A + ACC + ) =−BAC2 4

1025 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

56. Since B22−=−44AC B′′′ A C for any rotation θ (Problem 55), choose θ so that B′ = 0 . Then B2 −=−44AC A′′ C .

a. If BACAC2 −=−=440′′ then AC ′′ =0. Using the theorem for identifying conics without , the equation is a parabola.

b. If BACAC2 −=−<440′′ then AC ′′ >0. Thus, the equation is an ellipse (or circle).

c. If BACAC2 −=−>440′′ then AC ′′ <0. Thus, the equation is a hyperbola.

222=− +− 57. dyyxx()()21 21 =+−−+−−+()()′′′′θθθθ22 ′ θθθθ ′′′ xyxy2211sin cos sin cos x 2 cos yxy 211 sin cos sin =−()()′′θθ +− ′′22 +−() ′′ θθ −− () ′′ ()xx21sin yy 21 cos() xx 21 cos yy 21 sin =−′′2222θθθθ + ′′′′ − − +− ′′ ()xx21sin 2 ()() xxyy 2121 sin cos () yy 21 cos +−()′′2222θθθθ − ()()() ′′′′ − − +− ′′ xx21cos 2 xxyy 2121 sin cos yy 21 sin

=−()′′2222θθ + +−() ′′ 22 θθ + xx21()sin cos yy 21() cos sin

=−()()x′′xyy22+−′′()sin22θθ + cos 2 121 =−()()′′22 +− ′′ xx21 yy 21

58. xya1/ 2+= 1/ 2 1/2 yax1/ 2=− 1/ 2 1/ 2 2 ya=−()1/2 x 1/2

ya=−2 ax1/ 2 1/ 2 + x 1/ 2 1/ 2 =+− 2()ax ax y 4()2()ax=+ a x22 − y a ++ x y 4222ax=+ a22 ax +− x ay − xy + y 2 02=−xxyyaxaya22 +− 22 − + 2 BAC22−4 =− ( 2) − 4(1)(1) =−= 4 4 0 The graph of the equation is part of a parabola.

59 – 60. Answers will vary.

1026 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.6: Polar of Conics

Section 10.6 33 11. r == − θ 42cos − 1 θ 1. rrcosθθ ; sin 41 cos 2 2. r = 6cosθ 3 Begin by multiplying both sides of the equation = 4 ; 2 1 by r to get rr= 6cosθ . Since rxcosθ = and 1cos− θ 2 rxy222=+, we obtain xy22+=6 x. 31 3 Move all the variable terms to the left side of the ep==,; e p = equation and complete the square in x. 42 2 Ellipse; directrix is perpendicular to the polar xxy22−+60 = 3 22−++= axis and units to the left of the pole. ()xx69 y 9 2 ()−+=2 2 xy39 66 12. r == 82sin+ θ 1 3. conic; focus; directrix 81+ sinθ 4 4. 1; < 1; > 1 3 5. True = 4 1 1sin+ θ 6. True 4 31 7. ep==1; 1 ; parabola; directrix is ep==,;3 e p = perpendicular to the polar axis and 1 unit to the 44 right of the pole. Ellipse; directrix is parallel to the polar axis and 3 units above the pole. 8. ep==1; 3 ; parabola; directrix is parallel to 1 the polar axis and 3 units below the pole. 13. r = 1cos+ θ == = 44 ep1, e 1, p 1 9. r == 23sin− θ 3 Parabola; directrix is perpendicular to the polar 21− sinθ axis 1 unit to the right of the pole; vertex is 2 1 2 ,0 . = 2 3  1sin− θ 2 34 ep==2, e ; p = 23 Hyperbola; directrix is parallel to the polar axis 4 and units below the pole. 3

2 10. repep====;2,2;1 12cos+ θ Hyperbola; directrix is perpendicular to the polar axis and 1 unit to the right of the pole.

1027 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

3 10 14. r = 16. r = 1sin− θ 54cos+ θ ep===3, e 1, p 3 10 2 r == Parabola; directrix is parallel to the polar axis 3 4 + 4 θ 51+ cosθ 1cos 33π 5 5 units below the pole; vertex is , . 22 45 ep==2, e , p = 52 Ellipse; directrix is perpendicular to the polar 5 axis units to the right of the pole; vertices are 2 10 , 0 and () 10, π . 9 11050 Also: a =+=10 so the center is at 299 50 40 40 40 10−= ,ππ , , and c =−=0 so 99 99 that the second focus is at = 8 15. r 40 40 80 43sin+ θ +=,,ππ. 99 9 82  r == y 3 3  π  Directrix + θ + θ 2, 41 sin 1sin  2  4 4 38 ep==2, e , p = Polar Axis 43 (10, π) x 8  10  Ellipse; directrix is parallel to the polar axis ,0 3  9   3π  units above the pole; vertices are 2,  2  83ππ  , and 8, . 72 2  9 17. r = 1832 36cos− θ Also: a =+=8 so the center is at 277 93  r == −−θθ 32 3ππ 24 3 24 24 31() 2cos 1 2cos 8,−= ,, and c =−=0 72 72 77 3 ep==3, e 2, p = so that the second focus is at 2 24 24 3ππ 48 3 Hyperbola; directrix is perpendicular to the polar +=,, 772 72 3  axis units to the left of the pole; vertices are 8 π 2 , Directrix y 72 ()()−π3, 0 and 1, . 1 (2, π) (2, 0) Polar Also: a =−=()31 1 so the center is at x Axis 2 ()()11,+=ππ 2, [or ()−2,0 ], and c =−=20 2 so that the second focus is at ()()22,+=ππ 4, [or ()−4,0 ].  3π  8,  2 

1028 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.6: Polar Equations of Conics

8 19. r = 2sin− θ 84 r == 1 − 1 θ 21− sinθ 1sin 2 2 1 ep==4, e , p = 8 2 Ellipse; directrix is parallel to the polar axis 8 units below the pole; vertices are ππ 83  8, and  , . 232   = 12 18. r 1816 48sin+ θ Also: a =+=8 so the center is at 233 12 3  r == ()++θθ 16ππ 8 88 41 2sin 1 2sin 8,−= ,, and c =−=0 so that 32 32 33 == =3 ep3, e 2, p 88ππ 16 2 the second focus is at +=,,. Hyperbola; directrix is parallel to the polar axis 332 32 3 units above the pole; vertices are 2 ππ 3  1, and − 3, . 22   1 Also: a =−=()31 1 so the center is at 2 ππ3π 11,+= 2, [or −2, ], and 222 c =−=20 2 so that the second focus is at ππ3π 22,+= 4, [or −4, ]. 222

8 20. r = 24cos+ θ 84 r == 21()++ 2cosθθ 1 2cos ep==4, e 2, p = 2 Hyperbola; directrix is perpendicular to the polar axis 2 units to the right of the pole; vertices are 4 , 0 and ()−π 4, . 3 144 Also: a =−=4 so the center is at 233 44 8 8 +=,0 ,0 [or − ,π ], and 33 3 3 88 c =−=0 so that the second focus is at 33

1029 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

88 16 16 2 +=,0 ,0 [or − ,π ]. 22. rr()2cos−=θ 2 = 33 3 3 2cos− θ 21 r == 1 − 1 θ 21− cosθ 1cos 2 2 1 ep==1, e , p = 2 2 Ellipse; directrix is perpendicular to the polar axis 2 units to the left of the pole; vertices are 2 ()2, 0 and  , π . 3 124 Also: a =+=2 so the center is at 233 42 22 2,0,0−= , and c =−=0 so that 6 33 33 21. rr()32sin−=θ 6 = 32sin− θ 22 4 the second focus is at +=,0 ,0 . 62 33 3 r == 2 − 2 θ 31− sinθ 1sin 3 3 2 ep==2, e , p = 3 3 Ellipse; directrix is parallel to the polar axis 3 units below the pole; vertices are ππ 63  6, and  , . 252   1618 Also: a =+=6 so the center is at 255 18ππ 12 12 12 6,−= ,, and c =−=0 so 52 52 55 1 6 that the second focus is at 6 6secθ cosθ cosθ 12 12ππ 24 23. r == = +=,,. 2secθ − 1 1 2cos− θ 552 52 21− cosθ cosθ 6cos6θ == cosθθ 2−− cos 2 cos θ 63 r == 1 − 1 θ 21− cosθ 1cos 2 2 1 ep==3, e , p = 6 2 Ellipse; directrix is perpendicular to the polar axis 6 units to the left of the pole; vertices are ()6, 0 and () 2, π . 1 Also: a =+=()62 4 so the center is at 2

1030 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.6: Polar Equations of Conics

()()64,0−= 2,0, and c =−=20 2 so that the 3 26. r = − θ second focus is at ()()22,0+= 4,0. 1sin rr−=sinθ 3 rr=+3sinθ rr22=+(3 sinθ )

xy22+=+(3 y ) 2 xy22+=++96 yy 2 xy2 −−=690

8 27. r = 43sin+ θ 43sin8rr+=θ 483sinrr=− θ

16rr22=− (8 3 sinθ )

1 3 16(xy22+=− ) (8 3 y ) 2 3 3cscθ sinθ θ 22 2 24. r == =sin 16xy+=−+ 16 64 48 yy 9 cscθ − 1 11sin− θ −1 22++−= sinθθ sin 16xy 7 48 y 64 0 3sin3θ == 10 sinθθ 1−− sin 1 sin θ 28. r = 54cos+ θ ep===3, e 1, p 3 54cos10rr+=θ Parabola; directrix is parallel to the polar axis 3 5104cosrr=− θ 33π units below the pole; vertex is , . 22=− θ 22 25rr (10 4 cos )  25(xy22+=− ) (10 4 x ) 2 25xy22+=−+ 25 100 80 xx 16 2 925801000xyx22++−=

9 29. r = 36cos− θ 36cos9rr−=θ 396cosrr=+ θ rr=+32cosθ rr22=+(3 2 cosθ ) 22 2 1 xy+=+(3 2 x ) 25. r = 1cos+ θ xy22+=++9124 xx 2 rr+=cosθ 1 22 31290xy−+ x += rr=−1cosθ rr22=−(1 cosθ )

xy22+=−(1 x ) 2 xy22+=−+12 xx 2 yx2 +−=210

1031 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

12 34. r(2−= cosθ ) 2 30. r = 48sin+ θ 2cos2rr−=θ 48sin12rr+=θ 22cosrr=+ θ =− θ 4128sinrr 4(2cos)rr22=+ θ rr=−32sinθ 4(xy22+=+ ) (2 x ) 2 rr22=−(3 2 sinθ ) 44xy22+=++ 44 xx 2 xy22+=−(3 2 y ) 2 34440xyx22+−−= xy22+=−+9124 yy 2 22 6secθ xy−+−=31290 y 35. r = 2secθ − 1 8 6 31. r = r = 2sin− θ 2cos− θ 2sin8rr−=θ 2cos6rr−=θ 28sinrr=+ θ 26cosrr=+ θ 4(8sin)rr22=+ θ 4(6cos)rr22=+ θ

4(xy22+=+ ) (8 y ) 2 4(xy22+=+ ) (6 x ) 2 44xy22+=++ 6416 yy 2 44xy22+=++ 3612 xx 2 4316640xy22+−−= y 3xy22+−−= 4 12 x 36 0

8 3cscθ 32. r = 36. r = 24cos+ θ cscθ − 1 24cos8rr+=θ 3 r = 284cosrr=− θ 1sin− θ −=θ rr=−42cosθ rrsin 3 22 rr=+3sinθ rr=−(4 2 cosθ ) 22 22 2 rr=+ θ xy+=−(4 2 x ) (3 sin ) 22 2 22 2 xy+=+ y xy+=−16 16 xx + 4 (3 ) 22 2 22 xy+=++ yy 316160xy−− x += 96 xy2 −−=690 33. r(3−= 2sinθ ) 6 −=θ ep 32sin6rr 37. r = 362sinrr=+ θ 1sin+ e θ ep==1; 1 9(62sin)rr22=+ θ 1 22 2 = 9(xy+=+ ) (6 2 y ) r 1sin+ θ 99xy22+=++ 36244 yy 2 ep 9524360xy22+−−= y 38. r = 1sin− e θ ep==1; 2 2 r = 1sin− θ

1032 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.6: Polar Equations of Conics

ep 44. Consider the following diagram: 39. r = Directrix D 1cos− e θ 4 dD(, P) ep==;3 Pr = ( , θ ) 5 p r 12 12 r ==5 θ Polar 4 54cos− θ 1cos− θ Pole O Q Axis 5 (Focus F)

ep 40. r = 1sin+ e θ =⋅ 2 dFP(, ) edDP (, ) ep==;3 =− θ 3 dDP(,) p r sin 26 repr=−(sin)θ r == 2 + θ reper=−sinθ 1sin+ θ 32sin 3 rer+=sinθ ep re(1+= sinθ ) ep ep 41. r = ep 1sin− e θ r = + θ ep==6; 2 1sine 12 r = 45. Consider the following diagram: 16sin− θ Pr = ( , θ )

ep r 42. r = dD(, P) + θ 1cose θ Polar == ep5; 5 Pole O Q Axis 25 (Focus F) p r = 15cos+ θ Directrix D 43. Consider the following diagram: dFP(, )=⋅ edDP (, ) Pr = ( , θ ) Directrix D dDP(,)=+ p r sinθ r dD(, P) repr=+(sin)θ reper=+sinθ θ Polar rer−=sinθ ep Pole O Q Axis (Focus F) re(1−= sinθ ) ep p ep r = 1sin− e θ dFP(, )=⋅ edDP (, ) dDP(,)=− p r cosθ repr=−(cos)θ reper=−cosθ

rer+=cosθ ep re(1+= cosθ ) ep ep r = 1cos+ e θ

1033 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

=− = + ≤≤ ()3.442 107 8. xt3, y 2 t 4, 0 t 2 46. r = =++ 10.206cos− θ yx2( 3) 4 At aphelion, the greatest distance from the sun, yx=++264

cosθ = 1 . yx=+210 (3.442)1077 (3.442)10 2100xy−+ = r == 1− 0.206(1) 0.794 ≈×4.335 107 miles At perihelion, the shortest distance from the sun, cosθ =− 1. (3.442)1077 (3.442)10 r == 1−− 0.206( 1) 1.206 ≈×2.854 107 miles 410× 7

−×6107 610× 7 9. xt=+2, y = tt , ≥ 0 yx=−2 −×4107 Section 10.7

2ππ 1. 33;== 42 2. curve; parameter 3. ellipse 4. cycloid 5. False; for example: xt= 2cos , yt= 3sin define the same curve as in problem 3.

6. True 10. xtytt==≥2, 4, 0 7. xt=+32, yt =+ 1,0 ≤≤ t 4 2 xy=−+3( 1) 2 ==x 2 ≥ yx42 , x 0 xy=−+332 2

xy=−31 xy−+=310

1034 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.7: Plane and Parametric Equations

11. xt=+224, yt =− 4, −∞<<∞ t 14. xt=−24,4, yt =2 −∞<<∞ t yx=−−(4)4 + 2 x 4 2 =− yx==+44() yx8 2 For −∞

15. xey==+≥2,tt 1 et , 0 12. xt=+4, yt =− 4, t ≥ 0 x yx=−−=−44 x 8 y =+1 2 22yx=+

13. xtyt==+−∞<<∞3,2 1, t ==tt− ≥ xy=−3( 1)2 16. xe,,0 ye t − 1 yx==1 x

1035 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

17. xtytt==,,03/2 ≥ 20. xtytt==≤≤π2cos , 3sin , 0 3/2 xy yx= ()2 ==costt sin 23 yx= 3 22 xy  22 +=+=  costt sin 1 23  xy22 +=10y ≥ 49

18. xt=+3/2 1, y = t , t ≥ 0

2 3/2 xy=+() 1 ==−π≤≤ 21. xtyt2cos , 3sin , t 0 xy=+3 1 xy ==costt sin 23 22 xy  22 +=+=  costt sin 1 23  xy22 +=1 y ≤ 0 49

19. xtytt==≤≤π2cos , 3sin , 0 2 xy ==costt sin 23 22 xy  22 +=+=  costt sin 1 23  xy22 +=1 49

1036 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.7: Plane Curves and Parametric Equations

π 25. xtytt==sin22 , cos , 0 ≤≤π 2 22. xtytt==≤≤2cos , sin , 0 2 sin22tt+= cos 1 x ==costy sin t xy+=1 2 2 x 2 22 +=()yttcos + sin = 1 2 x2 +=yxy2 1 ≥ 0, ≥ 0 4

26. xt==2 ,ln,0 y tt > yx= ln 1 yx= ln 2 π 23. xtyt==≤≤sec , tan , 0 t 4 sec22tt=+ 1 tan xy22=+1 xy22−=1 1 ≤≤ x 2, 0 ≤≤ y 1

27. xty==−,41 t xt =+=+ 1,43 y t

28. xty==−+,83 t xt =+=−− 1,85 y t

29. xtyt==+,122 xt =−=−+ 1,22 yt t

30. xty==−+,21 t2

xt=−1, y =− 2 t2 + 4 t − 1 ππ 24. xtytt==csc , cot , ≤≤ 42 31. xtyt==,,3 x =3 tyt = csc22tt=+ 1 cot 4 4 xy22=+1 32. xtyt==+,1 x =−= t 1, yt xy22−=1 1 ≤≤ x 2, 0 ≤≤ y 1 33. xt==3/2 ,, yt x == ty3 t y 34. xtyt==,, xtyt ==24

2,1 1 35. xt=+2, yt = ; 0 ≤≤ t 5

36. xty==−+−≤≤,1;13 t t (1, 0) 0 x 1 37. xtytt==≤≤π3cos , 2sin ; 0 2

1037 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

ππ   43. C1 38. xtytt=−=−≤≤cos() 1 , 4sin () 1 ; 0 2 22 

39. Since the motion begins at (2, 0), we want x = 2 and y = 0 when t = 0. For the motion to be clockwise, we must have x positive and y negative initially. xtyt==−2cos()ωω , 3sin () 2π = 2  ω =π ω =π=−π≤≤() () xtytt2cos , 3sin , 0 2 C2

40. Since the motion begins at (0, 3), we want x = 0 and y = 3 when t = 0. For the motion to be counter-clockwise, we need x negative and y positive initially. xtyt=−2sin()ωω , = 3cos () 2π = 12 ω =π ω xtytt=−2sin() 2 π , = 3cos () 2 π , 0 ≤ ≤ 1

41. Since the motion begins at (0, 3), we want x = 0 C3 and y = 3 when t = 0. For the motion to be clockwise, we need x positive and y positive initially. xtyt==2sin()ωω , 3cos () 2π = 12 ω =π ω xtytt=π=π≤≤2sin2,3cos2,0() () 1

42. Since the motion begins at (2, 0), we want x = 2 and y = 0 when t = 0. For the motion to be counter-clockwise, we need x positive and y

positive initially. C4 xtyt==2cos()ωω , 3sin () 22ππ = 3  ω = ω 3 22ππ  xtytt==≤≤2cos , 3sin  , 0 3 33 

1038 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.7: Plane Curves and Parametric Equations

=+ =− 44. C1 46. xttyttsin cos , sin cos 2

–3 3

–2

47. xt=−4sin 2sin(2 t ) =− C2 yt4cos 2cos(2 t ) 3.5

–7.5 7.5

–6.5 48. xt=+4sin 2sin(2 t )

C3 yt=+4cos 2cos(2 t ) 6

–9 9

–6 49. a. Use equations (1): ==() C4 xt50cos90º 0 1 yt=−()322 + ( 50sin 90º ) t + 6 2 =−16tt2 + 50 + 6

b. The ball is in the air until y = 0 . Solve: −++=16tt2 50 6 0 −±50 502 −− 4( 16)(6) t = 2(− 16) == −± 45. xtsin tyt , cos t = 50 2884 7 −32 ≈−0.12 or 3.24 The ball is in the air for about 3.24 seconds. –7 11 (The negative solution is extraneous.)

–5 1039 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

c. The maximum height occurs at the vertex of d. We use x = 3 so that the line is not on top the . of the y-axis. b 50 40 t =− =− =1.5625 seconds 22(16)a − Evaluate the function to find the maximum height: −++=16(1.5625)2 50(1.5625) 6 45.0625 The maximum height is 45.0625 feet. 0 5 0 d. We use x = 3 so that the line is not on top = = of the y-axis. 51. Let y1 1 be the train’s path and y2 5 be 50 Bill’s path.

a. Train: Using the hint, 1 xtt==(2) 22 1 2 y = 1 0 5 1 0 Bill: 50. a. Use equations (1): xt=−5( 5) 2 ==() = xt40cos90º 0 y2 5 =−1 2 + + yt()32 ( 40sin 90º ) t 5 b. xx= 2 Bill will catch the train if 12. 2 =−16tt2 + 40 + 5 tt=−5( 5) tt2 =−525 b. The ball is in the air until y = 0 . Solve: tt2 −+5250 = −++=16tt2 40 5 0 Since 2 −±40 40 −− 4( 16)(5) 22−=−− t = bac4 ( 5) 4(1)(25) , − 2( 16) =−25 100 =−< 75 0 −±40 1920 the equation has no real solution. Thus, Bill =≈−0.12 or 2.62 −32 will not catch the train. The ball is in the air for about 2.62 seconds. c. (The negative solution is extraneous.) y

c. The maximum height occurs at the vertex of 10 the quadratic function. t = 7 b 40 t =− =− =1.25 seconds t = 10 22(16)a − 5 Bill

Evaluate the function to find the maximum t = 5 t = 10 height: Train 2 0 x −++=16(1.25) 40(1.25) 5 30 50 100 The maximum height is 30 feet.

1040 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.7: Plane Curves and Parametric Equations

= = 52. Let y1 1 be the bus’s path and y2 5 be c. Find the horizontal displacement: Jodi’s path. x =≈()()145cos 20º 3.20 436 feet a. Bus: Using the hint, 1 d. The maximum height occurs at the vertex of xtt==(3)22 1.5 the quadratic function. 1 2 b 145sin 20º y = 1 t =− =− ≈1.55 seconds 1 22(16)a − Jodi: Evaluate the function to find the maximum xt=−5( 2) height: 2 2 = −+16()( 1.55 145sin 20º ) (1.55) +≈ 5 43.43 y2 5 The maximum height is about 43.43 feet. b. Jodi will catch the bus if xx= . 12 e. 2 1.5tt=− 5( 2) 250 1.5tt2 =− 5 10 1.5tt2 −+= 5 10 0 Since bac22−=−−4 ( 5) 4(1.5)(10) , 0 440 =−=−<25 60 35 0 –50 the equation has no real solution. Thus, Jodi will not catch the bus. 54. a. Use equations (1): xt= ()125cos 40º c. y yt=−162 +() 125sin 40º t + 3 = 10 The ball is in the air until y 0 . Solve: −+16tt2 () 125sin 40º += 3 0 t = 5 = t 8 2 5 Jodi −−125sin 40º±−() 125sin 40º 4( 16)(3) t = − t = 5 t = 8 2( 16) 1 Bus ≈−0.037 or 5.059 0 x 50 100 b. The ball is in the air for about 5.059 sec. 53. a. Use equations (1): (The negative solution is extraneous.) = () xt145cos 20º c. Find the horizontal displacement: 1 x =≈()()125cos 40º 5.059 484.41 feet yt=−()322 + ( 145sin 20º ) t + 5 2 d. The maximum height occurs at the vertex of b. The ball is in the air until y = 0 . Solve: the quadratic function. b 125sin 40º −+16tt2 () 145sin 20º += 5 0 t =− =− ≈2.51 seconds 22(16)a − 2 −±145sin 20º ()145sin 20º−− 4( 16)(5) Evaluate the function to find the maximum t = 2(− 16) height: 2 ≈−0.10 or 3.20 −+16()( 2.51 125sin 40º )() 2.51 +≈ 3 103.87 The ball is in the air for about 3.20 seconds. The maximum height is about 103.87 feet. (The negative solution is extraneous.)

1041 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

e. xt= ()125cos 40º 56. a. Use equations (1): xtt==()40cos 45º 20 2 yt=−162 +() 125sin 40º t + 3 11 120 yt=− ⋅()9.82 + ( 40sin 45º ) t + 300 26 4.9 =−tt2 +20 2 + 300 0 500 6

b. The ball is in the air until y = 0 . Solve: −60 4.9 2 55. a. Use equations (1): −+tt20 2 += 300 0 6 xtt==()40cos 45º 20 2 2 4.9 1 −±20 2() 20 2 −− 4 (300) yt=−()9.82 + ( 40sin 45º ) t + 300 6 2 t = − 4.9 =−4.9tt2 + 20 2 + 300 2 6 b. The ball is in the air until y = 0 . Solve: ≈−8.51 or 43.15 −+4.9tt2 20 2 += 300 0 The ball is in the air for about 43.15 seconds. (The negative solution is 2 −±20 2() 20 2 −− 4( 4.9)(300) extraneous.) t = 2(− 4.9) c. Find the horizontal displacement: ≈−5.45 or 11.23 x =≈()20 2() 43.15 1220.5 meters The ball is in the air for about 11.23 seconds. (The negative solution is d. The maximum height occurs at the vertex of extraneous.) the quadratic function. b 20 2 c. Find the horizontal displacement: t =− =− ≈17.32 seconds 2a − 4.9 x =≈()20 2() 11.23 317.6 meters 2 6 Evaluate the function to find the maximum d. The maximum height occurs at the vertex of height: the quadratic function. 4.9 2 b 20 2 −+()17.32 20 2() 17.32 + 300 t =− =− ≈2.89 seconds 6 − 22(4.9)a ≈ 544.9 meters Evaluate the function to find the maximum height: e. −+()2 () + 700 4.9 2.89 20 2 2.89 300 = 340.8 meters

e. 350 0 1300

–100

0 400 0

1042 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.7: Plane Curves and Parametric Equations

57. a. At t = 0 , the Camry is 5 miles from the 58. a. At t = 0 , the Boeing 747 is 550 miles from intersection (at (0, 0)) traveling east (along the intersection (at (0, 0)) traveling west the x-axis) at 40 mph. Thus, xt=−40 5 , (along the x-axis) at 600 mph. y = 0 , describes the position of the Camry Thus, xt=−600 + 550 , y = 0 , describes the as a function of time. The Impala, at t = 0 , position of the Boeing 747 as a function of is 4 miles from the intersection traveling time. The Cessna, at t = 0 , is 100 miles north (along the y-axis) at 30 mph. Thus, from the intersection traveling south (along x = 0 , yt=−30 4 , describes the position of the y-axis) at 120 mph. Thus, x = 0 , the Impala as a function of time. yt=−120 + 100 describes the position of the Cessna as a function of time. P (0,0)

C d d

(0,0) B B b. Let d represent the distance between the planes. Use the Pythagorean Theorem to b. Let d represent the distance between the find the distance: cars. Use the Pythagorean Theorem to find dt=−( 600 + 550)22 +−+ ( 120 t 100) . the distance: dt=−+−(40 5)22 (30 t 4) . c. Note this is a function graph not a c. Note this is a function graph not a parametric graph. parametric graph.

d. The minimum distance between the cars is d. The minimum distance between the planes 0.2 miles and occurs at 0.128 hours (7.68 is 9.8 miles and occurs at 0.913 hours (54.8 min). min). e. e. y y Impala t = 0 200 t = 5 5 6 t = 5

t = 0.25 6 t = 0 t = 0 .125 = x t 0 .2 5 Camry Boeing 747 500 x t = 0 5 t = 0.1 25 = Cessna t 0

1043 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

1 59. a. We start with the parametric equations xv= ()cosθ t and ygtvth=−2 +()sinθ + . 0 2 0 We are given that θ =°45 , g = 32 ft/sec2 , and h = 3 feet. This gives us 2 12 xv=⋅()cos 45 °= t vt; ytvttvt=−()3222 +() sin 45 ° + 3 =− 16 + + 3 002 2200

where v0 is the velocity at which the ball leaves the bat. == b. Letting v0 90 miles/hr 132 ft/sec , we have 2 yt=−1622 +() 132 t + 3 =− 16 t + 66 2 t +3 2 66 2 33 2 The height is maximized when t =− = ≈2.9168 sec 21616()− y =−16() 2.91682 + 66 2() 2.9168 + 3 ≈ 139.1 The maximum height of the ball is about 139.1 feet. c. From part (b), the maximum height is reached after approximately 2.9168 seconds. 2 xv==⋅≈()cosθ t 132() 2.9168 272.25 0  2 The ball will reach its maximum height when it is approximately 272.25 feet from home plate. d. The ball will reach the left fence when x = 310 feet. ==θ xv()0 cos t 66 2 t 310= 66 2 t 310 t =≈3.3213 sec 66 2 Thus, it will take about 3.3213 seconds to reach the left field fence. y =−16() 3.32132 + 66 2() 3.3213 + 3 ≈ 136.5 Since the left field fence is 37 feet high, the ball will clear the Green Monster by 136.5−= 37 99.5 feet.

==−θθ2 60. a. xv()00cos t , yv () sin t 16 t x t = θ v0 cos 2 xx yv=−sinθ 16 0 θθ vv00cos cos 16 yx=−(tanθ ) x2 22θ v0 cos −16 y is a quadratic function of x; its graph is a parabola with a = , b = tanθ , and c = 0 . 22θ v0 cos b. y = 0 θ −=2 ()vtt0 sin 16 0 θ −= tv()0 sin 16 t 0 =−=θ tv0or 0 sin 16 t 0 v sinθ t = 0 16

1044 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10 Review Exercises

vvsinθθ2 sin 2 ==()()θθ00 = c. xv00cos tv cos feet 16 32 d. xy= ()()θθ=−2 vtvtt00cos sin 16 2 +−=()()θθ 16tv00 cos tv sin t 0 +−()()θθ = ttv()1600 cos v sin 0 =+−=θθ ttvv0 or 16()()00 cos sin 0 vvsinθθ− cos v t ==−00 0()sinθθ cos 16 16 v At t =−0 ()sinθθ cos : 16 vv2 =−=−θθθ00() θθθ() xv0 cos sin cos cos sin cos 16 16 v 2 y =−0 ()cos2 θθθ + sin cos (recall we want xy= ) 16 2 vv22 xy22+=200 cosθθ() sin − cos θ = 2 cos θθ() sin − cos θ  16 16 Note: since θ must be greater than 45° ( sinθθ−> cos 0 ), thus absolute value is not needed. ______

=− + 61. xxxtx()21 1, =− + −∞<<∞ yyyty()21 1, t xx− Chapter 10 Review Exercises 1 = t − xx21 1. yx2 =−16 xx− yyy=−()1 + y 21xx− 1 This is a parabola. 21 a = 4 yy− yy−=21() xx − Vertex: (0, 0) 11− xx21 Focus: (–4, 0) This is the two-point form for the equation of a Directrix: x = 4

line. Its orientation is from ()()xy11, to x 2, y 2. 1 2. 16xy22=→ x = y 62. a. xt()==≤≤π cos33 t , yt () sin t , 0 t 2 16 1 This is a parabola. 1 a = 64 –1.5 1.5 Vertex: (0, 0) 1 Focus: 0, 64 –1 1 Directrix: y =− 22 64 b. cos221/31/3ttxy+= sin ()() +

xy2/3+= 2/3 1

63 – 64. Answers will vary. 1045 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

x2 7. xy2 +=44 3. −=y2 1 25 This is a parabola. This is a hyperbola. Write in standard form: ab==5, 1 . xy2 =−44 +

Find the value of c: xy2 =−4( − 1) cab222=+=+=25 1 26 = a 1 c = 26 Vertex: (0, 1) = Center: (0, 0) Focus: (0, 0) Directrix: y 2 Vertices: (5, 0), (–5, 0) 22−= Foci: ()26,0 , ()− 26,0 8. 39yx This is a hyperbola. 11 Asymptotes: yxyx==−; Write in standard form: 55 yx22 −=1 2 39 y 2 4. −=x 1 == 25 ab3, 3 This is a hyperbola. Find the value of c: ab==5, 1 . cab222=+=+=3912

Find the value of c: c ==12 2 3 cab222=+=+=25 1 26 Center: (0, 0) c = 26 Vertices: ()0, 3 , ()0,− 3 Center: (0, 0) Vertices: (0, 5), (0, –5) Foci: ()0, 2 3 , ()0,− 2 3 − Foci: ()0, 26 , ()0, 26 33 Asymptotes: yxyx==−; Asymptotes: yxyx==−5; 5 33

yx22 22−= 5. +=1 9. 48xy 25 16 This is a hyperbola. This is an ellipse. Write in standard form: == ab5, 4 . xy22 −=1 Find the value of c: 28 cab222=−=−=25 16 9 ab===2, 8 2 2 . = c 3 Find the value of c: Center: (0, 0) cab222=+=+=2810 Vertices: (0, 5), (0, –5) Foci: (0, 3), (0, –3) c = 10 Center: (0, 0) xy22 6. +=1 Vertices: ()− 2, 0 ,() 2, 0 916 This is an ellipse. Foci: ()− 10,0 , ()10,0 ab==4, 3 . Asymptotes: yxyx==−2; 2 Find the value of c: cab222=−=−=16 9 7 c= 7 10. 94xy22+= 36 Center: (0, 0) This is an ellipse. Vertices: (0, 4), (0, –4) Write in standard form: Foci: ()0, 7 ,() 0,− 7

1046 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10 Review Exercises

xy22 13. yyxx22−−44 += 84 +=1 49 This is a hyperbola. ab==3, 2 . Write in standard form: 22 Find the value of c: (44)4(21)444yy−+− xx −+=+− cab222=−=−=945 (2)4(1)4yx−−−=22

c = 5 (2)(1)yx−−22 −=1 Cnter: (0, 0) 41 Vertices: (0, 3), (0, –3) ab==2, 1 . Foci: ()0, 5 , ()0,− 5 Find the value of c: cab222=+=+=41 5 11. xxy2 −=42 c = 5 This is a parabola. Center: (1, 2) Write in standard form: Vertices: (1, 0), (1, 4) xx2 −+=+4424 y Foci: ()1, 2−+ 5 , ()1, 2 5 (2)2(2)xy−=+2 Asymptotes: yxyx−=22(1); − −=−22(1) − 1 a = 2 14. 48440xy22++−+= xy Vertex: (2, –2) This is an ellipse. 3 Focus: 2, − Write in standard form: 2 4(xx22+++−+=−++ 2 1)( yy 4 4)444 5 Directrix: y =− ++−=22 2 4(xy 1) ( 2) 4 (1)(xy+−22 2) 2 +=1 12. 24yyx−=− 2 14 This is a parabola. ab==2, 1 Write in standard form: Find the value of c: 2 −+=−+ 222 22122()yy x cab=−=−=→=41 3 c 3 1 Center: (–1, 2) (1)yx−=2 2 Vertices: (–1, 0), (–1, 4) 1 Foci: ()−−1, 2 3 , () −+1, 2 3 a = 8 Vertex: (0, 1) 15. 4xy22+−−= 9 16 xy 18 11 1 This is an ellipse. Focus: ,1 8 Write in standard form: 1 4xy22+−−= 9 16 xy 18 11 Directrix: x =− 8 4(xx22−++ 4 4) 9( yy −+=++ 2 1) 11 16 9 4(xy−+−= 2)22 9( 1) 36 (2)(1)xy−−22 +=1 94 ab==3, 2 . Find the value of c: cab222=−=−=→94 5 c = 5 Center: (2, 1); Vertices: (–1, 1), (5, 1) Foci: ()25,1,25,1−+()

1047 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry

16. 4xy22+−+= 9 16 xy 18 11 19. 9418823xy22+−+= xy This is an ellipse. This is an ellipse. Write in standard form: Write in standard form: 4xy22+−+= 9 16 xy 18 11 9(xx22−++ 2 1) 4( yy ++=++ 2 1) 23 9 4 4(xx22−++ 4 4) 9( yy ++=++ 2 1) 11 16 9 9(xy−+ 1)22 4( += 1) 36 4(xy−++= 2)22 9( 1) 36 (1)(1)xy−+22 +=1 −+22 49 (2)(1)xy+= 1 ab==3, 2 . 94 ab==3, 2 . Find the value of c: cab222=−=−=945 Find the value of c: cab222=−=−=945 c = 5

Center: (1, –1) c = 5 Vertices: (1, –4), (1, 2) Center: (2, –1); Vertices: (–1, –1), (5, –1) Foci: ()1,−− 1 5 ,() 1, −+ 1 5 Foci: ()25,1,25,1−−() +−

20. xy22−−−=221 xy 2 −++= 17. 41616320xxy This is a hyperbola. This is a parabola. Write in standard form: Write in standard form: (xx22−+−++=+− 2 1) ( yy 2 1) 1 1 1 4(xx2 −+=−−+ 4 4) 16 y 32 16 (1)(1)1xy−−+=22 −=−+2 4(xy 2) 16( 1) ab==1, 1 . 2 (2)4(1)xy−=−+ Find the value of c: a = 1 cab222=+=+=11 2

Vertex: (2, –1); Focus: ()2,− 2 ; c = 2 Directrix: y = 0 Center: (1, –1) Vertices: (0, –1), (2, –1) 2 18. 4yxy+− 3 16 += 19 0 Foci: ()12,1,12,1+−() −− This is a parabola. yxy+= − +=−− x Write in standard form: Asymptotes: 11; 1(1) 2 4(yy−+=−−+ 4 4) 3 x 19 16 21. Parabola: The focus is (–2, 0) and the directrix is = 4(yx−=−+ 2)2 3( 1) x 2 . The vertex is (0, 0). a = 2 and since (–2, 0) is to the left of (0, 0), the parabola opens 3 (2)(1)yx−=−+2 to the left. The equation of the parabola is: 4 yax2 =−4 3 a = yx2 =−42 ⋅ ⋅ 16 yx2 =−8 Vertex: (–1, 2); Focus: ()−19 ,2 16 Directrix: x =−13 16

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22. Ellipse: The center is (0, 0), a focus is (0, 3), and 24. Parabola: Vertex: (0, 0); Directrix: y =−3 ; = a vertex is (0, 5). The major axis is x 0 . a = 3 ; the focus is the point ()0,3 ; the graph ac==5, 3 . Find b: opens up. The equation of the parabola is: 222=−=−= = bac25 9 16 . So, b 4 . The xay2 = 4 equation of the ellipse is: 2 = xy22 xy4(3) +=1 2 ba22 xy= 12 xy22 +=1 4522 xy22 +=1 16 25

25. Ellipse: Foci: (–3, 0), (3, 0); Vertex: (4, 0); Center: (0, 0); Major axis is the x-axis; ac==4; 3 . Find b:

bac222=−=−=16 9 7

23. Hyperbola: Center: (0, 0); = Focus: (0, 4); Vertex: (0, –2); b 7 Transverse axis is the y-axis; ac==2; 4 . xy22 Write the equation: +=1 Find b: 16 7 bca222=−=−=16 4 12

b ==12 2 3 yx22 Write the equation: −=1 412

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26. Hyperbola: Vertices: (–2, 0), (2, 0); Focus: (4, 0); Center: (0, 0); Transverse axis is the x-axis; ac==2; 4 . Find b: bca222=−=−=16 4 12

b ==12 2 3 xy22 Write the equation: −=1 412

29. Hyperbola: Center: (–2, –3); Focus: (–4, –3); Vertex: (–3, –3); Transverse axis is parallel to the x-axis; ac==1; 2 . Find b: bca222=−=−=41 3

b = 3 (2)(3)xy++22 Write the equation: −=1 13

27. Parabola: The focus is (2, –4) and the vertex is (2, –3). Both lie on the vertical line x = 2 . a = 1 and since (2, –4) is below (2, –3), the parabola opens down. The equation of the parabola is: ()xh−=−−2 4() ayk (2)41((3))xy−=−⋅⋅−−2 2 (2)4(3)xy−=−+ 30. Parabola: Focus: (3, 6); Directrix: y = 8 ; Parabola opens down. Vertex: (3, 7) a = 1 . The equation of the parabola is: ()xh−=−−2 4() ayk (3)4(1)(7)xy−=−−2 (3)4(7)xy−=−−2

28. Ellipse: Center: (–1, 2); Focus: (0, 2); Vertex: (2, 2). Major axis: y = 2 . ac==3; 1 . Find b: bac222=−=−=91 8

b ==822

(1)(xy+−22 2) Write the equation: +=1 98

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31. Ellipse: Foci: (–4, 2), (–4, 8); Vertex: (–4, 10); 34. Hyperbola: Center: (4, –2); ac==1; 4 ; Center: (–4, 5); Major axis is parallel to the y- Transverse axis parallel to the y-axis. axis; ac==5; 3 . Find b: Fnd b: bac222=−=−=25 9 16 →= b 4 bca222=−=−=→=16 1 15 b 15 (4)(5)xy+−22 (2)(4)yx+−22 Write the equation: +=1 Write the equation: −=1 16 25 115

35. Hyperbola: Vertices: (0, 1), (6, 1); Asymptote: 32. Hyperbola: Vertices: (–3, 3), (5, 3); 3290yx+−=; Center: (3, 1); Transverse axis Focus: (7, 3); Center: (1, 3); Major axis is = parallel to the x-axis; ac==4; 6 . Find b: is parallel to the x-axis; a 3 ; The slope of the 2 bca222=−=−=36 16 20 → b = 20 = 2 5 asymptote is − ; Find b: 3 −−22 (1)(xy−= 3) −−−bb 2 Write the equation: 1 ==→−=−→=36bb 2 16 20 a 33 (3)(1)xy−−22 Write the equation: −=1 94

33. Hyperbola: Center: (–1, 2); ac==3; 4 ; Transverse axis parallel to the x-axis; 36. Hyperbola: Vertices: (4, 0), (4, 4); Asymptote: yx+−=2100; Center: (4, 2); Find b: bca222=−=−=→16 9 7 b = 7 Transverse axis is parallel to the y-axis; a = 2 ; (1)(xy+−22 2) Write the equation: −=1 The slope of the asymptote is − 2 ; Find b: 97 −−a 2 ==−→−=−→=222bb 1 bb

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(2)(4)yx−−22 41. 9124xxyyxy22−+++= 8120 Write the equation: −=1 41 AB==−=9, 12, C 4

BAC22−=−−4 ( 12) 4(9)(4) = 0 Parabola

42. 44xxyy22++− 851650 x + y = ABC===4, 4, 1 BAC22−=−=444(4)(1)0 Parabola

43. 410490xxyy22++−= AB==4, 10, C = 4

BAC22−=−4 10 4(4)(4) => 36 0 37. yxy2 ++−=4380 Hyperbola ACAC====0 and 1; (0)(1) 0. Since 22−+−= AC = 0 , the equation defines a parabola. 44. 4104xxyy 90 AB==−=4, 10, C 4 2 38. 280xyx−+ = BAC22−=−−4 ( 10) 4(4)(4) => 36 0 ACAC====2 and 0; (2)(0) 0 . Since Hyperbola AC = 0 , the equation defines a parabola. 45. xxyyxy22−+++−=232410 39. xyxy22++−+=24820 AB==−=1, 2, C 3

ACAC====1 and 2; (1)(2) 2 . Since BAC22−=−−=−<4 ( 2) 4(1)(3) 8 0 ACAC>≠0 and , the equation defines an Ellipse ellipse. 46. 41210xxyyxy22+−++−= 100 22−−−= 40. xyxy820 AB==4, 12, C =− 10 ACAC==−=−=−1 and 8; (1)( 8) 8 . Since BAC22−=−−=>4124(4)(10)3040 AC < 0 , the equation defines a hyperbola. Hyperbola ______

9 47. 252xxyy22++−= 0 2 AC−−22 π π AB==2, 5, and C =2; cot() 2θθθ = = =→=→= 0 2 B 524 22 2 xx=−=−=−'cosθθ y 'sin x ' y '() xy ' ' 22 2 22 2 yx=+'sinθθ y 'cos =+=+ x ' y '() x ' y ' 22 2

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22 22229 ()xy−+() xy −() xy ++() xy +−= 2 '' 5 '' '' 2 '' 0 22222 59 ()x'222222−++−+++−= 2'' xyy '()( x '' y ' x ' 2'' xyy ' ) 0 22 91 9xy ''22 xy''22−=→−=→−= 9''9 xy 22 1 22 2 19 Hyperbola; center at (0, 0), transverse axis is the x'-axis, vertices at ()()xy', '=± 1, 0 ; foci at ()xy', '=±() 10, 0 ; asymptotes: yx'3'=± .

9 48. 252xxyy22−+−= 0 2 AC−−22 π π AB==−=2, 5, and C2; cot() 2θθθ = = =→=→= 0 2 B −524 22 2 xx=−=−=−'cosθθ y 'sin x ' y '() xy ' ' 22 2 22 2 yx=+'sinθθ y 'cos =+=+ x ' y '() x ' y ' 22 2 22 22229 ()xy−−() xy −() xy ++() xy +−= 2''5''''2''0 22222 59 ()x'2'''222222−+−−+++−= xyy()( x ''' y x '2''' xyy ) 0 22 19 9yx ''22 −+xy''22 =→−+=→−= xy '9'9 22 1 22 2 19 Hyperbola; center at (0, 0), transverse axis is the y'-axis, vertices at ()()xy', '=± 0, 1 ; foci at ()xy', '=±() 0, 10 ; 1 asymptotes: yx''=± . 3

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49. 649200xxyy22++−= AC−−69 3 3 AB==6, 4, and C =9; cot() 2θθ = = =−→ cos() 2 =− B 44 5

−−33  +−  11   55425  1 5 sinθθθ======→≈ ; cos  63.4º 255 255 5255 xx=−=−=−'cosθθ y 'sin x ' y '() x ' 2 y ' 55 5

25 5 5 yx=+'sinθθ y 'cos =+=+ x ' y '() 2 x ' y ' 555 22 5555 ()x−+ y() x − y() xy ++() xy +−= 6'2'4'2'2''92''200 5555 64 9 ()xxyy'4''4'222222−+ +() 2'3''2' xxyy −−+() 4'4''' xxyy ++−= 200 55 5 62424812836369 xxyyxxyyxxyy''''''''''''20222222−++−−+++= 55 555 55 5 5 xy''22 10x '22+=→+=5'y 20 1 24 Ellipse; center at the origin, major axis is the y'-axis, vertices at ()()xy', '=± 0, 2 ; foci at ()xy', '=±() 0, 2 .

50. xxyy22+++44165850 x − y = AC−−14 3 3 AB==1, 4, and C =4; cot() 2θθ = = =− ; cos() 2 =− B 44 5 33  11−− +−  55425 1 5 sinθθθ======→= ; cos  63.4º 255 255 5255 xx=−=−=−'cosθθ y 'sin x ' y '() x ' 2 y ' 55 5 25 5 5 yx=+'sinθθ y 'cos =+=+ x ' y '() 2 xy ' ' 555 2 2 5555 ()xy'2'−+ 4() xy '2' −() 2' xy ++ ' 4() 2' xy + ' 5555 55 +−−+=16 5()xy ' 2 ' 8 5() 2 xy ' ' 0 55 14 4 ()xxyy'222222−++ 4''4'() 2' xxyy −−+ 3''2'() 4' xxyyxyxy +++−−−= 4'' ' 16'32'16'8' 0 55 5

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14 4812816164 x'222222−++−−++ xyyx '''' xyy ''' x ' xyyy '''40'0 +−= 55 555 5 5 5 5 5'xy22−=→= 40' 0 xy ' 8' Parabola; vertex at the origin, focus at ()()xy', '= 0, 2 .

51. 41291280xxyyxy22−+++= AC−−49 5 5 AB==−4, 12, and C =9; cot() 2θθ = = =→ cos() 2 = B −12 12 13 55  11−+  134 2 13 13 9 3 13 sinθθθ======→≈ ; cos  33.7º 2 13 13 2 13 13 3 13 2 13 13 xx=−=−'cosθθ y 'sin x ' y ' =() 3 x ' − 2 y ' 13 13 13 2 13 3 13 13 yx=+'sinθθ y 'cos = x ' + y ' =() 2 x ' + 3 y ' 13 13 13 2 2 13 13 13 13 4() 3'2'xy−− 12() 3'2' xy −() 2'3' xy ++ 9() 2'3' xy + 13 13 13 13  13 13 +−++=()xy() xy 12 3 ' 2 ' 8 2 ' 3 ' 0 13 13 4129 ()9'x222222−+− 12''4' xyy() 6' x +−+ 5''6' xyy() 4' x ++ 12''9' xyy 13 13 13

36 13 24 13 16 13 24 13 +−++xyxy''''0 = 13 13 13 13 36 48 16 72 60 72 36 108 81 xxyyxxyyx'''''''''222222−+−−+++++= xyy '''413'0 x 13 13 13 13 13 13 13 13 13 413 13yx '22+= 4 13 ' 0  y '=− x ' 13 Parabola; vertex at the origin, focus at ()xy', '=−13 , 0 . ()13

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52. 9241680600xxyyxy22−+++= AC−−9167 7 AB==−9, 24, and C =16; cot() 2θθ = = =→ cos() 2 = B − 24 24 25 77  11−+  2593 25 164 sinθθθ======→≈ ; cos  36.9 2255 2255 43 1 xx=−=−=−'cosθθ y 'sin x ' y '() 4 x ' 3 y ' 55 5 34 1 yx=+'sinθθ y 'cos =+=+ x ' y '() 3 x ' 4 y ' 55 5 22 94'3'244'3'3'4'163'4'804'3'603'4'0()111111()xy−−()() xy −()() xy ++()() xy ++()() xy −+()() xy += 555555 92416 ()16'xxyy222222−+− 24'' 9'() 12'7' xxyy +−+ '12'() 9' xxyyxyxy +++−++= 24' '16' 64'48' 36'48' 0 25 25 25 144 216 81 288 168 288 144 384 256 xxyyxxyyxxyyx''''''''''''100'0222222−+−−+++++= 25 25 25 25 25 25 25 25 25 25yx '22+=→=− 100 ' 0 yx ' 4 ' Parabola; vertex at the origin; focus at ()()xy', '=− 1, 0 .

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4 53. r = 1cos− θ ep===4, e 1, p 4 Parabola; directrix is perpendicular to the polar axis 4 units to the left of the pole; vertex is ()2, π .

2  2 3 56. r == 32cos+ θ 2 1cos+ θ 3 22 ep==,,1 e p = 33 6 Ellipse; directrix is perpendicular to the polar 54. r = axis 1 unit to the right of the pole; vertices are 1sin+ θ === 2 4 ep6, e 1, p 6 ,0 and () 2,π . Center at ,π ; other 5 5 Parabola; directrix is parallel to the polar axis 6  8 π focus at ,π . units above the pole; vertex is 3, . 5 2 

63 55. r == 2sin− θ 1 1sin− θ 2 1 ep==3, e , p = 6 2 Ellipse; directrix is parallel to the polar axis 6 units below the pole; vertices are ππ 3  π 6, and  2, . Center at 2, ; other 22   2 π focus at 4, . 2

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82 4 57. r == 59. r = 48cos++θθ 12cos 1cos− θ ep==2, e 2, p = 1 rr−=cosθ 4 Hyperbola; directrix is perpendicular to the polar rr=+4cosθ axis 1 unit to the right of the pole; vertices are rr22=+(4 cosθ ) 2 4 , 0 and ()−π 2, . Center at ,0 ; other 22+=+ 2 3 3 xy(4 x ) 22 2 8 xy+=++16 8 xx focus at − ,π . 3 yx2 −−=8160

6 60. r = 2sin− θ 2sin6rr−=θ 26sinrr=+ θ 4(6sin)rr22=+ θ

4(6)()xy22+=+ y 2

44xy22+=++ 3612 yy 2

4312360xy22+−−= y 10 2 58. r == 520sin++θθ 14sin 8 61. r = 1 48cos+ θ ep==2, e 4, p = 2 48cos8rr+=θ Hyperbola; directrix is parallel to the polar axis 488cosrr=− θ 1 unit above the pole; vertices are rr=−22cosθ 2 22=− θ πππ rr(2 2 cos ) 2232−=  , and ,  ,. Center at 22+=− 2 52 3 2  32 xy(2 2 x ) 8 π 16 3ππ 16 xy22+=−+48 xx 4 2 , ; other focus at −=,,. 15 2 15 2 15 2 3840xy22−−+= x

2 62. r = 32cos+ θ 32cos2rr+=θ 322cosrr=− θ 9(22cos)rr22=− θ

9(22)()xy22+=− x 2

99xy22+=−+ 484 xx 2

59840xyx22++−=

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63. xt=−42,1, y =−−∞<<∞ t t

=−− xy4(1 ) 2 66. xtytt==>ln ,3 , 0 xy=−44 − 2 xy+=42

64. xt=+26,5,2 y =−−∞<<∞ t t

lnyt= ln 3 lnyt= 3ln lnyx=→= 3 ye3x

2 xy=−+2(5 ) 6 π 67. xtytt==≤≤sec22 , tan , 0 xyy=−++22510()2 6 4

xyy=−++50 20 22 6 xy=−+220562 y

65. xtyt==+≤≤π3sin , 4cos 2, 0 t 2 xy− 2 ==sintt , cos 34 sin22tt+= cos 1 22 xy − 2  +=   1 34   tan22ttyx+= 1 sec → += 1 xy22(2)− +=1 916

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68. xt==+≥3/2 ,24,0 y t t 73. Write the equation in standard form: xy22 49xy22+=→+= 36 1 94 The center of the ellipse is (0, 0). The major axis is the x-axis. ab==3; 2; cab222=−=−=→=94 5 c 5. For the ellipse: Vertices: (–3, 0), (3, 0); Foci: ()− 5, 0 ,() 5, 0

yx=+242/3 For the hyperbola: Foci: (–3, 0), (3, 0); 69. Answers will vary. One example: − yx=−24 + Vertices: ()5, 0 ,() 5, 0 ; xty==−+,24 t Center: (0, 0) 4 − t ac==5; 3; xyt==, 2 bca222=−=−=→=95 4 b 2 xy22 70. Answers will vary. One example: The equation of the hyperbola is: −=1 yx=−282 54 2 xty==−,28 t 74. Write the equation in standard form: xty==2, 22() t2 −=− 8 8 t2 8 xy22 xy22−=→−=416 1 16 4 xy The center of the hyperbola is (0, 0). The 71. ==cos()ωω tt; sin() 43 transverse axis is the x-axis. == 22πππ ab4; 2; = 44 ωπω= 2 == ω 42 cab222=+=+=→=16 4 20 c 20 = 2 5. ππ For the hyperbola: ∴=xy =  costt ; sin  Vertices: (–4, 0), (4, 0); 4232  ()− () ππ  Foci: 25,0,25,0  xtytt==≤≤4cos ; 3sin  , 0 4 22  For the ellipse: xy Foci: (–4, 0), (4, 0); 72. ==sin()ωω tt; cos() 43 Vertices: ()− 25,0,25,0(); 22ππ = 55 ωπω= 2 = Center: (0, 0) ω 5 ac==25; 4; ππ xy22  222 ∴= sintt ; = cos  bac=−=−=→=20 16 4 b 2 4535  xy22 22ππ  The equation of the ellipse is: += 1  xtytt==4sin ; 3cos  , 0 ≤≤ 5 20 4 55 

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75. Let (,xy ) be any point in the collection of 77. Locate the parabola so that the vertex is at (0, 0) points. and opens up. It then has the equation: The distance from xay2 = 4 . Since the light source is located at the (,xy ) to (3,0)=−+ ( x 3)22 y . focus and is 1 foot from the base, a = 1 . Thus, 2 The distance from xy= 4 . The width of the opening is 2, so the 16 16 point (1, y) is located on the parabola. Solve for (xy , ) to the line x=− is x . 33 y: 2 Relating the distances, we have: 14=→=→=yyy 14 0.25 feet 316 The mirror is 0.25 feet, or 3 inches, deep. (3)xyx−+=22 − 43 78. Set up the problem so that the vertex of the 2 22916 parabola is at (0, 0) and it opens down. Then the (3)xy−+= x − 16 3 equation of the parabola has the form: xcy2 = . The point (30, –20) is a point on the parabola. 222−++=932256 − + xx69 y x x Solve for c and find the equation: 16 3 9 302 =−cc ( 20) →=− 45 16xx222−++ 96 144 16 yxx =−+ 9 96 256 xy2 =−45 7xy22+= 16 112 22 716xy +=1 y 112 112 xy22 +=1 x 16 7 20 The set of points is an ellipse. (30,–20) 76. Let (,xy ) be any point in the collection of

points. The distance from To find the height of the bridge, 5 feet from the (,xy ) to (5,0)=−+ ( x 5)22 y . center, the point (5, y) is a point on the parabola. Solve for y: The distance from 54525450.562 =−yyy → =− → ≈− 16 16 (xy , ) to the line x=− is x . The height of the bridge, 5 feet from the center, 55 is 20 – 0.56 = 19.44 feet. Relating the distances, we have: To find the height of the bridge, 10 feet from the 516 (5)xyx−+=22 − center, the point (10, y) is a point on the 45 parabola. Solve for y:

2 102 =− 45yyy → 100 =− 45 → ≈− 2.22 −+=2225 − 16 (5)xy x The height of the bridge, 10 feet from the center, 16 5 is 20 – 2.22 = 17.78 feet. 22225 32 256 To find the height of the bridge, 20 feet from the xx−++=10 25 y x − x + 16 525 center, the point (20, y) is a point on the 16xx222−++= 160 400 16 yxx 25 −+ 160 256 parabola. Solve for y: 202 =− 45yyy → 400 =− 45 → ≈− 8.89 916144xy22−= The height of the bridge, 20 feet from the center, 916xy22 −=1 is 20 – 8.89 = 11.11 feet. 144 144 xy22 −=1 16 9 The set of points is a hyperbola.

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79. Place the semi-elliptical arch so that the x-axis 81. First note that all points where an explosion coincides with the water and the y-axis passes could take place, such that the time difference through the center of the arch. Since the bridge would be the same as that for the first detonation, has a span of 60 feet, the length of the major axis would form a hyperbola with A and B as the is 60, or 2aa== 60 or 30 . The maximum foci. height of the bridge is 20 feet, so b = 20 . The Start with a diagram: 22 xy+= N equation is: 1. D2 900 400 (1000, y) The height 5 feet from the center: 22 5 +=y 1 − D 900 400 (1000, 0) 1 (1000, 0) y2 25 A (0, 0) (,a 0) B =−1 400 900 2000 feet 2 875 Since A and B are the foci, we have yy=⋅→≈400 19.72 feet 900 2c = 2000 The height 10 feet from the center: c = 1000 1022y Since D is on the transverse axis and is on the +=1 1 900 400 hyperbola, then it must be a vertex of the y2 100 hyperbola. Since it is 200 feet from B, we have =−1 a = 800 . Finally, 400 900 bca222=−=1000 2 − 800 2 = 360,000 2 800 yy=⋅→≈400 18.86 feet Thus, the equation of the hyperbola is 900 x2 y2 The height 20 feet from the center: −=1 640,000 360,000 2022y +=1 () 900 400 The point 1000, y needs to lie on the graph of y2 400 the hyperbola. Thus, we have =−1 2 400 900 ()1000 y2 −=1 500 640,000 360,000 yy2 =⋅→≈400 14.91 feet 900 2 −=−y 9 360,000 16 80. The major axis is 80 feet; therefore, 280aa== or 40. The maximum height is 25 y2 = 202,500 feet, so b = 25 . To locate the foci, find c: y = 450 cab222=−=1600 − 625 = 975 →≈ c 31.22 The second explosion should be set off 450 feet The foci are 31.22 feet from the center of the due north of point B. room or 8.78 feet from each wall. 82. a. Train: xt= 3 2 ; 1 2 = Let y1 1 for plotting convenience. Mary: xt=−6( 2); 2 = Let y2 3 for plotting convenience.

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= b. Mary will catch the train if xx12. d. Find the horizontal displacement: =≈()() 3 tt2 =−6( 2) x 80cos 35º (2.99) 196 feet 2 (or roughly 65.3 yards) 3 tt2 =−612 2 e. tt2 −+=480 50 Since bac22−=−−4 ( 4) 4(1)(8) 0 220 =−=−<16 32 16 0 the equation has no real solution. Thus, Mary will not catch the train. −50 c. y 84. Answers will vary.

5

= = = t 3 t 4 t 2 Mary Chapter 10 Test

= t = 2 t 3 t = 4 2 2 Train ()x +1 y 1. −=1 49 x 10 20 Rewriting the equation as 2 83. a. Use equations (1) in section 9.7. ()x −−()1 ()y − 0 2 −=1, we see that this is the = () 22 xt()80cos 35º 23 1 equation of a hyperbola in the form yt=−(32)2 +() 80sin() 35º t + 6 2 ()xh−−22() yk −=. Therefore, we have 221 b. The ball is in the air until y = 0 . Solve: ab =− = = = 2 = −+16tt2 () 80sin() 35º += 6 0 h 1, k 0 , a 2 , and b 3. Since a 4 and b2 = 9 , we get cab222=+=+=4913, or 2 −−80sin() 35º±−() 80sin () 35º 4( 16)(6) = ()− t = c 13 . The center is at 1, 0 and the 2(− 16) transverse axis is the x-axis. The vertices are at −± ()()hak±=−±,12,0, or ()−3, 0 and ()1, 0 . ≈ 45.89 2489.54 − 32 The foci are at ()hck±=−±,113,0(), or ≈−0.13 or 2.99 The ball is in the air for about 2.99 seconds. ()−−113,0 and ()−+113,0. The asymptotes (The negative solution is extraneous.) 3 3 are yx−=±01() −−(), or yx=−() +1 and c. The maximum height occurs at the vertex of 2 2 the quadratic function. =+3 () −b −80sin() 35º yx1 . t == ≈1.43 seconds 2 22(16)a − Evaluate the function to find the maximum height: −+16(1.43)2 () 80sin() 35º (1.43) +≈ 6 38.9 ft

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2. 814yx=−()2 − 4. The vertex ()−1, 3 and the focus ()−1, 4.5 both Rewriting gives lie on the vertical line x =−1 (the axis of (1)8xy−=+2 4 symmetry). The distance a from the vertex to the focus is a = 1.5 . Because the focus lies above 2 1 the vertex, we know the parabola opens upward. (1)8xy−= −− 2 As a result, the form of the equation is −=2 − 2 1 ()xh4 ayk () ()xy−=14(2) −− 2 where ()()hk,1,3=− and a = 1.5 . Therefore, This is the equation of a parabola in the form the equation is 2 ()xh−=4 ayk () −. Therefore, the axis of ()xy+=141.532 ()() − symmetry is parallel to the y-axis and we have ()xy+=1632 ( − ) 1 ()hk,1,=− and a = 2 . The vertex is at ()± ()− 2 The points hak2, , that is 4, 4.5 and 1 ()2, 4.5 , define the lattice rectum; the line ()hk,1,=−, the axis of symmetry is x = 1 , 2 y = 1.5 is the directrix. 13 the focus is at ()hk,1,21,+=−+= a , 22 and the directrix is given by the line yka=−, 5 or y =− . 2

3. 234613xyxy22++−= Rewrite the equation by completing the square in x and y. 234613xyxy22++−= 5. The center is ()()hk,0,0= so h = 0 and k = 0 . 243613xxyy22++ −= Since the center, focus, and vertex all lie on the 223213()()xx22++ yy −= line x = 0 , the major axis is the y-axis. The distance from the center ()0, 0 to a focus ()0,3 2213211323()()xx22+++ yy −+=++ is c = 3 . The distance from the center ()0, 0 to a 213118()()xy++22 −= vertex ()0,− 4 is a = 4 . Then, 2 2 ()x −−()1 ()y −1 22222=−=−=−= +=1 bac431697 96 The form of the equation is This is the equation of an ellipse with center at ()xh−−22() yk ()−1,1 and major axis parallel to the x-axis. +=1 ba22 2 2 Since a = 9 and b = 6 , we have a = 3 , where h = 0 , k = 0 , a = 4 , and b = 7 . Thus, b = 6 , and cab222=−=−=96 3, or we get c = 3 . The foci are ()hck±=−±,13,1() or xy22 +=1 716 −− −+ ()13,1 and ()13,1. The vertices are at To graph the equation, we use the center ()()hak±=−±,13,1, or ()−4,1 and ()2,1 . ()()hk,0,0= to locate the vertices. The major axis is the y-axis, so the vertices are a = 4 units above and below the center. Therefore, the = =− vertices are V1 ()0, 4 and V2 ()0, 4 . Since c = 3 and the major axis is the y-axis, the foci

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are 3 units above and below the center. Therefore, the foci are c = 23 units above and = () Therefore, the foci are F1 0,3 and =+ below the center. The foci are F1 ()2, 2 2 3 =− = F2 ()0, 3 . Finally, we use the value b 7 and F =−()2, 2 2 3 . The asymptotes are given to find the two points left and right of the center: 2 ()− 7,0 and ()7,0 . by the lines a yk−=±() xh − . Therefore, the asymptotes b are 2 yx−=±22() − 22 2 yx=±() −22 + 2

6. The center ()()hk,2,2= and vertex ()2, 4 both lie on the line x = 2 , the transverse axis is parallel to the y-axis. The distance from the center ()2, 2 to the vertex ()2, 4 is a = 2 , so the other vertex must be ()2, 0 . The form of the equation is

()yk−−22() xh −=1 ab22 7. 253370xxyyx22+++−= is in the form = = = where h 2 , k 2 , and a 2 . This gives us Ax22+++++= Bxy Cy Dx Ey F 0 where 22 ()yx−−22() A = 2 , B = 5 , and C = 3 . −=1 4 b2 BAC22−=−45423()() Since the graph contains the point =−25 24 ()=+ xy,210,5(), we can use this point to = 1 determine the value for b. Since BAC2 −>40, the equation defines a 2 2 hyperbola. ()52− ()2102+− −=1 2 4 b 8. 32310xxyy22−+ ++= y is in the form 910 −=1 Ax22+++++= Bxy Cy Dx Ey F 0 where 4 b2 A = 3 , B =−1 , and C = 2 . 510 = 2 2 BAC−=−−41432() ()() 4 b2 =− b2 = 8 124 =−23 b = 22 2 −< Therefore, the equation becomes Since BAC40, the equation defines an ellipse. ()yx−−2222() −=1 48 Since cab222=+=+=4812, the distance from the center to either focus is c = 23.

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9. xxyyxy22−++−−= 34 43 69 2320 is in the form xxy=−' ' and yxy =+ ' ' Ax22+++++= Bxy Cy Dx Ey F 0 where 55 55 Substituting these values into the original A = 1, B =−6 , and C = 9 . equation and simplifying, we obtain 2 −=−−()2 ()() BAC46419 41xxyy22−+−= 24 34 25 0 =−36 36 2 34 34 43 = 0 41xy '−− ' 24 xy ' − ' xy ' + ' 55 55 55 2 −= Since BAC40, the equation defines a 2 parabola. 43 ++=34xy ' ' 25 55 10. 41xxyy22−+−= 24 34 25 0 Multiply both sides by 25 and expand to obtain Substituting xx=−'cosθθ y 'sin and 41()() 9xxyy '2222−+− 24 ' ' 16 ' 24 12 xxyy ' −− 7 ' ' 12 ' yx=+'sinθθ y 'cos transforms the equation +++=34() 16xxyy '22 24 ' ' 9 ' 625 into one that represents a rotation through an angle θ . To eliminate the xy'' term in the new 625xy '22+= 1250 ' 625 A − C xy'2'122+= equation, we need cot() 2θ = . That is, we B xy′′22 need to solve +=1 1 1 41− 34 cot() 2θ = 2 −24 12 7 Thus:ab== 1 1 and = = cot() 2θ =− 22 24 This is the equation of an ellipse with center at Since cot() 2θ < 0 , we may choose ()0, 0 in the xy''plane− . The vertices are at 90°< 2θ < 180 °, or 45°<θ < 90 °. ()−1, 0 and ()1, 0 in the xy''plane− . (7,− 24) 222 11 2 22+= cab=−=−=1 → c = 72425 22 2  24 ± 2 2θ The foci are located at ,0 in the 2 − 7 xy''plane. In summary: x′′ y−− plane xy plane center (0,0) (0,0)

7 34 3 4  We have cot() 2θ =− so it follows that vertices (1,0)±−− , , ,  24 55 5 5  cos() 2θ =− 7 . 25 2232 ,, 7 minor axis 2 510 −− 0, ± − θ 1   1cos2() 25 16 4 intercepts 2 22 32 sinθ ==== −−, 22255  510 7 +− 3222 + ()θ 1  1cos2 25 93 ,, cosθ ==== 2 10 5 22255 ± foci ,0 2 32 22 −1 3 −−, θ =≈°cos 53.13  5 10 5 With these values, the rotation formulas are

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The graph is given below.

To find the rectangular equation for the curve, we need to eliminate the variable t from the equations. ep 3 t 11. r == We can start by solving equation (1) for . 1cos12cos−−e θθ xt=−32 3 32tx=+ Therefore, e = 2 and p = . Since e > 1 , this is 2 x + 2 t = the equation of a hyperbola. 3 3 r = Substituting this result for t into equation (2) 12cos− θ gives r ()12cos−=θ 3 x + 2 y =−1 , −≤225x ≤ rr−=2cosθ 3 3 Since xr= cosθ and xyr222+=, we get 13. −=θ We can draw the parabola used to form the rr2cos 3 reflector on a rectangular coordinate system so rr=+2cosθ 3 that the vertex of the parabola is at the origin and 2 2 its focus is on the positive y-axis. rr=+()2cosθ 3 y 4 ft xy22+=()23 x +2 2 (− 2, 1.5) xy22+=4129 x 2 + x + (2, 1.5) 312xxy22+−=− 9 x − 0 34()xxy22+−=− 9 2 2

344()xx22++−=−+ y 912 −2

32()xy+−=2 2 3 The form of the equation of the parabola is 2 2 xay= 4 and its focus is at ()0, a . Since the ()x + 2 y2 −=1 () 13 point 2,1.5 is on the graph, we have 241.52 = a() 12. xt=−3 2 (1)  46= a  yt=−1 (2) a = 2 tx y() xy, 3 The microphone should be located 2 feet (or 8 030221012,1xy=−=−=−=−() () 3 inches) from the base of the reflector, along its =−==−=() () 131211101,0xy axis of symmetry. 43421014110,1xy=−==−=−−() () 93922519225,2xy=−==−=−−() ()

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y Using synthetic division on the quotient and 4 ft x = 2: 2 (− 2, 1.5) (2, 1.5) 2 9−−− 12 11 2 18 12 2 x −2 0 2 9 6 1 0 Since the remainder is 0, 2 is a zero for f. So − x − 2 is a factor; thus, 2 F = 2 0, 3 fx()=+ ( x 5)() x − 2() 9 x2 ++ 6 x 1 . =+(5)23131xx()()() − x + x + 1 Chapter 10 Cumulative Review Therefore, x =− is also a zero for f (with 3

multiplicity 2). Solution set: {−−5,1 , 2} . fx()()+− h fx 3 1. h 3. 6 −≥xx2 2 2 −+−−−−3()(xh + +5 x h ) 2() 3 x +5 x 2 06≥+−xx2 = h xx2 +−≤60 22 2 −++++−+−+32()xxhhxh 552352 xx ()()xx+−≤320 = h fx()=+− x2 x 6 −22 − − ++−+−+ 2 = 363552352xxhhxh xx xx=−3, = 2 are the zeros of f . h Interval()()()−∞ , − 3 − 3, 2 2, ∞ −−+635xh h2 h ==−−+635xh Test Value− 4 0 3 h Value of f 6− 6 6 2. 9xx43 +33−−− 71 x 2 57 x 10=0 Conclusion Positive Negative Positive There are at most 4 real zeros. The solution set is { xx−≤32 ≤} , or []−3, 2 . Possible rational zeros: pq=±1, ± 2, ± 5, ± 10; =± 1, ± 3, ± 9; ()=+x p 11 22 4. fx 32 =±1, ± , ± , ± 2, ± , ± , ± 5, q 39 39 a. Domain: ()−∞, ∞ ; Range: ()2,∞ . 5 5 10 10 ±±±± , , 10, , ± b. fx()=+32x 39 3 9 =+x =−−−43 2 y 32 Graphing yx1 9 +33 x 71 x 57 x 10 y indicates that there appear to be zeros at x = –5 x =+3 2 Inverse

and at x = 2. x −=23y Using synthetic division with x = –5: − log()xyfxx−= 2 1 ()()=− log 2 −−−−5 9 33 71 57 10 33 − Domain of f 1 = range of f = ()2,∞ . −45 60 55 10 −1 −∞ ∞ 912112−−− 0 Range of f = domain of f = (), . Since the remainder is 0, –5 is a zero for f. So xx−−(5) = + 5 is a factor. The other factor is the quotient: 32−−− 912112xxx. =+32 − − − Thus, fx( ) ( x 5)() 9 x 12 x 11 x 2 .

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()=− ( ) 2 5. fxlog4 x 2 ()xh−=4 ayk () − =−= 2 a. fx()log4 ( x 2 ) 2 ()xay−=14 −=2 2 x 24 ()01−= 4a () 2 −= x 216 18= a = x 18 1 a = The solution set is {18} . 8 221 b. fx()=−≤log ( x 2 ) 2 ()xyyx−=1 or = 2() − 1 4 2 xx−≤2 42 and −> 2 0 e. This graph is a hyperbola with center ()0, 0 xx−≤2 16 and > 2 and vertices ()0,± 1 , containing the point xx≤>18 and 2 ()3, 2 . 218<≤x 22 (2,18] ()yk−−() xh −=1 ab22 6. a. This graph is a line containing points 22 yx−= ()0,− 2 and ()1, 0 . 1 1 b2 Δy 02−−()2 22 slope== == 2 ()23() Δ−x 10 1 −=1 1 b2 −=() − using yy11 mxx 49 −=1 yx−=02() − 1 1 2 b yx=22 − or 2 xy −−= 20 9 41−= 2 b. This graph is a circle with center point (2, 0) b 9 and radius 2. 3 = 2 ()()xh−+−=22 yk r2 b 39b2 = ()()xy−+−=202222 b2 = 3 2 ()xy−+=242 The equation of the hyperbola is: yx22 c. This graph is an ellipse with center point −=1 (0, 0); vertices ()±3, 0 and y-intercepts 13 ()0,± 2 . f. This is the graph of an exponential function with y-intercept ()0,1 , containing the point ()xh−−22() yk +=1 ()1, 4 . ab22 x 22 yAb=⋅ ()xy−−00() 22 +=xy+= 1  1 () =⋅0 =⋅ = 3222 94 y-intercept 0,1  111Ab A A point ()1, 4  4 ==bb1 d. This graph is a parabola with vertex ()1, 0 = x and y-intercept ()0, 2 . Therefore, y 4 .

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7. sin() 2θ = 0.5 The domain is π ππ  ≠+3 π  22θθ=+πkk  =+π  xx k, where k is any integer . 612 4 , 55π π or 2θθ=+π 2kk =+π ()θθ=<

rr2 = 8sinθ Chapter 10 Projects r = 8sinθ Project I – Internet Based Project Project II

1. Figure:

6 37.1× 10 4458.0× 106

(0, 0) Sun 4495.1× 106

c =×37.2 106 3 222=− 10. fx()= bac sinxx+ cos 266=×−×22 f will be defined provided sinxx+≠ cos 0 . b ()()4495.1 10 37.1 10 sinxx+= cos 0 b =×4494.9 106 sinxx=− cos xy22 +=1 sin x xx62 62 =−1 (4495.1 10 ) (4494.9 10 ) cos x tanx =− 1 3π xkk=+π , is any integer 4

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2. Figure: The focal length is 0.03125 m. 11 zx=+22 y 1467.7× 106 4445.8× 106 88 2. (0, 0) Sun Targetxy Rθ Z 5913.5× 106 T 0 9.551−− 11.78 0.5 1.950 1 T 0 9.948−− 5.65 0.125 0.979 77381.2×+ 10666 4445.8 ×= 10 11827 × 10 2 T3 0 9.928 5.90 0.125 1.021 a =×=×0.5() 11827 1066 5913.5 10 T4 0 9.708 11.89 0.5 2.000 =×6 − c 1467.7 10 T5 9.510 9.722 11.99 11.31 0 22 T 9.865 9.917− 5.85 12.165 0 b26=×−×()()5913.5 10 1467.7 10 6 6 T 9.875 9.925 5.78 12.189 0 =×6 7 b 5728.5 10 T 9.350 9.551 11.78 10.928 0 22 8 xy+= 1 1122 (5913.5xx 1062 ) (5728.5 10 62 ) zx=+ y, y = Rsinθ, x = Rcosθ 88

3. The two graphs are being graphed with the same 4. T1 through T4 do not need to be adjusted. T5 center. Actually, the sun should remain in the must move 11.510 m toward the y-axis and the z same place for each graph. This means that the coordinate must move down 10.81 m. T6 must graph of Pluto needs to be adjusted. move 10.865 m toward the y-axis and the z coordinate must move down 12.04 m. T must 4. Shift=− Pluto's distance Neptune's distance 7 move 8.875 toward the y-axis and z must move =×−×1467.7 1066 37.1 10 down 12.064m. T8 must move 7.35 m toward the =×1430.6 106 y-axis and z must move down 10.425 m. (xx+ 1430.6 1062 ) y 2 +=1 62 62 (5913.5xx 10 ) (5728.5 10 ) Project IV Figure 1 5. Yes. One must adjust the scale accordingly to see it.

66 6. ()4431.6×× 10 ,752.6 10 , (0, 0) ()− x1,70 ()x ,70− ()4431.6×× 1066 ,752.6 10 2 280 ft 280 ft

7. No, The timing is different. They do not both (− 525, −350) (525, −350) pass through those points at the same time. 1050 ft

1. xay2 =−4 Project III (525)2 =− 4a ( − 350) 1. As an example, T will be used. (Note that any 1 272625= 1400a of the targets will yield the same result.) a = 196.875 zaxay=+4422 xy2 =−787.5 0.5=+− 4aa (0)22 4 ( 2) 0.5= 16a 1 a = 32

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2. Let y =−70 . (The arch needs to be 280 ft channel doesn’t shrink in width in a flood as high. Remember the vertex is at (0, 0), so we fast as a parabola. must measure down to the arch from the x-axis at the point where the arch’s height is Project V 280 ft.) π x2 =−787.5( − 70) −=2 ≤≤ 1. 42sec,ttt 0 2 =→=± 4 xx55125 234.8 π The channel will be 469.6 ft wide. 1tan,0−=tt2 ≤≤ t 4 3. Figure 2 For the x-values, t = 1.99, which is not in the (0, 350) domain [0, π/4]. Therefore there is no t-value that allows the two x values to be equal. 280 ft 280 ft 2. On the graphing utility, solve these in parametric form, using a t-step of π/32. (− 525, 0) (0, 0) x (525, 0) 2 It appears that the two graphs intersect at about (1.14, 0.214). However, for the first pair, 1050 ft t = 0.785 at that point. That t-value gives the xy22 point (2, 1) on the second pair. There is no +=1 22 intersection point. ab xy22 3. Since there were no solutions found for each +=1 275625 122500 method, the “solutions” matched. 4. xt=−42 y =− 1 t D=R; R = R x22(280) 11 4. +=1 1−=yt 275625 122500 1 xy=−−4(1 ) 2 (280)2 2 =− += x 1 275625 xy42 122500 xtyt==sec22 tan x2 = 99225 22 1tan+=22tt sec x =±315 D=[1,2], R=[0,1] The channel will be 630 feet wide. 1+=yx 5. If the river rises 10 feet, then we need to xy−=1 look for how wide the channel is when the xy+=41 height is 290 ft.  xy−=1 For the parabolic :  = x2 =−787.5( − 60) 50y y = 0 x2 = 47250 x = 1 x =±217.4 The t-values that go with those x, y values are There is still a 435 ft wide channel for the not the same for both pairs. Thus, again, there is ship. no solution. For the semi-ellipse: 3/2 x22(290) xt:ln= t +=1 5. 275625 122500 yt:243 =+ t 2 2 =−(290) Graphing each of these and finding the x 1 275625 122500 intersection: There is no intersection for the 2 = x-values, so there is no intersection for the x 86400 system. x =±293.9 Graphing the two parametric pairs: The The ship has a 588-ft channel. A semi- parametric equations show an intersection point. ellipse would be more practical since the However, the t-value that gives that point for

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each parametric pair is not the same. Thus there 7. Efficiency depends upon the equations. Graphing is no solution for the system. the parametric pairs allows one to see Putting each parametric pair into rectangular immediately whether the t-values will be the coordinates: same for each pair at any point of intersection. =→=x =∞ Sometimes, solving for t, as was done in the first xtte1 ln D = R, R (0, ) method is easy and can be quicker. It leads ==33x yt1 e straight to the t-values, so that allow the method 32 to be efficient. xt=→=23 tx DR=∞[0, ), =∞ [4, ) 2 If the two graphs intersect, one must be careful 2 =+=3 + yt2 242 x 4 to check that the t-values are the same for each Then solving that system: curve at that point of intersection. 3x  ye=

 2 yx=+243 This system has an intersection point at (0.56, 5.39). However, ln t = 0.56, gives t = 1.75 and t =≈()0.562/3 0.68 . Since the t-values are not the same, the point of intersection is false for the system. 6. x: 3 sin t = 2 cos t  tan t = 2/3  t = 0.588 y: 4 cos t +2 = 4 sin t  by graphing, the solution is t = 1.15 or t = 3.57. Neither of these are the same as for the x-values, thus the system has no solution. Graphing parametrically: If the graphs are done simultaneously on the graphing utility, the two graphs do not intersect at the same t-value. Tracing the graphs shows the same thing. This backs up the conclusion reached the first way. ==+ xtyt113sin 4cos 2 xy− 2 sintt== cos 34

sin22tt+= cos 1 xy22(2)− +=1 916 D=[-3, 3], R=[-2, 6] == xtyt222cos 4sin xy costt== sin 24

cos22tt+= sin 1 xy22 +=1 416 D=[-2, 2], R=[-4,4] Solving the system graphically: x = 1.3 , y =−3.05 . However, the t-values associated with these values are not the same. Thus there is no solution. (Similarly with the symmetric pair in the third quadrant.)