chapter 10 conic sections and analytic geometry · 7/31/2013 · conic sections and analytic...
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Chapter 10 Conic Sections and Analytic Geometry
Copyright © 2014 Pearson Education, Inc 1209
Section 10.1
Check Point Exercises
1. 2 2
136 9
x y
The center is 0, 0 . 2
2
36, 6
9, 3
a a
b b
2 2 2
36 9
27
27
3 3
c a b
c
The foci are located at ( 3 3, 0) and (3 3, 0) .
The vertices are 0 6, 0 which gives 6, 0 and
6, 0 .
The endpoints of the minor axis are 0, 0 3 which
gives 0, 3 and 0,3 .
2. 2 2
2 2
2 2
16 9 144
16 9 144
144 144 144
19 16
x y
x y
x y
The center is 0, 0 . 2
2
16, 4
9, 3
a a
b b
2 2 2
16 9
7
7
c a b
c
The foci are located at (0, 7) and (0, 7 ) .
The vertices are 0, 0 4 which gives 0, –4 and
0,4 .
The endpoints of the minor axis are 0 3, 0 which
gives 3,0 and 3,0 .
3. Because the foci are located at (–2, 0) and (2,0), on the x-axis, the major axis is horizontal. The center of the ellipse is midway between the foci, located at (0, 0).
Thus, the form of the equation is 2 2
2 21.
x y
a b
We need to determine the values for 2a and 2 .b The distance from the center, (0, 0), to either vertex is
3. Thus, 3a and 2 9.a The distance from the center, (0, 0), to either focus is
2. Thus, 2c and 2 4.c 2 2 2
9 4
5
b a c
The equation is 2 2
1.9 5
x y
4. 2 2( 1) ( 2)
19 4
x y
The center is –1, 2 . 2
2
9, 3
4, 2
a a
b b
2 2 2
9 4
5
5
c a b
c
The foci are located at
( 1 5, 2) and ( 1 5, 2) .
The vertices are –1 3, 2 which gives –4, 2 and
2, 2 .
Chapter 10 Conic Sections and Analytic Geometry
1210 Copyright © 2014 Pearson Education, Inc
The endpoints of the minor axis are –1, 2 2
which gives –1, 0 and 1, 4 .
5. 2 2
2 21
20 10
x y
2 2
1400 100
x y
Since the truck is 12 feet wide, substitute 6x into the equation to find y.
2 2
2
2
2
2
61
400 100
36400 400 1
400 100
36 4 400
4 364
91
91
9.54
y
y
y
y
y
y
y
6 feet from the center, the height of the archway is 9.54 feet. Since the truck’s height is 9 feet, it will fit under the archway.
Concept and Vocabulary Check 10.1
1. ellipse; foci; center
2. 25; 5; 5; ( 5, 0); (5, 0); 9; 3; 3;
(0, 3); (0, 3)
3. 25; 5; 5; (0, 5); (0, 5); 9; 3; 3;
( 3, 0); (3, 0)
4. 5; (0, 5); (0, 5)
5. ( 1, 4)
6. ( 2, 2); (8, 2)
7. (1, 9)
9. 4; 1; 16
Exercise Set 10.1
1. 2 2
116 4
x y
2
2
2 2 2
16, 4
4, 2
16 4 12
12 2 3
a a
b b
c a b
c
The foci are located at ( 2 3, 0) and (2 3, 0) .
2. 2 25, 5a a 2 16, 4b b 2 2 2 25 16 9c a b , c = 3
The foci are located at ( 3, 0) and (3, 0) .
3. 2
2
2 2 2
36, 6
9, 3
36 9 27
27 3 3
a a
b b
c a b
c
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1211
The foci are located at (0, 3 3) and (0, 3 3) .
4. 2 2
116 49
x y
2 49, 7a a 2 16, 4b b 2 2 2 49 16 33c a b , 33c
The foci are located at (0, 33) and (0, 33) .
5. 2
2
2 2 2
64, 8
25, 5
64 25 39
39
a a
b b
c a b
c
The foci are located at (0, 39) and (0, 39) .
6. 2 49, 7a a 2 36, 6b b 2 2 2 49 36 13c a b , 13c
The foci are located at ( 13, 0) and ( 13, 0) .
7. 2
2
2 2 2
81, 9
49, 7
81 49 32
32 4 2
a a
b b
c a b
c
The foci are located at (0, 4 2) and (0, 4 2) .
8. 2 100, 10a a 2 64, 8b b 2 2 2 100 64 36c a b
The foci are located at (0, 6) and (0, 6) .
Chapter 10 Conic Sections and Analytic Geometry
1212 Copyright © 2014 Pearson Education, Inc
9. 2 2
19 25
4 4
x y
2
2
2
25 9
4 416
4
4
2
c
c
c
c
The foci are located at (0, 2) and (0, −2).
10. 2 81 25
4 16c
2
2
324 25
16 16299
16
299
44.3
c
c
c
c
The foci are located at (4.3, 0) and (–4.3, 0).
11. 2 21 4x y 2 2
22
4 1
11
4
x y
yx
2
2
11
43
4
3
20.9
c
c
c
c
The foci are located at 3 3
,0 and ,0 .2 2
12. 2 21 4y x 2 24 1x y
2 2
11 1
4
x y
12 14
324
3
20.87
c
c
c
c
foci: (0, 0.87) (0, –0.87)
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1213
13. 2 2
2 2
2 2
25 4 100
25 4 100
100 100 100
14 25
x y
x y
x y
2
2
2 2 2
25, 5
4, 2
25 4 21
a a
b b
c a b
The foci are located at (0, 21) and (0, 21) .
14. 2 29 4 36x y 2 2
2 2
9 4 36
36 36 36
14 9
x y
x y
2 9, 3a a 2 4, 2b b 2 2 2 9 4 5c a b , 5c
The foci are located at (0, 5) and (0, 5) .
15. 2 24 16 64x y
2 2
2
2
2
2
116 4
16, 4
4, 2
16 4
12
12
2 3
3.5
x y
a a
b b
c
c
c
c
c
The foci are located at (2 3,0) and (-2 3,0).
16. 2 24 25 100x y
2 2
125 4
x y
2
2
25 4
21
21
4.6
c
c
c
c
The foci are located at (4.6, 0) and (–4.6, 0).
Chapter 10 Conic Sections and Analytic Geometry
1214 Copyright © 2014 Pearson Education, Inc
17. 2 27 35 5x y
2 2
2 2
7 5 35
15 7
x y
x y
2
2
2
2
7, 7
5, 5
7 5
2
2
1.4
a a
b b
c
c
c
c
The foci are located at (0, 2 ) and (0, 2).
18. 2 26 30 5x y
2 2
2 2
6 5 30
15 6
x y
x y
2
2
6 5
1
1
c
c
c
The foci are located at (0, 1) and (0, –1).
19. 2 24, 1,a b center at (0, 0) 2 2
14 1
x y
2 2 2 4 1 3
3
c a b
c
The foci are at ( 3, 0) and ( 3, 0) .
20. 2 216, 4,a b center at (0, 0) 2 2
116 4
x y
2 2 2 16 4 12c a b , 12 2 3c
The foci are at ( 2 3, 0) and (2 3, 0) .
21. 2 24, 1,a b
center: (0, 0) 2 2
11 4
x y
2 2 2 4 1 3
3
c a b
c
The foci are at (0, 3) and (0, 3) .
22. 2 216, 4a b , center: (0, 0) 2 2
14 16
x y
2 2 2 16 4 12c a b , 12 2 3c
The foci are at (0, 2 3) and (0, 2 3) .
23. 2 2( 1) ( 1)
14 1
x y
2 2
2
2
4, 1
4 1
3
3
a b
c
c
c
The foci are located at ( 1 3,1) and ( 1 3,1).
24. 2 24, 1a b center: (–1, –1)
2 2( 1) ( 1)
11 4
x y
2
2
4 1
3
3
1.7
c
c
c
c
(0 – 1, 1.7 – 1), (0 – 1, – 1.7 – 1) The foci are at ( –1, 0.7) and (–1, –2.7).
25. 2 2
2 2 2
2 2
25, 64
64 25 39
164 39
c a
b a c
x y
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1215
26. 2 24 36c a 2 2 2 36 4 32b a c 2 2
136 32
x y
27. 2 2
2 2 2
2 2
16, 49
49 16 33
133 49
c a
b a c
x y
28. 2 29, 16c a 2 2 2 16 9 7b a c 2 2
17 16
x y
29. 2 2
2 2 2
2 2
4, 9
9 4 13
113 9
c b
a b c
x y
30. 2 24, 4c b 2 2 2 4 4 8a b c 2 2
14 8
x y
31. 2
2
2 2
2 8, 4, 16
2 4, 2, 4
116 4
a a a
b b b
x y
32. 22 12, 6, 36a a a 2
2 2
2 6, 3, 9
136 9
b b b
x y
33. 22 10, 5, 25a a a
22 4, 2, 4
2 2( 2) ( 3)1
4 25
b b b
x y
34. 22 20, 10, 100a a a 22 10, 5, 25b b b
center (2, –3) 2 2( 2) ( 3)
125 100
x y
35. length of the major axis = 9 – 3 = 6 2a = 6, a = 3 major axis is vertical length of the minor axis = 9 – 5 = 4 2b = 4, b = 2 Center is at (7, 6).
2 2( 7) ( 6)1
4 9
x y
36. length of major axis = 8 – 2 = 6, 2a = 6, a = 3 length of minor axis = 5 – 3 = 2, 2b = 2, b = 1, center (5, 2)
2 2( 5) ( 2)1
9 1
x y
37. 2 9, 3a a 2 4, 2b b
center: (2, 1) 2 2 2 9 4 5
5
c a b
c
The foci are at (2 5, 1) and (2 5, 1) .
38. 2 16, 4a a 2 9, 3b b
center: (1, 2) 2 2 2 16 9 7c a b , 7c
The foci are at (1 7, 2) and (1 7, 2).
Chapter 10 Conic Sections and Analytic Geometry
1216 Copyright © 2014 Pearson Education, Inc
39. 2 2
2 2
( 3) 4( 2) 16
16 16 16
( 3) ( 2)1
16 4
x y
x y
2 16, 4a a 2 4, 2b b
center: ( 3, 2) 2 2 2 16 4 12
12 2 3
c a b
c
The foci are at ( 3 2 3, 2) and ( 3 2 3, 2) .
40. 2 2
2 2
( 3) 9( 2) 18
18 18 18
( 3) ( 2)1
18 2
x y
x y
2 18, 18 3 2a a 2 2, 2b b
center: (3, 2) 2 2 2 18 2 16, 4c a b c
The foci are at ( 1, 2) and (7, 2) .
41. 2 25, 5a a 2 9, 3b b
center: (4, 2) 2 2 2 25 9 16
4
c a b
c
The foci are at (4, 2) and (4, 6) .
42. 2 16, 4a a 2 9, 3b b
center: (3, 1) 2 2 2 16 9 7c a b , 7c
The foci are at (3, 1 7) and (3, 1 7) .
43. 2 36, 6a a 2 25, 5b b
center: (0, 2) 2 2 2 36 25 11
11
c a b
c
The foci are at (0, 2 11) and (0, 2 11) .
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1217
44. a2 = 25, a = 5 b2 = 4, b = 2 center: (4, 0)
c2 = a2 – b2 = 25 – 4 = 21, 21c
The foci are at (4, 21) and (4, 21) .
45. a2 = 9, a = 3
b2 = 1, b = 1 center: (–3, 2) c2 = a2 – b2 = 9 –1 = 8
8 2 2c
The foci are at (–3 – 2 2, 2) and
(–3 + 2 2, 2).
46. a2 = 16, a = 4 b2 = 1, b = 1 center: (–2, 3)
c2 = a2 – b2 = 16 – 1 = 15, 15c
The foci are at ( 2 15, 3) and ( 2 15, 3) .
47. 2 5 2c
2 3
3
1.7
c
c
c
The foci are located at (1, 3 3) and (1, 3 3).
48. 2 5 2c
2 3
3
1.7
c
c
c
(–1, 3 + 1.7) (–1 , 3 – 1.7) The foci are (– 1, 4.7) and (–1, 1.3).
49. 2 2
2 2
9( 1) 4( 3) 36
36 36 36
( 1) ( 3)1
4 9
x y
x y
a2 = 9, a = 3
b2 = 4, b = 2 center: (1, –3)
c2 = a2 – b2 = 9 – 4 = 5
5c
The foci are at (1, 3 5) and (1, 3 5) .
Chapter 10 Conic Sections and Analytic Geometry
1218 Copyright © 2014 Pearson Education, Inc
50. 2 236( 4) ( 3) 36
36 36 36
x y
22 ( 3)
( 4) 136
yx
a2 = 36, a = 6
b2 = 1, b = 1 center: (–4, –3)
c2 = a2 – b2 = 36 – 1 = 35, 35c
The foci are at (–4, –3 + 35) and ( 4, 3 35) .
51. 2 2
2 2
(9 36 ) (25 50 ) 164
9( 4 ) 25( 2 ) 164
x x y y
x x y y
2 29( 4 4) 25( 2 1)
164 36 25
x x y y
2 2
2 2
2 2
9( 2) 25( 1) 225
9( 2) 25( 1) 225
225 225 225
( 2) ( 1)1
25 9
x y
x y
x y
center: (2, –1) a2 = 25, a = 5 b2 = 9, b = 3 c2 = a2 – b2 = 25 – 9 = 16 c = 4 The foci are at (–2, –1) and (6, –1).
52. 2 2
2 2
(4 32 ) (9 36 ) 64
4( 8 ) 9( 4 ) 64
x x y y
x x y y
2 24( 8 16) 9( 4 4)x x y y 64 64 36
2 2
2 2
2 2
4 4 9 2 36
4 4 9 2 36
36 36 36
4 21
9 4
x y
x y
x y
center: (4, –2) 2 9, 3a a 2 4, 2b b 2 2 2 9 4 5c a b , 5c
The foci are at 4 5, 2 and 4 5, 2 .
53. (9x2 – 18x) + (16y2 + 64y) = 71 9(x2 – 2x) + 16(y2 + 4y) = 71 9(x2 – 2x + 1) + 16(y2+ 4y + 4) = 71 + 9 + 64 9(x – 1)2 + 16(y + 2)2 = 144
2 2
2 2
9( 1) 16( 2) 144
144 144 144
( 1) ( 2)1
16 9
x y
x y
center: (1, –2) a2 = 16, a = 4 b2 = 9, b = 3 c2 = a2 – b2 = 16 – 9 = 7
c = 7 The foci are at
(1 – 7 , –2) and (1+ 7 , –2).
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1219
54. 2 2
2 2
2 2
( 10 ) (4 8 ) 13
( 10 25) 4( 2 ) 13 25
( 10 25) 4( 2 1) 13 25 4
x x y y
x x y y
x x y y
2 2
2 2
2 2
( 5) 4( 1) 16
( 5) 4( 1) 16
16 16 16
( 5) ( 1)1
16 4
x y
x y
x y
center: (–5, 1) 2 16, 4a a 2 4, 2b b 2 2 2 16 4 12c a b , 12 2 3c
The foci are at ( 5 2 3, 1) and ( 5 2 3, 1) .
55. 2 2
2 2
2 2
2 2
2 2
2 2
(4 16 ) ( 6 ) 39
4( 4 ) ( 6 ) 39
4( 4 4) ( 6 9) 39 16 9
4( 2) ( 3) 64
4( 2) ( 3) 64
64 64 64
( 2) ( 3)1
16 64
x x y y
x x y y
x x y y
x y
x y
x y
center: (–2, 3) 2
2
2 2 2
64, 8
16, 4
64 16 48
48 4 3
a a
b b
c a b
c
The foci are at (–2, 3 + 4 3) and (–2, 3 – 4 3) .
56. 2 2
2 2
(4 24 ) (25 100 ) 36
4( 6 ) 25( 4 ) 36
x x y y
x x y y
2 24( 6 9) 25( 4 4) 36 36 100x x y y
2 2
2 2
2 2
4( 3) 25( 2) 100
4( 3) 25( 2) 100
100 100 100
( 3) ( 2)1
25 4
x y
x y
x y
center: (3, –2) 2 25, 5a a 2 4, 2b b 2 2 2 25 4 21c a b , 21c
The foci are at (3 21, 2) and (3 21, 2) .
57. 2 2 1x y 2 2
2 2
2 2
9 9
9 9
9 9 9
19 1
x y
x y
x y
The first equation is that of a circle with center at the origin and 1r . The second equation is that of an ellipse with center at the origin, horizontal major axis of length 6 units 3a , and vertical minor axis of
length 2 units 1b .
−5
5
−5 5
y
x
(0, 1)
(0, 1)−
Check each intersection point.
The solution set is 0, 1 , 0,1 .
Chapter 10 Conic Sections and Analytic Geometry
1220 Copyright © 2014 Pearson Education, Inc
58. 2 2 25x y 2 2
2 2
2 2
25 25
25 25
25 25 25
11 25
x y
x y
x y
The first equation is for a circle with center at the origin and 5r . The second is for an ellipse with center at the origin, vertical major axis of length 10 units 5b , and horizontal minor axis of length 2
units 1a .
−5
5
−5 5
y
x
(0, 5)
(0, 5)− Check each intersection point.
The solution set is 0, 5 , 0,5 .
59. 2 2
125 9
x y 3y
The first equation is for an ellipse centered at the origin with horizontal major axis of length 10 units and vertical minor axis of length 6 units. The second equation is for a horizontal line with a y-intercept of 3.
(0, 3)
−5
5
−5 5
y
x
Check the intersection point.
The solution set is 0,3 .
60. 2 2
14 36
x y 2x
The first equation is for an ellipse centered at the origin with vertical major axis of length 12 units and horizontal minor axis of length 4 units. The second equation is for a horizontal line with an x-intercept of
2.
−5
5
−5 5
y
x
( 2, 0)−
Check the intersection point.
The solution set is 2,0 .
61. 2 2
2 2
2 2
4 4
4 4
4 4 4
11 4
x y
x y
x y
2 2
2 2
2 2
x y
y x
y x
The first equation is for an ellipse centered at the origin with vertical major axis of length 4 units ( 2b ) and horizontal minor axis of length 2 units
1a . The second equation is for a line with slope
2 and y-intercept 2 .
(1, 0)
(0, 2)−
−5
5
−5 5
y
x
Check the intersection points.
The solution set is 0, 2 , 1,0 .
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1221
62. 2 2
2 2
2 2
4 4
4 4
4 4 4
11 4
x y
x y
x y
3
3
x y
y x
The first equation is for an ellipse centered at the origin with vertical major axis of length 4 units ( 2b ) and horizontal minor axis of length 2 units
1a . The second equation is for a line with slope
1 and y-intercept 3.
−5
5
−5 5
y
x
The two graphs never cross, so there are no intersection points. The solution set is or .
63. 22 2
2 2
2 2
2 2
16 4
16 4
4 16
14 16
y x
y x
x y
x y
We want to graph the bottom half of an ellipse centered at the origin with a vertical major axis of length 8 units ( 4b ) and horizontal minor axis of
length 4 units 2a .
64. 22 2
2 2
2 2
2 2
4 4
4 4
4 4
11 4
y x
y x
x y
x y
We want to graph the bottom half of an ellipse centered at the origin with a vertical major axis of length 4 units ( 2b ) and horizontal minor axis of
length 2 units 1a .
65. a = 15, b = 10 2 2
1225 100
x y
Let x = 4 2 2
2
2
2
41
225 100
16900 900(1)
225 100
64 9 900
9 836
8369.64
9
y
y
y
y
y
Yes, the truck only needs 7 feet so it will clear.
Chapter 10 Conic Sections and Analytic Geometry
1222 Copyright © 2014 Pearson Education, Inc
66. a = 25, b = 20 2 2
1625 400
x y
Let x = 10 2 2(10)
1625 400
y
10,000 2100
10,000(1)625 400
y
1600 + 25y2 = 10,000
25y2 = 8400
336 18.3y
Yes, the truck only needs 14 feet so it will clear.
67. a. a = 48, a2= 2304 b = 23, b2 = 529
2 2
12304 529
x y
b. c2 = a2 – b2 = 2304 – 529 = 1775
c = 1775 42.13 He situated his desk about 42 feet from the center of the ellipse, along the major axis.
68. a = 50, b = 30 2 2
2 21
50 30
x y
2 2 2
2 250 30 2500 900 1600
40
c a b
c
The focus is 40 feet from the center of the room so one person should stand at 10 feet along the 100 foot width and the other person should stand at 90 feet.
69. – 77. Answers will vary.
78. 2 186, 93a a
2 185.8, 92.9b a
Earth’s orbit: 2 2
2 2
2 2
1(93) (92.9)
18649 8630.41
x y
x y
22 2
2(92.9) 1
(93)
xy
2
292.9 1
(93)
xy
2 283.5, 141.75a a
2 278.5, 139.25b b
Mar’s orbit: 2 2
2 2
2 2
1(141.75) (139.25)
120,093.0625 19,390.5625
x y
x y
22 2
2(139.25) 1
(141.75)
xy
2
2139.25 1
(141.75)
xy
79. does not make sense; Explanations will vary. Sample explanation: The foci are on the major axis.
80. does not make sense; Explanations will vary. Sample explanation: An ellipse is symmetrical about both its major and minor axes.
81. does not make sense; Explanations will vary. Sample explanation: We must also know the other vertices.
82. makes sense
Section 10.1 The Ellipse
Copyright © 2014 Pearson Education, Inc. 1223
83. 26, 36a a 2 2
21
36
x y
b
When x = 2 and y = –4, 2 2
2
2
2
2
2
2 2
365
2 ( 4)1
364 16
1364 5
9
36 5
36
5
136
b
b
b
b
b
x y
84. a. The perigee is at the point (5000, 0). If the center of the earth is at (16, 0), and the radius is 4000 miles, the right endpoint of the earth along the major axis is (4016, 0). The perigee is 5000 – 4016 = 984 miles above the earth’s surface.
b. The apogee is at the point (–5000, 0). The left endpoint of the earth along the major axis is
(–3984, 0). The apogee is 5000 ( 3984) =
1016 miles above the earth’s surface.
85. The large circle has radius 5 with center
(0, 0). Its equation is x2 + y2 = 25. The small circle has radius 3 with center (0, 0). Its equation is
x2 + y2 = 9.
86. c
a is close to zero when c is very small. This happens
when a and b are nearly equal, or when the shape of the graph is nearly circular.
87. 2 2
2 2
2 2
4 9 36
4 9 36
36 36 36
19 4
x y
x y
x y
The terms are separated by subtraction rather than by addition.
88. 2 2
116 9
x y
a. Substitute 0 for y. 2 2
2
2
01
16 9
116
16
4
x
x
x
x
The x-intercepts are 4 and 4.
b. 2 2
2
2
01
16 9
19
9
y
y
y
The equation 2 9y has no real solutions.
89. 2 2
19 16
y x
a. Substitute 0 for x. 2 2
2
2
01
9 16
19
9
3
y
y
y
y
The y-intercepts are 3 and 3.
b. 2 2
2
2
01
9 16
116
16
x
x
x
The equation 2 16x has no real solutions.
Chapter 10 Conic Sections and Analytic Geometry
1224 Copyright © 2014 Pearson Education, Inc
Section 10.2
Check Point Exercises
1. a. 2 2
125 16
x y
2 25, 5a a
vertices: (5, 0) and (–5, 0) 2
2 2 2
16
25 16
41
41
b
c a b
c
The foci are at ( 41, 0) and ( 41, 0) .
b. 2 2
125 16
y x
2 25, 5a a
vertices: (0, 5) and (0, –5) 2
2 2 2
16
25 16
41
41
b
c a b
c
The foci are at (0, 41) and (0, 41) .
2. Because the foci are located at (0, –5) and (0, 5), on the y-axis, the transverse axis lies on the y-axis. The center of the hyperbola is midway between the foci, located at (0, 0).
Thus, the form of the equation is 2 2
2 21.
y x
a b
We need to determine the values for 2a and 2 .b The distance from the center, (0, 0), to either vertex is
3. Thus, 3a and 2 9.a The distance from the center, (0, 0), to either focus is
5. Thus, 5c and 2 25.c 2 2 2
25 9
16
b c a
The equation is 2 2
1.9 16
y x
3. 2 2
136 9
x y
2 36, 6a a
The vertices are (6, 0) and (–6, 0). 2 9, 3b b
asymptotes: 3 1
6 2
by x x x
a
2 2 2
36 9
45
45
3 5
c a b
c
The foci are at ( 3 5, 0) and (3 5, 0) .
4. 2 2
2 2
22
4 4
4 4
4 4 4
14
y x
y x
yx
2 4, 2a a
The vertices are (0, 2) and (0, –2). 2 1, 1b b
asymptotes: 2a
y x xb
2 2 2
4 1
5
5
c a b
c
The foci are at (0, 5) and (0, 5) .
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1225
5. 2 2( 3) ( 1)
14 4
x y
center at (3, 1) 2
2
4, 2
1, 1
a a
b b
The vertices are (1, 1) and (5, 1).
asymptotes: 1
1 ( 3)2
y x
2 2 2
4 1
5
5
c a b
c
The foci are at (3 5, 1) and (3 5, 1) .
6. 2 24 24 9 90 153 0x x y y
2 2
2 2
2 2
2 2
2 2
2 2
4 6 9 10 153
4 6 9 9 10 25 153 36 ( 225)
4 3 9 5 36
4 3 9 5 36
36 36 36
3 51
9 4
5 31
4 9
x x y y
x x y y
x y
x y
x y
y x
center at (3, –5)
2
2
4, 2
9, 3
a a
b b
The vertices are (3, –3) and (3, –7).
asymptotes: 2
5 ( 3)3
y x
2 2 2
4 9
13
13
c a b
c
The foci are at (3, 5 13) and (3, 5 13) .
7.
2 2 2 2 2
5280
2 3300, 1650
5280 1650 25,155,900
c
a a
b c a
The explosion occurred somewhere at the right branch of the hyperbola given by
2 2
1.2,722,500 25,155,900
x y
Concept and Vocabulary Check 10.2
1. hyperbola; foci; vertices; transverse
2. ( 5, 0); (5, 0); ( 34, 0); ( 34, 0)
3. (0, 5); (0, 5); (0, 34); (0, 34 )
4. asymptotes; center
5. dividing; 36
6. 3 3
; 2 2
y x y x
7. 2 ; 2y x y x
8. ( 3, 3); (7, 3)
9. (7, 2)
10. 16; 1; 128
Exercise Set 10.2
1. a2 = 4, a = 2 The vertices are (2, 0) and (–2, 0). b2 = 1
2 2 2 4 1 5
5
c a b
c
The foci are located at ( 5, 0) and ( 5, 0).
graph (b)
Chapter 10 Conic Sections and Analytic Geometry
1226 Copyright © 2014 Pearson Education, Inc
2. a2 = 1, a = 1 The vertices are (1, 0) and (–1, 0). b2 = 4
2 2 2 4 1 5c a b , 5c
The foci are at ( 5, 0) and ( 5, 0) .
graph (d)
3. a2 = 4, a = 2 The vertices are (0, 2) and (0, –2). b2 = 1
2 2 2 4 1 5
5
c a b
c
The foci are located at
(0, 5) and (0, 5) .
graph (a)
4. a2 = 1, a = 1 The vertices are (0, 1) and (0, –1). b2 = 4
2 2 2 1 4 5c a b , c = 5
The foci are at (0, 5) and (0, – 5) . graph (c)
5. a = 1, c = 3
b2 = c2 – a2 = 9 – 1 = 8 2
2 18
xy
6. a = 2, c = 6 b2 = c2 – a2 = 36 – 4 = 32
2 2
14 32
y x
7. a = 3, c = 4 b2 = c2 – a2 = 16 – 9 = 7
2 2
19 7
x y
8. a = 5, c = 7 b2 = c2 – a2 = 49 – 25 = 24
2 2
125 24
x y
9. 2a = 6 – (–6) 2a = 12 a = 6
2
62
6 2
3
a
b
bb
b
Transverse axis is vertical.
2 2
136 9
y x
10. a = 4
2
24
8
b
ab
b
Transverse axis is horizontal.
2 2
116 64
x y
11. a = 2, c = 7 – 4 = 3
2 2 2
2
2
2 3
4 9
5
b
b
b
Transverse axis is horizontal.
2 2( 4) ( 2)
14 5
x y
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1227
12. a = 4 – 1 = 3 c = 6 – 1 = 5
2 2 2
2
2
3 5
9 25
16
4
b
b
b
b
2 2( 1) ( 2)
19 16
y x
asymptotes: 3
1 ( 2)4
y x
Transverse axis is vertical.
13. a2 = 9, a = 3
b2 = 25, b = 5 vertices: (3, 0) and (–3, 0)
asymptotes: 5
3
by x x
a
c2 = a2 + b2 = 9 + 25 = 34
34c on x-axis
The foci are at ( 34, 0) and ( 34, 0) .
14. a2 = 16, a = 4 The vertices are (4, 0) and (–4, 0). b2 = 25, b = 5
asymptotes: 5
4
by x x
a
c2 = a2 + b2 = 16 + 25 = 41, 41c on x-axis
The foci are at ( 41,0) and ( 41,0) .
15. a2 = 100, a = 10 b2 = 64, b = 8 vertices: (10, 0) and (–10, 0)
asymptotes: 8
10
by x x
a
or 4
5y x
c2 = a2 + b2 = 100 + 64 = 164
164 2 41c on x-axis
The foci are at (2 41, 0) and ( 2 41, 0) .
16. a2 = 144, a = 12 b2 = 81, b = 9 The vertices are (12, 0) and (–12, 0).
asymptotes: 3
4
by x x
a
c2 = a2 + b2 = 144 + 81= 225 c = 15 on x-axis The foci are at (15, 0) and (–15, 0).
Chapter 10 Conic Sections and Analytic Geometry
1228 Copyright © 2014 Pearson Education, Inc
17. a2 = 16, a = 4 b2 = 36, b = 6 vertices: (0, 4) and (0, –4)
asymptotes: 4 2
6 3
ay x x x
b
or 2
3y x
c2 = a2 + b2 = 16 + 36 = 52
52 2 13c on y-axis
The foci are at (0, 2 13) and (0, 2 13) .
18. a2 = 25, a = 5 b2 = 64, b = 8 The vertices are (0, 5) and (0, –5).
asymptotes: y = 5
8
ax x
b
c2 = a2 + b2 = 25 + 64 = 89
89c on y-axis
The foci are at (0, 89) and (0, 89 ) .
19. 2
2 11
4
yx
2
2
2 2 2
2
1 1,
4 2
1, 1
11
4
a a
b b
c a b
c
2 5
4
5
21.1
c
c
c
The foci are located at 5 5
0, and 0, .2 2
asymptotes:
1
211
2
y x
y x
20. 2 2
11 19
y x
2
2
1, 1
31
1910
9
10
31.1
a b
c
c
c
c
The foci are (0, 1.1) and (0, –1.1).
Asymptote:
113
1 3y x x
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1229
21. 2 29 4 36
36 36 36
x y
2 2
14 9
x y
a2 = 4, a = 2 b2 = 9, b = 3 vertices: (2, 0) and (–2, 0)
asymptotes: 3
2
by x x
a
c2 = a2 + b2 = 4 + 9 = 13
13c on x-axis
The foci are at ( 13, 0) and ( 13, 0) .
22. 2 2
2 2
4 25 100
100 100 100
125 4
x y
x y
a2 = 25, a = 5 b2 = 4, b = 2 The vertices are (5, 0) and (–5, 0).
asymptotes: y = 2
5
bx x
a
c2 = a2 + b2 = 25 + 4 = 29
29c on x-axis
The foci are at ( 29, 0) and ( 29, 0) .
23. 2 2
2 2
9 25 225
225 225 225
125 9
y x
y x
a2 = 25, a = 5 b2 = 9, b = 3 vertices: (0, 5) and (0, –5)
asymptotes: 5
3
ay x x
b
c2 = a2 + b2 = 25 + 9 = 34
34c on y-axis
The foci are at (0, 34) and (0, 34) .
24. 2 216 9 144
144 144 144
y x
2 2
19 16
y x
a2 = 9, a = 3 b2 = 16, b = 4 The vertices are (0, 3) and (0, –3).
asymptotes: 3
4
ay x x
b
c2 = a2 + b2 = 9 + 16 = 25 c = 5 on y-axis The foci are at (0, 5) and (0, 5) .
Chapter 10 Conic Sections and Analytic Geometry
1230 Copyright © 2014 Pearson Education, Inc
25. 2 2 2y x
2 2
2 2
2
12 2
x y
x y
2
2
2
2
2, 2
2, 2
2 2
4
2
a a
b b
c
c
c
The foci are located at (2,0) and (–2, 0).
asymptotes: 2
2y x
y x
26. 2 2
2 2
2 2
3
3
13 3
y x
x y
x y
Vertices: ( 3,0) and ( 3,0).
Asymptotes: 3
3y x
y x
2
2
3 3
6
6
2.4
c
c
c
c
Foci: (2.4, 0) and (–2.4, 0).
27. a = 3, b = 5 2 2
19 25
x y
28. a = 3, b = 2 2 2
19 4
x y
29. a = 2, b = 3 2 2
14 9
y x
30. a = 5, b = 3 2 2
125 9
y x
31. Center (2, –3), a = 2, b = 3
2 2( 2) ( 3)
14 9
x y
32. Center (–1, –2) a = 2, b = 2
2 2( 1) ( 2)
14 4
x y
33. center: (–4, –3) a2 = 9, a = 3 b2 = 16, b = 4 vertices: (–7, –3) and (–1, –3)
asymptotes: y + 3 = 4
( 4)3
x
c2 = a2 + b2 = 9 + 16 = 25 c = 5 parallel to x-axis The foci are at (–9, –3) and (1, –3).
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1231
34. The center is located at (–2, 1). a2 = 9, a = 3 b2 = 25, b = 5 The vertices are (–5, 1) and (1, 1).
asymptotes: y – 1 = 5
( 2)3
x
c2 = a2 + b2 = 9 + 25 = 34
34c parallel to x-axis The foci are located at
( 2 34, 1) and ( 2 34, 1) .
35. center: (–3, 0) 2
2
25, 5
16, 4
a a
b b
vertices: (2, 0) and (–8, 0)
asymptotes: 4
( 3)5
y x
2 2 2 25 16 41
41
c a b
c
The foci are at ( 3 41, 0) and ( 3 41, 0) .
36. The center is located at (–2, 0). a2 = 9, a = 3 b2 = 25, b = 5 The vertices are (–5, 0) and (1, 0).
asymptotes: 5
( 2)3
y x
c2 = a2 + b2 = 9 + 25 = 34
34c parallel to x-axis
The foci are located at
( 2 34, 0) and ( 2 34, 0) .
37. center: (1, –2) a2 = 4, a = 2 b2 = 16, b = 4 vertices: (1, 0) and (1, –4)
asymptotes: 1
2 ( 1)2
y x
c2 = a2 + b2 = 4 + 16 = 20
c = 20 2 5 parallel to y-axis
The foci are at (1, 2 2 5) and (1, 2 2 5) .
38. The center is located at (–1, 2). a2 = 36, a = 6 b2 = 49, b = 7 The vertices are (–1, 8) and (–1, –4).
asymptotes: y – 2 = 6
( 1)7
x
c2 = a2 + b2 = 36 + 49 = 85
85c parallel to y-axis The foci are located at
( 1, 2 85) and ( 1, 2 85) .
Chapter 10 Conic Sections and Analytic Geometry
1232 Copyright © 2014 Pearson Education, Inc
39. 2 2
22
( 3) 4( 3) 4
4 4 4
( 3)( 3) 1
4
x y
xy
center: (3, –3) 2
2
4, 2
1, 1
a a
b b
vertices: (1, –3) and (5, –3)
asymptotes: 1
3 ( 3)2
y x
2 2 2 4 1 5
5
c a b
c
The foci are at (3 5, 3) and (3 5, 3).
40. 2 2( 3) 9( 4) 9
9 9 9
x y
22( 3)
( 4) 19
xy
The center is located at (–3, 4). a2 = 9, a = 3 b2 = 1, b = 1 The vertices are (–6, 4) and (0, 4).
asymptotes: y – 4 = 1
( 3)3
x
c2 = a2 + b2 = 9 + 1 = 10
10c parallel to x-axis
The foci are located at ( 3 10, 4) and
( 3 10, 4) .
41. 2 2( 1) ( 2)
13 3
x y
center: (1, 2)
a2 = 3, a = 3
b2 = , b = 3 vertices: (–1, 2) and (3, 2) asymptotes: y – 2 = (x – 1) c2 = a2 + b2 = 3 + 3 = 6
c = 6 parallel to y-axis
The foci are at (1 6,2) and (1 6,2).
42. 2 2( 2) ( 3)
15 5
y x
The center is located at (–3, 2).
a2 = 5, a = 5
b2 = 5, b = 5 The vertices are (–3, 0) and (–3, 4). asymptotes: 2 ( 3)y x
c2 = a2 + b2 = 5 + 5 = 10
10c parallel to y-axis The foci are located
at ( 3,2 10) and ( 3,2 10) .
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1233
43. (x2 – 2x) – (y2 + 4y) = 4 (x2 – 2x + 1) – (y2 + 4y + 4) = 4 + 1 – 4 (x – 1)2 – (y + 2)2 = 1 center: (1, –2) a2 = 1, a = 1 b2 = 1, b = 1 c2 = a2 + b2 = 1 + 1 = 2
c = 2 asymptotes: 2 ( 1)y x
The foci are at (1 2, 2) and (1 2, 2) .
44. (4x2 + 32x) – (y2 – 6y) = –39
4(x2 + 8x + 16) – (y2 – 6y + 9) = –39 + 64 – 9 4(x + 4)2 – (y – 3)2 = 16
2 2( 4) ( 3)
14 16
x y
center: (–4, 3) a2 = 4, a = 2 b2 = 16, b = 4 c2 = a2 + b2 = 4 + 16= 20
20 2 5c
The foci are at ( 4 2 5, 3) and ( 4 2 5, 3) .
Asymptotes: 4
3 ( 4)2
3 2( 4)
y x
y x
45. (16x2 + 64x) – (y2 + 2y) = –67 16(x2 + 4x + 4) – (y2 + 2y + 1) = –67 + 64 – 1
2 2
2 2
2 2
14
16( 2) ( 1) 4
16( 2) ( 1) 4
4 4 4
( 1) ( 2)1
4
x y
x y
y x
center: (–2, –1) a2 = 4, a = 2
2 1 1,
4 2b b
c2 = a2 + b2 = 4 + 1
4 =
17
4
174 4.25c
asymptotes:
2( 1) ( 2)
1
21 4( 2)
y x
y x
The foci are at 2, 1 4.25 and 2, 1 4.25 .
46. (9y2 – 18y) – (4x2 – 24x) = 63 9(y2 – 2y + 1) – 4(x2 – 6x + 9) = 63 + 9 – 36 9(y – 1)2 – 4(x – 3)2 = 36
2 2( 1) ( 3)
14 9
y x
The center is located at (3, 1). a2 = 4, a = 2 b2 = 9, b = 3
c2 = a2 + b2 = 4 + 9 = 13, 13c
The foci are at (3, 1 13) and (3, 1 13) .
Asymptotes: 2
1 ( 3)3
y x
Chapter 10 Conic Sections and Analytic Geometry
1234 Copyright © 2014 Pearson Education, Inc
47. (4x2 – 16x) – (9y2 – 54y) = 101
4(x2 – 4x + 4) – 9(y2 – 6y + 9) = 101 + 16 – 81
4(x – 2)2 – 9(y – 3)2 = 36
2 2( 2) ( 3)
19 4
x y
center: (2, 3) a2 = 9, a = 3 b2 = 4, b = 2 c2 = a2 + b2 = 9 + 4 = 13
13c
asymptotes: 2
3 ( 2)3
y x
The foci are at (2 13, 3) and (2 13, 3) .
48. (4x2 + 8x) – (9y2 + 18y) = 6 4(x2 + 2x + 1) – 9(y2 +2y + 1) = 6 + 4 – 9 4(x + 1)2 – 9(y + 1)2 = 1
2 2
1 14 9
( 1) ( 1)1
x y
The center is located at (–1, –1).
2 1 1,
4 2a a
2 1 1,
9 3b b
2 2 2 1 1 13
4 9 36c a b ,
13
6c
The foci are at
13 131 , 1 and 1 , 1
6 6
.
Asymptotes:
1
31 ( 1)1
22
1 ( 1)3
y x
y x
49. (4x2 – 32x) – 25y2 = –164 4(x2 – 8x + 16) – 25y2 = –164 + 64 4(x – 4)2 – 25y2 = –100
2 2
2 2
4( 4) 25 100
100 100 100
( 4)1
4 25
x y
y x
center: (4, 0) a2 = 4, a = 2 b2 = 25, b = 5 c2 = a2 + b2 = 4 + 25 = 29
29c
asymptotes: 2
( 4)5
y x
The foci are at (4, 29) and (4, 29) .
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1235
50. (9x2 – 36x) – (16y2 + 64y) = –116 9(x2 – 4x + 4) – 16(y2 + 4y + 4) = –116 + 36– 64 9(x – 2)2 – 16(y + 2)2 = –144
2 29( 2) 16( 2) 144
144 144 144
x y
2 2( 2) ( 2)
19 16
y x
The center is located at (2, –2). a2 = 9, a = 3 b2 = 16, b = 4 c2 = a2 + b2 = 9 + 16 = 25, c = 5 The foci are at (2, –7) and (2, 3).
Asymptotes: 3
2 ( 2)4
y x
51. 2 2
19 16
x y
The equation is for a hyperbola in standard form with the transverse axis on the x-axis. We have
2 9a and 2 16b , so 3a and 4b .
Therefore, the vertices are at ,0a or 3,0 .
Using a dashed line, we construct a rectangle using the 3 on the x-axis and 4 on the y-axis. Then use dashed lines to draw extended diagonals for the rectangle. These represent the asymptotes of the graph.
From the graph we determine the following: Domain: | 3 or 3x x x or
, 3 3,
Range: | is a real numbery y or ,
52. 2 2
125 4
x y
The equation is for a hyperbola in standard form with the transverse axis on the x-axis. We have
2 25a and 2 4b , so 5a and 2b .
Therefore, the vertices are at ,0a or 5,0 .
Using a dashed line, we construct a rectangle using the 5 on the x-axis and 2 on the y-axis. Then use dashed lines to draw extended diagonals for the rectangle. These represent the asymptotes of the graph.
From the graph we determine the following: Domain: | 5 or 5x x x or
, 5 5,
Range: | is a real numbery y or ,
53. 2 2
19 16
x y
The equation is for an ellipse in standard form with
major axis along the y-axis. We have 2 16a and 2 9b , so 4a and 3b . Therefore, the
vertices are 0, a or 0, 4 . The endpoints of
the minor axis are ,0b or 3,0 .
From the graph we determine the following: Domain: | 3 3x x or 3,3
Range: | 4 4y y or 4,4 .
Chapter 10 Conic Sections and Analytic Geometry
1236 Copyright © 2014 Pearson Education, Inc
54. 2 2
125 4
x y
The equation is for an ellipse in standard form with
major axis along the y-axis. We have 2 25a and 2 4b , so 5a and 2b . Therefore, the
vertices are ,0a or 5,0 . The endpoints of the
minor axis are 0, b or 0, 2 .
From the graph we determine the following: Domain: | 5 5x x or 5,5
Range: | 2 2y y or 2,2 .
55. 2 2
116 9
y x
The equation is in standard form with the transverse
axis on the y-axis. We have 2 16a and 2 9b , so 4a and 3b . Therefore, the vertices are at
0, a or 0, 4 . Using a dashed line, we
construct a rectangle using the 4 on the y-axis and 3 on the x-axis. Then use dashed lines to draw extended diagonals for the rectangle. These represent the asymptotes of the graph.
From the graph we determine the following: Domain: | is a real numberx x or ,
Range: | 4 or 4y y y or , 4 4,
56. 2 2
14 25
y x
The equation is in standard form with the transverse
axis on the y-axis. We have 2 4a and 2 25b , so 2a and 5b . Therefore, the vertices are at
0, a or 0, 2 . Using a dashed line, we
construct a rectangle using the 2 on the y-axis and 5 on the x-axis. Then use dashed lines to draw extended diagonals for the rectangle. These represent the asymptotes of the graph.
From the graph we determine the following: Domain: | is a real numberx x or ,
Range: | 2 or 2y y y or , 2 2,
57. 2 2
2 2
4
4
x y
x y
5
5
y
x(−2, 0)
(2, 0)
Check 2,0 :
2 22 0 4
4 0 4
4 4 true
2 22 0 4
4 0 4
4 4 true
Check 2,0 :
2 22 0 4
4 0 4
4 4 true
2 22 0 4
4 0 4
4 4 true
The solution set is 2,0 , 2,0 .
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1237
58. 2 2
2 2
9
9
x y
x y
5
5
y
x
(−3, 0)
(3, 0)
Check 3,0 :
2 23 0 9
9 0 9
9 9 true
2 23 0 9
9 0 9
9 9 true
Check 3,0 :
2 23 0 9
9 0 9
9 9 true
2 23 0 9
9 0 9
9 9 true
The solution set is 3,0 , 3,0 .
59.
2 2
2 2
9 9
9 9
x y
y x
or 2 2
2 2
11 9
19 1
x y
y x
5
5
y
x
(0, −3)
(0, 3)
Check 0, 3 :
2 29 0 3 9
0 9 9
9 9
true
2 23 9 0 9
9 0 9
9 9
true
Check 0,3 :
2 29 0 3 9
0 9 9
9 9
true
2 23 9 0 9
9 0 9
9 9
true
The solution set is 0, 3 , 0,3 .
60. 2 2
2 2
4 4
4 4
x y
y x
or 2 2
2 2
11 4
14 1
x y
y x
5
5
y
x
(0, −2)
(0, 2)
Check 0, 2 :
2 24 0 2 4
0 4 4
4 4
true
2 22 4 0 4
4 0 4
4 4
true
Check 0,2 :
2 24 0 2 4
0 4 4
4 4
true
2 22 4 0 4
4 0 4
4 4
true
The solution set is 0, 2 , 0,2 .
61. | d2 – d1 | = 2a = (2 s)(1100 ft / s) = 2200 ft
a = 1100 ft 2c = 5280 ft, c = 2640 ft
b2 = c2 – a2 = (2640)2 – (1100)2 = 5,759,600
2 2
2
2 2
15,759,600(1100)
11,210,000 5,759,600
x y
x y
If M1 is located 2640 feet to the right of the origin on the x-axis, the explosion is located on the right branch of the hyperbola given by the equation above.
Chapter 10 Conic Sections and Analytic Geometry
1238 Copyright © 2014 Pearson Education, Inc
62. a. 2c = 200 km, c = 100 km
2 1m
2 500 s 300s
d d a μμ
2a = 150,000 m = 150 km a = 75 km
2 22 2 2 100 75 4375b c a
2 2
21
437575
x y
2 2
15625 4375
x y
b. The x-coordinate of the ship is 100 km:
2 21001
5625 4375
y
2 10,0001
4375 5625
y
10,0004375 1
5625y 58.3
The ship is about 58.3 kilometers from the coast.
63. 2 2
2 2
2 2
625 400 250,000
625 400 250,000
250,000 250,000 250,000
1400 625
y x
y x
y x
a2 = 400, a = 400 = 20 2a = 40 The houses are 40 yards apart at their closest point.
64. a = 3 2 2
21
9
x y
b
To find b, use the equation of the slope of the asymptote,
1:
3 2
b b
a
Solving for b: 2b = 3, b = 3
.2
2 2
94
19
x y
65. a. ellipse
b. 2 24 4x y
66. a. hyperbola
b. 2 2 1x y
67. – 76. Answers will vary.
77. 2 2
04 9
x y
2 29
4y x
3
2y x
No; in general, the graph is two intersecting lines.
78. Answers will vary depending on the choice for a and b. For a= 2, b = 3, a graph is shown. The two graphs open right/left and up/down, sharing a common set of
asymptotes given by y = .b
xa
79. 4x2 – 6xy + 2y2 – 3x + 10y – 6 = 0 2y2 + (10 – 6x)y + (4x2 – 3x – 6) = 0
2 2
2
2
6 10 (10 6 ) 8(4 3 6)
4
6 10 4( 24 37)
4
3 5 24 37
2
x x x xy
x x xy
x x xy
The xy-term rotates the hyperbola. Separation of terms into ones containing only x or only y would not be possible.
Section 10.2 The Hyperbola
Copyright © 2014 Pearson Education, Inc. 1239
80. 2 2
116 9
x y
2 2
19 16
y x
22
2
2
169
16
169
16
316
4
xy
xy
y x
116 9
19 16
169
16
x x y y
y y x x
x xy y
If y ≥ 0, 2y y y
2 169
16
169
16
316 ( 4)
4
x xy
x xy
y x x x
If y < 0, 2y y y
2
2
169
16
169
16
169
16
3( 16)
4
x xy
x xy
x xy
y x x
3
164
y x x (x ≤ 4)
The second equation is a function with domain (–∞, ∞).
81. does not make sense; Explanations will vary. Sample explanation: This would change the ellipse to a hyperbola.
82. makes sense
83. makes sense
84. makes sense
85. false; Changes to make the statement true will vary. A sample change is: If a hyperbola has a transverse axis along the x–axis and one of the branches is removed, the remaining branch does not define a function of x.
86. false; Changes to make the statement true will vary. A sample change is: The points on the hyperbola’s asymptotes do not satisfy the hyperbola’s equation.
87. true
88. false; Changes to make the statement true will vary. A sample change is: It is possible for two different hyperbolas to share the same asymptotes. For
example 2 2
14 9
x y and 2 2
19 4
y x share the
same asymptotes.
89. c
a will be large when a is small. When this happens,
the asymptotes will be nearly vertical.
90. The center is at the midpoint of the line segment joining the vertices, so it is located at (5, 0). The standard form is:
2 2
2 2
( ) ( )1
y k x h
a b
(h, k) = (5, 0), and a = 6, so a2 = 36. 2 2
2
( 5)1.
36
y x
b
Substitute x = 0 and y = 9: 2 2
2
2
2
2
9 (0 5)1
3625 5
4
100 5
20
b
b
b
b
Standard form: 2 2( 5)
136 20
y x
Chapter 10 Conic Sections and Analytic Geometry
1240 Copyright © 2014 Pearson Education, Inc
91. If the asymptotes are perpendicular, then their slopes are negative reciprocals. For the hyperbola
2 2
2 21
x y
a b , the asymptotes are .
by x
a The
slopes are negative reciprocals when b a
a b (since
one is already the negative of the other). This
happens when b2 = a2, so a = b. Any hyperbola where a = b, such as
2 2
1,4 4
x y has perpendicular asymptotes.
92. 2 4 5y x x Since 1a is positive, the parabola opens upward. The x-coordinate of the vertex is
42.
2 2(1)
bx
a The y-coordinate of the
vertex is 2( 2) 4( 2) 5 9.y
Vertex: ( 2, 9).
93. 23 1 2y x
Since 3a is negative, the parabola opens downward. The vertex of the parabola is
, 1,2h k .
The y–intercept is 1.
94. 2
2
2
2
2 12 23 0
2 12 23
2 1 12 23 1
( 1) 12 24
y y x
y y x
y y x
y x
Section 10.3
Check Point Exercises
1. 2 8y x
4 8
2
p
p
focus: , 0 2, 0p
directrix: ; 2x p x
2. 2 12x y
4 12
3
p
p
focus: 0, 0, 3p
directrix: ; 3y p y
3. The focus is (8, 0). Thus, the focus is on the x-axis. We use the standard form of the equation in which
there is x-axis symmetry, namely, 2 4y px . The focus is 8 units to the right of the vertex, (0, 0). Thus, p is positive and 8.p
2
2
2
4
4 8
32
y px
y x
y x
Section 10.3 The Parabola
Copyright © 2014 Pearson Education, Inc. 1241
4. 2( 2) 4( 1)x y
From the equation we have 2h and 1.k vertex: (2, –1) Find p: 4 4
1
p
p
focus: ( , ) (2, 1 1)
(2,0)
h k p
directrix:
1 1
2
y k p
y
y
5. 2
2
2
2 4 7
2 1 4 7 1
( 1) 4( 2)
y y x
y y x
y x
From the equation we have 2h and 1.k vertex: (2, –1) Find p: 4 4
1
p
p
focus: ( , ) (2 1, 1)
(1, 1)
h p k
directrix:
2 ( 1)
3
x h p
x
x
6. 2 4x py
Let x = 3 and y = 4. 2
2
3 4 4
9 16
9
169
4
p
p
p
x y
The light should be placed at 9
0,16
or 9
16 inch
above the vertex.
Concept and Vocabulary Check 10.3
1. parabola; directrix; focus
2. a
3. ( 7, 0)
4. 7x
5. 28; ( 7, 14); ( 7, 14)
6. d
7. ( 2, 0)
8. 2y
9. 4; ( 4, 0); (0, 0)
Exercise Set 10.3
1. y2 = 4x 4p = 4, p = 1 vertex: (0, 0) focus: (1, 0) directrix: x = –1 graph (c)
2. x2 = 4y 4p = 4, p = 1 vertex: (0, 0) focus: (0, 1) directrix: y = –1 graph (a)
Chapter 10 Conic Sections and Analytic Geometry
1242 Copyright © 2014 Pearson Education, Inc
3. x2 = –4y 4p = –4, p = –1 vertex: (0, 0) focus: (0, –1) directrix: y = 1 graph (b)
4. y2 = –4x 4p = –4, p = –1 vertex: (0, 0) focus: (–1, 0) directrix: x = 1 graph (d)
5. 4p = 16, p = 4 vertex: (0, 0) focus: (4, 0) directrix: x = –4
6. 4p = 4, p = 1 vertex: (0, 0) focus: (1, 0) directrix: x = –1
7. 4p = –8, p = –2 vertex: (0, 0) focus: (–2, 0) directrix: x = 2
8. 4p = –12, p = –3 vertex: (0, 0) focus: (–3, 0) directrix: x = 3
9. 4p = 12, p = 3 vertex: (0, 0) focus: (0, 3) directrix: y = –3
10. 4p = 8, p = 2
vertex: (0, 0) focus: (0, 2) directrix: y = –2
ς
11. 4p = –16, p = –4 vertex: (0, 0) focus: (0, –4) directrix: y = 4
Section 10.3 The Parabola
Copyright © 2014 Pearson Education, Inc. 1243
12. 4p = –20, p = –5 vertex: (0, 0) focus: (0, –5) directrix: y = 5
13. y2 = 6x
4p = 6, p = 6 3
4 2
vertex: (0, 0)
focus: 3
, 02
directrix: x = 3
2
14. x2 = 6y
4p = 6, p = 6 3
4 2
vertex: (0, 0)
focus: 3
0,2
directrix: y = 3
2
15. 2
2
8 4
1
2
x y
x y
14
21
8
p
p
focus: 1
0,8
directrix: 1
8y
16. 2
2
8 4
1
2
y x
y x
14
21
8
p
p
vertex: (0, 0)
focus: 1
,08
directrix: 1
8x
17. p = 7, 4p = 28
y2 = 28x
18. p = 9, 4p = 36 y2 = 36x
19. p = –5, 4p = –20 y2 = –20x
20. p = –10, 4p = –40 y2 = –40x
Chapter 10 Conic Sections and Analytic Geometry
1244 Copyright © 2014 Pearson Education, Inc
21. p = 15, 4p = 60 x2 = 60y
22. p = 20, 4p = 80 x2 = 80y
23. p = –25, 4p = –100 x2 = –100y
24. p = –15, 4p = –60 x2 = –60y
25. 5 ( 3) 2p Vertex, (2, –3) 2( 2) 8( 3)x y
26. vertex: (5, –2 ); p = 7 – 5 = 2
2
2
2 2
4
8
( 2) 8( 5)
y px
y x
y x
27. vertex: (1, 2) p = 2 2( 2) 8( 1)y x
28. vertex: ( –1, 4) p = 3 2
2
2
2
4
4(3)
12
( 4) 12( 1)
y px
y x
y x
y x
29. vertex: (–3, 3), p = 1 2( 3) 4( 3)x y
30. vertex: (7, –5) p = 4 2( 7) 16( 5)x y
31. (y – 1)2 = 4(x – 1) 4p = 4, p = 1 vertex: (1, 1) focus: (2, 1) directrix: x = 0 graph (c)
32. (x + 1)2 = 4(y + 1) 4p = 4, p = 1 vertex: (–1, –1) focus: (–1, 0) directrix: y = –2 graph (a)
33. (x + 1)2 = –4(y + 1) 4p = –4, p = –1 vertex: (–1, –1) focus: (–1, –2) directrix: y = 0 graph (d)
34. (y – 1)2 = –4(x – 1) 4p = –4, p = –1 vertex: (1, 1) focus: (0, 1) directrix: x = 2 graph (b)
35. 4p = 8, p = 2 vertex: (2, 1) focus: (2, 3) directrix: y = –1
36. 4p = 4, p = 1 vertex: (–2, –1) focus: (–2, 0) directrix: y = –2
37. 4p = –8, p = –2
vertex: (–1, –1) focus: (–1, –3) directrix: y = 1
Section 10.3 The Parabola
Copyright © 2014 Pearson Education, Inc. 1245
38. 4p = –8, p = –2 vertex: (–2, –2) focus: (–2, –4) directrix: y = 0
39. 4p = 12, p = 3
vertex: (–1, –3) focus: (2, –3) directrix: x = –4
40. 4p = 12, p = 3 vertex: (–2, –4) focus: (1, –4) directrix: x = –5
41. (y + 1)2 = –8(x – 0) 4p = –8, p = –2 vertex: (0, –1) focus: (–2, –1) directrix: x = 2
42. (y – 1)2 = –8(x – 0) 4p = –8, p = –2 vertex: (0, 1) focus: (–2, 1) directrix: x = 2
43. x2 – 2x + 1 = 4y – 9 + 1 (x – 1)2 = 4y – 8 (x – 1)2 = 4(y – 2) 4p = 4, p = 1 vertex: (1, 2) focus: (1, 3) directrix: y = 1
44. x2 + 6x = –8y – 1 x2 + 6x + 9 = –8y – 1 + 9 (x + 3)2 = –8y + 8 = –8(y – 1) 4p = –8, p = –2 vertex: (–3, 1) focus: (–3, –1) directrix: y = 3
Chapter 10 Conic Sections and Analytic Geometry
1246 Copyright © 2014 Pearson Education, Inc
45. y2 – 2y + 1 = –12x + 35 + 1 (y – 1)2 = –12x + 36 (y – 1)2 = –12(x– 3) 4p = –12, p = –3 vertex: (3, 1) focus: (0, 1) directrix: x = 6
46. y2 – 2y = 8x – 1 y2 – 2y + 1 = 8x – 1 + 1 (y – 1)2 = 8x 4p = 8, p = 2 vertex: (0, 1) focus: (2, 1) directrix: x = –2
47. x2 + 6x = 4y – 1 x2 + 6x + 9 = 4y – 1 + 9 (x + 3)2 = 4(y + 2) 4p = 4, p = 1 vertex: (–3, –2) focus: (–3, –1) directrix: y = –3
48. x2 + 8x + 16 = 4y – 8 +16 (x + 4)2 = 4y+ 8 (x + 4)2 = 4(y + 2) 4p = 4, p = 1 vertex: (–4, –2) focus: (–4, –1) directrix: x = –3
49. The y-coordinate of the vertex is
6
32 2 1
by
a
The x-coordinate of the vertex is
23 6 3 5
9 18 5
4
x
The vertex is 4, 3 .
Since the squared term is y and 0a , the graph opens to the right. Domain: | 4x x or 4,
Range: | is a real numbery y or ,
The relation is not a function.
50. The y-coordinate of the vertex is 2
12 2 1
by
a
The x-coordinate of the vertex is
21 2 1 5
1 2 5
6
x
The vertex is 6,1 .
Since the squared term is y and 0a , the graph opens to the right. Domain: | 6x x or 6,
Range: | is a real numbery y or ,
The relation is not a function.
Section 10.3 The Parabola
Copyright © 2014 Pearson Education, Inc. 1247
51. The x-coordinate of the vertex is 4
22 2 1
bx
a
The y-coordinate of the vertex is
22 4 2 3
4 8 3
1
y
The vertex is 2,1 .
Since the squared term is x and 0a , the graph opens down. Domain: | is a real numberx x or ,
Range: | 1y y or ,1
The relation is a function.
52. The x-coordinate of the vertex is
42
2 2 1
bx
a
The y-coordinate of the vertex is
22 4 2 4
4 8 4
8
y
The vertex is 2,8 .
Since the squared term is x and 0a , the graph opens down. Domain: | is a real numberx x or ,
Range: | 8y y or ,8
The relation is a function.
53. The equation is in the form 2x a y k h
From the equation, we can see that the vertex is
3,1 .
Since the squared term is y and 0a , the graph opens to the left. Domain: | 3x x or ,3
Range: | is a real numbery y or ,
The relation is not a function.
54. The equation is in the form 2x a y k h
From the equation, we can see that the vertex is
2,1 .
Since the squared term is y and 0a , the graph opens to the left. Domain: | 2x x or , 2
Range: | is a real numbery y or ,
The relation is not a function.
55.
Check 4,2 :
24 2 2 4
4 0 4
4 4
true
12 4
22 2
true
Check 0,0 :
20 0 2 4
0 4 4
0 0
true
10 0
20 0
true
The solution set is 4,2 , 0,0 .
56.
Check 2,3 :
22 3 3 2
2 0 2
2 2
true
2 3 5
5 5
true
Check 3,2 :
23 2 3 2
3 1 2
3 3
true
3 2 5
5 5
true
The solution set is 2,3 , 3,2 .
Chapter 10 Conic Sections and Analytic Geometry
1248 Copyright © 2014 Pearson Education, Inc
57.
Check 2,1 :
22 1 3
2 1 3
2 2 true
22 1 3 1
2 1 3
2 2 true
The solution set is 2,1 .
58.
Check 5,0 :
25 0 5
5 0 5
5 5 true
2 25 0 25
25 0 25
25 25 true
Check 4, 3 :
24 3 5
4 9 5
4 4 true
2 24 3 25
16 9 25
25 25 true
Check 4,3 :
24 3 5
4 9 5
4 4 true
2 24 3 25
16 9 25
25 25 true
The solution set is 5,0 , 4, 3 , 4,3 .
59.
The two graphs do not cross. Therefore, the solution set is the empty set, or .
60.
The two graphs do not cross. Therefore, the solution set is the empty set, or .
61. x2 = 4py
22 = 4p(1) 4 = 4 p = 1 The light bulb should be placed 1 inch above the vertex.
62. x2 = 4py 42 = 4p(1 16 = 4p p = 4 The light bulb should be placed 4 inches above the vertex.
63. x2 = 4py 62 = 4p(2) 36 = 8p
p = 36 9
4.58 2
The receiver should be located 4.5 feet from the base of the dish.
64. x2 = 4py
32 = 4p(2) 9 = 8p
p = 9
8 = 1.125
The receiver should be placed 1.125 feet from the base of the smaller dish.
65. 2
2
2
2
2
4
(640) 4 (160)
(640)640
640640 200 440
(440) 4(640)
(440)75.625
4(640)
x py
p
p
x
y
y
The height is 76 meters.
Section 10.3 The Parabola
Copyright © 2014 Pearson Education, Inc. 1249
66. x2 = 4py (400)2 = 4p(160) 160,000 = 640p
p = 160,000
250640
x = 400 – 100 = 300
3002 = 4(250)y 2300
904(250)
y
The height is 90 feet.
67. 2 4x py 2
2
2
2004 ( 50)
2
10,0004
504 200
200
(30) 200
9004.5
200
p
p
p
x y
y
y
(height of bridge) = 50 – 4.5 = 45.5 feet. Yes, the boat will clear the arch.
68. 2 4y px
25 4 (6)
254
6
p
p
25 feet
6
69. – 76. Answers will vary.
77. y2 + 2y – 6x + 13 = 0 y2 + 2y + (–6x + 13) = 0
22 2 4( 6 13)
2
xy
2 24 48
2
1 6 12
xy
y x
78. 2
2
10 25 0
10 ( 25) 0
y y x
y y x
210 10 4( 25)
2
10 4
2
5
xy
xy
y x
79. 16x2 – 24xy + 9y2 – 60x – 80y + 100 = 0
9y2 – (24x + 80)y + (16x2 – 60x + 100) = 0 2 224 80 (24 80) 36(16 60 100)
18
24 80 6000 2800
18
24 80 20 15 7
18
12 40 10 15 7
9
x x x xy
x xy
x xy
x xy
80. x2 + 2 3 xy + 3y2 + 8 3 x – 8y + 32 = 0
3y2 + (2 3 x – 8)y + (x2 + 8 3 x + 32) = 0
2 2(2 3 8) (2 3 8) 12( 8 3 32)
6
2 3 8 128 3 320
6
2 3 8 8 2 3 5
6
3 4 4 2 3 5
3
x x x xy
x xy
x xy
x xy
Chapter 10 Conic Sections and Analytic Geometry
1250 Copyright © 2014 Pearson Education, Inc
81. does not make sense; Explanations will vary. Sample explanation: Horizontal parabolas will rise without limit.
82. does not make sense; Explanations will vary. Sample explanation: More information is necessary to determine how quickly it opens.
83. makes sense
84. makes sense
85. false; Changes to make the statement true will vary. A sample change is: Because a = −1, the parabola will open to the left.
86. true
87. false; Changes to make the statement true will vary. A sample change is: If a parabola defines y as a function of x, it will open up or down.
88. false; Changes to make the statement true will vary. A sample change is: x a y k h is not a
parabola. There is no squared variable.
89. 2 0Ax Ey
2
2
Ax Ey
Ex y
A
4
4
Ep y
AE
p yA
focus: 0,4
E
A
,
directrix: 4
Ey
A
90. y = 4 is the directrix and (–1, 0) is the focus. The
vertex must be located halfway between them at the point (–1, 2). p = –2 and the parabola opens down. (x + 1)2 = 4(–2)(y – 2) (x + 1)2 = –8(y – 2)
91. Answers will vary.
92.
2 2
2 2
2 2
2 21
2 2
2 21
2 22
( ) ( ) 14
12 2
2
x y x y
x y x y
x y
x y
x y
93. a.
b. 7
cos225
θ
c. 1 cos 2
sin2
71
25sin
2
16sin
254
sin5
θθ
θ
θ
θ
1 cos2cos
2
71
25cos
2
9cos
253
cos5
θθ
θ
θ
θ
d. Since 90 2 180 ,θ we have 45 90 .θ Both sinθ and cosθ are positive when
45 90 .θ
94. 2 24 ( 2 3) 4(3)(1)
12 12
0
B AC