Chapter 10 Conic Sections and Analytic Geometry · 7/31/2013 · Conic Sections and Analytic Geometry a . . . . (

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 .


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


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




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)



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


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


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).



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



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).



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


42. 2 16, 4a a 2 9, 3b b

center: (3, 1) 2 2 2 16 9 7c a b , 7c


43. 2 36, 6a a 2 25, 5b b

center: (0, 2) 2 2 2 36 25 11

11

c a b

c




44. a2 = 25, a = 5 b2 = 4, b = 2 center: (4, 0)

c2 = a2 – b2 = 25 – 4 = 21, 21c


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




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).



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


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


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 .



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.


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.


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.




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.



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



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.



Section 10.2


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


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


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


Section 10.2 The Hyperbola


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


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


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


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)



2. a2 = 1, a = 1 The vertices are (1, 0) and (–1, 0). b2 = 4

2 2 2 4 1 5c a b , 5c


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


(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



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


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).



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


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



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


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


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


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) .



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).



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


( 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).


( 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) .



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).


( 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) .



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


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


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


Asymptotes: 2

1 ( 3)3

y x



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


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




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,


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 .



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 .



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


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


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


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.



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


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.



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



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


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


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.


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)



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



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



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



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



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,


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,





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



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



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


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




57.

Check 2,1 :

22 1 3

2 1 3

2 2 true

22 1 3 1

2 1 3

2 2 true


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.



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


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



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

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