chapter 7nhmath.lonestar.edu/faculty/turnelle/1314/chpt7.pdf · chapter 7 ism: college algebra 744...

743

Chapter 7

Section 7.1

Check Point Exercises

1. 2

2

2 2 2

36, 69, 3

36 9 27

27 3 3

a ab bc a b

c

= =

= =

= − = − =

= =

The foci are located at ( 3 3, 0) and (3 3, 0)− .

2. 2 2

2 2

16 9 144144 144 144

19 16

x y

x y

+ =

+ =

2

2

2 2 2

16, 49, 3

16 9 7

7

a ab bc a b

c

= =

= =

= − = − =

=

The foci are located at (0, 7) and (0, 7)− .

3. 2 2

2 2 2

4, 99 4 5

c ab a c

= =

= − = − =

2 2

19 5x y

+ =

4. 2

2

9, 34, 2

a ab b

= =

= =

center at (–1, 2)

2 2 2

2

2

9 45

5

c a bcc

c

= −

= −

=

=

The foci are located at ( 1 5, 2) and ( 1 5, 2)− − − + .

5. 2 2

2 2 120 10x y

+ =

2 2

1400 100x y

+ =

Since the truck is 12 feet wide, substitute 6x = into the equation to find y.

( )

2 2

2

2

2

2

6 1400 100

36400 400 1400 100

36 4 4004 364

91

919.54

y

y

yyy

yy

+ =

⎛ ⎞+ =⎜ ⎟

⎝ ⎠+ =

=

=

=≈

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.

Chapter 7 ISM: College Algebra

744

Exercise Set 7.1

1. 2 2

116 4x y

+ =

2

2

2 2 2

16, 44, 2

16 4 12

12 2 3

a ab bc 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, 69, 3

36 9 27

27 3 3

a ab bc a b

c

= =

= =

= − = − =

= =

The foci are located at (0, 3 3)− and

(0, 3 3) .

4. 2 2

116 49x 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, 825, 5

64 25 39

39

a ab bc a b

c

= =

= =

= − = − =

=


(0, 39) .

ISM: College Algebra Section 7.1

745

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, 949, 7

81 49 32

32 4 2

a ab bc 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) .

9. 2 2

19 254 4

x y+ =

2

2

2

25 94 4

164

42

c

c

cc

= −

=

==

The foci are located at (0, 2) and (0, −2).

10. 2 81 254 16

c = −

2

2

324 2516 1629916

2994

4.3

c

c

c

c

= −

=

= ±

≈ ±

The foci are located at (4.3, 0) and (–4.3, 0).


746

11. 2 21 4x y= −

2 2

22

4 1

114

x yyx

+ =

+ =

2

2

114

34

32

0.9

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 14

x y+ =

12 14

324

32

0.87

c

c

c

c

= −

=

= ±

≈ ±

foci: (0, 0.87) (0, –0.87)

13. 2 2

2 2

2 2

25 4 10025 4 100100 100 100

14 25

x yx y

x y

+ =

+ =

+ =

2

2

2 2 2

25, 54, 2

25 4 21

a ab bc a b

= =

= =

= = = − =


(0, 21) .

14. 2 29 4 36x y+ =

2 2

2 2

9 4 3636 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) .


747

15. 2 24 16 64x y+ =

2 2

2

2

2

2

116 4

16, 44, 216 412

12

2 33.5

x y

a ab bcc

c

cc

+ =

= =

= =

= −

=

= ±

= ±≈ ±

The foci are located at (2 3,0) and (-2 3,0).

16. 2 24 25 100x y+ =

2 2

125 4x y

+ =

2

2

25 421

214.6

cc

cc

= −

=

=≈ ±

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 yx y

+ =

+ =

2

2

2

2

7, 7

5, 57 52

21.4

a a

b bcc

cc

= =

= =

= −

=

= ±≈ ±

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 yx y

+ =

+ =

2

2

6 51

1

ccc

= −

== ±

The foci are located at (0, 1) and (0, –1).

19. 2 24, 1,a b= = center at (0, 0) 2 2

14 1x 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 4x 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 4x y

+ = 2 2 2 4 1 3

3

c a b

c

= − = − =

=

The foci are at (0, 3) and (0, 3)− .


748

22. 2 216, 4a b= = , center: (0, 0) 2 2

14 16x 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) 1

4 1x y+ −

+ =

2 2

2

2

4, 14 13

3

a bcc

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) 1

1 4x y+ +

+ =

2

2

4 13

31.7

cc

cc

= −

=

= ±≈ ±

(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, 6464 25 39

164 39

c ab a cx y

= =

= − = − =

+ =

26. 2 24 36c a= = 2 2 2 36 4 32b a c= − = − = 2 2

136 32x y

+ =

27. 2 2

2 2 2

2 2

16, 4949 16 33

133 49

c ab a cx y

= =

= − = − =

+ =

28. 2 29, 16c a= = 2 2 2 16 9 7b a c= − = − = 2 2

17 16x y

+ =

29. 2 2

2 2 2

2 2

4, 99 4 13

113 9

c ba b cx y

= =

= + = + =

+ =

30. 2 24, 4c b= = 2 2 2 4 4 8a b c= + = + = 2 2

14 8x y

+ =

31. 2

2

2 2

2 8, 4, 162 4, 2, 4

116 4

a a ab b b

x y

= = =

= = =

+ =

32. 22 12, 6, 36a a a= = = 2

2 2

2 6, 3, 9

136 9

b b bx y

= = =

+ =

33. 22 10, 5, 25a a a= = =

22 4, 2, 42 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) 1

25 100x y− +

+ =


749

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) 14 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) 19 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) 1616 16 16

( 3) ( 2) 116 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) 1818 18 18

( 3) ( 2) 118 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)− .


750

41. 2 25, 5a a= = 2 9, 3b b= =

center: (4, 2)− 2 2 2 25 9 16

4c a bc

= − = − ==


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)− .

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)− + .


751

47. 2 5 2c = −

2 3

31.7

c

cc

=

= ±≈ ±

The foci are located at (1, 3 3)− + and

(1, 3 3).− −

48. 2 5 2c = −

2 3

31.7

c

cc

=

= ±≈ ±

(–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) 3636 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)− − .

50. 2 236( 4) ( 3) 36

36 36 36x y+ +

+ =

22 ( 3)( 4) 1

36yx +

+ + =

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)− − − .


752

51. 2 2

2 2

(9 36 ) (25 50 ) 1649( 4 ) 25( 2 ) 164x x y y

x x y y− + + =

− + + =

2 29( 4 4) 25( 2 1)164 36 25x x y y− + + + +

= + +

2 2

2 2

2 2

9( 2) 25( 1) 2259( 2) 25( 1) 225

225 225 225( 2) ( 1) 1

25 9

x yx 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 ) 644( 8 ) 9( 4 ) 64x 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 3636 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) 144144 144 144

( 1) ( 2) 116 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 yx 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 yx 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)− + .


753

55. 2 2

2 2

2 2

2 2

2 2

2 2

(4 16 ) ( 6 ) 394( 4 ) ( 6 ) 39

4( 4 4) ( 6 9) 39 16 94( 2) ( 3) 644( 2) ( 3) 64

64 64 64( 2) ( 3) 1

16 64

x x y yx x y y

x x y yx yx y

x y

+ + − =

+ + − =

+ + + − + = + +

+ + − =

+ −+ =

+ −+ =

center: (–2, 3) 2

2

2 2 2

64, 816, 4

64 16 48

48 4 3

a ab bc 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 ) 364( 6 ) 25( 4 ) 36

x x y yx 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) 1004( 3) 25( 2) 100

100 100 100( 3) ( 2) 1

25 4

x yx 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 99 9

9 9 9

19 1

x yx 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− .


754

58. 2 2 25x y+ = 2 2

2 2

2 2

25 2525 2525 25 25

11 25

x yx 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 9x 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 36x 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 44 44 4 4

11 4

x yx y

x y

+ =

+ =

+ =

2 22 2

2 2

x yy xy 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− .


755

62. 2 2

2 2

2 2

4 44 44 4 4

11 4

x yx y

x y

+ =

+ =

+ =

33

x yy 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 44 16

14 16

y x

y xx yx 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 44 4

11 4

y x

y xx yx 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 100x y

+ =

Let x = 4 2 2

2

2

2

4 1225 100

16900 900(1)225 100

64 9 9009 836

836 9.649

y

y

yy

y

+ =

⎛ ⎞+ =⎜ ⎟

⎝ ⎠+ =

=

= ≈

Yes, the truck only needs 7 feet so it will clear.


756

66. a = 25, b = 20 2 2

1625 400x y

+ =

Let x = 10 2 2(10) 1

625 400y

+ =

10,000 2100 10,000(1)

625 400y⎛ ⎞

+ =⎜ ⎟⎝ ⎠

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 2 150 30x y

+ =

2 2 2

2 250 30 2500 900 160040

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.

78. 2 186, 93a a= = 2 185.8, 92.9b a= = Earth’s orbit:

2 2

2 2 1(93) (92.9)

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 1(141.75) (139.25)

x y+ =

22 2

2(139.25) 1(141.75)

xy⎛ ⎞

= −⎜ ⎟⎝ ⎠

2

2139.25 1(141.75)

xy = ± −

79. 26, 36a a= = 2 2

2 136

x yb

+ =

When x = 2 and y = –4, 2 2

2

2

2

2

2

2 2

365

2 ( 4) 136

4 16 1364 5

936 5

365

136

b

b

bb

b

x y

−+ =

+ =

=

=

=

+ =


757

80. 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.

81. 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.

82. ca

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.

Section 7.2


1. a. 2 25, 5a a= = vertices: (5, 0) and (–5, 0)

2

2 2 2

1625 16 41

41

bc a b

c

=

= + = + =

=

The foci are at ( 41, 0) and ( 41, 0)− .

b. 2 25, 5a a= = vertices: (0, 5) and (0, –5)

2

2 2 2

1625 16 41

41

bc a b

c

=

= + = + =

=


2. a = 3, c = 5 2 2 2 25 9 16b c a= − = − = 2 2

19 16y x

− =

3. 2 36, 6a a= = The vertices are (6, 0) and (-6, 0).

2 9, 3b b= =

asymptotes: 3 16 2

by x x xa

= ± = ± = ±

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

22

4 44 4 4

14

y x

y x

− =

− =

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

2 1, 1b b= =

asymptotes: 2ay x xb

= ± = ±

2 2 2 4 1 5

5

c a b

c

= + = + =

=



758

5. center at (3, 1) 2

2

4, 21, 1

a ab b

= =

= =

The vertices are (1, 1) and (5, 1).

asymptotes: 11 ( 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 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 3636 36 36

3 51

9 45 3

14 9

x x y y

x x y y

x y

x y

x y

y x

− − + =

− + − + + = + + −

− − + = −

− + −− =

− − −

− +− + =

+ −− =

center at (3, –5) 2

2

4, 29, 3

a ab b

= =

= =

The vertices are (3, –3) and (3, –7).


y x+ = ± −

2 2 2 4 9 13

13

c a b

c

= + = + =

=

The foci are at (3, 5 13) and (3, 5 13)− − − + .


759

7.

2 2 2 2 2

52802 3300, 1650

5280 1650 25,155,900

ca 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− =

Exercise Set 7.2

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 = 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

22 1

8xy − =

6. a = 2, c = 6 b2 = c2 – a2 = 36 – 4 = 32

2 2

14 32y x

− =

7. a = 3, c = 4 b2 = c2 – a2 = 16 – 9 = 7

2 2

19 7x y

− =

8. a = 5, c = 7 b2 = c2 – a2 = 49 – 25 = 24

2 2

125 24x y

− =

9. 2a = 6 – (–6) 2a = 12 a = 6

2

6 2

6 23

ab

bb

b

=

=

==

Transverse axis is vertical.

2 2

136 9y x

− =

10. a = 4

2

24

8

bab

b

=

=

=

Transverse axis is horizontal.

2 2

116 64x y

− =

11. a = 2, c = 7 – 4 = 3

2 2 2

2

2

2 34 9

5

bbb

+ =

+ =

=

Transverse axis is horizontal.

2 2( 4) ( 2) 1

4 5x y− +

− =


760

12. a = 4 – 1 = 3 c = 6 – 1 = 5

2 2 2

2

2

3 59 25

164

bbbb

+ =

+ =

==

2 2( 1) ( 2) 1

9 16y x− +

− =


y x− = ± +

Transverse axis is vertical.

13. a2 = 9, a = 3

b2 = 25, b = 5 vertices: (3, 0) and (–3, 0)

asymptotes: 53

by x xa

= ± = ±

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: 54

by x xa

= ± = ±

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: 810

by x xa

= ± = ±

or 45

y 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: 34

by x xa

= ± = ±

c2 = a2 + b2 = 144 + 81= 225 c = 15 on x-axis The foci are at (15, 0) and (–15, 0).


761

17. a2 = 16, a = 4 b2 = 36, b = 6 vertices: (0, 4) and (0, –4)

asymptotes: 4 26 3

ay x x xb

= ± = ± = ±

or 23

y 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 = 58

a x xb

± = ±

c2 = a2 + b2 = 25 + 64 = 89 89c = on y-axis


19. 2

2 114

y x− =

2

2

2 2 2

2

2

1 1,4 21, 1

1 1454

52

1.1

a a

b bc a b

c

c

c

c

= =

= =

= +

= +

=

= ±

≈ ±

The foci are located at 5 50, and 0, .2 2

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

asymptotes:

12112

y x

y x

= ±

= ±

20. 2 2

11 19

y x− =

2

2

1 , 131 19109

103

1.1

a b

c

c

c

c

= =

= +

=

= ±

= ±

The foci are (0, 1.1) and (0, –1.1).


762

Asymptote:

113

1 3y x x= ± = ±

21. 2 29 4 36

36 36 36x y

− =

2 2

14 9x y

− =

a2 = 4, a = 2 b2 = 9, b = 3 vertices: (2, 0) and (–2, 0)

asymptotes: 32

by x xa

= ± = ±

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 100100 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 = 25

b x xa

± = ±

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 225225 225 225

125 9

y x

y x

− =

− =

a2 = 25, a = 5 b2 = 9, b = 3 vertices: (0, 5) and (0, –5)

asymptotes: 53

ay x xb

= ± = ±

c2 = a2 + b2 = 25 + 9 = 34 34c = on y-axis The foci are at (0, 34) and (0, 34)− .


763

24. 2 216 9 144

144 144 144y x

− =

2 2

19 16y x

− =

a2 = 9, a = 3 b2 = 16, b = 4 The vertices are (0, 3) and (0, –3).

asymptotes: 34

ay x xb

= ± = ±

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 yx y

= −

= −

2

2

2

2

2, 2

2, 22 24

2

a a

b bcc

c

= =

= =

= +

==

The foci are located at (2,0) and (–2, 0).

asymptotes: 22

y x

y x

= ±

= ±

26. 2 2

2 2

2 2

33

13 3

y xx yx y

= −

= −

= −

Vertices: ( 3,0) and ( 3,0).−

Asymptotes: 33

y x

y x

= ±

= ±

2

2

3 36

62.4

cc

cc

= +

=

= ±≈ ±

Foci: (2.4, 0) and (–2.4, 0).

27. a = 3, b = 5 2 2

19 25x y

− =

28. a = 3, b = 2 2 2

19 4x y

− =

29. a = 2, b = 3 2 2

14 9y x

− =

30. a = 5, b = 3 2 2

125 9y x

− =

31. Center (2, –3), a = 2, b = 3

2 2( 2) ( 3) 1

4 9x y− +

− =

32. Center (–1, –2) a = 2, b = 2

2 2( 1) ( 2) 1

4 4x y+ +

− =


764

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, 516, 4

a ab 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= ± +


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)


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)− − .


765

38. The center is located at (–1, 2). a2 = 36, a = 6 b2 = 49, b = 7 The vertices are (–1, 8) and (–1, –4).


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) 44 4 4

( 3) ( 3) 14

x y

x y

− +− =

−− + =

center: (3, –3) 2

2

4, 21, 1

a ab b

= =

= =

vertices: (1, –3) and (5, –3)


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 9x y+ −

− =

22( 3) ( 4) 1

9x y+

− − =

The center is located at (–3, 4). a2 = 9, a = 3 b2 = 1, b = 1 The vertices are (–6, 4) and (0, 4).


x± +


The foci are located at ( 3 10, 4)− − and

( 3 10, 4)− + .

41. 2 2( 1) ( 2) 1

3 3x 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).+ −


766

42. 2 2( 2) ( 3) 1

5 5y 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+ = ± −

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) 1

4 16x 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: 43 ( 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) 416( 2) ( 1) 4

4 4 4( 1) ( 2) 1

4

x yx y

y x

+ − + = −

+ + −− =

− − −+ +

− =

center: (–2, –1) a2 = 4, a = 2

2 1 1,4 2

b b= =

c2 = a2 + b2 = 4 + 14

= 174

174 4.25c = =

asymptotes:

2( 1) ( 2)12

1 4( 2)

y x

y x

+ = ± +

+ = ± +

The foci are at ( )2, 1 4.25− − + and

( )2, 1 4.25− − − .


767

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) 1

4 9y 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)+ − .


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) 1

9 4x y− −

− =

center: (2, 3) a2 = 9, a = 3 b2 = 4, b = 2 c2 = a2 + b2 = 9 + 4 = 13

13c =


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) 1x y+ +− =

The center is located at (–1, –1).

2 1 1,4 2

a a= =

2 1 1,9 3

b b= =

2 2 2 1 1 134 9 36

c a b= + = + = , 136

c =

The foci are at 13 131 , 1 and 1 , 16 6

⎛ ⎞ ⎛ ⎞− + − − − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠.

Asymptotes:

131 ( 1)1221 ( 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 100100 100 100

( 4) 14 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= ± −



768

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 144x y− + −

− =− − −

2 2( 2) ( 2) 1

9 16y 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).


y x+ = ± −

51. 2 2

19 16x 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 4x y

− =


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 16x 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− .


769

54. 2 2

125 4x 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 9y 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 25y 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

44

x yx y

− =

+ =

5

5

y

x(−2, 0)

(2, 0)

Check ( )2,0− :

( )2 22 0 44 0 4

4 4 true

− − =

− ==

( )2 22 0 44 0 4

4 4 true

− + =

+ ==

Check ( )2,0 :

( )2 22 0 44 0 4

4 4 true

− =

− ==

( )2 22 0 44 0 4

4 4 true

+ =

+ ==

The solution set is ( ) ( ){ }2,0 , 2,0− .


770

58. 2 2

2 2

99

x yx y

− =

+ =

5

5

y

x(−3, 0)

(3, 0)

Check ( )3,0− :

( )2 23 0 99 0 9

9 9 true

− − =

− ==

( )2 23 0 99 0 9

9 9 true

− + =

+ ==

Check ( )3,0 :

( )2 23 0 99 0 9

9 9 true

− =

− ==

( )2 23 0 99 0 9

9 9 true

+ =

+ ==

The solution set is ( ) ( ){ }3,0 , 3,0− .

59. 2 2

2 2

9 99 9

x yy 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 90 9 9

9 9true

+ − =

+ ==

( ) ( )2 23 9 0 99 0 9

9 9true

− − =

− ==

Check ( )0,3 :

( ) ( )2 29 0 3 90 9 9

9 9true

+ =

+ ==

( ) ( )2 23 9 0 99 0 9

9 9true

− =

− ==

The solution set is ( ) ( ){ }0, 3 , 0,3− .

60. 2 2

2 2

4 44 4

x yy 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.


771

62. a. 2c = 200 km, c = 100 km

( )2 1m2 500 s 300

sd d a μ

μ⎛ ⎞

− = = ⎜ ⎟⎝ ⎠

2a = 150,000 m = 150 km a = 75 km

( ) ( )2 22 2 2 100 75 4375b c a= − = − =

( )

2 2

2 1437575

x y− =

2 2

15625 4375

x y− =

b. The x-coordinate of the ship is 100 km: ( )2 2100

15625 4375

y− =

2 10,000 14375 5625

y= −

10,0004375 15625

y = ± − 58.3≈ ±

The ship is about 58.3 kilometers from the coast.

63. 2 2

2 2

2 2

625 400 250,000625 400 250,000

250,000 250,000 250,000

1400 625

y xy 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

2 19x y

b− =

To find b, use the equation of the slope of the asymptote,

1:3 2

b ba

=

Solving for b: 2b = 3, b = 3 .2

2 2

94

19x y

− =

75. 2 2

04 9x y

− =

2 294

y x=

32

y x= ±

No; in general, the graph is two intersecting lines.

76. Answers may 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

±

77. 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 24 372

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.


772

78. 2 2

116 9x y

− =

2 2

19 16y x

= −

22

2

2

16916

16916

3 164

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

3 16 ( 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 (–∞, ∞).

79. a. False; one branch of the hyperbola 2 2

2 2 1x ya b

− = will not pass the vertical line

test, so will not define y as a function of x. b. False; none of the points on the asymptotes

satisfy the hyperbola’s equation, since the hyperbola never touches its asymptotes.

c. True; 23

y x= − is one of the asymptotes of

the hyperbola and they will not intersect.

d. False; for example, 2 2

14 4x y

− = and

2 2

14 4y x

− = each have asymptotes

y = ± x, but are different hyperbolas.

(c) is true.

80. ca

will be large when a is small. When this

happens, the asymptotes will be nearly vertical.

81. 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

( ) ( ) 1y k x ha b− −

− =

(h, k) = (5, 0), and a = 6, so a2 = 36. 2 2

2

( 5) 1.36y x

b−

− =

Substitute x = 0 and y = 9: 2 2

2

2

2

2

9 (0 5) 136

25 54

100 520

b

bb

b

−− =

− = −

− = −

=

Standard form: 2 2( 5) 1

36 20y x −

− =

ISM: College Algebra Mid-Chapter 7 Check Point

773

82. If the asymptotes are perpendicular, then their slopes are negative reciprocals. For the

hyperbola 2 2

2 2 1x ya b

− = , the asymptotes are

.by xa

= ± The slopes are negative reciprocals

when b aa 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 4x y

− = has perpendicular asymptotes.

Mid-Chapter 7 Check Point 1. Center: ( )0,0

Because the denominator of the 2 termx − is greater than the denominator of the 2 term,y − the major axis is horizontal. Since 2 25a = , 5a = and the vertices are ( ) ( )5,0 and 5,0 .− Since 2 4b = ,

2b = and endpoints of the minor axis are ( ) ( )0, 2 and 0, 2 .−

Foci: ( )21,0±

2. Divide both sides by 36 to get the standard form: 2 2

14 9x y

+ =

Center: ( )0,0

Because the denominator of the 2 termy − is greater than the denominator of the 2 term,x − the major axis is vertical. Since 2 9a = , 3a = and the vertices are ( ) ( )0, 3 and 0,3 .− Since 2 4b = ,

2b = and endpoints of the minor axis are ( ) ( )2,0 and 2,0 .−

Foci: ( )0, 5±

3. Center: ( )2, 1−

Because the denominator of the 2 termy − is greater than the denominator of the 2 term,x − the major axis is vertical. We have 2 25a = and

2 16b = , so 5a = and 4b = . The vertices lie 5 units above and below the center. The endpoints of the minor axis lie 4 units to the left and right of the center. Vertices: ( )2,4 and ( )2, 6−

Minor endpoints: ( )2, 1− − and ( )6, 1−

Foci: ( )2, 2 , ( )2, 4−


774

4. Center: ( )2,1−

Because the denominator of the 2 termx − is greater than the denominator of the 2 term,y − the major axis is horizontal. We have 2 25a = and

2 16b = , so 5a = and 4b = . The vertices lie 5 units to the left and right of the center. The endpoints of the minor axis lie 4 units above and below the center. Vertices: ( )7,1− and ( )3,1

Minor endpoints: ( )2,5− and ( )2, 3− −

Foci: ( )5,1− , ( )1,1

5.

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )

2 2

2 2

2 2

2 2

2 2

2 2

4 9 54 49

4 9 6 49

4 4 9 6 9 49 4 81

2 9 3 36

2 9 3 3636 36 36

2 31

36 4

x x y y

x x y y

x x y y

x y

x y

x y

− + + = −

− + + = −

− + + + + = − + +

− + + =

− ++ =

− ++ =

Center: ( )2, 3−

Foci: ( )2 4 2, 3± −

6. The equation is for a hyperbola in standard form with the transverse axis on the x-axis. We have

2 9a = and 2 1b = , so 3a = and 1b = . 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 1± on the y-axis. Then use dashed lines to draw extended diagonals for the rectangle. These represent the asymptotes of the graph. Foci: ( )10,0±

7. The equation is in the form

2 2

2 2 1y xa b

− = with

2 29, and 1a b= = . We know the transverse axis lies on the y-axis and the vertices are ( ) ( )0, 3 and 0,3 .− Because 2 29 and 1,a b= =

3 and 1.a b= = Construct a rectangle using –1 and 1 on the x–axis, and –3 and 3 on the y–axis. Draw extended diagonals to obtain the asymptotes.

Foci: ( )0, 10±


775

8. 2 2

116 4y x

− =

The equation is in the form 2 2

2 2 1y xa b

− = with

2 216, and 4a b= = . We know the transverse axis lies on the y-axis and the vertices are ( ) ( )0, 4 and 0, 4 .− Because 2 216 and 4,a b= =

4 and 2.a b= = Construct a rectangle using –2 and 2 on the x–axis, and –4 and 4 on the y–axis. Draw extended diagonals to obtain the asymptotes. Foci: ( )0, 2 5±

9. 2 2

149 4x y

− =


2 49a = and 2 4b = , so 7a = and 2b = . Therefore, the vertices are at ( ),0a± or ( )7,0± . Using a dashed line, we construct a rectangle using the 7± 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. Foci: ( )53,0±

10. The equation is for a hyperbola in standard form with center (2, –2). We have 2 9a = and 2 16b = , so 3a = and 4b = .


y x+ = ± −

Foci: ( )3, 2− − , ( )7, 2−

11. Write the equation for the hyperbola in standard form:

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )

2 2

2 2

2 2

2 2

2 2

2 2

2 2

4 8 6 11 04 8 6 11

4 2 6 11

4 2 1 6 9 11 4 9

4 1 3 16

4 1 3 1616 16 16

3 11

16 4

x y x yx x y y

x x y y

x x y y

x y

x y

y x

− + + + =

+ − + = −

+ − − = −

+ + − − + = − + −

+ − − = −

+ − −− =

− − −

− +− =

Center (–1, 3). Asymptotes: 3 2( 1)y x− = ± +

Foci: ( )1,3 2 5− ± , ( )7, 2−


776

12. This is the equation of a circle centered at the origin with radius 4 2r = = . We can plot points that are 2 units to the left, right, above, and below the origin and then graph the circle. The points are ( )2,0− , ( )2,0 , ( )0, 2 ,

and ( )0, 2− .

13. 44

x yy x

+ == − +

This is the equation of a line with slope 1m = − and a y-intercept of 4. We can plot the point ( )0, 4 , use the slope to get an additional point, connect the points with a straight line and then extend the line to represent the graph of the equation.

14. 2 2

2 2

4

14 4

x yx y

− =

− =


2 4a = and 2 4b = , so 2a = and 2b = . Therefore, the vertices are at ( ),0a± or ( )2,0± . Using a dashed line, we construct a rectangle using the 2± 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.

Graph the hyperbola.

15. 2 2

2 2

4 4

14 1

x yx y

+ =

+ =

Center: ( )0,0

Because the denominator of the 2 termx − is greater than the denominator of the 2 term,y − the major axis is horizontal. We have 2 4a = and

2 1b = , so 2a = and 1b = . The vertices lie 2 units to the left and right of the center. The endpoints of the minor axis lie 1 unit above and below the center. Vertices: ( )2,0− and ( )2,0

Minor endpoints: ( )0, 1− and ( )0,1

16. Center: ( )1,1−

Radius: 4 2r = = We plot the points that are 2 units to the left, right, above and below the center. These points are ( )3,1− , ( )1,1 , ( )1,3− , and

( )1. 1− − .


777

17. ( )( )

22

22

4 1 4

11

4 1

x y

yx

+ − =

−+ =

18. ( ) ( )( ) ( )

2 2

2 2

1 1 4

1 11

4 4

x y

x y

− − − =

− −− =

The equation is for a hyperbola in standard form centered at (1, 1). We have 2 4a = and 2 4b = , so

2a = and 2b = .

19. The foci and vertices show that c is 4 and a is 5. 2 2 2

2 2 2

2

2

4 525 169

c a bb

bb

= −

= −

= −

=

2 2

125 9x y

+ =

20. The endpoints show that the center is (1, 2). Since 22 18, 9 and 81.a a a= = = Since 22 10, 5 and 25.c c c= = =

2 2 2

2

2

2

25 8181 2556

c a bb

bb

= −

= −

= −

=

( ) ( )2 21 21

81 56x y− −

+ =

21. The foci and vertices show that c is 3 and a is 2. 2 2 2

2 2 2

2

2

3 29 45

b c abbb

= −

= −

= −

=

2 2

14 5x y

− =

22. The endpoints show that the center is (–1, 5). Since 22 4, 2 and 4.a a a= = = Since 22 6, 3 and 9.c c c= = =

2 2 2

2

2

9 45

b c abb

= −

= −

=

( ) ( )2 21 51

4 5x y+ −

+ =

23. a = 15, b = 10 2 2

2 2 115 10x y

+ =

2 2

1225 100x y

+ =

Since the truck is 10 feet wide, substitute 5x = into the equation to find y.

( )

2 2

2

2

2

2

2

2

5 1225 10025 1225 100

1 19 100

1900 900 19 100

100 9 9009 800

88.8889

88.88899.43

y

y

y

y

yyy

yy

+ =

+ =

+ =

⎛ ⎞+ =⎜ ⎟

⎝ ⎠+ =

=

=

=≈

5 feet from the center, the height of the archway is 9.43 feet. Since the truck’s height is 9.5 feet, it will not fit under the archway.


778

24. Find the distance between the foci. Since 22 40, 20 and 400.a a a= = = Since 22 20, 10 and 100.b b b= = =

2 2 2

2

2

400 100300

300

10 3

2 20 32 34.64

c a bcc

c

cc

= −

= −

=

=

=

=≈

The kidney stone should be 34.64 cm from the electrode that sends the ultrasound waves.

25. a. Since 22 6, 3 and 9.c c c= = = The ranger at the primary station heard the explosion 6 seconds before the other ranger. This means that the explosion occurred 6 0.35 2.1× = miles closer to the primary station. Since 22 2.1, 1.05 and 1.1025.a a a= = =

2 2 2

2

2

9 1.10257.8975

b c abb

= −

= −

=

2 2

11.1025 7.8975

x y− =

b.

Section 7.3


1. 4 8, 2p p= = focus: (2, 0) directrix: x = –2

2. 2 124 12, 3x yp p

= −= − =

focus: (0, –3) directrix: y = 3

3. 2

2

84 832

py xy x

=

= ⋅

=

4. 4 4, 1p p= = vertex: (2, –1) focus: (2, 0) directrix: y = –2


5.

vertex: (2, –1) focus: (1, –1) directrix: x = 3

2

2

2

2 4 72 1 4 7 1

( 1) 4( 2)

y y xy y x

y x

+ = − +

+ + = − + +

+ = − −4 4,p p= − = −1

6. 2 4x py= Let x = 3 and y = 4.

2

2

3 4 49 16

91694

pp

p

x y

= ⋅=

=

=

The light should be placed at 90,16

⎛ ⎞⎜ ⎟⎝ ⎠

or 916

inch above the vertex.

Exercise Set 7.3

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


7. 4p = –8, p = –2 vertex: (0, 0) focus: (–2, 0) directrix: x = 2

779


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 34 2=

vertex: (0, 0)

focus: 3 , 02

⎛ ⎞⎜ ⎟⎝ ⎠

directrix: x = 32

−

14. x2 = 6y

4p = 6, p = 6 34 2=

vertex: (0, 0)

focus: 30,2

⎛ ⎞⎜ ⎟⎝ ⎠

directrix: y = 32

−

780


15.

2

2

8 412

x y

x y

= −

= −

14218

p

p

= −

= −

focus: 10,8

⎛ ⎞−⎜ ⎟⎝ ⎠

directrix: 18

y =

16. 2

2

8 412

y x

y x

= −

= −

14218

p

p

= −

= −

vertex: (0, 0)

focus: 1 ,08

⎛ ⎞−⎜ ⎟⎝ ⎠

directrix: 18

x =

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( 3x y )− = − +

26. vertex: (5, –2 ); p = 7 – 5 = 2

2

2

2 2

48

( 2) 8( 5)

y pxy xy x

=

=

+ = −

27. vertex: (1, 2) p = 2 2( 2) 8( 1y x )− = −

28. vertex: ( –1, 4) p = 3 2

2

2

2

44(3)12

( 4) 12( 1)

y pxy xy xy x

=

=

=

− = +

29. vertex: (–3, 3), p = 1 2( 3) 4( 3x 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)

781


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

782


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

783


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 3

2 2 1bya

= − = − = −

The x-coordinate of the vertex is ( ) ( )23 6 3

9 18 54

x = − + − +

= − += −

5

The vertex is ( ) . 4, 3− −

Since the squared term is y and , the graph opens to the right.

0a >

Domain: { } or [ | 4x x ≥ − )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 1bya

−= − = − =

The x-coordinate of the vertex is ( ) ( )21 2 1

1 2 56

x = − −

= − −= −

5

The vertex is ( ) . 6,1−

Since the squared term is y and , the graph opens to the right.

0a >

Domain: { } or [ | 6x x ≥ − )6,− ∞

Range: { }| is a real numbery y or ( ),−∞ ∞ The relation is not a function.

51. The x-coordinate of the vertex is ( )( )4

22 2 1bxa

= − = − =−

The y-coordinate of the vertex is ( ) ( )22 4 24 8 3

1

y 3= − + −

= − + −=

The vertex is ( )2,1 . Since the squared term is x and , the graph opens down.

0a <

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 1bxa

−= − = − = −

−

The y-coordinate of the vertex is ( ) ( )22 4 24 8 4

8

y 4= − − − − +

= − + +=

The vertex is ( )2,8− . Since the squared term is x and , the graph opens down.

0a <

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 , the graph opens to the left.

0a <

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 , the graph opens to the left.

0a <

Domain: { }| 2x x ≤ − or ( ], 2−∞ −

Range: { }| is a real numbery y or ( ) ,−∞ ∞The relation is not a function.

784


55.

Check : ( )4, 2−

( )24 2 2 44 0 44 4

true

− = − −

− = −− = −

( )12 42

2 2true

= − −

=

Check : ( )0,0

( )20 0 20 4 40 0

true

= − −

= −=

4 ( )10 02

0 0true

= −

=

The solution set is . ( ) ({ }4,2 , 0,0− )

56.

Check : ( )2,3

( )22 3 32 0 22 2

true

= − +

= +=

2

2

2 3 55 5

true

+ ==

Check : ( )3,2

( )23 2 33 1 23 3

true

= − +

= +=

3 2 55 5

true

+ ==

The solution set is . ( ) ( ){ }2,3 , 3, 2

57.

Check ( )2,1− :

( )22 1 32 1 32 2 true

− = −

− = −− = −

( ) ( )22 1 3 12 1 32 2 true

− = −

− = −− = −

The solution set is . ( ){ }2,1−

58.

Check ( )5,0− :

25 0 55 0 55 5 true

− = −− = −− = −

( ) 2 25 0 2525 0 25

25 25 true

− + =

+ ==

Check ( )4, 3− :

( )24 34 9 54 4 true

5= − −

= −=

( ) ( )2 24 3 2516 9 25

25 25 true

+ − =

+ ==

Check ( )4,3 :

( )24 34 9 54 4 true

5= −

= −=

( ) ( )2 24 3 2516 9 25

25 25 true

+ =

+ ==

The solution set is ( ) ( ) ( ){ }5,0 , 4, 3 , 4,3− − .

785


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 = 98

= 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) 640640

640 200 440(440) 4(640)

(440) 75.6254(640)

x pyp

p

xy

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

2

2

2

4

200 4 ( 50)2

10,000 450

4 200200

(30) 200900 4.5200

x py

p

p

px y

y

y

=

⎛ ⎞ = −⎜ ⎟⎝ ⎠

=−

= −

= −

= −

= = −−

(height of bridge) = 50 – 4.5 = 45.5 feet. Yes, the boat will clear the arch.

68. 2 4y p= x

25 4 (6)25 46

p

p

=

=

25 feet6

786


77. y2 + 2y – 6x + 13 = 0 y2 + 2y + (–6x + 13) = 0

22 2 4( 6 13)2

xy

− ± − − +=

2 24 482

1 6 12

xy

y x

− ± −=

= − ± −

78. 2

2

10 25 010 ( 25) 0

y y xy y x

+ − + =

+ + − + =

210 10 4( 25)2

10 42

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)

1824 80 6000 2800

1824 80 20 15 7

1812 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 3206

2 3 8 8 2 3 56

3 4 4 2 3 53

x x x xy

x xy

x xy

x xy

− − ± − − + +=

− + ± − −=

− + ± − −=

− + ± − −=

787


81. a. False; it opens to the left.

b. True; it opens to the right and has a domain [3, ∞).

c. False; any parabola that opens to the right will not be a function of x because at least one x-value will be paired with more than 1 y-value.

d. False; the graph is a line.

(b) is true.

82. 2 0Ax Ey+ =

2

2

Ax EyEx yA

= −

= −

4

4

Ep yAEp yA

= −

= −

focus: 0,4EA

⎛ ⎞−⎜ ⎟⎝ ⎠

,

directrix: 4EyA

=

83. 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)

Chapter 7 Review Exercises

1. a2 = 36, a = 6 b2 = 25, b = 5 c2 = a2 – b2 = 36 – 25 = 11

11c = The foci are at ( 11, 0) and ( 11, 0)−

2. 2

2

25, 516, 4

a ab b

= =

= =

2 2

2

2

25 1693

c a bccc

2= −

= −

==

The foci are (0, 3) and (0, –3).

3. 2 2

2 2

4 116 16 16

14 16

x y

x y

+ =

+ =

6

b2 = 4, b = 2 a2 = 16, a = 4 c2 = a2 – b2 = 16 – 4 = 12 c = 12 2 3= The foci are at (0, 2 3) and (0, 2 3)− .

788

ISM: College Algebra Chapter 7 Review Exercises

4. 2 2

2 2

4 9 336 36 36

19 4

x y

x y

+ =

+ =

6

a2 = 9, a = 3 b2 = 4, b = 2 c2 = a2 – b2 = 9 – 4 = 5, c = 5 The foci are at ( 5 , 0) and ( 5, 0)− .

5. a2 = 16 a = 4 b2 = 9 b = 3 c2 = 16 – 9 = 7, c = 7 center: (1, –2) The foci are at (1 7, 2) and (1 7, 2)+ − − − .

6. a2 = 16, a = 4

b2 = 9, b = 3 c2 = a2 – b2 = 16 – 9 = 7, 7c = center: (–1, 2) The foci are at (–1, 2 + 7) and ( 1, 2 7)− − .

7. 4x2 + 24x + 9y2 – 36y = –36 4(x2 + 6x + 9) + 9(y2 – 4y + 4) = –36 +36 +36 = 4(x + 3)2 + 9(y – 2)2 = 36

2 2( 3) ( 2) 19 4

x y+ −+ =

c2 = a2 – b2 = 5, 5c = center: (–3, 2) The foci are at ( 3 5, 2)− + and ( 3 5, 2)− − .

8. 9x2 – 18x + 4y2 + 8y = 23

9(x2 – 2x + 1) + 4(y2 + 2y + 1) = 23 + 9 + 4 9(x – 1)2 + 4(y + 1)2 = 36

2 2( 1) ( 1) 14 9

x y− ++ =

c2 = a2 – b2 = 9 – 4 = 5, 5c = center: (1, –1) The foci are at (1, –1 + 5 ) and (1, 1 5)− − .

9. c = 4, c2 = 16

a = 5, a2 = 25 b2 = a2 – c2 = 25 – 16 = 9

2 2

125 9x y

+ =

10. c = 3, c2 = 9 a = 6, a2 = 36 b2 = a2 – c2 = 36 – 9 = 27

2 2

127 36x y

+ =

789


11. 2a = 12, a = 6, a2 = 36 2b = 4, b = 2, b2 = 4

2 2( 3) ( 5) 136 4

x y+ −+ =

12. 2a = 20, a = 10, a2 = 100 b = 6, b2 = 36

2 2

1100 36x y

+ =

13. 2a = 50, a = 25 b = 15

2 2

1625 225x y

+ =

Let x = 14. 2 2(14) 1

625 225y

+ =

y2 = 225 1961625

⎛ ⎞−⎜ ⎟⎝ ⎠

y ≈ 15(0.8285) ≈ 12.4 > 12 Yes, the truck can drive under the archway.

14. The hit ball will collide with the other ball.

15. c2 = a2 + b2 = 16 + 1 = 17, c = 17 The foci are at ( 17 , 0) and ( 17, 0)− .

Asymptotes: 14

y x= ±

16. c2 = a2 + b2 = 16 + 1 = 17 17c =

The foci are at (0, 17 ) and (0, 17)− . Asymptotes: 4y x= ±

17. 2 2

116 9x y

− =

c2 = a2 + b2 = 16 + 9 = 25, c = 5 The foci are at (5, 0) and (–5, 0).

Asymptotes: 34

y x= ±

18. 2 2

14 16y x

− =

c2 = a2 + b2 = 4 + 16 = 20 20 2 5c = =

The foci are at (0, 2 5 ) and (0, 2 5)− .

Asymptotes: 12

y x= ±

19. c2 = a2 + b2 = 25 + 16 = 41, c = 41

center: (2, –3) The foci are at (2 41, 3) and (2 41, 3)+ − − − .

Asymptotes: 43 (5

y x 2)+ = ± −

790

ISM: College Algebra Chapter 7 Review Exercises

20. c2 = a2 + b2 = 25 + 16 = 41, 41c = center: (3, –2) The foci are at (3, 2 41) and (3, 2 41)− + − − .

Asymptotes: 52 (4

y x+ = ± −3)

21. y2 – 4y – 4x2 + 8x – 4 = 0

(y2 – 4y + 4) – 4(x2 – 2x + 1) = 4 + 4 – 4 (y – 2)2 – 4(x – 1)2 = 4

22( 2) ( 1)

4y x−

− − =1

c2 = a2 + b2 = 4 + 1 = 5, 5c = center: (1, 2) The foci are at (1, 2 5) and (1, 2 5)+ − .

Asymptotes: 2 2( 1y x− = ± − )

22. x2 – 2x – y2 – 2y = 1 (x2 – 2x + 1) – (y2 + 2y + 1) = 1 + 1 – 1 (x – 1)2 – (y + 1)2 = 1

2 2 2 1 1 2, 2c a b c= + = + = = center: (1, –1) The foci are at (1 2, 1) and (1 2, 1)+ − − − . asymptotes: 1 ( 1y x+ = ± −

23. c = 4, c2 = 16

a = 2, a2 = 4 b2 = c2 – a2 = 16 – 4 = 12

2 2

14 12y x

− =

24. c = 8, c2 = 64 a = 3, a2 = 9 b2 = c2 – a2 = 64 – 9 = 55

2 2

19 55x y

− =

25. If the foci are at (0, –2) and (0, 2), then c = 2. If the vertices are at (0, –3) and (0, 3) then a = 3. This is not possible since c must be greater than a.

26. foci: (±100, 0), c = 100

(1 2mi0.186 500 s

sd d )μ

μ⎛ ⎞

− = ⎜ ⎟⎝ ⎠

= 93 mi = 2a

932

a =

( )2

22 2 2 931002

b c a ⎛ ⎞= − = − ⎜ ⎟⎝ ⎠

= 7837.75

( )

2 2

2932

17837.75

x y− =

2 2

12162.25 7837.75

x y− =


)

791


31. x2 = –4y + 4 x2 = –4 (y – 1) 4p = –4, p = –1 vertex: (0, 1) focus: (0, 0) directrix: y = 2

28.

2 16 0x y+ =2 16x y= −

4p = –16 p = –4 vertex: (0, 0) focus: (0, –4) directrix: y = 4

32.

( ) ( )

2

2

2

10 4 2110 25 4 21 25

5 4 1

y y xy y x

y x

− = −

− + = − +

− = +

4p = 4, p = 1 vertex: (–1, 5) focus: (0, 5) directrix: x = –2

29. 4p = –16

p = –4 vertex: (0, 2) focus: (–4, 2) directrix: x = 4

30. 4p = 4, p = 1

vertex: (4, –1) focus: (4, 0) directrix: y = –2

33. x2 – 4x – 2y = 0

x2 – 4x = 2y (x2 – 4x + 4) = 2y + 4 (x – 2)2 = 2(y + 2)

4p = 2, p = 12

vertex: (2, –2)

focus: 32,2

⎛ ⎞−⎜ ⎟⎝ ⎠

directrix: 52

y = −

792

ISM: College Algebra Chapter 7 Test

34. p = 12 y2 = 48x

2. 4p = –8, p = –2 parabola vertex: (0, 0) focus: (0, –2) directrix: y = 2

35. p = –11 x2 = –44y

36.

( ) ( )

2

2

2

4

6 4 3312

x py

pp

x y

=

=

=

=

Place the light 3 inches from the vertex at (0, 3).

37. 2 4x py=

4p ≈ 9691 ( ) (21750 4 316p=

3. The center is at (–2, 5). 2 2 2 25 9 16c a b= − = − = , c = 4

ellipse The foci are at (–6, 5) and (2, 5).

)

2 9691x y= Let x = 1750 – 1000 = 750.

( )22 75058

9691 9691xy = = ≈

The height is approximately 58 feet.

38. x2 = 4py (150)2 = 4p(44) 22,500 = 176p p ≈ 128 The receiver should be placed approximately 128 feet from the base of the dish. 4.

( ) ( )( ) ( )

( ) ( )( ) ( )( ) ( )

2 2

2 2

2 2

2 2

2 2

22

4 8 2 7 0

4 8 2 7

4 2 1 2 1 7 4

4 1 1 4

1 4 1 4

11 1

4

x y x y

x x y y

x x y y

x y

y x

yx

− + + + =

+ − − = −

1+ + − − + = − + −

+ − − = −

− − + =

−− + =

2 2 2 4 1 5c a b= + = + = , 5c = The center is at (–1, 1).

Chapter 7 Test

1. 2 2

14 9x y

− =

, 2 2 2 4 9 13c a b= + = + = 13c = hyperbola The foci are at ( ) ( )13,0 and 13, 0− .

Asymptotes: 32

y x= ±

Asymptotes: 1 2( 1y x )− = ± + hyperbola The foci are at ( ) ( )1, 1 5 and 1, 1 5− + − − .

793


5. 4p = 8, p = 2 parabola vertex: (–5, 1) focus: (–5, 3) directrix: y = –1

6. c = 7, c2 = 49

a = 10, a2 = 100 b2 = a2 – c2 = 100 – 49 = 51

2 2

1100 51x y

+ =

7. c = 10, c2 = 100 a = 7, a2 = 49 b2 = c2 – a2 = 100 – 49 = 51

2 2

149 51y x

− =

8. p = 50 y2 = 4px y2 = 200x

9. b = 24, b2 = 576 2a = 80, a = 40, a2 = 1600 c2 = a2 – b2 = 1600 – 576 = 1024

1024 32c = = The two people should each stand 32 feet from the center of the room, along the major axis.

10. a. 2 4x py= when x = ±3, y = 3 9 4 (3)3 434

pp

p

==

=

2 3x y=

b. focus: 30,4

⎛ ⎞⎜ ⎟⎝ ⎠

The light is placed 34

inch above the

vertex.

Cumulative Review Exercises (Chapters 1–7)

1. 2(x – 3) + 5x = 8(x – 1) 2x – 6 + 5x = 8x – 8 7x – 6 = 8x – 8 –x = –2 x = 2 The solution set is { }2 .

2. –3(2x – 4) > 2(6x – 12) –6x +12 > 12x – 24 –18x > –36 x < 2 The solution set is { }2x x < .

3. x – 5 = 7x + (x – 5)2 = x + 7 x2 – 10x + 25 = x + 7 x2 – 11x + 18 = 0 (x – 2)(x – 9) = 0 x = 2 or x = 9 The solution x = 2 is extraneous, so the only solution is x = 9. The solution set is { }9 .

4. (x – 2)2 = 20 x – 2 = 20± x – 2 = 2 5± x = 2 2 5± The solution set is { }2 2 5, 2 2 5+ − .

5. 2 1 7x − ≥ 2 1 7 or 2 1 7

2 8 2 64 or 3

x xx xx x

− ≥ −≥ ≤≥ ≤

≤ −−−

The solution set is { }3 or 4x x x≤ − ≥

794

ISM: College Algebra Cumulative Review Exercises (Chapters 1−7)

6.

3 23 4 – 7 2x x x+ + 0=: 1, 2p ± ±: 1, 3q ± ±

1 2: 1, 2, ,3 3

pq

± ± ± ±

Let Evaluate f at the possible rational zeros to find

3 2( ) 3 4 – 7 2.f x x x x= + +

2 0.3

f ⎛ ⎞ =⎜ ⎟⎝ ⎠

23

3 4 –7 2

2 4 –2

3 6 –3 0

22– (3 6 – 3) 03

x x x⎛ ⎞ + =⎜ ⎟⎝ ⎠

2(3 – 2)( 2 –1) 0x x x+ =

23

x = or 2–2 (2) – 4(1)(–1)2

x±

=

–2 82

x ±=

1 2x = − ±

The solution set is 2 , –1 2, 1 23

⎧ ⎫+ − −⎨ ⎬⎩ ⎭

.

7. 2 22

22 3

2

log ( 1) log ( –1) 3

log ( –1) 3

–1 29

3

x x

x

xxx

+ + =

=

=

== ±

x = –3 is not a solution of the original equation. The solution set is {3}.

8. 3 4 22 5

x yx y+ =+ = −

9. 2 2

2 2 2

2 866

6( 6) 12 36

x yx yx y

x yx y y y

− = −− =− =

= +

= + = + +

Substitute into first equation. 2(y2 + 12y +36) – y2 = –8 2y2 + 24y + 72 – y2 = –8 y2 + 24y + 80 = 0 (y + 4)(y + 20) = 0 y = –4 or y = –20 x = 2 x = –14 The solution set is {(2, –4), (–14, –20)}.

10. Set up the augmented matrix and use Gauss-Jordan reduction.

1 –1 1 17–4 1 5 22 3 1 8

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

1 2

1 3

1 1 1 170 3 9 66 40 5 1 26 2

R RR R

−⎡ ⎤⎢ ⎥− +⎢ ⎥⎢ ⎥− − − +⎣ ⎦

2

1 1 1 1710 1 3 223

0 5 1 26R

−⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥− −⎣ ⎦

2 1

2 3

1 0 2 50 1 3 220 0 14 84 5

R R

R R

− − +⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥ − +⎣ ⎦

3

1 0 2 50 1 3 220 0 1 6 1

14R

− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦

3 1

3 2

1 0 0 7 20 1 0 4 30 0 1 6

R RR R

+⎡ ⎤⎢ ⎥− +⎢ ⎥⎢ ⎥⎣ ⎦

x = 7, y = –4, z = 6 The solution set is {(7, –4, 6)}.

16 8 46 15 3

7 71

3 4( 1) 23 6

2

x yx y

yy

xxx

+ =− − =

− == −

+ − ===

The solution set is {(2, –1)}.

795


11. Parabola with vertex at (1, –4).

12. Ellipse with center at (0, 0) and vertices at (3, 0)

and (–3, 0).

13. 15 104

5 10

x y y x

y x

+ ≤ ≥ +

≤ − +

2

Graph with solid line and 5 10y x= − +

1 2.4

y x= + Shade the region that is below the

line and above the line 5 10y x= − +

1 24

y x= + . Then dash the solid lines that do

not contain the solution set.

14. a.

: 1, 3p ± ±: 1, 2, 4, 8, 16, 32q ± ± ± ± ± ±

1 3 1 3 1: 1, 3, , , , ,2 2 4 4 8

pq

± ± ± ± ± ± ±

b. x = 1 appears to be a root.

1 32 –52 17 3

32 –20 –3

32 –20 –3 0 3 232 – 52 17 3 0x x x+ + =

(x – 1)(4x – 3)(8x + 1) = 0

x = 1 or

2( –1)(32 – 20 – 3) 0x x x =

34

x = or 18

x = −

The solution set is 1 3– , , 18 4

⎧ ⎫⎨ ⎬⎩ ⎭

.

15. a. domain: ( 2,2)− range: [ )3,− ∞

b. the relative minimum of –3 occurs at 0.x =

c. increasing: ( )0, 2

d. ( 1) (0) 0 ( 3) 3f f− − = − − =

e. ( ) ( )( )(1) (1) 0f f f f f 3= = = −

f. ( ) as 2 or as 2f x x x+ −→ ∞ → − →

g.

h.

,

3 1 3 1 3, , , ,8 16 16 32 32

± ± ± ± ±

796

ISM: College Algebra Cumulative Review Exercises (Chapters 1−7)

16.

( )2( ) 4, 2f x x g x x= − = +

( ) ( )2 2 2( )( ) 4 4 2g f x g x x x= − = − + = − 2

17. 3

35 5

35 5

5 5

log log log 125125

log log 313log log 32

x yx y

x y

x y

= −

= + −

= +

5

−

18. ( )2 1

2 1

8 4 12 25 1 6

y ym

x x− −−

= = = =− − − −

−

5

y + 4 = –2(x – 1) y = –2x – 2

1 1( )y y m x x− = −

19. Let R = the cost of a rental at Rent-a-Truck and let A = the cost of a rental at Ace Truck Rentals. R = 39 + 0.16m A = 25 + 0.24m where m is the number of miles. 39 + 0.16m = 25 + 0.24m 14 = 0.08m m = 175 R = 39 + 0.16(175) = 67 The cost will be the same when the number of miles driven is 175 miles. The cost will be $67.

20. Let x = cost of basic cable, Let y = cost of movie channel.

Multiply the first equation by –1 and then add the two equations.

352 4

x yx y+ =+ =

35 2 45

10

x yx y

y

− − = −+ =

=

Use back-substitution to find x.

10 35

25x

x+ =

=Basic cable costs $25 and one movie channel costs $10.

797

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