P. 623-625

P. 623-625
Chapter 9,
continued
When da
N
W
}}
20Ï 9t 2 30t 1 29 a50
100
E
S
2
2. x2 5 2y
y
60 mph
Car
(−100, 40) d
}}
Ï9t 2 2 30t 1 29 a}52
25
Tower
(0, 0)
25
9t 2 30t 1 29a}
4
2
x
1
Axis of symmetry: vertical (x 5 0)
6t 2 13a0
13
ta}
6
7
13
So, the car is in range of the tower when }6 ata}
, or
6
(0, 21)
Directrix: y 5 2p, or y 5 1
0 5 3x2 2 12x 2 15
Axis of symmetry: vertical (x 5 0)
0 5 (3x 1 3)(x 2 5)
or
x2550
or
x55
1
4. x 5 } y 2
3
y
y 2 5 3x
The x-interepts of the function are 21 and 5.
( , 0)
3
4
Focus: ( p, 0)
59. F;
x
3x 5 4px
Find the height of one of the triangles.
3
x 5 24
3
4
}5p
Focus: 1 }4, 0 2
3
Directrix: x 5 2p, or x 5 2}4
4.5
Axis of symmetry: horizontal ( y 5 0)
20.25 1 h2 5 81
5. Directrix: y 5 2p 5 2
h2 5 60.75
p 5 22
h ø 7.8
Equation of parabola: x 2 5 4py
S 5 area of square 1 4 + area of triangle
x 2 5 4(22)y
ø 81 1 41 }2 2(9)(7.8) 5 81 1 140.4 5 221.4
1
x 2 5 28y
The surface area of the figure is about 221 square inches.
6. Directrix: x 5 2p 5 4
p 5 24
Equation of parabola: y 2 5 4px
Lesson 9.2
y 2 5 4(24) x
9.2 Guided Practice (pp. 622– 623)
1. y2 5 26x
Focus: ( p, 0)
26x 5 4px
26 5 4p
7. Focus: (22, 0)
3
(2 , 0 )
3
2
1
21
x52
p 5 22
x
Equation of parabola: y 2 5 4px
y 2 5 4(22) x
3
2}2 5 p
3
Focus: 2}2, 0
y 2 5 216x
y
y 2 5 28x
2
3
Directrix: x 5 2p, or x 5 }2
Axis of symmetry: horizontal ( y 5 0)
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
9
(4.5)2 1 h2 5 92
480
x
22
Focus: (0, 21)
y 5 3x2 2 12x 2 15
1
y51
x2 5 24y
21 5 p
58. A;
4.5
y
2
24y 5 4py
Mixed Review for TAKS
h
1
3. y 5 2} x2
4
Focus: (0, p)
between about 1.17 hours and 2.17 hours.
9
y5
1
2
Directrix: y 5 2p, or y 5 2}2
tq} and
x 5 21
x
1
22
1
(6t 2 7)(6t 2 13)a0
3x 1 3 5 0
1
Focus: 1 0, }2 2
36t 2 2 120t 1 91a0
and
1
2
}5p
36t 2 2 120t 1 116a25
6t 2 7q0
y
(0, )
Focus: (0, p)
x2 5 4py
2y 5 4py
Chapter 9,
continued
8. Focus: (0, 3)
5.
p53
Equation of parabola: x 2 5 4py
2
x 5 4(3)y
x 2 5 12y
9. Vertex: (0, 0)
6.
Focus: (10, 0)
Directrix: x 5 2p 5 27
(7, 0)
2
22
x
Axis of symmetry: y 5 0
y 5 4(10)x
7.
2
y 5 40x
y 2 5 210x
y
4px 5 210x
The antenna’s depth is the x-value at the outside edge.
10
1
(2
5
p 5 2}
5 2}2
4
Radius 5 }2 (16) 5 8. The antenna extends 8 feet to either
side of the vertex (0, 0).
5
Focus: 2}2, 0
1
When y 5 8:
)
,0
1
1
x
2
x
5
52
Directrix: x 5 2p 5 }2
Axis of symmetry: y 5 0
64 5 40x
8.
1.6 5 x
x 2 5 30y
y
4 py 5 30y
The antenna is 1.6 feet deep.
(0 , )
15
2
30
15
p5}
5}
4
2
9.2 Exercises (pp. 623–625)
4
Focus: 1 0, }
22
15
Skill Practice
1. A parabola is the set of all points in a plane equidistant
2. Sample answer: x 2 5 4py opens up or down (depending
2
on the value of p) while y 5 4px opens left or right
(depending on the value of p).
y 2 5 16x
x
10.
3
y 52
x
23
(0, )
3
22
Axis of symmetry: x 5 0
2
21
1
x
x 52
x 2 5 236y
4py 5 236y
1
3
)
Axis of symmetry: y 5 0
4py 5 26y
Directrix: y 5 2p 5 }2
,0
Directrix: x 5 2p 5 }2
y
2
1
2
1
x 5 24
x 2 5 26y
3
(2
1
Axis of symmetry: y 5 0
p 5 2}4 5 2}2
y
Focus: 1 2}2, 0 2
(4, 0)
1
Directrix: x 5 2p 5 24
y 2 5 22x
4px 5 22x
4px 5 16x
Focus: (4, 0)
x
Axis of symmetry: x 5 0
9.
1
p 5 2}2
1
4
15
2
15
y
p54
y52
Directrix: y 5 2p 5 2}
2
from a point called the focus and a line called the
directrix.
6
5
2
5
(8)2 5 40x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
x 5 27
p57
y 2 5 4px
1
2
y 5 25
y
Focus: (7, 0)
3
Focus: 0, 2}2
2
y 2 5 28x
The parabola opens right, so its equation has the
form y 2 5 4px.
2
4.
y
(0, 5)
4px 5 28x
p 5 10
3.
x 2 5 20y
4py 5 20y
p55
Focus: (0, 5)
Directrix: y 5 2p 5 25
Axis of symmetry: x 5 0
y
y59
3
36
p 5 2}
5 29
4
3
Focus: (0, 9)
x
(0, 29)
Directrix: y 5 2p 5 9
Axis of symmetry: x 5 0
11. x 2 5 12y
y
4py 5 12y
2
p53
Focus: (0, 3)
(0, 3)
2
x
y 5 23
Directrix: y 5 2p 5 23
Axis of symmetry: x 5 0
Algebra 2
Worked-Out Solution Key
481
Chapter 9,
continued
12. 22y 5 x 2
1
22y 5 4py
y
17. 224x 5 3y 2
1
y 52
1
y 5 28x
x
1
2}2 5 p
(0,
Focus: 1 0, 2}2 2
1
1
x
x52
Directrix: x 5 2p 5 2
Axis of symmetry: y 5 0
18. 14x 5 6y 2
Axis of symmetry: x 5 0
2
(
0.1
1
4px 5 }4 x
1
16
)
,0
x
0.2
1
p5}
16
(
1
7
4px 5 }3 x
7
12
,0
)
1
x
7
x 5 2 12
7
p5}
12
x 5 2 16
1
y
14
y2 5 }
x
6
y
1
y 2 5 }4 x
7
Focus: 1 }
,0
12 2
Focus: 1 }
,0
16 2
1
7
Directrix: x 5 2p 5 2}
12
1
Directrix: x 5 2p 5 2}
16
Axis of symmetry: y 5 0
1
19. } x 2 2 y 5 0
8
Axis of symmetry: y 5 0
2
14. 2x 5 48y
y
y 5 12
x 2 5 248y
4py 5 248y
1
8
x
5
48
(0, 212)
Focus: (0, 212)
1
1
x 2 5 8y
4py 5 8y
y 5 22
Focus: (0, 2)
15. 5x 2 5 215y
y
x 2 5 23y
Axis of symmetry: x 5 0
3
y 54
20. 4x 2 11y 2 5 0
0.5
x
1.5
4py 5 23y
( 0, )
211y 5 24x
0.1
4
y2 5 }
x
11
3
24
3
p 5 2}4
y
2
x5
4
Focus: 1 0, 2}4 2
3
x
4px 5 }
11
3
1
Directrix: y 5 2p 5 }4
p5}
11
Axis of symmetry: x 5 0
Focus: 1 }
,0
11 2
1
2
16. 2y 5 18x
y
(2
4px 5 218x
9
2
,0
)
2
18
9
5 2}2
p 5 2}
4
Focus: 1 2}2 , 0 2
9
9
Directrix: x 5 2p 5 }2
Axis of symmetry: y 5 0
1
9
Directrix: x 5 2p 5 2}
11
x 52
2
x
Axis of symmetry: y 5 0
1
2 11
(
0.1
1
,
11
0
)
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Directrix: y 5 2p 5 22
Axis of symmetry: x 5 0
y 2 5 218x
x
p52
Directrix: y 5 2p 5 12
Algebra 2
Worked-Out Solution Key
y
(0, 2)
} x2 5 y
5
5 212
p 5 2}
4
482
4px 5 28x
Focus: (22, 0)
1
x 5 4y
1
(22, 0)
p 5 22
1
22)
Directrix: y 5 2p 5 }2
13.
y
2
Chapter 9,
continued
21. 5x 2 1 12y 5 0
25. D; 15y 1 3x 2 5 0
y
2
5x 5 212y
y5
1
12
x 2 5 2}
y
5
3
5
3x 2 5 215y
2
(0, 2 )
3
5
x 2 5 25y
x
4py 5 25y
12
y
4py 5 2}
5
5
p 5 2}4
3
5
p 5 2}5
Directrix: y 5 2p 5 }4 5 1.25
Focus: 1 0, 2}5 2
3
26. Focus: (2, 0)
p52
3
p 5 25
Directrix: y 5 2p 5 }5
y 2 5 4px
y 2 5 4(25)x
Axis of symmetry: x 5 0
y 2 5 4(2)x
y 2 5 220x
2
y 5 8x
1
22. 25x 1 } y 2 5 0
3
y
1
} y 2 5 5x
3
( 3 , 0)
3
4
1
y 2 5 15x
x
21
4px 5 15x
3
4
x 5 23
15
3
5 3 }4
p5}
4
28. Focus: (3, 0)
29. Focus: (0, 24)
p53
p 5 24
2
x 2 5 4py
2
y 5 4(3)x
x 2 5 4(24)y
y 2 5 12x
x 2 5 216y
30. Focus: (0, 8)
31. Focus: (0, 210)
y 5 4px
p58
Focus: 1 3}4 , 0 2
3
3
Directrix: x 5 2p 5 23 }4
x 2 5 4(8)y
x 2 5 4(210)y
x 2 5 240y
32. Focus: (0, 26)
26x 1 y 2 5 0
y
y2 5 6x
(
1
4px 5 6x
3
2
)
,0
x
1
3
x 5 22
3
x 2 5 4py
x 5 32y
23. The parabola should open to the right rather than up.
p 5 }2
p 5 210
x 2 5 4py
2
Axis of symmetry: y 5 0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
27. Focus: (25, 0)
p 5 26
x 2 5 4py
x 2 5 4(26)y
x 2 5 224y
7
34. Focus: 0, }
4
1
2
7
Focus: 1 }2 , 0 2
3
24. The parabola should open to the left rather than to
the right.
0.5y 2 1 x 5 0
0.5y 5 2 x
x5
y2 5 22x
(
)
1
2 ,0
2
4px 5 22x
x 2 5 41 }4 2 y
x 2 5 41 2}8 2 y
x 2 5 7y
x 2 5 2}2 y
5
36. Focus: } , 0
2
1
y
2
1
22
1
2
3
3
2
5
x
Focus: 1 2}2 , 0 2
1
1
Directrix: x 5 2p 5 }2
Axis of symmetry: y 5 0
1
9
37. Focus: 2} , 0
16
2
9
p 5 }2
p 5 2}
16
y 2 5 4px
y 2 5 4px
y 2 5 41 }2 2 x
y 2 5 41 2}
x
16 2
y 2 5 10x
y 2 5 2}4 x
5
1
p 5 2}2
2
x 2 5 4py
7
Axis of symmetry: y 5 0
1
3
35. Focus: 0, 2}
8
p 5 2}8
x 2 5 4py
3
p 5 29
y 2 5 4px
y 2 5 4(29)x
y 2 5 236x
3
p 5 }4
Directrix: x 5 2p 5 2}2
33. Focus: (29, 0)
9
9
38. A; Focus: (28, 0)
39. Directrix: x 5 3
p 5 28
y 2 5 4px
y 2 5 4(28)x
y 2 5 232x
p 5 23
y 2 5 4px
y 2 5 4(23)x
y 2 5 212x
Algebra 2
Worked-Out Solution Key
483
continued
40. Directrix: y 5 27
41. Directrix: x 5 25
p57
x 2 5 4py
x 2 5 4(7)y
x 2 5 28y
42. Directrix: y 5 12
p 5 212
x 2 5 4py
x 2 5 4(212)y
x 2 5 248y
44. Directrix: x 5 22
p52
y 2 5 4px
y 2 5 4(2)x
y 2 5 8x
p55
y 2 5 4px
y 2 5 4(5)x
y 2 5 20x
43. Directrix: y 5 24
p54
x 2 5 4py
x 2 5 4(4)y
x 2 5 16y
45. Directrix: y 5 6
p 5 26
x 2 5 4py
x 2 5 4(26)y
x 2 5 224y
46. Directrix: x 5 11
y 5 4px
y 2 5 4(211)x
x 2 5 4 1 2}
y
12 2
1
located at 1 2}8 , 0 2 rather than 1 }2 , 0 2. The new directrix
1
3
3
1
will be x 5 }8 rather that x 5 2}2. The parabola will be
narrower and will open left rather than right.
1
52. The equation x 2 5 4py can be rewritten y 5 } x 2. So, if
4p
x 2 5 4py and y 5 ax 2 represent the same parabola, then
1
1 2
x 5 ax 2 and a 5 }
.
y5 }
4p
4p
Distance from (x, y) to (0, p):
5
2
y 5 6x
}}
1
50. Directrix: x 5 2}
18
11
p5}
6
1
p5}
18
x 2 5 4py
y 2 5 4px
y
x2 5 4 1 }
62
y2 5 4 1 }
x
18 2
22
y
x 5}
3
2
y 5 }9 x
x 2 5 4py
Problem Solving
55. Focus is (0, 6)
p56
x 2 5 4py
y
x 2 5 4(6)y 5 24y
4p 5 a
a
1
x 2 5 4y
An equation for the cross section of the trough is
x 2 5 24y.
x2 5 y
21
x
As a changes from 1 to 4, the new focus will be
1
2
x 2 1 ( y 2 p) 2 5 ( y 1 p) 2
x 1 y 2 2 2py 1 p 2 5 y 2 1 2py 1 p 2
2
51. a. x 2 5 ay
The point (x, y) is equidistant from the focus and
directrix, so the above distances are equal.
2
1
2
}
d 5 Ï (x 2 x) 2 1 ( y 2 (2p)) 2 5 Ï ( y 1 p) 2
x 5 2} y
11
}}
Distance from (x, y) to directrix y 5 2p:
5
3
2
1
located at (0, 1) rather than 0, }4 . The new directrix
1
will be y 5 21 rather than y 5 2}4. The parabola will
Diameter 5 17 ft
Radius 5 8.5 ft
When x 5 8.5:
(8.5) 2 5 24y
72.25
24
}5y
3.01 ø y
The trough is about 3 feet deep.
Algebra 2
Worked-Out Solution Key
x
As a changes from 6 to2}2, the new focus will be
}}
y 2 5 4 1 }2 2 x
3
484
y 2 5 6x
d 5 Ï (x 2 0) 2 1 ( y 2 p) 2 5 Ï x 2 1 ( y 2 p) 2
x 5 4py
be wider.
22
Directrix: y 5 2p
2
y 5 4px
p 5 }4
a
p 5 }4
54. Focus: (0, p)
5
48. Directrix: y 5 }
12
5
p 5 2}
12
2
11
49. Directrix: y 5 2}
6
y
1
directrix move further away from the vertex, and the
graph becomes wider. Because each point on a parabola
is equidistant from the focus and the directrix, this has
the effect of making the parabola wider and wider as
{p{ increases.
2
3
47. Directrix: x 5 2}
2
3
p 5 }2
1
y2 5 22 x
4p 5 a
53. As {p{ increases, the focus of the parabola and the
p 5 211
y 2 5 244x
b. y 2 5 ax
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,
Chapter 9,
continued
56. y 2 5 4px, p > 0
b. Original reflector: x 2 5 10.5y.
When x 5 3 and y 5 4:
For a wider reflector, 4p > 10.5.
2
(4) 5 4p(3)
Sample answer:
Let p 5 12.5: x 2 5 4py
16 5 12p
x 2 5 4(12.5)y
4
3
}5p
x 2 5 50y
4
The focal length is }3 inches.
57. a.
When y 5 9.5: x 2 5 50(9.5)
y
x 2 5 475
y
48 in.
x ø 21.8
(0, 48)
The wider reflector’s beam is about
2(21.8) 5 43.6 inches wide.
48 in.
146 in.
x
(248, 0)
x
146 in.
c. For a narrower reflector, 4 p < 10.5.
Sample answer:
Let p 5 2: x 2 5 4py
b. x 2 5 4py
x 2 5 4(2)y
y 2 5 4px
x 2 5 4(48)y
y 2 5 4(248)
x 2 5 192y
y 2 5 2192x
x 2 5 8y
When y 5 9.5: x 2 5 8(9.5)
x 2 5 76
146
c. Using x 5 192y and x 5 } 5 73:
2
2
x ø 8.7
The narrower reflector’s beam is about 2(8.7) 5
17.4 inches wide.
x 2 5 192y
2
73 5 192y
27.8 ø y
The dish is about {27.8{ 5 27.8 inches deep.
146
Using y 5 2192 x and y 5 }
5 73:
2
2
Find the x-values where y 5 p:
x 2 5 4p 2
x 5 62p
2
The length of the latus rectum is the distance from
(22p, p) to (2p, p):
227.8 ø x
d 5 Ï (2p 2 (22p)) 2 1 ( p 2 p) 2 5 Ï (4p) 2 5 4p
y 2 5 2192x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
60. x 2 5 4py
73 5 2192x
The dish is about {227.8{ 5 27.8 inches deep.
No, you get the same answer using each equation.
58. Because the parabola opens up, p is positive.
Focal distance 5 {p{ 5 p
p 5 0.36(diameter)
}}}
The length of the latus rectum is 4 p.
Mixed Review for TAKS
61. C;
x2 1 752 5 852
x 1 5625 5 7225
x 5 4py
25
When x 5 }
5 12.5:
2
(12.5)2 5 36y
4.3 ø y
Each dish is about 4.3 meters deep.
59. a. When y 5 9.5:
75 yd
x2 5 1600
2
An equation for the cross section of the dish is x 2 5 36y.
85 yd
x
2
p 5 0.36(25) 5 9
x 2 5 4(9)y 5 36y
}
x 5 40 yd
40 1 75 1 85 5 200
The perimeter of the park is 200 yards.
62. H; Points needed for an A 5 423 1 9 5 432
432
x
90
100
}5}
90x 5 43,200
x 5 480
There are 480 points possible in the course.
x 2 5 10.5y
x 2 5 10.5(9.5)
x 2 5 99.75
x ø 9.99
The diameter is about 2(9.99) ø 20 inches.
Algebra 2
Worked-Out Solution Key
485
P. 655-656
Chapter 9,
continued
13. 4x2 2 y2 2 16x 2 4y 2 4 5 0
3. (x 1 4)2 5 28( y 2 2)
h 5 24, k 5 2
A 5 4, B 5 0, C 5 21
Parabola
B2 2 4AC 5 0 2 4(4)(21) 5 16
Because B2 2 4AC > 0, the conic is a hyperbola.
Directrix: y 5 4
4(x 2 2)2 2 ( y 1 2)2 5 16
4. (x 2 2)2 1 ( y 2 7)2 5 9
( y 1 2)2
16
}2 }51
h 5 2, k 5 22
Center: (2, 22)
2
a52
2
b54
a 54
b 5 16
y
2
Center: (2, 7)
(4, 22)
x
(2, 22)
Hyperbola
B2 2 4AC 5 0 2 4(4)(6.25) 5 2100
Center: (6, 21)
Because A Þ C and B2 2 4AC < 0, the path is an ellipse.
4x2 1 6.25y2 2 12x 2 16 5 0
4(x2 2 3x) 1 6.25y2 5 16
y2
4
b2 5 1
b51
(6
, )0
(6
, 21)
22
x
(11, 21)
21
(1, 21)
(6, 22)
}
h 5 1.5, k 5 0
Co-vertices: (1.5, 2) and
(1.5, 22)
a55
c 5 Ï 26 ø 5.1
Foci: (0.9, 21) and (11.1, 21)
y
Center: (1.5, 0)
Vertices: (21, 0) and (4, 0)
a 5 25
5 26
}1}51
b52
y
c2 5 a2 1 b2 5 25 1 1
4(x 2 1.5)2 1 6.25y2 5 25
b2 5 4
2
Vertices: (1, 21) and
(11, 21)
4 (x2 2 3x 1 1.52) 1 6.25y2 5 16 1 9
a 5 2.5
x
22
h 5 6, k 5 21
A 5 4, B 5 0, C 5 6.25
a2 5 6.25
2
}
Radius: Ï9 5 3
(x 2 6)2
5. } 2 ( y 1 1)2 5 1
25
(2, 26)
14. 4x2 1 6.25y2 2 12x 2 16 5 0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(2, 7)
Circle
(2, 2)
(0, 22)
Vertices: (0, 22) and (4, 22)
(x 2 1.5)2
6.25
y
h 5 2, k 5 7
4
x
Focus: (24, 0)
4(x2 2 4x 1 4) 2 ( y2 1 4y 1 4) 5 4 1 4(4) 2 4
(x 2 2)2
4
4
(24, 0)
p 5 22
4(x2 2 4x) 2 ( y2 1 4y) 5 4
y
(24, 2)
4p 5 28
4x2 2 y2 2 16x 2 4y 2 4 5 0
6
y54
Vertex: (24, 2)
(21, 0)
1
(1.5, 2)
2
(1.5, 0) (4, 0)
x
1
1
11
1
and y 5 2}5 x 1 }5
Asymptotes: y 5 }5 x 2 }
5
(1.5, 22)
9.6 Exercises (pp. 655–657)
Skill Practice
1. Circles, ellipses, parabolas and hyperbolas are
called conic sections because they are formed by the
intersection of a plane and a double-napped cone.
2. If the discriminant of a general second-degree equation
is less than 0 and B 5 0 and A 5 C, the conic is a circle.
If it is less than 0 and either B Þ 0 or A Þ C, the conic is
an ellipse. If the discriminant is equal to 0, the conic is a
parabola. If it is greater than 0, the conic is a hyperbola.
(x 1 8)2
(y 1 4)2
6. } 2 } 5 1
9
49
y
(28, 3)
h 5 28, k 5 24
2
Hyperbola
22
Center: (28, 24)
(211, 24)
a2 5 49
(28, 24)
(25, 24)
2
b 59
a57
b53
Vertices: (28, 211) and
(28, 3)
2
2
x
(28, 211)
2
c 5 a 1 b 5 49 1 9
5 58
}
c 5 Ï 58 ø 7.6
Foci: (28, 211.6) and (28, 3.6)
7
44
7
68
and y 5 2}3 x 2 }
Asymptotes: y 5 }3 x 1 }
3
3
Algebra 2
Worked-Out Solution Key
509
Chapter 9,
continued
2
( y 2 2)
(x 1 2)
7. } 1 } 5 1
36
16
(y 2 4)2
(x 1 3)2
11. } 2 } 5 1
16
9
h 5 22, k 5 2
Co-vertices: (26, 2) and (2, 2)
h 5 23, k 5 4
Hyperbola
Center: (23, 4)
a2 5 9
a53
2
b 5 16 b 5 4
c2 5 a2 1 b2 5 9 1 16 5 25
c55
Vertices: (26, 4) and (0, 4)
Foci: (28, 4) and (2, 4)
Foci: (22, 6.5) and (22, 22.5)
Asymptotes: y 5 }3 x 1 8 and y 5 2}3 x
Ellipse
Center: (22, 2)
a2 5 36
a56
b2 5 16
b54
c2 5 a2 2 b2 5 36 2 16 5 20
}
c 5 Ï20 ø 4.5
Vertices: (22, 8) and (22, 24)
4
y
6
(26, 2)
(22, 2)
(23, 8)
(23, 4)
(26, 4)
(0, 4)
2
(23, 0)
4 x
4
( y 2 1)2
(x 2 4)2
12. C; } 1 } 5 1
4
16
(22, 8)
(2, 2)
4
x
(22, 24)
8. (x 2 5) 2 1 ( y 1 1) 2 5 64
y
h 5 5, k 5 21
2
Circle
24
Center: (5, 21)
x
(5, 21)
}
Radius: Ï 64 5 8
9. ( y 2 1)2 5 4(x 1 6)
8
h 5 26, k 5 1
y
x 5 27
Parabola
(26, 1)
(25, 1)
Vertex: (26, 1)
x
22
4p 5 4
p51
Focus: (25, 1)
Directrix: x 5 27
( y 2 2) 2
x2
10. } 1 } 5 1
4
25
h 5 0, k 5 2
Ellipse
(0, 4) y
a2 5 25, a 5 5
b2 5 4, b 5 2
c2 5 a2 2 b2 5 25 2 4 5 21
c ø 4.6
Vertices: (25, 2) and (5, 2)
Co-vertices: (0, 4) and (0, 0)
Foci: (24.6, 2) and (4.6, 2)
(0, 2)
(25, 2)
Center: (0, 2)
(5, 2)
1
21
(0, 0)
x
h 5 4, k 5 1
Center: (4, 1)
a2 5 16
a54
b2 5 4
b52
The co-vertices are 2 units above and below the center, at
(4, 3) and (4, 21).
13. Circle
14. Circle
Center: (25, 1)
Center: (9, 21)
r 5 6, r2 5 36
r52
r2 5 4
h 5 25, k 5 1
h 5 9, k 5 21
2
2
(x 1 5) 1 ( y 2 1) 5 36
(x 2 9)2 1 (y 1 1)2 5 4
15. Parabola
16. Parabola
Vertex: (24, 23)
Vertex: (5, 3)
Focus: (1, 23)
Directrix: y 5 6
h 5 24, k 5 23
h 5 5, k 5 3
The parabola opens
The parabola opens down.
to the right.
p 5 2(6 2 3) 5 2 3
p 5 1 2 (24) 5 5
(x 2 h)2 5 4p( y 2 k)
( y 2 k)2 5 4p(x 2 h)
(x 2 5)2 5 212( y 2 3)
2
( y 1 3) 5 20(x 1 4)
17. Ellipse
Vertices: (23, 4) and (5, 4)
Foci: (21, 4) and (3, 4)
Then center is the midpoint of the vertices.
23 1 5 4 1 4
,}
5 (1, 4)
1}
2
2 2
h 5 1, k 5 4
The vertices are 4 units from the center, so a 5 4 and
a2 5 16.
The foci are 2 units from the center, so c 5 2 and c2 5 4.
c2 5 a2 2 b2
4 5 16 2 b2
b2 5 16 2 4 5 12
The major axis is horizontal.
(x 2 h)2
( y 2 k)2
}
1}
51
2
2
b
a
( y 2 4)2
(x 2 1)2
}1}51
12
16
510
y
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
Chapter 9,
continued
18. Ellipse
21. In writing the equation, the h and k values should be
Vertices: (22, 1) and (22, 9)
Co-vertices (24, 5) and (0, 5)
The center is the midpoint of the vertices:
119
,}
5 (22, 5)
1}
2
2 2
22 1 (22)
h 5 22, k 5 5
(y 2 k)2
}
1}
51
2
2
a
b
(y 2 5)2
(x 1 2)2
}1}51
16
4
19. Hyperbola
Vertices: (6, 23) and (6, 1)
Foci: (6, 26) and (6, 4)
The center is the midpoint of the vertices:
6 1 6 23 1 1
,}
5 (6, 21)
1}
2
2 2
h 5 6, k 5 21
The vertices are 2 units from the center, so a 5 2 and
a2 5 4.
The foci are 5 units from the center, so c 5 5 and
c2 5 25.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
2
2
c 5a 1b
25 5 4 1 b2
b2 5 25 2 4 5 21
The transverse axis is vertical.
( y 2 k)
2
2
(x 2 h)
}
2}
51
2
2
b
a
(x 2 6)2
( y 1 1)2
}2}51
21
4
20. Hyperbola
Vertices: (1, 7) and (7, 7)
Foci: (21, 7) and (9, 7)
The center is the midpoint of the vertices:
1
2
117 717
}, } 5 (4, 7)
2
2
h 5 4, k 5 7
The vertices are 3 units from the center, so a 5 3 and
a2 5 9.
The foci are 5 units from the center, so c 5 5 and
c2 5 25.
c2 5 a2 1 b2
25 5 9 1 b2
b2 5 25 2 9 5 16
The transverse axis is horizontal.
(x 2 h)2
(x 1 2)2
( y 2 3)2
equation is }
1}
5 1.
9
25
( y 2 2)2
(x 1 5)2
22. } 1 } 5 1
16
49
Ellipse with center (25, 2)
The vertices are 4 units from the center, so a 5 4 and
a2 5 16.
The co-vertices are 2 units from the center, so b 5 2 and
b2 5 4.
The major axis is vertical.
(x 2 h)2
subtracted from x and y, not added. The correct
(y 2 k)2
}
2}
51
2
2
b
a
(y 2 7)2
(x 2 4)2
}2}51
16
9
Lines of symmetry: x 5 25 and y 5 2
23. ( y 2 4)2 5 6(x 1 60)
Parabola with vertex (26, 4)
Line of symmetry: y 5 4
( y 2 2)
(x 2 1)2
24. } 2 } 5 1
9
36
Hyperbola with center (1, 2)
Lines of symmetry: x 5 1 and y 5 2
(x 2 3)2
25. ( y 2 5)2 2 } 5 1
9
Hyperbola with center (3, 5)
Lines of symmetry: x 5 3 and y 5 5
26. (x 1 3)2 5 10( y 2 1)
Parabola with vertex (23, 1)
Line of symmetry: x 5 23
27. (x 1 2)2 1 ( y 1 1)2 5 121
Circle with center (22, 21)
Any line passing through (22, 21) is a line of symmetry.
28. 6x2 2 2y2 1 24x 1 2y 2 1 5 0
A 5 6, B 5 0, C 5 22
B2 2 4AC 5 0 2 4(6)(22) 5 48
The conic is a hyperbola, because B2 2 4AC > 0.
29. x2 1 y2 2 10x 2 6y 1 18 5 0
A 5 1, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(1)(1) 5 24
The conic is a circle because B2 2 4AC < 0, B 5 0, and
A 5 C.
30. y 2 210y 2 5x 1 57 5 0
A 5 0, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(0)(1) 5 0
The conic is a parabola because B2 2 4AC 5 0.
31. 4x2 1 y2 2 48x 2 14y 1 189 5 0
A 5 4, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(4)(1) 5 216
The conic is an ellipse because B2 2 4AC < 0 and A Þ C.
32. 9x2 1 4y2 1 8y 1 18x 2 41 5 0
A 5 9, B 5 0, C 5 4
B2 2 4AC 5 0 2 4(9)(4) 5 2144
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
33. x2 2 18x 1 6y 1 99 5 0
A 5 1, B 5 0, C 5 0
B2 2 4AC 5 0 2 4(1)(0) 5 0
The conic is a parabola because B2 2 4AC 5 0.
Algebra 2
Worked-Out Solution Key
511
Chapter 9,
continued
34. x2 1 y2 2 6x 1 8y 2 24 5 0
39. x2 2 16x 2 8y 1 80 5 0
A 5 1, B 5 0, C 5 1
A 5 1, B 5 0, C 5 0
2
B2 2 4AC 5 0 2 4(1)(0) 5 0
B 2 4AC 5 0 2 4(1)(1) 5 24
2
The conic is a circle because B 2 4AC < 0, B 5 0, and
A 5 C.
The conic is a parabola because
B2 2 4AC 5 0.
35. 8x2 2 9y2 2 40x 1 4y 1 145 5 0
x2 2 16x 2 8y 1 80 5 0
(x2 2 16x 1 64) 5 8y 2 80 1 64
A 5 8, B 5 0, C 5 29
2
B 2 4AC 5 0 2 4(8)(29) 5 288
(x 2 8)2 5 8y 2 16
2
The conic is a hyperbola because B 2 4AC > 0.
2
(x 2 8)2 5 8(y 2 2)
2
36. B; 4x 1 y 1 32x 2 10y 1 85 5 0
h 5 8, k 5 2
y
A 5 4, B 5 0, C 5 1
Vertex: (8, 2)
B2 2 4AC 5 0 2 4(4)(1) 5 216
4p 5 8
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
p52
Focus: (8, 4)
37. x2 1 y2 2 14x 1 4y 2 11 5 0
(8, 4)
2
(8, 2)
y50
22
x
A 5 1, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(1)(1) 5 24
40. 9y2 2 x2 2 54y 1 8x 1 56 5 0
2
The conic is a circle because B 2 4AC < 0, B 5 0 and
A 5 C.
A 5 21, B 5 0, C 5 9
B2 2 4AC 5 0 2 4(21)(9) 5 36
x2 1 y2 2 14x 1 4y 2 11 5 0
The conic is a hyperbola because B2 2 4AC > 0.
2 14x 1 49) 1 ( y 1 4y 1 4) 5 11 1 49 1 4
2
9y2 2 x2 2 54y 1 8x 1 56 5 0
(x 2 7)2 1 (y 1 2)2 5 64
h 5 7, k 5 22
9( y 2 6y 1 9) 2 (x2 2 8x 1 16) 5 256 1 81 2 16
2
9( y 2 3)2 2 (x 2 4)2 5 9
y
Center: (7, 22)
(x 2 4)2
}
Radius: Ï64 5 8
( y 2 3)2 2 }
51
9
2
x
22
h 5 4, k 5 3
(7, 22)
y
Center: (4, 3)
a2 5 1, a 5 1
(1, 3) (4, 4)
(7, 3)
3
b2 5 9, b 5 3
2
Vertices: (4, 4) and (4, 2)
2
38. x 1 4y 2 10x 1 16y 1 37 5 0
2
B 2 4AC 5 0 2 4(1)(4) 5 216
The conic is an ellipse because B 2 4AC < 0 and AÞC.
2
B2 2 4AC 5 0 2 4(9)(4) 5 2144
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
x 1 4y 2 10x 1 16y 1 37 5 0
(x2 2 10x 1 25) 1 4( y2 1 4y 1 4) 5 237 1 25 1 16
2
9x2 1 4y2 2 36x 2 24y 1 36 5 0
9(x 2 4x 1 4) 1 4( y2 2 6y 1 9) 5 236 1 36 1 36
2
2
(x 2 5) 1 4( y 1 2) 5 4
2
9(x 2 2)2 1 4(y 2 3)2 5 36
(x 2 5)
4
} 1 ( y 1 2)2 5 1
h 5 5, k 5 22
Center: (5, 22)
(x 2 2)2
4
a 5 4, a 5 2
b2 5 1, b 5 1
( y 2 3)2
9
}1}51
y
21
2
1
(5, 21)
x
(3, 22)
(5, 22) (5, 23)
(7, 22)
h 5 2, k 5 3
y (2, 6)
Center: (2, 3)
a2 5 9
2
a53
Vertices: (3, 22) and (7, 22)
b 54
Co-vertices: (5, 21) and (5, 23)
Vertices: (2, 6) and (2, 0)
b52
Co-vertices: (0, 3) and (4, 3)
Algebra 2
Worked-Out Solution Key
x
A 5 9, B 5 0, C 5 4
2
512
(4, 3)
1
41. 9x 2 1 4y2 2 36x 2 24y 1 36 5 0
A 5 1, B 5 0, C 5 4
2
(4, 2)
(0, 3)
(2, 3)
(4, 3)
1
21
(2, 0)
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(x
2
Chapter 9,
continued
42. y2 1 14y 1 16x 1 33 5 0
45. If A 5 C, then the conic will be a circle. If AÞC but
A and C have the same sign, the conic will be an ellipse.
If either A 5 0 or C 5 0, the conic will be a parabola.
If A and C have opposite signs, the conic will be a
hyperbola.
A 5 0, B 5 0, C 5 1
2
B 2 4AC 5 0 2 4(0)(1) 5 0
The conic is a parabola because B2 2 4AC 5 0.
y2 1 14y 1 16x 1 33 5 0
46.
( y2 1 14y 1 49) 5 216x 2 33 1 49
xy 5 a
( y 1 7)2 5 216x 1 16
A 5 0, B 5 1, C 5 0
( y 1 7)2 5 216(x 2 1)
h 5 1, k 5 27
2
Vertex: (1, 27)
B2 2 4AC 5 1 2 4(0)(0) 5 1
y
The conic is a hyperbola because B2 2 4AC > 0.
47. The foci are c units above and below the center, at
x
22
4p 5 216
a
y 5 }x
(h, k 1 c) and (h, k 2 c).
p 5 24
Focus: (23, 27)
a
The asymptotes are y 5 6}b x shifted horizontally h units
and vertically k units:
(23, 27)
(1, 27)
a
a
( y 2 k) 5 }b(x 2 h)
x55
( y 2 k) 5 2}b (x 2 h)
a
ah
a
bk 2 ah
y 5 }bx 2 }
1k
b
43. x 2 1 y 2 1 16x 2 8y 1 16 5 0
y 5 }b x 1}
b
A 5 1, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(1)(1) 5 24
Problem Solving
The conic is a circle because B2 2 4AC < 0, B 5 0,
and A 5 C.
48.
(x
a
bk 1 ah
y 5 2}b x 1 }
b
y
(0, 4)
1 16x 1 64) 1 ( y 2 2 8y 1 16) 5 216 1 64 1 16
2
(x 1 8)2 1 ( y 2 4)2 5 64
x
6
h 5 28, k 5 4
y
(0, 24)
Center: (28, 4)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
ah
8 ft
x2 1 y2 1 16x 2 8y 1 16x 5 0
2
a
y 5 2}b x 1 }
1k
b
}
Radius: Ï 64 5 8
Top circle:
(28, 4)
4
Center: (0, 4)
x
24
Radius: 4
Equation: (x 2 0)2 1 ( y 2 4)2 5 42
x2 1 (y 2 4)2 5 16
44. x2 2 4y2 1 8x 2 24y 2 24 5 0
Bottom circle:
A 5 1, B 5 0, C 5 24
Center: (0, 24)
B2 2 4AC 5 0 2 4(1)(24) 5 16
Radius: 4
The conic is a hyperbola because B2 2 4AC > 0.
Equation: (x 2 0)2 1 (y 1 4)2 5 42
x2 2 4y2 1 8x 2 24y 2 24 5 0
x2 1 (y 1 4) 5 16
(x2 1 8x 1 16) 2 4( y 2 1 6y 1 9) 5 24 1 16 2 36
49. x2 2 10x 1 4y 5 0
(x 1 4)2 2 4( y 1 3)2 5 4
x2 2 10x 5 24y
(x 1 4)2
} 2 ( y 1 3)2 5 1
4
h 5 24, k 5 23
Center: (24, 23)
a2 5 4, a 5 2
b2 5 1, b 5 1
Vertices: (26, 23) and
(22, 23)
x2 2 10x 1 25 5 24y 1 25
2
(24, 22)
210
(26, 23)
(24, 23)
(24, 24)
(x 2 5)2 5 241 y 2 }
42
25
y
x
(22, 23)
An equation for the path of the leap is
(x 2 5)2 5 241 y 2 }
.
42
25
Vertex: 1 5, }
4 2
25
25
The person’s jump is }
feet high and
4
2(5) 5 10 feet wide.
Algebra 2
Worked-Out Solution Key
513
continued
50. 21y 2 2 210y 2 4x2 5 2441
53. a. The intersection is not a circle when the plane crosses
the point where the cones meet. It is a point.
A 5 24, B 5 0, C 5 21
b. The intersection is not a hyperbola when the plane
2
B 2 4AC 5 0 2 4(24)(21) 5 336
crosses the point where the cones meet. It is a pair
of lines.
The shape of the path is a hyperbola because
B2 2 4AC > 0.
c. The intersection is not a parabola when the plane lies
21y2 2 210y 2 4x2 5 2441
along the edge of the cone. It is a line.
21( y2 2 10y 1 25) 2 4x2 5 2441 1 525
Mixed Review for TAKS
21( y 2 5)2 2 4x2 5 84
2
54. D;
( y 2 5)
x
}2}51
21
4
2
Center: (0, 5)
(0, 7)
}
a 5 2, b 5 Ï21
Vertices: (0, 3) and (0, 7)
(2
The expression 3x 2 39,000 represents the population
of Texas.
y
55. H;
Let n 5 number of sides.
(
21, 5 )
(0, 5)
21, 5 )
(n 2 2) + 180
5 135
n
}}
(0, 3)
1
180n 2 360 5 135n
x
45n 5 360
22
51. a. For the hotel, h 5 100, k 5 260 and r 5 150.
You will be in range of the transmitter when
(x 2 100)2 1 ( y 1 60)2a1502.
For the café, h 5 280, k 5 270, and r 5 100.
You will be in range of the transmitter when
(x 1 80)2 1 ( y 1 70)2a1002.
b. At point (0, 0):
2
(0 2 100)
The polygon has 8 sides.
Lesson 9.7
9.7 Guided Practice (pp. 659–661)
1. x 2 1 y 2 5 13 and y 5 x 2 1
y2 5 13 2 x2
(0 1 80) 1 (0 1 70) a100
1002
6400 1 4900a
2
n58
}
2
11,300 µ 10,000
1502
2
1 (0 1 60)2a
22,500
10,000 1 3600 a
y 5 6Ï 13 2 x2
Using the calculator’s intersect feature, the solutions are
(22, 23) and (3, 2).
2. x 2 1 8y 2 2 4 5 0 and y 5 2x 1 7
8y2 5 4 2 x2
y2 5 }2 2 }
8
y 5 6Î
y
}
You
(0, 0)
x
(100, 260)
Hotel
(280, 270)
Cafe
At the origin, you are in range of the hotel’s transmitter
because its inequality is satisfied. You are not in
range of the café’s transmitter because its inequality
is not satisfied.
range is 150 yards, if the hotel’s distance from the café
is less than their combined range of 250 yards, there is
an overlap.
52. a. An ellipse is formed by cutting the cone-shaped tip,
because the cut enters the cone diagonally and exits the
other side.
b. Hyperbolas are formed by each flat side and the
cone-shaped tip, because the flat sides are parallel
to the axis of the cone.
Algebra 2
Worked-Out Solution Key
1
2
x2
8
}2}
The graphs do not intersect. There is no solution.
3. y2 1 6x 2 1 5 0 and y 5 20.4x 1 2.6
y2 5 26x 1 1
}
y 5 6Ï 26x 1 1
Using the calculator’s intersect feature, the solutions are
approximately (21.57, 3.23) and (222.9, 11.8).
4. y 5 0.5x 2 3
x2 1 4y2 2 4 5 0
x2 1 4(0.5x 2 3)2 2 4 5 0
c. Because the café’s range is 100 yards and the hotel’s
514
x2
1
13,600a22,500 x 1 4(0.25x2 2 3x 1 9) 2 4 5 0
2
x2 1 x2 2 12x 1 36 2 4 5 0
2x2 2 12x 1 32 5 0
2(x2 2 6x 1 16) 5 0
There is no solution.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,