P.645-647

P.645-647
Chapter 9,
continued
3. 4y2 2 9x2 5 36
6. G;
2
O
B
A
(0,
6 x
a2 5 9, a 5 3
(3, − 2)
b2 5 4, b 5 2
(0, 2
}
c 5 Ï 13
x
(0, 23)
13 )
Vertices: (0, 63);
3
}
Foci: (0, 6Ï13 ); Asymptotes: y 5 6}2 x
1
2
5 2}3
Slope of tangent line at point A: 2}
3
}
4. Foci: (63, 0)
2
2
5 2}3
Slope of BO: }
320
} 22 2 0
5. Foci: (0, 610)
c 5 3, c2 5 9
3
1
Slope of tangent line at point B: 2}2 5 }2
2}3
Because the slopes of the two tangent lines are negative
reciprocals, the two lines are perpendicular.
x2 5 4py
c 5 10, c 2 5 100
Vertices: (61, 0)
Vertices: (0, 66)
a 5 1, a2 5 1
a56
c2 5 a2 1 b2
c2 5 a2 1 b2
2
100 5 36 1 b2
8 5 b2
64 5 b2
y
2.25 5 p
a2 5 36
9511b
2
y2
36
x2 2 }
51
8
(6)2 5 4p(4)
x2
64
}2}51
6. a 5 3 cm, c 5 5 cm
The focus is at (0, 2.25), so the wire should be placed
about 2.3 inches from the bottom.
b2 5 c2 2 a2
b 2 5 25 2 9 5 16
b54
Lesson 9.5
y2
9.5 Guided Practice (p. 644)
3
(0, 7)
a2 5 16
a54
b2 5 49
b57
y2
y
(24, 0)
3
y2
9
}2}51
16
9
(4, 0)
(2 65, 0)
c 5 Ï65
(
65, 0)
(0, 27)
y ø 23.75
}
Foci: (6Ï 65 , 0); Vertices: (64, 0)
The mirror has a height of about 23 2 (23.75) 5
0.75 centimeters.
7
Asymptotes: y 5 6}4 x
9.5 Exercises (pp. 645–648)
(0,
a 5 36
a56
b2 5 1
b51
y
37 )
4
(21, 0)
2
Skill Practice
(0, 6)
}
x
(0, 2
The transverse axis
is vertical.
}
Foci: (0, 6Ï 37 ); Vertices: (0, 66);
6
Asymptotes: y 5 6}1 x 5 66x
Algebra 2
Worked-Out Solution Key
the vertices of the hyperbola. The line segment joining
these two points is the transverse axis.
2. All ellipse is the set of all points such that the sum of the
(0, 26)
c 5 Ï37
1. The points (22, 0) and (2, 0) in the graph at the right are
(1, 0)
24
c 5 36 1 1 5 37
25
16
}5}
y 2 ø 14.06
The transverse axis is
horizontal.
y2
2. } 2 x2 5 1
36
y2
9
x
22
}
32
4
When x 5 3: }2 2 }2 5 1
4
c2 5 16 1 49 5 65
2
x2
4
}2 2 }2 5 1
y2
x2
1. } 2 } 5 1
49
16
500
(0, 3)
(2, 0)
21
c 5 9 1 4 5 13
7.
2
(22, 0)
2
23 2 0
3
}
Slope of AO: 5 }
5 }2
22 2 0
13 )
37 )
distances between any point and the foci is a constant. A
hyperbola is the set of all points such that the difference
of the distances between any point and the foci is
a constant.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(− 2, −3)
y
y2
x2
}2}51
4
9
y
Chapter 9,
continued
2
y2
x2
4. } 2 } 5 1
36
9
y
x2
3. } 2 } 5 1
4
25
a2 5 25
a2 5 9
a55
y2
x2
8. } 2 } 5 1
49
121
a2 5 49
a53
Vertices: (63, 0)
Vertices: (0, 67)
b2 5 4
b2 5 36
b2 5 121, b 5 11
b52
b56
c2 5 9 1 36 5 45
}
2
}
}
Foci: (6Ï 29 , 0)
(0, 6)
y
(2
29, 0 )
( 29, 0)
(0, 2) (5, 0)
a2 5 64
(3, 0)
5, 0)
(3
x
5, 0)
a 2 5 144
a59
Vertices: (0, 69)
b 5 25
b55
b 5 36
}
Foci: (68Ï5 , 0)
Foci: (6Ï53 , 0)
Asymptotes:
Asymptotes: y 5 6}2 x
7
16
b56
y
8
y
(0, 7)
(0, 16)
Foci: (66Ï5 , 0)
9
(25, 0)
(26
(0, 9)
( 6 5, 0)
(0, 6)
8
y2
x2
7. } 2 } 5 1
100
196
Vertices: (0, 614)
b2 5 100, b 5 10
c 5 196 1 100 5 296
}
(210, 0)
}
Foci: (0, 62Ï 74 )
7
14
Asymptotes: y 5 6}
x 5 6}5 x
10
(0, 27)
x
y2
64
x
9
a2 5 64
a55
Vertices: (0, 68)
b2 5 9
b2 5 25
b53
c 5 64 1 25 5 89
}
}
c 5 Ï 34
c 5 Ï89
}
Foci: (0, 6Ï34 )
Foci: (0, 6Ï89 )
5
8
Asymptotes: y 5 6}3 x
74 )
8
( 0,
34 )
(23, 0)
24
b55
2
}
x
a58
Vertices: (0, 65)
c 5 25 1 9 5 34
(0, 14)
x2
25
}2}51
2
(10, 0)
24
(0, 214)
(0,22
c 5 2Ï74
8
y
25
a2 5 25
y
74 )
x
12. 25y2 2 64x2 5 1600
2
}2}51
x
(0, 26)
(212, 0)
(12, 0)
(0, 2
53, 0)
(0, 216)
11. 9y2 2 25x2 5 225
2
x
a 5 14
(
53, 0)
y
5, 0)
4
(0, 29)
(2, 0)
24
(2
(8, 0)
1
6
29
2
5, 0)
(22, 0)
5, 0)
15
(28
y56}
x 5 6}2 x
12
(5, 0)
106 )
(28, 0)
Asymptotes:
y
6
(8
10
}
Asymptotes: y 5 6}5 x
a 5 196
}
}
Foci: (0, 6Ï106 )
b57
c2 5 4 1 49 5 53
c 5 Ï53
c 5 6Ï 5
}
2
b2 5 49
b 5 16
y 5 6}
x 5 62x
8
c2 5 144 1 36 5 180
}
(0, 2
a 5 12
2
c 5 Ï106
106 )
b2 5 256
}
Vertices: (612, 0)
c2 5 81 1 25 5 106
(0,
Vertices: (62, 0)
}
y
x2
6. } 2 } 5 1
36
144
a52
Vertices: (68, 0)
c 5 8Ï 5
2
y
x
5. } 2 } 5 1
25
81
2
a2 5 4
a58
c 5 64 1 256 5 320
(0, 26)
y2
49
x2
4
}2}51
2
2
a 2 5 81
y
256
x2
64
(25, 0) (0, 22)
2
10. 49x2 2 4y2 5 196
2
} 2 } 5 11
24
(23
(0, 27)
7
9. 4x2 2 y2 5 256
y
(23, 0)
x
28
170)
}
4
6
x
x
Foci: (0, 6Ï170 ); Asymptotes: y 5 6}
11
Asymptotes: y 5 62x
2
Asymptotes: y 5 6}5 x
(0, 2
c 5 Ï 170
}
Foci: (63Ï5 , 0)
(11, 0)
28
}
c 5 Ï 45 5 3Ï 5
c 5 25 1 4 5 29
c 5 Ï29
4
(211, 0)
c2 5 49 1 121 5 170
}
170 ) y
(0, 7)
a57
Vertices: (65, 0)
c2 5 a2 1 b2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(0,
Asymptotes: y 5 6}5 x
(0, 8) y
y
(0, 5)
(3, 0)
(0,2
6
(25, 0)
x
34 )
26
(0, 2
89 )
(0,
89 )
(5, 0)
x
(0, 28)
(0, 25)
Algebra 2
Worked-Out Solution Key
501
continued
13. 81x 2 2 16y 2 5 1296
14. 49y 2 2 100x2 5 4900
2
a2 5 15
a2 5 100
a54
Vertices: (0, 610)
b2 5 81
b2 5 49
b59
Vertices: (0, 62)
a 5 10
Vertices: (64, 0)
2
c 5 Ï149
}
Foci: (0, 6Ï149 )
9
Asymptotes: y 5 6}
x
7
(0, 9) y
(
(24, 0)
(0,
97, 0)
(0, 10)
(4, 0)
(27, 0)
x
26
97, 0)
149 )
6
(0, 2
(7, 0)
x
149 )
c 5 5, c2 5 25
36 5 4 1 b2
25 5 1 1 b2
24 5 b2
x2
4
y2
32
y2
x2 2 }
51
24
}2}51
21. Foci: (0, 612)
c 5 12
x2
9
}2}51
22. Foci (610, 0)
Vertices: (0, 67)
a57
15. D; 45y2 2 200x2 5 1800
y2
40
c 5 6, c 5 36
32 5 b
(0, 210)
(0, 29)
a 5 1, a2 5 1
2
2
29
c2 5 144
c 5 10
95 5 b 2
2
c 5 40 1 9 5 49
up and down instead of left and right. The value of a2
is 36, so the value of a is 6 and the vertices should be
located at (0, 66).
a 5 4, a2 5 16
}
c 5 4Ï 5 , c2 5 80
2
x
64
}
Verticies: (62, 0)
2
a 5 2, a 5 4
}
x2
17. In the equation } 2 y 2 5 4, the right side of the equation
4
should have been set equal to 1:
x2
16
2
y
4
}2}51
2
Therefore, a 5 16 and a 5 4. The vertices are located
at (64, 0). Similarly, b2 5 4 and b 5 2.
y
1
22
2
y2
8
} 2 x2 5 1
}
26. C; Foci: (0, 66Ï 3 )
Vertices: (0, 68)
a 5 8, a2 5 64
}
c 5 6Ï 3 , c2 5 108
54 5 4 1 b2
108 5 64 1 b2
2
44 5 b2
50 5 b
x2
4
y2
50
}2}51
y2
64
x2
44
}2}51
y2
x2
27. } 2 } 5 1
49
25
y
9 (0, 7)
The equation is a hyperbola.
a2 5 25
a55
b2 5 49
b57
(2
74, 0)
(
c 5 25 1 49 5 74
x
Vertices: (65, 0)
74, 0)
(5, 0) x
29
(25, 0)
2
}
Algebra 2
Worked-Out Solution Key
c 5 3, c2 5 9
c 5 3Ï 6 , c 5 54
Foci: 1 6Ï74 , 0 2
502
a 5 2Ï 2 , a2 5 8
1 5 b2
2
25. Foci: (63Ï 6 , 0)
x
}
64 5 b2
}2}51
24
}
Vertices: (0, 62Ï2 )
9 5 8 1 b2
y
16
y
24. Foci: (0, 63)
80 5 16 1 b
2
12
y2
25
x
75
Vertices: (0, 64)
y2
x2
16. In the equation } 2 } 5 1, the hyperbola should open
4
36
2
}2}51
}
Foci: (0, 67)
c2 5 100
100 5 75 1 b
2
23. Foci: (0, 64Ï 5 )
c57
a2 5 75
25 5 b 2
y2
x2
}2}51
95
49
b 59
}
a 5 5Ï 3
144 5 49 1 b
2
}
Vertices: (65Ï3 , 0)
a2 5 49
2
a2 5 40
Vertices: (61, 0)
2
a 5 2, a 5 4
y
6
20. Foci: (65, 0)
Vertices: (62, 0)
10
Asymptotes: y 5 6}4 x
c2 5 16
19. Foci: (66, 0)
}
Foci: (6Ï 97 , 0)
c54
y2
x2
}2}51
12
4
}
c 5 Ï97
a2 5 4
12 5 b2
c 5 100 1 49 5 149
}
a52
16 5 4 1 b2
b57
2
c 5 16 1 81 5 97
(2
18. Foci: (0, 64)
y2
x2
}2}51
49
100
y
x2
}2}51
81
16
(0, 27)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,
Chapter 9,
continued
2
28. y 2 5 18x
y
x2
b. } 2 } 5 1
4
16
y
The equation is a parabola.
9
2
b2 5 4
9
x 5 22
9
9
Focus: }2, 0 ; Directrix: x 5 2}2
1
2
2
29. 48x 1 12y 5 48
Foci: 1 0, 62Ï 5 2
(0, 2) y
(21, 0)
(0, 2
The equation is an ellipse.
( 0,
1
3)
By changing b2 to 25, c2 5 41 and the foci move to
3)
(1, 0)
}
(0, 6Ï41 ). The vertices are the same, but the foci
4 x
change and the hyperbola becomes wider.
y2
x2
34. a. } 2 } 5 1
30
15
(0, 22)
2
a 54
a52
b51
b2 5 1
c2 5 4 2 1 5 3
y2
15
x2
30
}511}
}
Foci: (0, 6Ï 3 )
x2
y2 5 15 1 }
2
Vertices: (0, 62); Co-vertices: (61, 0)
y2
x2
30. } 1 } 5 1
256
144
Î
}
(0, 16) y
(0, 4
The equation is an ellipse.
a2 5 256 a 5 16
b2 5 144 b 5 12
c2 5 256 2 144 5 112
4
7)
y2
x2
b. } 2 } 5 1
5.5
8.4
(12, 0)
(212, 0)
4
y2
x
Foci: (0, 64Ï 7 )
(0, 24
(0, 216)
y2
5.5
7)
5.5x2
2 5.5
y2 5 }
8.4
y 5 6Î
}
32. 18x2 1 18y 2 5 288
The equation is a hyperbola.
2
x 2 1 y 2 5 16
r 2 5 16
r54
The equation is a circle.
a55
a 5 25
b 2 5 121 b 5 11
c2 5 25 1 121 5 146
Vertices: (0, 65)
27.5y2 5 12 2 5x 2
7.5y 2 5 5x2 2 12
5x 2 2 12
y2 5 }
7.5
Î
}
Foci: 1 0, 6Ï146 2
}
5
Asymptotes: y 5 6}
x
11
(24, 0)
1
(4, 0)
x
9
(0, 5) (11, 0)
x
(0, 24)
(0, 25)
y
x2
33. a. } 2 } 5 1
36
9
b 5 2a
Sample answer:
a 5 1, a2 5 1; b 5 2, b2 5 4
y2
4
}2}51
a53
Vertices: (63, 0)
b2 5 36
b
35. y 5 62x 5 6} x
a
b
}52
a
x2
1
2
a2 5 9
5x2 2 12
y56 }
7.5
21
y
(211, 0)
5.5x2
8.4
} 2 5.5
c. 5x2 2 7.5y 2 5 12
(0, 4) y
26
x2
8.4
}5}21
Vertices: (0, 616); Co-vertices: (612, 0)
y2
x2
31. } 2 } 5 1
121
25
x2
2}
512}
5.5
8.4
}
}
x2
y 5 6 15 1 }
2
c 5 4Ï7
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b52
c2 5 20
}
2
y2
x2
}1}51
4
1
a54
Vertices: (0, 64)
x
2
9
p 5 }2
a2 5 16
( , 0)
2
4p 5 18
b56
c2 5 36 1 9 5 45
}
Foci: (63Ï 5 , 0)
By changing b2 to 4, b 5 2, c2 5 13, and the foci
}
move to (6Ï 13 , 0). The vertices remain the same, but
the foci change and the hyperbola becomes narrower.
a 5 2, a2 5 4; b 5 4, b2 5 16
y2
x2
}2}51
16
4
a 5 3, a 2 5 9; b 5 6, b 2 5 36
y2
x2
}2}51
36
9
All three hyperbolas approach the same asymptotes, but
as a and b get larger the foci and vertices move outward.
Algebra 2
Worked-Out Solution Key
503
Chapter 9,
continued
36. Sample answer:
b. a 5 4.1
}}
625 5 16.81 1 b2
d1 5 Ï(c 2 a)2 1 (0 2 0)2 5 c 2 a
608.19 5 b2
}}
d2 5 Ï(a 2 (2c))2 1 (0 2 0)2 5 a 1 c
An equation of the hyperbola that models the
{d2 2 d1{ 5 {(a 1 c) 2 (c 2 a){ 5 {2a{
Because a is positive, {2a{ 5 2a.
37. Foci: (62, 0)
1
2
{d2 2 d1{ 5 2a 5 2
} (170) 5 85; point B is located at (85, 240).
a51
b. Vertices (630.5, 0)
2
c 5 a 1 b2
Horizontal transverse axis
4 5 1 1 b2
x2
a2
y2
} 2 }2 5 1
2
b
y2
x2
} 2 }2 5 1
2
b
30.5
y2
x2 2 }
51
3
Using point B(85, 240):
Problem Solving
38. a 5 33
852
30.52
(240)2
}2}
51
2
c 5 56
b
1600
7225
}2}
51
930.25
b2
c2 5 a2 1 b 2
56 2 5 332 1 b2
6294.75
930.25
2047 5 b 2
b2 ø 236.5
y2
x2
}2}51
2047
1089
}
1
Ï2
39. Vertices: 6}, 0
2
An equation that models the cross section is
1
}
Ï3
y2
236.5
x2
930.25
} 2 } 5 1.
2
1
c. At the top outside edge of the roof, x 5 } (84) 5 42:
2
y2
422
}2}51
236.5
930.25
2
1
a2 5}
5 }2
4
Foci: 6}
,0
2
2
833.75
930.25
3
y 2 ø 212
c2 5 a2 1 b2
y ø 14.6
The total height of the roof is h ø 40 1 14.6 5
54.6 feet.
1
3
} 5 } 1 b2
2
4
2
4
1
} 5 b2
4
42. a.
} 2 } 5 b2
2
x
y2
236.5
}5}
c2 5 }4
3
4
1600
b
}5}
2
3136 2 1089 5 b2
x
y2
}2}51
1
1
}
}
4
2
An equation that models the hyperbola is 2x 2 2 4y 2 5 1.
40. a. a 5 13.3
28.22 5 13.32 1 b2
2
795.24 2 176.89 5 b
618.35 5 b2
An equation of the hyperbola that models the June 21
x2
y2
path is }
2}
5 1.
618.35
176.89
504
y
c 5 28.2
Algebra 2
Worked-Out Solution Key
Area of
outer
circle
:y 2
Area of
Area of
2 inner
5
walkway
circle
2
:x2
5
600
An equation is :y 2 2 :x2 5 600.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
35b
y2
2
x
September 1 path is }
2}
5 1.
608.19
16.81
1
41. a. } (61) 5 30.5; point A is located at (30.5, 0).
2
c 52
2
c 5 25
252 5 4.12 1 b2
Choose (a, 0) for (x, y).
Chapter 9,
continued
b. :y 2 2 :x 2 5 600
44. Microphone A receives the sound 2 seconds
after microphone B, so microphone A is
:(y2 2 x2) 5 600
(2 sec)1 }
sec 2 5 2200 feet farther away from the
elk than microphone B. The set of all points that are
2200 feet closer to B than to A is one branch of the
1100 ft
600
y2 2 x2 5 }
:
600
y2 5 x2 1 }
:
Ï
y2
x2
a
hyperbola }2 2 }2 5 1.
}
600
y 5 x2 1 }
:
b
5280
The microphones are 1 mile apart, so c 5 }
5 2640.
2
Sample answer:
d2 2 d1 5 2200
x
1
3
6
9
(c 1 a) 2 (c 2 a) 5 2200
2a 5 2200
}
Ï
600
:
y 5 x2 1 }
13.9 14.1 15.1 16.5
c. :y2 2 :x 2 5 600
y2
}2
600
}
:
a 5 1100
b2 5 c2 2 a2 5 26402 2 11002 5 5,759,600
x2
So, the elk is located somewhere along the right branch
}51
600
}
:
The equation represents a hyperbola.
600
a 5}
:
enough information to determine its exact location.
600
b 5}
:
2
2
4000
}
1
Ï 600 2
Because the radii of the walkway, x and y, must be
positive, only the first quadrant of the graph represents
solutions that make sense.
y
600
600
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
B
c1a
x
4000
c2a
45. A;
( p , 0)
300
x
(0, 2 p )
600
d. As x and y increase, the width of the walkway,
y 2 x, decreases because the area of concrete
remains constant.
43. a. Foci (66, 0)
(22, 4), (26, 22)
22 2 4
3
26
5}
5 }2
m5}
24
26 2 ( 2 2)
y 2 y1 5 m(x 2 x1)
3
y 2 4 5 }2(x 2 (22))
3
y 2 4 5 }2(x 1 2)
3
2
c 5 36
Vertices: (64, 0)
a2 5 16
a54
d1
Mixed Review for TAKS
(0, p )
300
(2 p 0)
Elk
A
a
Asymptotes: y 5 6}b x 5 6 x
600 ,
y
d2
Vertices: 0, 6 }
:
c56
y2
2
x
2}
5 1. There is not
of the hyperbola }
5,759,600
1,210,000
c2 5 a2 1 b2
36 5 16 1 b2
b2 5 36 2 16 5 20
An equation for the hyperbola is:
y 2 4 5 }2x 1 3
3
2}2x 1 y 5 7
23x 1 2y 5 14
The graph represents the equation 23x 1 2y 5 14
46. H;
The center is at the midpoint of the two points.
1 25 1 9
8 1 (23)
2
y
20
5
,}
5 1 2, }2 2
M5 }
2
2
b. Sample answer:
The center of the circle is 1 2, }2 2.
x2
16
2
} 2 } 5 1.
By definition of a hyperbola, the difference of the
distances from any point to the two foci is a constant.
Because the graph shown lies along the intersection
points of the circular ripples, you can see that at any
point, the difference of the radii always equals 8.
5
Algebra 2
Worked-Out Solution Key
505
Chapter 9,
continued
y2
x2
7. } 2 } 5 1
64
25
Quiz 9.4–9.5 (p. 648)
y2
x
1. } 1 } 5 1
4
25
2
a2 5 25
a2 5 25
Co-vertices: (0, 62)
c2 5 a2 2 b2
c2 5 25 2 4 5 21
2
b 5 64
y
3
(2
21, 0)
(0, 2)
(
(5, 0)
x
21
y
x2
2. } 1 } 5 1
49
16
x
2
c 5a 1b
(0, 25)
(0, 2
89 )
}
c 5 Ï 89
5
y
a57
Vertices: (0, 67)
b2 5 16
b54
(0, 7)
(0,
2
x
(0, 2
(0, 2)
a54
b2 5 4
y
(4, 0)
x
26
(0, 22)
(24, 0)
(2
b52
5, 0)
c2 5 a2 1 b2
c2 5 16 1 4 5 20
}
c 5 2Ï 5
}
Foci: (62Ï5 , 0)
2
y2
36
(0, 6)
(0, 3 3)
(23, 0)
2
x
(0, 23 3)
}
a2 5 20
2
a 5 2Ï 5
c 5a 2b
Vertices: (0, 62Ï 5 )
}
b2 5 12
8
(0, 4 2)
y2
x2
}2}51
12
20
}
(0, 26)
Co-vertices: (63, 0)
9. 12y2 2 20x2 5 240
(3, 0)
24
Vertices: (0, 66)
1
Asymptotes: y 5 6}4 x 5 6}2 x
y
}1}51
b53
33)
(0, 27)
3. 36x2 1 9y2 5 324
a56
5, 0) 6
Vertices: (64, 0)
(4, 0)
(24, 0) 22
Co-vertices: (64, 0)
c2 5 a2 2 b2
c2 5 49 2 16 5 33
}
c 5 Ï33
}
Foci: 1 0, 6Ï33 2
a2 5 36
a2 5 16
33)
(22
(22
y
(0, 2 5)
3, 0)
(2 3, 0) x
(0, 24 2)
26
(0, 22 5)
b 5 2Ï 3
c2 5 a2 1 b2
2
c 5 36 2 9 5 27
c2 5 20 1 12 5 32
}
c 5 3Ï3
}
c 5 4Ï 2
}
Foci: (0, 63Ï 3 )
}
Foci: (0, 64Ï2 )
4. Vertex: (0, 5)
}
2Ï 5
2Ï3
Co-vertex: (24, 0)
a55
a2 5 25
b54
b2 5 16
10. Foci: (65, 0)
Focus: (28, 0)
}
6. Co-vertex: (2Ï 15 , 0)
Focus: (0, 25)
}
a2 5 100
b 5 Ï15
b2 5 15
c58
c2 5 64
c55
c2 5 25
c2 5 a2 2 b2
c2 5 a2 2 b2
64 5 100 2 b2
25 5 a2 2 15
b 5 100 2 64 5 36
2
y
36
}1}51
Algebra 2
Worked-Out Solution Key
a2 5 25 1 15 5 40
x2
15
y2
40
}1}51
c53
c2 5 9
2
a51
a2 5 1
a 54
a52
c2 5 a2 1 b2
2
25 5 4 1 b
b2 5 25 2 4 5 21
2
x
4
2
y
21
}2}51
Vertices: (0, 61)
2
c 5 25
c55
a 5 10
2
11. Foci: (0, 63)
Vertices: (62, 0)
y2
x2
}1}51
25
16
5. Vertex: (10, 0)
}
Ï15
Asymptotes: y 5 6}
} x 5 6} x
3
c2 5 a2 1 b2
9 5 1 1 b2
b2 5 8
x2
y2 2 }
51
8
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
a2 5 49
506
(8, 0)
6
8. 4x 2 2 16y 2 5 64
y2
x2
}2}51
4
16
2
x2
100
3
Asymptotes: y 5 6}8 x
}
2
2
Foci: (0, 6Ï89 )
(0, 22)
Foci: (6Ï 21 , 0)
2
b58
(0, 5)
}
(25, 0)
}
b 59
2
89 )
(28, 0)
c2 5 25 1 64 5 89
21, 0)
c 5 Ï21
2
a55
Vertices: (0, 65)
a55
Vertices: (65, 0)
b2 5 4
b52
x2
9
y
(0,
P. 655-656
Chapter 9,
continued
( y 2 4)2
3. (x 1 3)2 2 } 5 1
4
}
12. Foci: (63Ï 6 , 0)
The equation represents a hyperbola with center at
(h, k) 5 (23, 4).
Vertices: (63, 0)
}
2
c 5 3Ï6
c 5 54
a53
a2 5 9
2
2
a2 5 1
a51
The vertices are 1 unit to the left and right of the centers
at (24, 4) and (22, 4).
2
c 5a 1b
54 5 9 1 b2
b2 5 4
b2 5 54 2 9 5 45
2
b52
c 5 a 1 b2
y2
x2
}2}51
45
9
2
c2 5 1 1 4 5 5
}
13. 2a 5 2.98 1 2.55
c 5 Ï 5 ø 2.23
y
The foci are about 2.23 units left and right of the center,
at (25.23, 4) and (20.77, 4).
2a 5 5.53
a 5 2.765
sun
(22, 4)
c 5 2.765 2 2.55
2.98
c 5 0.215
2
2
2.55
(24, 4)
2
c 5a 2b
(25.23, 4)
(0.215) 2 5 (2.765)2 2 b2
(23, 4)
b2 ø 7.645 2 0.046 5 7.599
x2
2
Lesson 9.6
Investigating Algebra Activity 9.6 (p. 649)
1. No, the equations will not all be the same. The equations
will vary as the distance of the flashlight from the wall
changes.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2. By holding the flashlight parallel to and against the wall,
the beam will form a branch of a hyperbola on the wall.
( y 2 1)
(x 2 2)
4. } 1 } 5 1
9
16
The equation is an ellipse with center at (h, k) 5 (2, 1).
a2 5 16
a54
The vertices are 4 units left and right of the center, at
(22, 1) and (6, 1).
b2 5 9
b53
The co-vertices are 3 units above and below the center at
(2, 4) and (2, 22).
y
(2, 4)
2
1. (x 1 1) 1 ( y 2 3) 5 4
The equation represents a circle with
center at
}
(h, k) 5 (21, 3) and radius r 5 Ï 4 5 2.
y
(21, 5)
(21, 3)
x
2
2
9.6 Guided Practice (pp. 651–654)
2
1
22
y
1}
5 1.
An equation for the orbit of 1 ceres is }
7.599
7.645
(23, 3)
y
(20.77, 4)
x
c 5 a 2 2.55
(22, 1)
(6, 1)
2
(2, 1)
x
22
(1, 3)
(2, 22)
2
5. Parabola with vertex (3, 21) and focus (3, 2)
(21, 1)
y
x
21
2. (x 2 2)2 5 8( y 1 3)
The equation represents a parabola with vertex at
(h, k) 5 (2, 23).
2
(3, 2)
(3, 21)
8
x
4p 5 8
p52
The form of the equation is (x 2 h)2 5 4p( y 2 k) where
p > 0.
The focus lies 2 units above the vertex, at (2, 21).
The directrix lies 2 units below the vertex, at y 5 25.
The vertex is (3, 21), so h 5 3 and k 5 21.
y
1
21
x
(2, 21)
(2, 23)
The focus is 3 units away from the vertex, so p 5 3.
4p 5 4(3) 5 12
An equation is ( x 2 3)2 5 12( y 1 1).
y 5 25
Algebra 2
Worked-Out Solution Key
507
Chapter 9,
continued
6. Hyperbola with vertices at (27, 3) and (21, 3) and foci
12
y
1
at (29, 3) and (1, 3)
(1, 22)
(21, 3)
(29, 3)
(27, 3)
x
4
y
(1, 3)
11. 2x 2 1 y 2 2 4x 2 4 5 0
x
3
(24, 3)
A 5 2, B 5 0, C 5 1
The foci lie on the transverse axis, so the transverse
axis is horizontal. The form of the equation is
2
(x 2 h)
( y 2 k)
a
b
2x 2 1 y 2 2 4x 2 4 5 0
}
2}
5 1.
2
2
The center is the midpoint of the vertices:
1
Because AÞC and B2 2 4AC < 0, the conic is an ellipse.
2
27 1 (21) 3 1 3
}, } 5 (24, 3)
2
2
2(x2 2 2x) 1 y 2 5 4
2(x 2 2x 1 1) 1 y 2 5 4 1 2
2
2(x 2 1) 2 1 y 2 5 6
}1}51
The vertices are 3 units from the center, so a 5 3 and
a2 5 9.
h 5 1, k 5 0
The foci are 5 units from the center, so c 5 5 and
c 2 5 25.
Center: (1, 0)
a2 5 6
a 5 Ï 6 ø 2.45
c2 5 a2 1 b2
b2 5 3
b 5 Ï3 ø 1.73
25 5 9 1 b2
Vertices: (1, Ï6 ) and (1, 2Ï6 )
(x 1 4)2
( y 2 3)2
}
y
3
(1 2
(1,
6)
3, 0)
(1, 0)
x
4
k50
The center of the ellipse is (5, 0).
Lines of symmetry: x 5 5 and y 5 0
8. (x 1 5)2 5 8( y 2 2)
k52
The vertex of the parabola is (25, 2).
Line of symmetry: x 5 25.
( y 2 2)2
(x 2 1)2
9. } 2 } 5 1
121
49
h51
}
Co-vertices: (1 1 Ï 3 , 0) and (1 2 Ï3 , 0)
y2
(x 2 5)2
7. } 1 } 5 1
16
64
h 5 25
}
}
An equation is }
2}
51
16
9
h55
}
}
b2 5 25 2 9 5 16
k52
The center of the hyperbola is (1, 2).
Lines of symmetry: x 5 1 and y 5 2
10. x 2 1 y 2 2 2x 1 4y 1 1 5 0
A 5 1, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(1)(1) 5 24
Because B 2 2 4AC < 0, B 5 0, and A 5 C, the conic is
a circle.
x 2 1 y 2 2 2x 1 4y 1 1 5 0
(x 2 2 2x) 1 ( y 2 1 4y) 5 21
(x 2 2x 1 1) 1 ( y 2 1 4y 1 4) 5 21 1 1 1 4
2
(x 2 1) 2 1 ( y 1 2) 2 5 4
h 5 1, k 5 22, r 2 5 4, r 5 2
508
y2
6
(x 2 1)2
3
h 5 24, k 5 3
Algebra 2
Worked-Out Solution Key
(1, 2 6 )
(1 1
3, 0)
12. y 2 2 4y 2 2x 1 6 5 0
A 5 0, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(0)(1) 5 0
Because B2 2 4AC 5 0, the conic is a parabola.
y 2 2 4y 2 2x 1 6 5 0
y 2 2 4y 5 2x 2 6
y 2 2 4y 1 4 5 2x 2 6 1 4
( y 2 2)2 5 2(x 2 1)
h 5 1, k 5 2
y
4p 5 2
(1, 2)
1
p 5 }2
3
2
1
Vertex: (1, 2)
1
3
Focus: }2, 2
( , 2)
2
1
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
B 2 2 4AC 5 0 2 4(2)(1) 5 28
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,