P. 629-630

P. 629-630
continued
Lesson 9.3
9.3 Exercises (pp. 629–632)
9.3 Guided Practice (pp. 627–628)
Skill Practice
2
2
1. x 1 y 5 9
r53
1. The radius of a circle is the distance from any point on
y
(0, 3)
r2 5 9
the circle to a fixed point called the circle’s center.
1
(3, 0)
1
x
(23, 0)
(0, 23)
y 2 5 2x 2 1 49
2.
(0, 7
)
x 2 1 y 2 5 49
2. A line tangent to a circle is perpendicular to the radius
at the point of tangency, so their slopes are negative
reciprocals of one another.
3. C; x 2 1 y 2 5 9
4. E; x 2 1 y 2 5 36
r2 5 9
r 2 5 36
y
r53
2
r56
2
2
6. D; x 1 y 2 5 6
2
r2 5 6
5. A; x 1 y 5 4
r 2 5 49
2
r57
r 54
(7, 0)
(27
, 0)
r ø 2.4
r52
x
22
7. F; x 2 1 y 2 5 16
8. B; x 2 1 y 2 5 3
r 2 5 16
r2 5 3
(0, 27
)
r ø 1.7
r54
3. x 2 2 18 5 2y 2
x 2 1 y 2 5 18
y
(23
r 2 5 18
2
(0, 3
2 , 0)
2
2
r 5 3Ï 2
2
10. x 1 y 5 81
r2 5 1
r 2 5 81
r51
r59
x
(3
(0, 23
2
y
2 , 0)
and (5, 21).
(0, 9)
0.5
3
(1, 0)
x
0.5
r 5 Ï 5 2 1 (21) 2
(0, 29)
11. x 2 1 y 2 5 25
12. x 2 1 y 2 5 12
r 2 5 25
r 2 5 12
}
r 5 Ï26
}
r 5 2Ï 3
r55
r 2 5 26
An equaton for the circle is x 2 1 y 2 5 26.
120
1
5. Slope of radius to point (6, 1): m 5 } 5 }
620
6
1
Slope of tangent line: 2}
5 26
m
(0, 5)
y
(0,
2
(25, 0)
2 3)
(5, 0)
(22
13. y 2 5 27 2 x 2
2
x
3, 0)
(0, 22
2
x 1 9 5 10
486
Algebra 2
Worked-Out Solution Key
}
r 5 3Ï 3
(0, 3
3)
x ø 64.4
You will be in the tower’s range from (4, 9) to (24.4, 9),
a distance of {4 2 (24.4){ 5 8.4 miles.
r 2 5 40
}
x 2 5 19
r 5 2Ï 10
y
(3
, 0)
3
(0, 2
(2
10, 0 )
2
x
22
, 0)
3
y
10 )
2
(23
)
x 2 1 y 2 5 40
r 2 5 27
2
3
14. x 2 5 2y 2 1 40
2
x 1 y 5 27
x 2 1 y 2 5 10 2
2
3, 0)
(0, 25)
y 2 1 5 26x 1 36
x 2 1 y 2 5 10 2.
(2
21
y 2 1 5 26(x 2 6)
y 5 26x 1 37
y
1
x
22
Equation of tangent line:
6. Find the point (x , 9) where x < 0 on the circle.
x
(29, 0)
(0, 21)
}}
(9, 0)
23
(21, 0)
r 5 Ï (5 2 0) 2 1 (21 2 0) 2
}
y
(0, 1)
)
4. The radius is the distance between the center (0, 0)
2
9. x 1 y 5 1
)
2
}
2
( 0, 23
3)
x
22
(22
10, 0 )
(0, 22
10 )
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,
Chapter 9,
continued
15. x 2 5 15 2 y 2
}
16. y 2 5 2x 2 1 9
}
2
}
2
x 2 1 y 2 5 15
x2 1 y 2 5 9
x 2 1 y 2 5 (5Ï 2 )
x 2 1 y 2 5 (4Ï6 )
r 2 5 15
r2 5 9
x 2 1 y 2 5 25(2)
x 2 1 y 2 5 16(6)
}
r 5 Ï15
(0,
2
r53
15 ) y
(
Center: (0, 0); Radius: 12
1
(3, 0)
x
(23, 0)
15, 0 )
2
17. 15x 1 15y 5 60
}
x 2 1 y 2 5 (6) 2
2
18. 7x 1 7y 5 112
x2 1 y 2 5 4
x 2 1 y 2 5 16
r2 5 4
r 2 5 16
r52
r54
y
x 2 1 y 2 5 36
32. r 5
}
x 2 1 y 2 5 52
x 2 1 y 2 5 25
1
(2, 0)
(24, 0)
x
21
1
33. r 5
(4
, 0)
x
21
(0, 22)
19. 4x 2 1 4y 2 5 128
(4
(0,
}
2, 0 )
35. r 5
(2
2
2
x
22
2
)
x
6, 0)
(0, 22
2
}
2
}}
(4 2 0) 2 1 (210 2 0) 2
Ï}
}
2
x 2 1 y 2 5 (Ï 116 )
2
x 2 1 y 2 5 144
x 2 1 y 2 5 64
x 2 1 y 2 5 116
38. r 5
25. r 5 16
x 2 1 y 2 5 22
x 2 1 y 2 5 16 2
x2 1 y2 5 4
x 2 1 y 2 5 256
}}
Ï(28 2 0)2 1 (25 2 0)2
}
}
r 5 Ï 64 1 25 5 Ï 89
}
2
x 2 1 y 2 5 (Ï 89 )
x 2 1 y 2 5 89
}
27. r 5 Ï 15
2
}}
Ï(29 2 0)2 1 (2 2 0)2
}
2
x 1y 58
}
36. r 5
r 5 Ï 16 1 100 5 Ï 116
x 1 y 5 12
26. r 5 Ï 2
x 2 1 y 2 5 100
37. r 5
23. r 5 8
}
x 2 1 y 2 5 10 2
x 2 1 y 2 5 85
}
24. r 5 2
}
x 2 1 y 2 5 (Ï 85 )
r 5 3Ï2
2
)
}
r 5 Ï 36 1 64 5 Ï100 5 10
}
r 2 5 18
2
6
}}
Ï(26 2 0)2 1 (8 2 0)2
}
x 2 1 y 2 5 18
2
2
r 5 Ï 81 1 4 5 Ï85
21. A; 3x 2 1 3y 2 5 54
22. r 5 12
6, 0)
22
(22
}
x 2 1 y 2 5 20
y
2 6)
}
x 2 1 y 2 5 (2Ï 5 )
r 5 2Ï6
y
}}
Ï(2 2 0)2 1 (24 2 0)2
}
}
r 5 4Ï 2
( 0, 24
}
r 5 Ï 4 1 16 5 Ï20 5 2Ï 5
r 2 5 24
}
2, 0 )
34. r 5
x 2 1 y 2 5 24
r 2 5 32
(24
}
r 5 Ï 16 1 9 5 Ï25 5 5
x 2 1 y 2 5 25
20. 8x 2 1 8y 2 5 192
x 2 1 y 2 5 32
)
}}
Ï(24 2 0)2 1 (3 2 0)2
x 2 1 y 2 5 52
(0, 24)
2
}}
Ï(0 2 0)2 1 (5 2 0)2
r 5 Ï 25 5 5
y
(0, 4)
(0, 2)
( 0, 4
Ï(26 2 0)2 1 (0 2 0)2
r56
2
(22, 0)
}}
r 5 Ï 36
(0,2 15 )
2
x 2 1 y 2 5 144
31. r 5
(0, 23)
(2
Equation: x 2 1 y 2 5 12 2
x
21
21
x 2 1 y 2 5 96
30. The radius should be squared.
(0, 3) y
15, 0 )
2
x 1 y 5 50
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
29. r 5 4Ï 6
28. r 5 5Ï 2
}
2
x 2 1 y 2 5 (Ï2 )
x 2 1 y 2 5 (Ï 15 )
x2 1 y2 5 2
x 2 1 y 2 5 15
}}
39. r 5 Ï (28 2 0) 2 1 (14 – 0) 2
}
}
r 5 Ï 64 1 196 5 Ï 260
}
x 2 1 y 2 5 (Ï 260 )2
x 2 1 y 2 5 260
Algebra 2
Worked-Out Solution Key
487
Chapter 9,
40. r 5
continued
}}
Ï(5 2 0)2 1 (212 2 0)2
}
50. 12x 2 1 12y 2 5 192
}
2
51. 2x 2 1 2y 2 5 16
x 1 y 5 16
x2 1 y2 5 8
x 2 1 y 2 5 13 2
r 2 5 16
r2 5 8
x 2 1 y 2 5 169
r54
r 5 Ï25 1 144 5 Ï169 5 13
41. r 5
}}}
Ï(211 2 0)2 1 (211 2 0)2
}
2
}
r 5 2Ï2
(0, 4) y
(0, 2
2
)
r 5 Ï121 1 121 5 Ï 242
}
2
(24, 0)
2
x 1 y 5 242
42. r 5
1
(22
(0, 22
2
)
(0, 24)
x 2 1 y 2 5 412
2
y
52. 6x 1 6y 5 0
x 2 1 y 2 5 1681
6y 2 5 26x
}}
Ï(4 2 0)2 1 (26 2 0)2
2
2
y 5 2x
}
r 5 Ï16 1 36 5 Ï52
}
2 , 0)
}
r 5 Ï81 1 1600 5 Ï1681 5 41
}
x
x
}}
}
2 , 0)
1
(4, 0)
1
Ï(9 2 0)2 1 (40 2 0)2
43. C; r 5
(2
1
x 2 1 y 2 5 (Ï242 )
2
y
}
1
x
2
x 1 y 5 (Ï52 )
2
2
x 2 1 y 2 5 52
45. 4x 2 1 y 5 0
2
r 5 49
4x 5 2y
1
x 2 5 2}4 y
r57
(0, 7)
420
53. Slope of radius to point (1, 4): m 5 } 5 4
120
1
1
Slope of tangent line: 2}
5 2}4
m
2
y
Equation of tangent line:
y
1
1
x
21
2
2
1
1
1
17
y 2 4 5 2}4 x 1 }4
(7, 0)
(27, 0)
y 2 4 5 2}4 (x 2 1)
x
y 5 2}4 x 1 }
4
(0, 27)
23 2 0
3
54. Slope of radius to point (2, 23): m 5 } 5 2}
220
2
46. 7x 2 1 7y2 5 63
2
(0, 3) y
1
x 1y 59
1
r2 5 9
(3, 0)
1
r53
Equation of tangent line:
x
2
y 2 (23) 5 }3 (x 2 2)
(23, 0)
(0, 23)
47. y 2 2 121 5 2x 2
2
2
5 }3
Slope of tangent line: 2}
m
2
y
(0, 11)
2
x 1 y 5 121
4
2
13
y 5 }3 x 2 }
3
(211, 0)
r 2 5 121
2
y 1 3 5 }3 x 2 }3
3
320
3
55. Slope of radius to point (25, 3): m 5 } 5 2}
5
25 2 0
(11, 0)
r 5 11
3
x
1
5
Slope of tangent line: 2}
5 }3
m
(0, 211)
2
48. x 1 16y 5 0
49. 3x 5 2y
x 2 5 216y
1
5
y 2 3 5 }3 (x 2 (25))
y 2 5 23x
x
2
25
5
34
y 5 }3 x 1 }
3
1
Algebra 2
Worked-Out Solution Key
5
y 2 3 5 }3 x 1 }
3
y
y
1
488
Equation of tangent line:
2
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
44. y 2 1 x 2 5 49
Chapter 9,
continued
22 2 0
1
56. Slope of radius to point (26, 22): m 5 } 5 }
26 2 0
3
Slope of segment from (x , y) to (r , 0):
}
Ïr 2 2 x2
y20
1
Slope of tangent line: 2}
5 23
m
m2 5 }
x 2 r 5}
x2r
Equation of tangent line:
}
m1m2 5 }
x1r + x2r
}
y 2 (22) 5 23(x 2 (26))
r 2 2 x2
5 }}
(x 1 r)(x 2 r)
y 1 2 5 23x 2 18
y 5 23x 2 20
(r 1 x)(r 2 x)
(x 1 r)(x 2 r)
5 }}
920
9
57. Slope of radius to point (25, 9): m 5 } 5 2}
5
25 2 0
2(x 1 r)(x 2 r)
5 }}
(x 1 r)(x 2 r)
5
1
Slope of tangent line: 2}
5 }9
m
5 21
Equation of tangent line:
Because m1 and m2 are negative reciprocals, the line
segments meet at a right angle.
5
y 2 9 5 }9 (x 2 (25))
61.
y
5
25
y 2 9 5 }9 x 1 }
9
5
(x, y)
106
y 5 }9 x 1 }
9
1
Let x 2 1 y 2 5 r 2 represent the equation of the lefthand
}
}
circle. Because AB bisects OS, the x-coordinate of
the point
Slope of tangent line: 2}
5 23
m
Equation of tangent line:
1
y 2 5 5 23(x 2 15)
(x , y) is equal to }2 r.
y 2 5 5 23x 1 45
1
When x 5 }2 r:
y 5 23x 1 50
59. The circles must lie between the circles that pass through
2
1 }12 r 2
(23, 5) and (26, 2).
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
r
520
1
58. Slope of radius to point (15, 5): m 5 } 5 }
15 2 0
3
Circle that passes through (23, 5):
* 5 2y
1 y2 5 r2
}
1 Ï3 2
1
4
*52 }
r
2
3
9 1 25 5 r 2
}
* 5 Ï3 r
y 2 5 }4 r 2
}
Ï34 5 r
}
Ï3
When y 5 }
r:
2
} r2 1 y2 5 r2
(23) 2 1 5 2 5 r 2
}
Ï3
Circle that passes through (26, 2):
r
y 5 6}
2
(26) 2 1 22 5 r 2
Problem Solving
36 1 4 5 r 2
62. Perimeter of tower’s range: x 2 1 y 2 5 15 2
}
Ï40 5 r
Region covered by the tower: x 2 1 y 2 a 15 2
15 2
9 2 1 11 2a
So, circles between these two circles have radii r such
that 34 < r 2 < 40.
225
81 1 121a
Sample answers:
x 2 1 y 2 5 35
202a225 x 2 1 y 2 5 36
x 2 1 y 2 5 38
Because the point (9, 11) falls within the radius of the
circle, you are in the region served by the tower.
63. Perimeter of feeding range: x 2 1 y 2 5 50 2
60. x 2 1 y 2 5 r 2
y 2 5 r 2 2 x2
}
y 5 Ïr 2 2 x2
Slope of segment from (2r, 0) to (x , y):
y20
x 2 (2r)
}
Ïr 2 2 x2 Ïr 2 2 x2
y
}
Ïr 2 2 x2
}
m1 5 } 5 }
x1r5 x1r
Feeding range: x 2 1 y 2 a 50
(225) 2 1 40 2a50 2
625 1 1600a2500
2225 a0 Because the point (225, 40) is within the radius of the
circle, the location is within the bats’ feeding range.
Algebra 2
Worked-Out Solution Key
489
Chapter 9,
continued
64. D; Perimeter of delivery area: x 2 1 y 2 5 100 2
67. The sprinklers should be placed so that their circular
regions intersect at the edge of the flower bed, when
y 5 4.5.
Free delivery area: x 2 1 y 2 a 1002
1002
752 1 702a
x 2 5 36 2 20.25
Because the point (75, 70) is not within the radius of the
circle, the house located at (75, 70) is located outside the
free delivery area.
65. a. Points A and F :
x 2 5 209
20 x
x ø 614.5
AF 5 {14.5 2 (214.5){
A B
y 5 24
5 29
C D
220
y 4.5
3
x
3
x ø 63.97
x2 y2 62
x 2 1 16 5 225
x 2 5 15.75
(x, 4.5)
y
x 2 1 y 2 5 225 x 2 1 y 2 5 25
x 2 1 (24) 2 5 15 2
y
x 2 1 (4.5) 2 5 62
10,525 ÷ 10,000
The distance between the sprinklers should be about
2(3.97) 5 7.9 feet.
68. a.
40
E F
y
x 2 1 y 2 5 100
40 x
The plane will be in the top most layer for about
29 miles.
b. Points B and E:
x 2 1 (24) 2 5 10 2
b. Rating of at least 6.0: x 2 1 y 250
x 2 1 16 5 100
Rating of at least 5.7: x 2 1 y 2a225
x 2 5 84
Rating of at least 5.4: x 2 1 y 2a625
x ø 69.2
Rating of at least 5.1: x 2 1 y 2a1225
BE 5 {9.2 2 (29.2){ 5 18.4
c. When x 5 212 and y 5 216:
The plane will be above the middle layer for about
18.4 miles.
(212) 2 1 (216) 2 5 r 2
144 1 256 5 r 2
c. Points C and D:
Because 225 < 400 < 625, the rating will be between
5.4 and 5.7.
x 2 1 16 5 25
2
x 59
69. 40-mile transmitter, when y 5 10:
x 5 63
Find (x, 10) where x > 0 on the circle x 2 1 10 2 5 40 2.
CD 5 {3 2 (23){ 5 6
x 2 5 1600 2 100
The plane will be above the lowest layer for 6 miles.
x 2 5 1500
66. The tunnel’s diameter is 8 feet, so its radius is 4 feet, and
an equation for its cross-section is x 2 1 y 2 5 4 2.
Half the width of the walkway is 3 feet, so let x 5 3.
60-mile transmitter, when y 5 10:
Find (x, 10) where x < 0 on the circle x 2 1 10 2 5 60 2.
32 1 y 2 5 42
x 2 5 3600 2 100
y
2
y 5 16 2 9 5 7
x 2 5 3500
y 5 62.65
x ø 259.2
4 ft
x
y ft
Walkway
(x, y)
6 ft
The height of the tunnel at the center of the walkway is
about 4 1 2.65 5 6.65 feet, or about 79.8 inches. Yes, a
worker who is 6 feet 2 inches, or 74 inches, could walk
down the center of the walkway without ducking.
490
x ø 38.7
Algebra 2
Worked-Out Solution Key
The 60-mile transmitter
is 80 miles away from
40-mile transmitter.
When you enter the
60-mile transmitter range,
you will be at a point
whose x-coordinate is
80 2 59.2 5 20.8.
y
(20.8, 10)
20
(38.7, 10)
60
You will be in range of both transmitters for about
38.7 2 20.8 5 17.9 miles.
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
400 5 r 2
x 2 1 (24) 2 5 5 2
Chapter 9,
continued
3. (25, 1), (24, 8)
At a speed of 60 miles per hour:
}}}
d 5 Ï (25 2 (24)) 2 1 (1 2 8) 2
d 5 rt
17.9 5 60 t
}}
5 Ï (21) 2 1 (27) 2 5 Ï1 1 49
60 min
t ø 0.3 h + } 5 18 min
h
}
}
}
5 Ï 50 5 5Ï2
You will be in range of both transmitters for about
18 minutes.
1
2
25 1 (24)
118
9
M }
,}
5 1}
, } 5 (24.5, 4.5)
2
2 22
2
29
4. (1, 2), (7, 1)
Mixed Review for TAKS
}}
d 5 Ï (1 2 7) 2 1 (2 2 1) 2
70. D;
y
}}
5 Ï (26) 2 1 12 5 Ï36 1 1 5 Ï37
2
1
}
}
3
211
,}
5 4, } 5 (4, 1.5)
M1 }
2
2 2 1 22
117
x
8x 2 2y 5 18
5. (26, 25), (21, 8)
}}}
212x 1 3y 5 227
d 5 Ï (26 2 (21)) 2 1 (25 2 8) 2
}}
Because the graphs of the equations are the same line,
there is an infinite number of solutions.
71. H;
1
2
26 1 (21)
}
25 1 8
3
M }
,}
5 1}
, } 5 (23.5, 1.5)
2
2 22
2
27
}}
d 5 Ï (3 2 6) 2 1 (22 2 5) 2
}
5 Ï52 122
}}
5 Ï (23) 2 1 (27) 2 5 Ï9 1 49 5 Ï58
}
5 Ï 29
}}
}
}}
RS 5 Ï(21 2 4)2 1 (24 2 2)2
}}
5 Ï(25) 1 (26)
2
2
}
5 Ï 61
PS 5 Ï(21 2 (23)) 1 (24 2 1)
2
7. Focus: (0, 3); p 5 3
5 Ï22 1 (25)2
y 2 5 4px
x 2 5 4(3) y 5 12 y
y 2 5 4(22) x 5 28x
x 2 5 4py
2
x 2 5 4(24)y 5 216y
11. Focus: (0, 5); p 5 5
}
}
}
}
}
Perimeter P 5 Ï 29 1 Ï5 1 Ï 61 1 Ï 29
ø 5.4 1 2.2 1 7.8 1 5.4
5 20.8 units
Quiz 9.1–9.3 (p. 632)
y 2 5 4px
x 2 5 4(5) y 5 20y
y 2 5 4(21) x 5 24x
2
2
13. x 1 y 5 4
14. x 2 1 y 2 5 64
r2 5 4
r 2 5 64
r52
r58
(0, 8) y
(0, 2) y
}}}
d 5 Ï (4 2 8) 1 (23 2 (27))
2
2
}}
}
12. Focus: (21, 0); p5 21
x 2 5 4py
1. (4, 23), (8, 27)
5 Ï (24) 1 (4)
10. Focus: (0, 24); p 5 24
2
y 5 4(6) x 5 24x
5 Ï 29
2
8. Focus: (22, 0); p 5 22
x 2 5 4py
y 5 4px
}
1
(22, 0)
2
}
}
5 Ï16 1 16 5 Ï 32 5 4Ï 2
23 1 (27)
}
9 3
9. Focus: (6, 0); p 5 6
}}}
2
}
M 1}
,}
5 }, } 5 (4.5, 1.5)
2
2 2 12 22
3 1 6 22 1 5
}
QR 5 Ï(4 2 2)2 1 (2 2 3)2 5 Ï22 1 (21)2 5 Ï5
1 418
}
6. (3, 22), (6, 5)
}}
PQ 5 Ï(2 2 (23))2 1 (3 2 1)2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5 Ï (25) 2 1 (213) 2 5 Ï25 1 169 5 Ï194
(2, 0)
1
x
(28, 0)
2
(0, 22)
(8
, 0)
2
x
2
M }
,}
5 (6, 25)
2
2
2. (22, 5), (4, 9)
(0, 28)
}}
d 5 Ï (22 2 4) 2 1 (5 2 9) 2
}}
5 Ï (26) 2 1 (24) 2
}
}
}
5 Ï36 1 16 5 Ï 52 5 2Ï13
M 1}
,}
5 (1, 7)
2
2 2
22 1 4 5 1 9
Algebra 2
Worked-Out Solution Key
491
Chapter 9,
continued
15. x 2 1 y 2 5 20
16. x 2 1 y 2 5 75
2
r 5 20
r 5 75
}
}
r 5 5Ï 3
r 5 2Ï 5
(0, 2
5
)
y
(0, 5
(2
2
5 , 0)
5
(5
x
(25
3 , 0)
(0, 25
x 2 1 y 2 5 18
2
2
r 5 16
(0, 3
2
5. 7x 2 1 7y 2 5 105
4. 0.5x 2 1 0.5y 2 5 12
x 2 1 y 2 5 24
x 2 1 y 2 5 15
y
)
1
(4, 0)
x
(23
2 , 0)
(3
2 , 0)
1
x
(0, 23
(0, 24)
19. Jupiter’s orbit:
r 5x 1y
3
2
144
96
}
1
2
Xmax 2 Xmin
Ymax 2 Ymin
}} 5 } 5 }
)
r 5 3Ï 2
(0, 4) y
2
272axa72, 248aya48
r 5 18
r54
1
3
18. 6x 2 1 6y 2 5 108
x 2 1 y 2 5 16
(24, 0)
3 , 0)
23
)
17. 3x 2 1 3y 2 5 48
y
)
3
x
(0, 22
3
5 , 0)
2
(22
3. x 2 1 y 2 5 576
2
2
)
29axa9, 26aya6
29axa9, 26aya6
Xmax 2 Xmin
18
3
}} 5 } 5 }
12
2
Ymax 2 Ymin
}} 5 } 5 }
Xmax 2 Xmin
Ymax 2 Ymin
18
12
3
2
6. 16x 2 1 16y 2 5 9
9
x2 1 y2 5 }
16
2
r 2 5 (350) 2 1 (370) 2
r 2 5 122,500 1 136,900
r ø 509
Because 509 < 650, KY Cygni would contain
Jupiter’s orbit.
23axa3, 22aya2
Graphing Calculator Activity 9.3 (p. 633)
}} 5 } 5 }
1. x 2 1 y 2 5 144
Xmax 2 Xmin
Ymax 2 Ymin
6
4
3
2
Lesson 9.4
9.4 Guided Practice (pp. 635–636)
y2
x2
1. } 1 } 5 1
9
16
y
a 2 5 16
(0, 3)
b2 5 9
(4, 0)
}
224axa24, 216aya16
Vertices: (6Ï16 , 0) 5 (64, 0)
Xmax 2 Xmin
Ymax 2 Ymin
Co-vertices:
}
(0, 6Ï9 ) 5 (0, 63)
48
32
3
2
}} 5 } 5 }
c 2 5 a 2 2 b 2 5 16 2 9 5 7
2. x 2 1 y 2 5 80
}
c 5 Ï7
}
Foci: (6Ï7 , 0)
215axa15, 210aya10
Xmax 2 Xmin
Ymax 2 Ymin
30
20
3
2
}} 5 } 5 }
492
Algebra 2
Worked-Out Solution Key
1
(2
7, 0)
1
(
7, 0)
(24, 0)
(0, 23)
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
r 2 5 259,400
P. 637-639
Chapter 9,
continued
2
y
x2
2. } 1 } 5 1
49
36
(0, 7)
a 2 5 49
b 2 5 36
y
(26, 0)
2 (0, 13 )
Vertices: (0, 67)
Co-vertices: (66, 0)
2
2
2
(6, 0)
x
2
c 5 a 2 b 5 49 2 36 5 13
}
c 5 Ï13
(0, 2
}
Foci: (0, 6Ï13 )
13 )
Skill Practice
y
9y 2
225
1. An ellipse is the set of all points P such that the sum of
(0, 5)
}1}51
the distances between P and two fixed points, called the
foci, is a constant.
(0, 4)
y2
25
x2
9
2. If the vertices of an ellipse are located at (6a, 0) or
}1}51
1
a 2 5 25
b2 5 9
(23, 0)
Vertices: (0, 65)
c 2 5 25 2 9 5 16
(0, 25)
}
c 5 Ï16 5 4
Foci: (0, 64)
4. Vertex: (7, 0); Co-vertex: (0, 2)
a57
a 2 5 49
b52
b2 5 4
y2
4
}1}51
5. Vertex: (0, 6); Co-vertex: (25, 0)
a 2 5 36
a56
b 2 5 25
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b 5 25
2
y
x
}1}51
36
25
2
6. Vertex: (0, 8); Focus: (0, 23)
a58
a 2 5 64
c 5 {23{ 5 3
c2 5 9
c2 5 a2 2 b2
9 5 64 2 b 2
7. Vertex: (25, 0); Focus: (3, 0)
9 5 25 2 b
a 2 5 25
2
c 59
c2 5 a2 2 b2
2
2
b 5 25 2 9 5 16
y2
x2
}1}51
16
25
y2
x2
3. } 1 } 5 1
4
16
a 2 5 16
a54
b2 5 4
b52
c 2 5 16 2 4 5 12
}
c 5 2Ï3
Vertices: (64, 0)
Co-vertices: (0, 62)
}
Foci: (62Ï3 , 0)
y
(0, 2)
(24, 0)
1
(22
3, 0)
y2
x2
4. } 1 } 5 25
1
4
x2
100
(2
(4, 0)
3, 0)
1
x
(0, 22)
y
y2
25
(0, 5)
(210, 0)
}1}51
a 2 5 100
a 5 10
b 2 5 25
b55
c 2 5 100 2 25 5 75
}
}
c 5 Ï75 5 5Ï3
Vertices: (610, 0)
Co-vertices: (0, 65)
2
(10, 0)
(5 3, 0)
(25 3, 0)
2
x
(0, 25)
y2
x2
5. } 1 } 5 1
49
9
y2
64
c53
(0, 6a) and the co-vertices are located at (0, 6b) or
(6b, 0), you know that c2 5 a2 2 b2, so you can
determine the value of c. The foci will then be located
at (6c, 0) or (0, 6c).
}
}1}51
a 5 {25{ 5 5
x
Foci: 1 65Ï3 , 0 2
b 2 5 64 2 9 5 55
x2
55
(3, 0)
1
(0, 24)
Co-vertices: (63, 0)
x2
49
The area is A 5 :(175)(125) ø 68,722 square meters.
9.4 Exercises (pp. 637–639)
(0, 27)
3. 25x 2 1 9y 2 5 225
25x 2
225
350
8. The major axis is vertical, with a 5 } 5 175 and
2
250
}
b 5 2 5 125.
y2
y2
x2
x2
}2 1 }2 5 1, or } 1 } 5 1
30,625
15,625
175
125
(0, 7)
2
a 5 49
a57
b2 5 9
b53
y
(0, 2
2
c 2 5 49 2 9 5 40
10 )
(3, 0)
2
x
}
c 5 2Ï10
(23, 0)
Vertices: (0, 67)
Co-vertices: (63, 0)
(0, 22
10 )
(0, 27)
}
Foci: 1 0, 62Ï10 2
Algebra 2
Worked-Out Solution Key
493
continued
2
y
x2
6. } 1} 5 1
144 64
11. 16x 2 1 9y 2 5 144
y
a 2 5 144
a 5 12
b 2 5 64
b58
(0, 8)
(24 5 , 0)
(12, 0)
3
2
c 5 144 2 64 5 80
(4 5 , 0)
}
c 5 4Ï5
(212, 0)
(0, 28)
Vertices: (612, 0)
Co-vertices: (0, 68)
}
Foci: 1 64Ï5 , 0 2
y2
x2
7. } 1 } 5 1
81
400
y
2
a 5 400
a 5 20
b 2 5 81
b59
4
2
c 5 400 2 81 5 319
c 5 Ï319
Vertices: (620, 0)
(
319, 0)
x
(2
319, 0)
(0, 29)
Co-vertices: (0, 69)
}
Foci: (6Ï 319 , 0)
y2
x2
8. } 1 } 5 1
225
36
(0, 3
a 2 5 225
y
21 )
(0, 15)
2
b 5 36
c 5 225 2 36 5 189
}
c 5 3Ï21
3
(26, 0)
(6, 0)
Vertices: (0, 615)
Co-vertices: (66, 0)
}
Foci: (0, 63Ï 21 )
(0, 23
21 )
(0, 215)
9. 4x 2 1 y 2 5 36
x2
9
y
y2
36
(0, 6)
}1}51
2
a 5 36
2
b 59
a56
b53
2
c 5 36 2 9 5 27
(27, 0)
(0, 3
2
(23, 0)
3)
a 5 49
a57
b 2 5 25
b55
c 2 5 49 2 25 5 24
}
c 5 2Ï 6
Vertices: (67, 0)
Co-vertices: (0, 65)
}
Foci: (62Ï6 , 0)
2
(0,23
(0, 26)
c 5 3Ï3
a 2 5 81
a59
b2 5 9
b53
c 2 5 81 2 9 5 72
Co-vertices: (63, 0)
c 5 6Ï 2
Foci: (0, 63Ï 3 )
y2
9
(0, 2
a 59
a53
b51
c2 5 9 2 1 5 8
}
c 5 2Ï2
Vertices:(0, 63)
(21, 0)
1
2)
(1, 0)
x
22
(0, 22
2)
(0, 23)
}
x
2)
y2
x2
15. In the equation } 1 } 5 1, the value of a2 is 16
16
4
and corresponds to the y-axis, not the x-axis.
Therefore, the vertices should be located at (0, 64)
and the co-vertices should be located at (62, 0).
y
1
(22, 0)
(2, 0)
x
2)
(3, 0)
Foci: 1 0, 66Ï 2 2
(0, 24)
Algebra 2
Worked-Out Solution Key
(0, 6
(0, 29)
21
494
2
y
24
(0, 26
(0, 4)
x
(0, 28)
(23, 0)
Co-vertices: (61, 0)
Foci: (0, 62Ï 2 )
(6, 0)
(26, 0) 22
}
(0, 3)
}1}51
(10, 0)
(210, 0) 2
Co-vertices: (63, 0)
y
b2 5 1
(0, 25)
(0, 9)
Vertices: (0, 69)
10. 9x 2 1 y 2 5 9
x
y
}
}
6, 0)
(0, 8)
Vertices: (0, 66)
2
6, 0)
y2
x
}1}51
81
9
3)
}
x2
1
(22
2
x
7)
(7, 0)
(2
22
14. 72x 2 1 8y 2 5 648
(3, 0)
(0, 2
2
2
a 2 5 100
a 5 10
2
b 5 64
b58
c 2 5 100 2 64 5 36
c56
Vertices: (610, 0)
Co-vertices: (0, 68)
Foci: (66, 0)
x
23
x
y
y2
x2
}1}51
64
100
b56
2
(3, 0)
(0, 5)
13. 16x 2 1 25y 2 5 1600
a 5 15
7)
1
(0, 24)
y2
x
}1}51
25
49
(20, 0)
(0,
(23, 0)
2
24
}
1
12. 25x 2 1 49y 2 5 1225
(0, 9)
(220, 0)
(0, 4)
a 2 5 16
a54
2
b 59
b53
c 2 5 16 2 9 5 7
}
c 5 Ï7
Vertices: (0, 64)
Co-vertices: (63, 0)
}
Foci: (0, 6Ï7 )
x
3
y
y2
x2
}1}51
16
9
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,
Chapter 9,
continued
2
y
}
x2
16. In the equation } 1 } 5 1, a2 5 3 and a 5 Ï 3 .
3
2
23. Vertex: (0, 8)
}
Therefore, the vertices should be located at (0, 6}Ï3 )
instead of (0, 63). Likewise, b2 5 2 and}b 5 Ï2 , so
the co-vertices should be located at (6Ï 2 , 0) instead
of (62, 0).
( 0,
y
3)
a58
a 2 5 64
c56
c 2 5 36
36 5 64 2 b 2
b 2 5 64 2 36 5 28
y2
64
x2
28
1
(
Focus: (0, 6)
}1}51
2, 0 )
1
x
24. Vertex: (4, 0)
25. Vertex: (0, 9)
}
}
Focus:(Ï 7 , 0)
(2
2, 0 )
( 0,2
}
c 57
7 5 16 2 b 2
b 5 16 2 7 5 9
a55
a 2 5 25
b53
b2 5 9
y2
9
}1}51
y2
9
x2
16
c53
}
a 2 5 25
2
c 59
a 2 5 100
9 5 25 2 b 2
b56
b 2 5 36
b 5 25 2 9 5 16
x2
36
y2
100
b59
b 2 5 81
y2
81
}1}51
b54
b 2 5 16
y2
x2
}1}51
36
16
a 2 5 144
2
b 5 121
x2
16
30. Co-vertex: (23Ï 5 , 0)
Focus: (0, 6)
}
b 5 3Ï 5
x2
45
2
b 5 256
b 2 5 175
2
c 5 15
c 2 5 225
y2
81
x2
400
}
33. Co-vertex: (2Ï 15 , 0)
Focus: (0, 14)
}
2
b 5 2Ï15
b 2 5 60
2
c 5 14
c 2 5 196
b 5 225
c 5 64
2
y2
225
y2
175
}1}51
64 5 a 2 2 225
x2
289
225 5 a 2 2 175
400 5 a 2
} 1} 5 1
289 5 a
}
b 5 5Ï7
b 5 45
2
c58
a 2 5 400
}
31. Co-vertex: (0, 25Ï 7 )
Focus: (215, 0)
c 5 36
b 5 15
Co-vertex: (0, 216)
y2
7
2
Focus: (28, 0)
22. Vertex: (20, 0)
16 5 a 2
}1}51
32. Co-vertex: (0, 15)
y2
x
}1}51
144
121
y2
256
y2
121
81 5 a
2
x2
400
9 5 a2 2 7
36 5 a 2 2 45
Co-vertex: (11, 0)
}1}51
c2 5 9
48 5 169 2 b 2
c56
21. Vertex: (0, 12)
b 5 16
c53
c 5 48
}
a 2 5 36
a 5 20
b2 5 7
2
}1}51
a56
b 5 11
c 5 4Ï 3
}
b 5 Ï7
b 5 169 2 48 5 121
Co-vertex: (4, 0)
a 5 12
}
Focus: (23, 0)
2
a 5 169
a 5 13
x2
169
y2
16
}
2
20. Vertex: (0, 26)
b 2 5 16 2 12 5 4
29. Co-vertex: (0, Ï 7 )
}
a 2 5 196
c 2 5 12
c 5 2Ï 3
x2
4
Focus: (24Ï3 , 0)
a 5 14
}
}1}51
28. Vertex: (13, 0)
Co-vertex: (0, 29)
x2
196
y2
16
} 1} 5 1
19. Vertex: (14, 0)
a 2 5 16
a54
12 5 16 2 b 2
2
x2
25
y2
81
Focus: (0, 22Ï3 )
a 5 10
}1}51
b 2 5 81 2 32 5 49
27. Vertex: (0, 24)
Focus: (3, 0)
Co-vertex: (6, 0)
c 2 5 32
c 5 4Ï 2
}1}51
26. Vertex: (25, 0)
a55
}
x2
49
}1}51
18. Vertex: (0, 210)
a 2 5 81
a59
32 5 81 2 b 2
2
Co-vertex: (0, 23)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
c 5 Ï7
17. Vertex: (5, 0)
x2
25
a 2 5 16
a54
3)
Focus: (0, 24Ï2 )
}1}51
196 5 a 2 2 60
256 5 a 2
x2
60
y2
256
}1}51
Algebra 2
Worked-Out Solution Key
495
continued
34. Co-vertex: (232, 0)
Focus: (0, 24)
Co-vertex: (28, 0)
b 2 5 1024
b 5 32
2
c 5 24
4y
x2
41. } 1 } 5 0
25
75
35. B; Vertex: (0, 212)
c 5 576
576 5 a 2 2 1024
1600 5 a 2
a 2 5 144
a 5 12
2
b58
2
36. x 1 y 5 64
Focus: (0, 23)
Directrix: y 5 3
(0, 8) y
4y 2
3x 2
42. } 1 } 5 1
400
48
The equation is a circle.
r 2 5 64
(28, 0)
r58
2
x2
16
(0, 10) y
2
y
100
}1}51
(8, 0)
x
22
The equation is an ellipse.
a 2 5 100
2
(0, 28)
37. 25x 2 1 81y 2 5 2025
y2
25
x2
81
(0, 5)
}1}51
The equation is an ellipse.
b 5 16
x
b54
y2
x2
43. } 1 } 5 4
64
64
(0, 16) y
x 2 1 y 2 5 256
The equation is a circle.
y
y59
The equation is a parabola.
3
(0, 216)
44. 16 x 2 1 10 y 2 5 160
Directrix: y 5 9
x5
The equation is a parabola.
1
22
(0, 4) y
y2
x2
}1}51
16
10
y
y 2 5 2x
(
1,
2
0
)
2
a 5 16
x
21
1
The equation is an ellipse.
4p 5 2
b 2 5 10
(
10, 0)
x
21
a54
}
b 5 Ï 10
(2
1
p 5 }2
10, 0)
(0, 24)
2
y
x2
45. } 1 } 5 1
25
9
2
The conic changes from an
ellipse elongated along the
y-axis to a circle to an
ellipse elongated along
the x-axis.
1
Directrix: x 5 2}2
40. 30x 2 1 30y 2 5 480
x 2 1 y 2 5 16
(0, 4) y
The equation is a circle.
2
r 5 16
r54
(24, 0)
1
(4, 0)
x
21
(0, 24)
496
x
(0, 29)
39. 65y 2 5 130x
(16, 0)
24
r 5 16
Focus: (0, 29)
1
4
x
3
4p 5 236
1
Focus: }2, 0
(216, 0)
r 2 5 256
p 5 29
x
(0, 210)
(0, 25)
x 5 236y
(4, 0)
22
(9, 0)
22
38. 36y 1 x 5 0
2
(24, 0)
a 5 10
y
2
(29, 0)
2
2
(0, 23)
p 5 23
}1}51
2
x
1
4p 5 212
y
1600
x
1024
1
The equation is a parabola.
}1}51
2
2
y53
4
x2
} 5 2} y
25
75
x 2 5 212y
b 5 64
y2
144
x2
64
y
Algebra 2
Worked-Out Solution Key
y
x2
y2
1
51
9
25
1
1
x2
y2
1
51
9
4
x2
y2
1
51
9
9
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 9,
Chapter 9,
continued
y2
x2
46. Sample answer: } 1 } 5 1
49
64
49. Largest field:
Rewrite the equation as two functions:
y2
49
x2
64
}512}
49x2
49x
A viewing window that does not distort the shape of the
ellipse is 212axa12 and 28aya8. The length of
2
Sum of distances from (a, 0) to the two foci:
}}
5 (a 2 c) 1 (a 1 c) 5 2 a
}}
}}
5 Ï c 1 b 1 Ï c 1 b 5 2Ï c 1 b
2
a 5 14
b58
a2 5 196
b2 5 64
x2
196
0) 2 1 (0 2 b) 2 1 Ï (c 2 0) 2 1 (0 2 b) 2
Ï(c 2}
}
}
2
2b 5 16
y2
64
}1}51
Sum of the distances from (0, b) to the two foci:
c 2 5 a2 2 b2
c 2 5 196 2 64 5 132
2
}
c 5 2Ï 33
By definition of an ellipse, the two sums of the distances
are equal.
}
Foci: 1 62Ï 33 , 0 2
}
2a 5 2Ï c 2 1 b 2
}
}
Distance between the foci 5 21 2Ï33 2 5 4Ï 33
}
}
a 5 Ïc2 1 b2
a 5c 1b
b 5 55
50. 2a 5 28
}}
Ï(a 2 c)2 1 (0 2 0)2 1 Ï(a 1 c)2 1 (0 2 0)2
2
2b 5 110;
An inequality for the possible areas (in square meters) of
the fields is 11,700aAa22,500.
Foci: (c, 0) and (2c, 0)
2
a 5 67.5
A 5 :(67.5)(55) ø 11,663
y2
x2
47. }2 1 }2 5 1
b
a
2
2a 5 135;
y2
x2
} 1 }2 5 1
2
(67.5)
(55)
the y-axis is }3 the length of the x-axis.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b 5 77.5
Smallest field:
2
y 5 6 49 2 }
64
2
2b 5 155;
A 5 :(77.5)(92.5) ø 22,521
}
2
a 5 92.5
y2
x2
} 1 }2 5 1
2
(92.5)
(77.5)
y2 5 49 2 }
64
Î
2a 5 185;
The foci are about 4Ï33 centimeters apart.
0.59
y
51.
2
Sun
a2 2 b2 5 c2
35.3
y
(0, b)
x
(2a, 0)
Comet
(a, 0)
(2c, 0)
(c, 0)
x
(0, 2b)
The sum of the two distances is 2a.
2a 5 35.3 1 0.59
a 5 17.945
Problem Solving
48. Major axis: 81 km; minor axis: 12 km
y
(− 40.5, 0)
c 2 5 a2 2 b2
(40.5, 0)
(0, 6)
(0, − 6)
x
The distance from the origin to a focus c is about
17.945 2 0.59 5 17.355.
12 km
81 km
2
a 5 40.5
a 5 1640.25
b56
b 2 5 36
y2
x2
}1}51
36
1640.25
A 5 :F
ab 5 :F
(40.5)(6) ø 763
b2 5 a2 2 c 2
b2 5 (17.945) 2 2 (17.355) 2
b2 ø 322.023 2 301.196
b2 ø 20.827
Sample answer:
x2
y2
1}
5 1.
An equation for the orbit is }
322
20.8
The area of the ellipse is about 763 square kilometers.
Algebra 2
Worked-Out Solution Key
497
Chapter 9,
52.
continued
55. G;
y
1
A 5 }2bh
A
(− 225, 0)
B
(225, 0)
1
45m7n13 5 }2b(15m10n9)
x
15
45m7n13 5 }
m10n9b
2
450 mi.
2 + 45m7n13
15 + m n
6n4
}
5b
m3
}
5b
10 9
a. The sum of the distances from any point on the ellipse
to each airport remains constant, in this case 600 miles.
}
b. The length of AB is 450 miles.
Foci: (6225, 0)
Problem Solving Workshop 9.4 (p. 640)
The airports’ coordinates are (6225, 0).
This is equal to the length of the major axes, 2a.
2a 5 600
a 5 300
Vertices: (6300, 0)
c 2 5 a2 2 b2
d.
(225) 2 5 (300) 2 2 b2
b2 5 90,000 2 50,625
b2 5 39,375
y2
x2
An equation for the ellipse is }
1}
5 1.
39,375
90,000
53.
As the painting slides down the
wall, the two triangles remain similar.
So, corresponding side lengths
are proportional.
y
2
x
(x, y)
6
y
x
d
d
6
x
2
}5}
1. y 5
2
x1 5 0
x2 5 20
x3 5 40
x4 5 60
x5 5 80
x6 5 100
x7 5 120
x8 5 140
x9 5 160
x10 5 180
x2
4
}
y1 5 100
y2 ø 99.5
y3 ø 98.0
y4 ø 95.4
y5 ø 91.7
y6 ø 86.6
y7 5 80
y8 ø 71.4
y9 5 60
y10 ø 43.6
A ø 20( y1 1 y2 1 y3 1 y4 1 y5 1 y6 1 y7 1
y8 1 y9 1 y10) ø 20(826.2) 5 16,524
An estimate for the area of the ellipse is about
4(16,524) 5 66,096 m2.
This is a better estimate; more rectangles means less area
that extends beyond the ellipse.
2. Use rectangles whose widths are smaller than the
previous set of rectangles to obtain a closer and closer
approximation of the ellipse’s area.
2d 5 6x
d 5 3x
3. a. 2a 5 250; a 5 125
Using the Pythagorean Theorem:
2b 5 200; b 5 100
d 2 1 y 2 5 62
2
2
x2
4
y2
36
x2
125
y2
An equation of the ellipse is }2 1 }2 5 1.
(3x)2 1 y 2 5 36
9x 1 y 5 36
}1}51
x2
y2
An equation for the path of point (x, y) is }
1}
5 1.
36
4
Mixed Review for TAKS
54. C;
Volume of cylinder 5 :r 2h
5 :(8.5)2(18)
ø 4086 in.3
Volume of 1 candle 5 *wh 5 3(2)(5) 5 30
4086
Number of candles 5 }
ø 136
30
She can make 136 candles.
498
Î100 2
}
c. By definition of an ellipse, the plane flies 600 miles.
Algebra 2
Worked-Out Solution Key
y2
16x2
b. } 1 } 5 1002
1
25
16x 2
2
y 5 1002 2 }
25
Î
100
}
16x2
y 5 1002 2 }
25
Sample answer:
x1 5 0
y1 5 100
x2 5 25
y2 ø 98.0
x3 5 50
y3 ø 91.7
x4 5 75
y4 5 80
x5 5 100 y5 5 60
A ø 25 ( y1 1 y2 1 y3 1 y4 1 y5)
ø 25(429.7) ø 10,743
An estimate for the area of the ellipse is about
4(10,743) 5 42,972 square meters.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
2
} (450) 5 225
6n4
m
The length of the triangle’s base is }3 units.
Chapter 9,
continued
Mixed Review for TEKS (p. 641)
4. G;
e
m
o
H
1. A;
x2 5 4py
i
m
8
(11.2)2 5 4p(6)
5.2 ø p
i
m
6
1
4m
i
The focus is about 5.2 inches from the vertex.
ea
B
ch
Total driving: 16 miles east, 12 miles south
2. G;
}
x 2 1 y 2a162
2
5. C;
y
An inequality describing the region covered by the radar
is x 2 1 y 2a256.
The coordinates of the second boat are (4, 6).
B (16, 2)
1
A (− 10, 1)
O (0, 0)
y
(0, 16)
(x, 6)
(16, 0)
1
x
2}
5 10
m
} 210 1 0 1 1 0
1
Midpoint of AO: 1 }
,}
5 25, }2 2
2
2 2 1
(0, −16)
}
Perpendicular bisector of AO:
x 2 1 (6)2 5 256
1
x 2 5 220
y 2 }2 5 10(x 2 (25))
}
x 5 62Ï55
1
y 2 }2 5 10x 1 50
}
The boat will be in range until (22Ï55 , 6).
}
Distance from (4, 6) to (22Ï55 , 6):
}}}
} 2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Ï (4 2 (22Ï55 ))
101
}}
} 2
1 (6 2 6)2 5 Ï (4 1 2Ï55 ) ø 18.8
The second boat will be in radar range of the fishing boat
for about 18.8 miles.
3. D;
y 5 10x 1 }
2
220
1
}
Slope of BO: m 5 }
5 }8
16 2 0
1
2}
5 28
m
} 16 1 0 2 1 0
Midpoint of BO: 1 }
,}
5 (8, 1)
2
2 2
}
Perpendicular bisector of BO:
y 2 1 5 28(x 2 8)
y 2 1 5 28x 1 64
y 5 28x 1 65
Mercury
29
44
Sun
44 1 29 5 2a
Intersection of perpendicular bisectors:
73 5 2a
101
10x 1 }
5 28x 1 65
2
36.5 5 a
29
c 5 a 2 29 5 36.5 2 29 5 7.5
2
18x 5 }
2
2
c 5a 2b
29
x5}
36
7.52 5 36.522 2 b2
101
527
y 5 101 }
1}
5}
2
9
36 2
29
b2 5 1332.25 2 56.25
b2 5 1276
b ø 35.7
x
2
120
1
1
}
Slope of AO: m 5 }
5}
5 2}
210 2 0
210
10
(4, 6)
(−16, 0)
2
}
The straight-line distance is 20 miles.
x 1 y a256
2
}
5 Ï 162 1 122 5 Ï256 1 144 5 Ï400 5 20
x2 1 y 2 ar 2
The center of the circle is 1 }
,} .
36 9 2
29 527
x2
36.5
y2
An equation for Mercury’s orbit is }2 1 }2 5 1.
35.7
Distance from 1 }
, } to (0, 0):
36 9 2
29 527
}}
d5
Ï1 2936 2 0 2 1 1 9
}
2
527
}20
2
2
ø 58.6
The radius of the skid mark is about 58.6 meters.
}
}}
v 5 Ï 9.8mr ø Ï9.8(0.7)(58.6) ø 20.05
The car was traveling about 20 meters per second.
Algebra 2
Worked-Out Solution Key
499
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
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,