14. 4x2 1 6.25y2 2 12x 2 16 5 0
h 5 1, k 5 0
Center: (1, 0)
A 5 4, B 5 0, C 5 6.25
}
a2 5 6 B2 2 4AC 5 0 2 4(4)(6.25) 5 2100
a 5 ​Ï 6 ​ ø 2.45
}
b2 5 3 Because A Þ C and B2 2 4AC < 0, the path is an ellipse.
b5Ï
​ 3 ​ ø 1.73
}
}
Vertices: (1, ​Ï 6 ​)  and (1, 2​Ï 6 ​ )
4x2 1 6.25y2 2 12x 2 16 5 0
}
}
Co-vertices: (1 1 Ï
​ 3 ​,  0) and (1 2 Ï
​ 3 ​,  0)
4 (x 2 3x 1 1.52) 1 6.25y2 5 16 1 9
3
y
6)
(1,
4(x 2 1.5)2 1 6.25y2 5 25
3, 0)
(1 2
4(x2 2 3x) 1 6.25y2 5 16
2
(x 2 1.5)2
(1, 0)
3
h 5 1.5, k 5 0
(1, 2 6 )
(1 1
3, 0)
Center: (1.5, 0)
12. y 2 2 4y 2 2x 1 6 5 0
a2 5 6.25 Vertices: (21, 0) and (4, 0)
Because B2 2 4AC 5 0, the conic is a parabola.
Co-vertices: (1.5, 2) and (1.5, 22)
y 2 4y 1 4 5 2x 2 6 1 4
( y 2 2)2 5 2(x 2 1)
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.
y
4p 5 2
(1, 2)
1
2
( , 2)
3
2
p 5 ​ } ​
1
Vertex: (1, 2)
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 Þ0 or AÞC, 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
1  2
3
Focus: ​ }
​ 2 ​, 2  ​
13. 4x2 2 y2 2 16x 2 4y 2 4 5 0
3. (x 1 4)2 5 28( y 2 2)
A 5 4, B 5 0, C 5 21
h 5 24, k 5 2
B2 2 4AC 5 0 2 4(4)(21) 5 16
Parabola
Because B2 2 4AC > 0, the conic is a hyperbola.
4x2 2 y2 2 16x 2 4y 2 4 5 0
4(x 2 4x) 2 ( y 1 4y) 5 4
Vertex: (24, 2)
4p 5 28
2
2
Center: (2, 22)
a2 5 4 2
b 5 16
a52
4. (x 2 2)2 1 ( y 2 7)2 5 9
4
4
(0, 22)
(2, 7)
2
Circle
(4, 22)
(2, 22)
x
Center: (2, 7)
2
}
Radius: Ï
​ 9 ​ 5 3
x
22
(2, 26)
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 395
x
y
h 5 2, k 5 7
y
(2, 2)
b54
Vertices: (0, 22) and (4, 22)
(24, 2)
Directrix: y 5 4
( y 1 2)2
(x 2 2)2
​ }
   
 
​2 }
​  16   
​ 5 1
4
h 5 2, k 5 22
y
Focus: (24, 0)
4(x 2 2)2 2 ( y 1 2)2 5 16
6
y54
(24, 0)
p 5 22
4(x 2 4x 1 4) 2 ( y 1 4y 1 4) 5 4 1 4(4) 2 4
2
(1.5, –2)
Skill Practice
h 5 1, k 5 2
2
b52
Exercises for the lesson “Translate and
­Classify Conic Sections”
y 2 2 4y 5 2x 2 6
2
(1.5, 0) (4, 0)
x
b 5 4 (1.5, 2)
(21, 0)
B2 2 4AC 5 0 2 4(0)(1) 5 0
2
y
a 5 2.5
A 5 0, B 5 0, C 5 1
y 2 2 4y 2 2x 1 6 5 0
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
y2
​ }
   
​ 1 }
​ 4  ​5 1
6.25
x
4
395
7/13/11 12:26:32 PM
(x 2 6)2
5.​ }
   
​ 2 ( y 1 1)2 5 1
25
8. (x 2 5) 2 1 ( y 1 1)2 5 64
h 5 5, k 5 21
h 5 6, k 5 21
Center: (6, 21)
a55
b2 5 1
b51
Vertices: (1, 21) and
(11, 21)
x
9. ( y 2 1)2 5 4(x 1 6)
(6, 22)
Focus: (25, 1)
2
(x 1 8)
Directrix: x 5 27
y
   
​ 2 }
​  9   
​ 5 1
6.​ }
49
(28, 3)
x2
h 5 28, k 5 24
Hyperbola
x
22
Center: (28, 24)
(211, 24)
a2 5 49 (28, 24)
(25, 24)
Vertices: (28, 211) and
(28, 3)
(0, 4) y
(0, 2)
(25, 2)
(5, 2)
1
Center: (0, 2)
21
(0, 0)
x
a2 5 25, a 5 5
b2 5 4, b 5 2
(28, 211)
c2 5 a2 2 b2 5 25 2 4 5 21
c 5 a 1 b 5 49 1 9
c ø 4.6
5 58
Co-vertices: (0, 4) and (0, 0)
Foci: (28, 211.6) and (28, 3.6)
Foci: (24.6, 2) and (4.6, 2)
7
7
68
44
​ 3  ​and y 5 2​ }3 ​ x 2 }
​ 3  ​
Asymptotes: y 5 }
​ 3 ​ x 1 }
(x 1 3)2
( y 2 2)
(x 1 2)
   
​ 1 }
​  36   
​ 5 1
7.​ }
16
h 5 22, k 5 2
Ellipse
y
Center: (22, 2)
a2 5 36 a56
b2 5 16 b54
(22, 8)
6
(26, 2)
(22, 2)
c2 5 a2 2 b 2 5 36 2 16 5 20
}
 
c 5 Ï
​ 20 ​ø 4.5
Vertices: (22, 8) and (22, 24)
Co-vertices: (26, 2) and (2, 2)
Foci: (22, 6.5) and (22, 22.5)
(2, 2)
4
(22, 24)
(y 2 4)2
   
​ 2 }
​  16   
​ 5 1
11.​ }
9
2
x
h 5 23, k 5 4
Hyperbola
Center: (23, 4)
a2 5 9 a 5 3
b2 5 16 b 5 4
c2 5 a2 1 b2 5 9 1 16 5 25
c 5 5
Vertices: (26, 4) and (0, 4)
Foci: (28, 4) and (2, 4)
4
y
(23, 8)
(23, 4)
(26, 4)
(0, 4)
2
(23, 0)
4
Asymptotes: y 5 }
​ 3 ​ x 1 8 and y 5 2​ }3 ​ x
4 x
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
Vertices: (25, 2) and (5, 2)
}
c 5 Ï
​ 58 ​ ø 7.6
396
h 5 0, k 5 2
Ellipse
2
2
( y 2 2) 2
 ​1 }
​  4   
​ 5 1
10.​ }
25
2
b53
x
22
p 5 1
1
1
11
1
​ 5  ​and y 5 2​ }5 ​ x 1 }
​ 5 ​
Asymptotes: y 5 }
​ 5 ​ x 2 }
(25, 1)
4p 5 4
Foci: (0.9, 21) and (11.1, 21)
2
(26, 1)
Vertex: (26, 1)
a57
y
x 5 27
Parabola
c 5 Ï
​ 26 ​ø 5.1
2
8
h 5 26, k 5 1
}
 
b 5 9 Radius: Ï
​ 64 ​ 5 8
(11, 21)
21
(1, 21)
5 26
2
x
(5, 21)
}
(6, 0)
(6, 21)
22
c2 5 a2 1 b2 5 25 1 1
(y 1 4)
24
Center: (5, 21)
y
a2 5 25 2
2
Circle
Hyperbola
y
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 396
7/13/11 12:26:33 PM
( y 2 1)2
(x 2 4)2
12. C; ​ }
   
​ 1 }
​  4   
​ 5 1
16
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, r 2 5 36 r 5 2 r2 5 4
h 5 25, k 5 1 h 5 9, k 5 21
(x 1 5) 2 1 ( y 2 1)2 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)
( y 1 3)2 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.
​1 }
​  2   
​, }
​  2   
​  2​5 (1, 4)
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
23 1 5 4 1 4
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 }
​  2   
​ 5 1
2
b
a
( y 2 4)2
(x 2 1)2
   
​ 1 }
​  12   
​ 5 1
​ }
16
18. Ellipse
Vertices: (22, 1) and (22, 9)
Co-vertices (24, 5) and (0, 5)
The center is the midpoint of the vertices:
22 1 (22)
1 
119
2
(x 2 h)2
(y 2 k)2
​ }
   
​ 1 }
​  2   
​ 5 1
2
a
b
(y 2 5)2
(x 1 2)2
   
​ 1 }
​  16   
​ 5 1
​ }
4
19. Hyperbola
Vertices: (6, 23) and (6, 1)
Foci: (6, 26) and (6, 4)
The center is the midpoint of the vertices:
​1 }
​  2   
​,  }
​  2   
​  2​5 (6, 21)
6 1 6 23 1 1
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.
c2 5 a2 1 b2
25 5 4 1 b2
b2 5 25 2 4 5 21
The transverse axis is vertical.
( y 2 k) 2
(x 2 h)2
​ }
   
​ 2 }
​  2   
​ 5 1
2
b
a
(x 2 6)2
( y 1 1)2
   
​ 2 }
​  21   
​ 5 1
​ }
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   
​,  ​ }
   
​  ​5 (4, 7)
2 2
117 717
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
(y 2 k)2
​ }
   
​ 2 }
​  2   
​ 5 1
2
b
a
(y 2 7)2
(x 2 4)2
   
​ 2 }
​  16   
​ 5 1
​ }
9
21.In writing the equation, the h and k values should be
subtracted from x and y, not added. The correct
(x 1 2)2
( y 2 3)2
​ }
​ 
 
 
​, }
​  2   
 
​  ​5 (22, 5)
2
equation is ​ }
   
​ 1 }
​  9   
​ 5 1.
25
h 5 22, k 5 5
   
​ 1 }
​  16   
​ 5 1
22.​ }
49
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 1 5)2
( y 2 2)2
Ellipse with center (25, 2)
Lines of symmetry: x 5 25 and y 5 2
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 397
397
7/13/11 12:26:33 PM
23. ( y 2 4)2 5 6(x 1 60)
35. 8x2 2 9y2 2 40x 1 4y 1 145 5 0
Parabola with vertex (26, 4)
A 5 8, B 5 0, C 5 29
Line of symmetry: y 5 4
B2 2 4AC 5 0 2 4(8)(29) 5 288
2
(x 2 1)
( y 2 2)
   
​ 2 }
​  9   
​ 5 1
24.​ }
36
The conic is a hyperbola because B2 2 4AC > 0.
36. B; 4x2 1 y2 1 32x 2 10y 1 85 5 0
Hyperbola with center (1, 2)
A 5 4, B 5 0, C 5 1
Lines of symmetry: x 5 1 and y 5 2
B2 2 4AC 5 0 2 4(4)(1) 5 216
(x 2 3)2
   
​ 5 1
25. ( y 2 5)2 2 ​ }
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
2
B 2 4AC 5 0 2 4(6)(22) 5 48
The conic is a hyperbola, because B2 2 4AC > 0.
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
37. x2 1 y2 2 14x 1 4y 2 11 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.
x2 1 y2 2 14x 1 4y 2 11 5 0
(x 2 14x 1 49) 1 ( y2 1 4y 1 4) 5 11 1 49 1 4
2
(x 2 7)2 1 (y 1 2)2 5 64
h 5 7, k 5 22
y
Center: (7, 22)
}
Radius: Ï
​ 64 ​ 5 8
2
x
22
(7, 22)
29. x2 1 y2 2 10x 2 6y 1 18 5 0
A 5 1, B 5 0, C 5 1
The conic is a circle because B2 2 4AC < 0, B 5 0, and
A 5 C.
2
30. y 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.
34. x2 1 y2 2 6x 1 8y 2 24 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.
398
38. x2 1 4y2 2 10x 1 16y 1 37 5 0
A 5 1, B 5 0, C 5 4
B2 2 4AC 5 0 2 4(1)(4) 5 216
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
x2 1 4y2 2 10x 1 16y 1 37 5 0
(x 2 10x 1 25) 1 4( y2 1 4y 1 4) 5 237 1 25 1 16
2
(x 2 5)2 1 4( y 1 2)2 5 4
(x 2 5)2
​ }
   
​ 1 ( y 1 2)2 5 1
4
h 5 5, k 5 22
Center: (5, 22)
y
21
2
a 5 4, a 5 2
b2 5 1, b 5 1
Vertices: (3, 22) and (7, 22)
Co-vertices: (5, 21) and (5, 23)
1
(5, 21)
x
(3, 22)
(5, 22) (5, 23)
(7, 22)
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
B2 2 4AC 5 0 2 4(1)(1) 5 24
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 398
7/13/11 12:26:34 PM
39. x2 2 16x 2 8y 1 80 5 0
42. y2 1 14y 1 16x 1 33 5 0
A 5 1, B 5 0, C 5 0
A 5 0, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(1)(0) 5 0
B2 2 4AC 5 0 2 4(0)(1) 5 0
The conic is a parabola because
The conic is a parabola because B2 2 4AC 5 0.
2
B 2 4AC 5 0.
y2 1 14y 1 16x 1 33 5 0
x2 2 16x 2 8y 1 80 5 0
(x
2
2 16x 1 64) 5 8y 2 80 1 64
( y 1 7)2 5 216x 1 16
(x 2 8)2 5 8y 2 16
( y 1 7)2 5 216(x 2 1)
2
(x 2 8) 5 8( y 2 2)
h 5 8, k 5 2
h 5 1, k 5 27
p 5 24
(8, 4)
p 5 2
2
Focus: (8, 4)
(8, 2)
y50
22
x
(23, 27)
Focus: (23, 27)
(1, 27)
x
40. 9y2 2 x2 2 54y 1 8x 1 56 5 0
x55
43. x 2 1 y2 1 16x 2 8y 1 16 5 0
A 5 21, B 5 0, C 5 9
A 5 1, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(21)(9) 5 36
B2 2 4AC 5 0 2 4(1)(1) 5 24
2
The conic is a hyperbola because B 2 4AC > 0.
The conic is a circle because B2 2 4AC < 0, B 5 0,
and A 5 C.
9y2 2 x2 2 54y 1 8x 1 56 5 0
9( y2 2 6y 1 9) 2 (x2 2 8x 1 16) 5 256 1 81 2 16
9( y 2 3)2 2 (x 2 4)2 5 9
( y 2 3)2 2 }
​  9   
​ 5 1
h 5 4, k 5 3
y
Center: (4, 3)
(x2 1 16x 1 64) 1 ( y 2 2 8y 1 16) 5 216 1 64 1 16
(x 1 8)2 1 ( y 2 4)2 5 64
h 5 28, k 5 4
(1, 3) (4, 4)
(4, 2)
b2 5 9, b 5 3
y
Center: (28, 4)
(7, 3)
3
Vertices: (4, 4) and (4, 2)
x2 1 y2 1 16x 2 8y 1 16x 5 0
(x 2 4)2
a2 5 1, a 5 1
y
22
4p 5 216
4p 5 8
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
2
Vertex: (1, 27)
y
Vertex: (8, 2)
( y2 1 14y 1 49) 5 216x 2 33 1 49
}
Radius: Ï
​ 64 ​ 5 8
(28, 4)
4
(4, 3)
1
x
x
24
41. 9x 2 1 4y2 2 36x 2 24y 1 36 5 0
A 5 9, B 5 0, C 5 4
B2 2 4AC 5 0 2 4(9)(4) 5 2144
44. x2 2 4y2 1 8x 2 24y 2 24 5 0
The conic is an ellipse because B2 2 4AC < 0 and AÞC.
A 5 1, B 5 0, C 5 24
2
B2 2 4AC 5 0 2 4(1)(24) 5 16
2
9x 1 4y 2 36x 2 24y 1 36 5 0
The conic is a hyperbola because B2 2 4AC > 0.
9(x2 2 4x 1 4) 1 4( y2 2 6y 1 9) 5 236 1 36 1 36
(x 2 2)2
(x 1 8x 1 16) 2 4( y 2 1 6y 1 9) 5 24 1 16 2 36
2
( y 2 3)2
​ }
   
 
​1 }
​  9   
​ 5 1
4
h 5 2, k 5 3
a53
2
b52
b 5 4 (0, 3)
​ }
   
​ 2 ( y 1 3)2 5 1
4
h 5 24, k 5 23
(4, 3)
Center: (24, 23)
a2 5 4, a 5 2
1
Vertices: (2, 6) and (2, 0)
Co-vertices: (0, 3) and (4, 3)
(2, 3)
21
(x 1 4)2 2 4( y 1 3)2 5 4
(x 1 4)2
y (2, 6)
Center: (2, 3)
a2 5 9 x2 2 4y2 1 8x 2 24y 2 24 5 0
9(x 2 2)2 1 4(y 2 3)2 5 36
(2, 0)
x
b2 5 1, b 5 1
Vertices: (26, 23) and
(22, 23)
2
(24, 22)
210
(26, 23)
y
x
(22, 23)
(24, 23)
(24, 24)
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 399
399
7/13/11 12:26:35 PM
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
46. y 5 }
​ x ​
50. 21y 2 2 210y 2 4x2 5 2441
A 5 24, B 5 0, C 5 21
B2 2 4AC 5 0 2 4(24)(21) 5 336
The shape of the path is a hyperbola because
B2 2 4AC > 0.
21y2 2 210y 2 4x2 5 2441
xy 5 a
21( y 2 10y 1 25) 2 4x2 5 2441 1 525
A 5 0, B 5 1, C 5 0
2
B2 2 4AC 5 1 2 4(0)(0) 5 1
21( y 2 5)2 2 4x2 5 84
( y 2 5)2
2
The conic is a hyperbola because B 2 4AC > 0.
x2
​ }
   
​ 2 }
​ 21 ​ 5 1
4
47.The foci are c units above and below the center, at
(h, k 1 c) and (h, k 2 c).
Center: (0, 5)
The asymptotes are y 5 6​ }b ​ x shifted horizontally h units
and vertically k units:
Vertices: (0, 3) and (0, 7) (2
a
(0, 7)
}
 
a 5 2, b 5 Ï
​ 21 ​
(
21, 5 )
(0, 5)
21, 5 )
a
a
( y 2 k) 5 ​ }b ​(x 2 h) ( y 2 k) 5 2​ }b ​(x 2 h)
a
ah
a
ah
y 5 ​ }b ​x 2 }
​ b  ​1 k y 5 2​ }b ​x 1 }
​ b  ​1 k
a
bk 2 ah
a
bk 1 ah
y 5 ​ }b ​ x 1​ }
   
 
​
y 5 2​ }b ​x 1 }
​  b   
​ 
b
Problem Solving
48.
(0, 3)
1
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)2 ≤ 1502.
y
8 ft
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)2 ≤ 1002.
(0, 4)
2
b.At point (0, 0):
x
6
(0, 24)
?
(0 1 80)2 1 (0 1 70)2 ≤ 1002
6400 1 4900 ?
≤ 1002
Top circle:
Center: (0, 4)
(0 2 100)
2
2
2
Equation: (x 2 0) 1 ( y 2 4) 5 4
13,600 ≤ 22,500 ✓
y
x2 1 (y 2 4)2 5 16
You
(0, 0)
x
Center: (0, 24)
(100, 260)
Hotel
Radius: 4
Equation: (x 2 0)2 1 (y 1 4)2 5 42
2
x 1 (y 1 4) 5 16
2
49. x 2 10x 1 4y 5 0
x2 2 10x 5 24y
x2 2 10x 1 25 5 24y 1 25
11,300 µ 10,000
1 (0 1 60)2 ?
≤ 1502
10,000 1 3600 ?
≤ 22,500
Bottom circle:
2
Radius: 4
x
22
(x 2 5) 2 5 24​1 y 2 }
​ 4  ​ 2​
25
An equation for the path of the leap is
(x 2 5) 2 5 24​1 y 2 }
​ 4  ​ 2​.
25
Vertex: 1​ 5, }
​  4   ​ 2​
25
(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.
c. Because the café’s range is 100 yards and the hotel’s
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.
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
y
25
The person’s jump is ​ }
  ​feet high and
4
2(5) 5 10 feet wide.
400
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 400
7/13/11 12:26:35 PM
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.
53. a. The intersection is not a circle when the plane crosses
the point where the cones meet. It is a point.
b. The intersection is not a hyperbola when the plane
crosses the point where the cones meet. It is a pair
of lines.
c. The intersection is not a parabola when the plane lies
along the edge of the cone. It is a line.
y2 2 2x 2 10 5 0
5.
y 5 2x 2 1
(2x 2 1)2 2 2x 2 10 5 0
x2 1 2x 1 1 2 2x 2 10 5 0
x2 2 9 5 0
(x 1 3)(x 2 3) 5 0
x 5 63
When x 5 3: When x 5 23:
y 5 2(3) 2 1 y 5 2(23) 2 1
y 5 24 y 5 2
The solutions are (3, 24) and (23, 2).
6. y 5 4x 2 8
Lesson 8.7 Solve Quadratic Systems
Guided Practice for the lesson “Solve
­Quadratic Systems”
9x 2 (16x2 2 64x 1 64) 2 36 5 0
1. x 2 1 y2 5 13 and y 5 x 2 1
9x2 2 16x2 1 64x 2 64 2 36 5 0
2
y 5 13 2 x
2
}
y 5 6​Ï 13 2 x2 ​ 
Using the calculator’s intersect feature, the solutions are
(22, 23) and (3, 2).
2. x 2 1 8y2 2 4 5 0 and y 5 2x 1 7
8y 2 5 4 2 x2
y2 5 }
​ 2 ​2 }
​ 8  ​
Î
}
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
x2
The graphs do not intersect. There is no solution.
2
3. y 1 6x 2 1 5 0 and y 5 20.4x 1 2.6
y2 5 26x 1 1
}
y 5 6​Ï26x 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
x2 1 4(0.25x2 2 3x 1 9) 2 4 5 0
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.
27x2 1 64x 2 100 5 0
2(7x2 2 64x 1 100) 5 0
2(7x 2 50)(x 2 2) 5 0
7x 2 50 5 0 or x2250
50
x5}
​ 7  ​ or x52
  ​: When x 5 2:
When x 5 ​ }
7
1
​ 2 ​2 }
​ 8  ​ ​ 
y 5 6​ }
9x2 2 (4x 2 8)2 2 36 5 0
2
50
x2
1
9x2 2 y2 2 36 5 0
​ 7  ​ 2​2 8 y 5 4(2) 2 8
y 5 4​1 }
50
144
 
y50
y 5 }
​  7   ​
The solutions are (2, 0) and 1​ }
​ 7  ​, }
​  7   ​  2​.
50 144
7. 22y2 1 x 1 2 5 0
x2 1 y2 2 1 5 0
22y2 1 x 1 2 5 0
2
2x 1 2y2
2x2
32
2250
1x
50
x(2x 1 1) 5 0
x 5 0 or 2x 1 1 5 0
1
x 5 2​ }2 ​
1
When x 5 0: When x 5 2​ }2 ​:
0 2 1 y2 2 1 5 0​1 }
​ 2 ​ 2​ 1 y2 2 1 5 0
1 2
y2 5 1
y 61
3
y2 5 }
​ 4 ​
Î3
}
1 
​ 2 ​ ​ 
y 5 6​ }
1
}
2
​Ï 3 ​ 
​ 2 ​, 6​ }
   ​  ​.
The solutions are (0, 61) and ​ 2}
2
Algebra 2
Worked-Out Solution Key
CS10_CC_A2_MEWO710648_C8.indd 401
401
7/13/11 12:26:36 PM