1. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
(For help, go to Lessons 2-2, 5-2, and 5-5.)
Find the x- and y-intercepts of the graph of each function.
1. y = 3x y = –x – x – 4y = –12
4. y = x2 – y = (x – 3)2 6. y = –4x
10-1

2. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Solutions
1. y = 3x + 6
x-intercept (let y = 0):
0 = 3x
–3x = 6
x = –2
y-intercept (let x = 0):
y = 3(0) + 6
y =
y = 6
2. 2y = –x – 3
x-intercept (let y = 0):
2(0) = –x – 3
0 = –x – 3
x = –3
y-intercept (let x = 0):
2y = –0 – 3
2y = –3
y = – , or –1.5 3 2
3. 3x – 4y = –12
x-intercept (let y = 0):
3x – 4(0)= –12
3x = –12
x = –4
y-intercept (let x = 0):
3(0) – 4y = –12
–4y = –12
y = 3
4. y = x2 – 4
x-intercept (let y = 0):
0 = x2 – 4
4 = x2
x = ± 2
y-intercept (let x = 0):
y = 02 – 4
y = –4
10-1

3. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Solutions (continued)
5. y = (x – 3)2
x-intercept (let y = 0):
0 = (x – 3)2
0 = x – 3
3 = x
y-intercept (let x = 0):
y = (0 – 3)2
y = (–3)2
y = 9
6. y = –4x
x-intercept (let y = 0):
0 = –4x
4x2 = 1
x2 =
x = ± = ±
y-intercept (let x = 0):
y = –4(0)
y =
y = 1 1 4 1 4 1 2
10-1

4. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Graph the equation x2 + y2 = 16. Describe the graph and its lines of symmetry. Then find the domain and range.
x y –
–3 ± ± 2.6
–2 ± ± 3.5
–1 ± ± 3.9 ±4
1 ± ± 3.9
2 ± ± 3.5
3 ± ± 2.6
Make a table of values.
Plot the points and connect them with a smooth curve.
10-1

5. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
(continued)
The graph is a circle of radius 4. Its center is at the origin. Every line through the center is a line of symmetry.
Recall from Chapter 2 that you can use set notation to describe a domain or a range. In this Example, the domain is {x| – x }. The range is {y| – y }.
< –
10-1

6. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Graph the equation 9x y2 = 36. Describe the graph and the lines of symmetry. Then find the domain and range.
x y –
–1 ± 2.6
0 ± 3
1 ± 2.6
Make a table of values.
Plot the points and connect them with smooth curves.
The graph is an ellipse. The center is at the origin. It has two lines of symmetry, the x-axis and the y-axis.
The domain is {x| – x }.
The range is {y| – y }.
< –
10-1

7. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Graph the equation x2 – y2 = 4. Describe the graph and its lines of symmetry. Then find the domain and range.
x y
–5 ± 4.6
–4 ± 3.5
–3 ± 2.2 –
– — — —
3 ± 2.2
4 ± 3.5
5 ± 4.6
Make a table of values.
Plot the points and connect them with smooth curves.
10-1

8. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
(continued)
The graph is a hyperbola that consists of two branches. Its center is at the origin. It has two lines of symmetry, the x-axis and the y-axis.
The domain is {x| x –2 or x }. The range is all real numbers.
> –
<
10-1

9. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Identify the center and intercepts of the conic section. Then find the domain and range.
The center of the ellipse is (0, 0).
The x-intercepts are (–5, 0) and (5, 0), and the y-intercepts are (0, –4) and (0, 4).
The domain is {x| – x }. The range is {y| – y }.
< –
10-1

10. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Describe each Moiré pattern as a circle, an ellipse, or a hyperbola. Match it with one of these possible equations,
2x2 – y2 = 16, 25x y2 = 100, or x2 + y2 = 16.
a. b. c.
a. The equation 25x y2 = 100 represents a conic section with two sets of intercepts, (±2, 0) and (0, ±5). Since the intercepts are not equidistant from the center, the equation models an ellipse.
10-1

11. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
(continued)
a. b. c.
b. The equation x2 + y2 = 16 represents a conic section with two sets of intercepts, (±4, 0) and (0, ±4). Since each intercept is 4 units from the center, the equation models a circle.
c. The equation 2x2 – y2 = 16 represents a conic section with one set of intercepts, (±2 2, 0), so the equation must be a hyperbola.
10-1

12. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
pages 538–541  Exercises 1.
Hyperbola: center (0, 0),
y-intercepts at ± , no
x-intercepts, the lines of
symmetry are the x- and
y-axes; domain: all real
numbers, range: y
or y 2.
Ellipse: center (0, 0), x-intercepts at ±3 2,
y-intercepts at ±6, the lines of symmetry
are the x- and y-axes; domain:
– x , range –6 y 6.
5 3 3
< –
>
10-1

13. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1 3.
Circle: center (0, 0), radius 4,
x-intercepts at ±4, y-intercepts
at ±4, there are infinitely many
lines of symmetry; domain:
–4 x 4, range: –4 y 4. 4.
Hyperbola: center (0, 0), y-intercepts
at ± 3, no x-intercepts, the lines of
symmetry are the x- and y-axes; domain:
all real numbers, range: y or y 5.
Ellipse: center (0, 0), y-intercepts
at ±2, x-intercepts at ±5, the lines
of symmetry are the x- and y-axes;
domain: –5 x 5, range: –2 y 2. 6.
Circle: center (0, 0), radius 7, x- and
y-intercepts at ±7, there are infinitely
many lines of symmetry; domain:
–7 x 7, range: –7 y 7.
< –
>
10-1

14. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1 7.
Hyperbola: center (0, 0), y-intercepts
at ±1, the lines of symmetry are the
x- and y-axes; domain: all real numbers,
range: y –1 or y 1. 8.
Hyperbola: center (0, 0), x-intercepts
at ±2, the lines of symmetry are the
x- and y-axes; domain: x –2 or x 2,
range: all real numbers. 9.
Circle: center (0, 0), radius 10, x- and
y-intercepts at ±10, there are infinitely
many lines of symmetry; domain:
–10 x 10, range: –10 y
10.
Circle: center (0, 0), radius 2, x- and
y-intercepts at ±2, there are infinitely
–2 x 2, range: –2 y 2.
> –
<
10-1

15. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
11.
Ellipse: center (0, 0), x-intercepts
at ±4, y-intercepts at ±2, the lines
of symmetry are the x- and y-axes;
domain: –4 x 4, range: –2 y 2.
12.
Circle: center (0, 0), radius 5, x-
and y-intercepts at ± 5, there are
infinitely many lines of symmetry;
domain: – x ,
range: – y
13.
Ellipse: center (0, 0), x-intercepts at ±1,
y-intercepts at ± , the lines of symmetry
are the x- and y-axes; domain:
–1 x 1, range: y 1 3
< –
10-1

16. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
14.
Hyperbola: center (0, 0), x-intercepts
at ±6, the lines of symmetry are the
x- and y-axes; domain: x –6 or x 6,
range: all real numbers.
15.
Hyperbola: center (0, 0), y-intercepts at ± ,
the lines of symmetry are the x- and y-axes;
domain: all real numbers, range: y – or y
16.
Ellipse: center (0, 0), x-intercepts
at ±2, y-intercepts at ±6, the lines of
symmetry are the x- and y-axes;
domain: –2 x 2, range: –6 y 6.
17. center (0, 0), x-intercepts at ±3,
y-intercepts at ±2; domain: –3 x 3,
range: –2 y 2 1 2
< –
>
10-1

17. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
18. center (0, 0), no x-intercepts,
y-intercepts at 42; domain:
all real numbers, range: y –2 or y 2
19. center (0, 0), x-intercepts at 43,
no y-intercepts; domain: x –3 or x 3,
range: all real numbers
20. center (0, 0), x-intercepts at 48,
y-intercepts at 44; domain: –8 x 8,
range: –4 y 4
21. center (0, 0), x-intercepts at 43,
y-intercepts at 45; domain: –3 x 3,
range: –5 y 5
22. center (0, 0), no x-intercepts,
y-intercepts at 43; domain:
all real numbers; range: y –3 or y 3
23. 19
24. 17
25. 18
26. 20
27. 21
28. 22
29.
Hyperbola: center (0, 0), x-intercepts
±4, the lines of symmetry are the x-
and y-axes; domain: x –4 or x 4,
range: all real numbers.
> –
<
10-1

18. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
30.
Circle: center (0, 0), radius 2,
x- and y-intercepts at ±2, there
are infinitely many lines of
symmetry; domain: –2 x 2,
range: –2 y 2.
31.
Hyperbola: center (0, 0), y-intercepts
at ±2, the lines of symmetry are the
x- and y-axes; domain: all real numbers,
range: y –2 or y 2.
8 5 5
< –
32.
Ellipse: center (0, 0), x-intercepts at
± , y-intercepts at ±2 5, the
lines of symmetry are the x- and
y-axes; domain: – x ,
range: – y
< –
> –
10-1

19. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
33. a. All lines in the plane that pass through
the center of a circle are axes of
symmetry of the circle.
b. The axes of symmetry of an ellipse
intersect at the center of the ellipse.
The same is true for a hyperbola.
This can be confirmed using, for example,
4x2 + 9y2 = 36 and 4x2 – 9y2 = 36.
34. a. Let the lamp sit in a normal, upright position,
but close enough to the wall for the bottom rim
of the shade to almost touch the wall.
b. Hold the lamp so that the shade contacts
the wall along a vertical line.
c. Hold the lamp at an angle so that the light from
the top of the shade gives a closed, curved
oblong area of light on the wall.
d. Hold the lamp so that the circular top rim of
the shade is parallel to the wall.
35–40.  Answers may vary.
Samples are given.
35. (2, 4)
36. ( 2, 1)
37. (–2, )
38. (2, 0)
39. (–3, )
40. (0, – 7)
41.
x2 + y2 = 36
10-1

20. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
42.
x2 + y2 =
43.
x2 + y2 = 16
44.
x2 + y2 =
45. Check students’ work.
46. a.
b. hyperbola
c. No intercepts, but y = x
and y = –x are lines of
symmetry.
d. yes; ƒ(x) =
47. The curve is a small piece of a hyperbola.
Each side of the unsharpened pencil is a
portion of a plane parallel to the line
represented by the lead in the pencil. The
plane would intersect both parts of the cone
if the pencil were a double cone. 1 4 16 x
10-1

21. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
48. a. b.
49. a parabola, or part of a hyperbola
50. D
51. F
52. D 1 2
< –
53. I
54. B
55. [2] It is a circle with center at (0, 0)
and a radius of 11. The x- and
y-intercepts are at 411. The lines
of symmetry are all lines through
the origin. The domain is –11 x 11
and the range is –11 y
[1] includes either description OR domain
and range, but not both
56.
57.
58.
10-1

22. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
69. y = (2)x
70. x3 – 3x2y + 3xy2 – y3
71. p6 + 6p5q + 15p4q2+ 20p3q3 + 15p2q4+ 6pq5 + q6
72. x4 – 8x3 + 24x2 – 32x + 16
– 405x + 270x2 – 90x3 + 15x4 – x5
59.
60. z = –5xy; 30
61. z = –4xy; 24
62. z = 40xy; –240
63. z = – xy; 3
64. y = 4(2)x
65. y = (4)x
66. y = 3(2)x
67. y = (3)x
68. y = 2(3)x 1 2 3 4
10-1

23. Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Graph each equation. Identify the conic section and determine its intercepts. Then, give the domain and range of each graph.
1. 2y2 – x2 = 8
hyperbola, (0, ±2);
domain: all real numbers
range: {y| – < y –2
or y < }
< –
2. 16x y2 = 400
ellipse, x-int: (±5, 0),
y-int: (0, ±4);
domain: {x| – x },
range: {y| – y }
< –
3. 2x y2 = 50
circle, x-int: (±5, 0),
y-int: (0, ±5);
domain: {x| – x },
range: {y| – y }
< –
10-1

24. Find the distance between the given points.
Parabolas
ALGEBRA 2 LESSON 10-2
(For help, go to Skills Handbook pages 844 and 856.)
Solve for c.
= = = = 1 8
Find the distance between the given points.
5. (2, 3) and (4, 1) 6. (4, 6) and (3, –2) 7. (–1, 5) and (2, –3) c 2c 12 4c
10-2

25. 5. (2, 3) and (4, 1); distance = (2 – 4)2 + (3 – 1)2 =
Parabolas
ALGEBRA 2 LESSON 10-2
= ; c = 8
= ; 2c = 8; c = 4
= ; 4c = 12; c = 3
4. 2(4c) = 1; 8c = 1; c =
5. (2, 3) and (4, 1); distance = (2 – 4)2 + (3 – 1)2 =
(–2) = = = • =
Solutions 1 2c 12 4c 8 c
10-2

26. Solutions (continued)
Parabolas
ALGEBRA 2 LESSON 10-2
Solutions (continued)
6. (4, 6) and (3, –2); distance = (4 – 3)2 + (6 – (–2))2 =
= =
7. (–1, 5) and (2, –3); distance = (–1 – 2)2 + (5 – (–3))2 =
(–3) = =
10-2

27. Parabolas
ALGEBRA 2 LESSON 10-2
Write an equation for a graph that is the set of all points in the plane that are equidistant from point F(0, 1) and the line y = –1.
You need to find all points P(x, y) such that FP and the distance from P to the given line are equal.
10-2

28. (x – 0)2 + (y – 1)2 = (x – x)2 + (y – (–1))2
Parabolas
ALGEBRA 2 LESSON 10-2
(continued)
FP = PQ
(x – 0)2 + (y – 1)2 = (x – x)2 + (y – (–1))2
x2 + (y – 1)2 = (y + 1)2
x2 + y2 – 2y = y y + 1
x2 = 4y
y = x2 1 4
An equation for a graph that is the set of all points in the plane that are equidistant from the point F (0, 1) and the line y = –1 is y = x2. 1 4
10-2

29. Step 1: Determine the orientation of the parabola.
Parabolas
ALGEBRA 2 LESSON 10-2
Write an equation for a parabola with a vertex at the origin and a focus at (0, –7).
Step 1: Determine the orientation of the parabola.
Make a sketch.
Since the focus is located below the vertex, the parabola must open downward. Use y = ax2.
Step 2: Find a.
10-2

30. = Since the focus is a distance of 7 units from the vertex, c = 7.
Parabolas
ALGEBRA 2 LESSON 10-2
(continued)
| a | =
= Since the focus is a distance of 7 units from the vertex, c = 7. = 1
4(7) 28 4c
Since the parabola opens downward, a is negative.
So a = – 1 28
An equation for the parabola is y = – x2. 1 28
10-2

31. The distance from the vertex to the focus is 4 in., so
Parabolas
ALGEBRA 2 LESSON 10-2
A parabolic mirror has a focus that is located 4 in. from the vertex of the mirror. Write an equation of the parabola that models the cross section of the mirror.
The distance from the vertex to the focus is 4 in., so
c = 4. Find the value of a.
a = 1 4c = 1
4(4)
= Since the parabola opens upward, a is positive. 1 16
The equation of the parabola is y = x2. 1 16
10-2

32. Identify the focus and directrix of the graph of the
Parabolas
ALGEBRA 2 LESSON 10-2
Identify the focus and directrix of the graph of the
equation x = – y2. 1 8
The parabola is of the form x = ay2, so the vertex is at the origin and the parabola has a horizontal axis of symmetry. Since a < 0, the parabola opens to the left.
| a | = 1 4c
| – | = 1 4c 8
4c = 8
c = 2
The focus is at (–2, 0). The equation of the directrix is x = 2.
10-2

33. 8y = –x2 – 4x + 4 Solve for y, since y is the only term.
Parabolas
ALGEBRA 2 LESSON 10-2
Identify the vertex, the focus, and the directrix of the graph of the equation x x + 8y – 4 = 0. Then graph the parabola.
x x + 8y – 4 = 0
8y = –x2 – 4x Solve for y, since y is the only term.
8y = –(x x + 4) Complete the square in x.
y = – (x + 2) vertex form 1 8
The parabola is of the form y = a(x – h)2 + k, so the vertex is at (–2, 1) and the parabola has a vertical axis of symmetry. Since a < 0, the parabola opens downward.
10-2

34. Parabolas | a | = | – | = Substitute – for a. 4c = 8 Solve for c.
ALGEBRA 2 LESSON 10-2
(continued)
| – | = Substitute – for a. 1 8
| a | =
4c = 8 Solve for c.
c = 2 4c
The vertex is at (–2, 1) and the focus is at (–2, –1). The equation of the directrix is y = 3.
Locate one or more points on the parabola. Select a value for x such as –6.
The point on the parabola with an x-value of –6 is (–6, –1). Use the symmetric nature of a parabola to find the corresponding point (2, –1).
10-2

35. b. The light produced by the bulb will reflect off the
Parabolas
ALGEBRA 2 LESSON 10-2
pages 546–548  Exercises
1. y = x2
2. y = – x2
3. x = – y2
4. y = – x2
5. y = x2 + 2
6. x = y2
7. x = y2
8. y = – x2
9. y = x2 1 8 4 12 32 2 24 16 28
10. x = – y2
11. x = y2
12. y = – x2
13. y = x2
14. y = 2x2
15. a. Answers may vary.
Sample: y = x2
b. The light produced by the
bulb will reflect off the
parabolic mirror and will
be emitted in parallel rays.
16. (0, 1), y = –1
17. 0, , y = –
18. (0, –2), y = 2
, 0 , x = –
20. 0, , y = –
21. (9, 0), x = –9
22. – , 0 , x =
23. 0, – , y = 20 6 9 1 4 1 4 1 4
10-2

36. Parabolas 24. (0, 0), (6, 0), x = –6 25. (0, 0), (0, –1), y = 1
ALGEBRA 2 LESSON 10-2 25 4
24. (0, 0), (6, 0), x = –6
25. (0, 0), (0, –1), y = 1
26. (0, 0), (3, 0), x = –3
27. (0, 0), , 0 , x = –
28. (0, 0), (0, 1), y = –1
29. (0, 0), (0, –1), y = 1
10-2

37. Parabolas 30. (2, 0), (2, 1), y = –1 31. (0, 0), (–2, 0), x = 2
ALGEBRA 2 LESSON 10-2
30. (2, 0), (2, 1), y = –1
31. (0, 0), (–2, 0), x = 2
32. (–2, 4), –2, , y =
33. (–3, 0), – , 0 , x = –
34. (4, 0), (4, –6), y = 6
35. (3, –1), (6, –1), x = 0 17 4 15 3 2 9
10-2

38. Parabolas 36. x = y2 43. x = – y2 48. 37. y = x2 44. y = x2
ALGEBRA 2 LESSON 10-2
36. x = y2
37. y = x2
38. y = – x2
39. x = – y2
40. x = – y2
41. y = – x2
42. 1 12
400 20 28 36 5 56
43. x = – y2
44. y = x2
45. x = y2
46.
47. 8 4
48.
49.
50.
10-2

39. 56. Answers may vary. Sample: Write the equation in the form
Parabolas
ALGEBRA 2 LESSON 10-2
51.
52. y = (x – 1)2 + 1
53. x = – (y – 1)2 + 1
54. y = – (x – 1)2 + 1
55. Check students’ work.
56. Answers may vary. Sample:
Write the equation in the form
x = y2. The distance from
the focus to the directrix is , or . 1 4 8
57. The directrix will have equation y = k – c.
A point (x, y) is on the parabola if and
only if the distance from (x, y) to the
directrix is equal to the distance from
(x, y) to the focus. So (x, y) is on the
parabola if and only if
|y – (k – c)| = (x – h)2 + (y – k – c)2.
Square and simplify to get the
equivalent equation 4cy – 4kc = (x – h)2,
or (x – h)2 = 4c(y – k).
58. 6 2
10-2

40. b. domain: all non-negative real numbers, range: all
Parabolas
ALGEBRA 2 LESSON 10-2
59. a. the bottom half of the
parabola y2 = x
b. domain: all non-negative
real numbers, range: all
non-positive real numbers
60. a. The vertex moves up, and
the parabola widens.
b. The vertex moves down, and
the parabola narrows.
c. The parabola would degenerate
into the ray with endpoint (0, –2)
that passes through the origin.
61. B
62. F
63. A
64. H
65. [2] |a| =
|a| =
so since the focus is at (0, 4) and
the directrix is y = –4, a is positive.
The equation is y = x2.
[1] answer correct, without work shown
66. [2] |a| = =
36 = 4c
c = 9
so the focus is at (0, 9) since a > 0
and the equation of the directrix
is y = –9. 1
4.4 16 4c 36 1 4c
10-2

41. Parabolas 67. 68. 69. 70. the x- and y-axes 71. x = 0 and y = 4
ALGEBRA 2 LESSON 10-2
67.
68.
69.
70. the x- and y-axes
71. x = 0 and y = 4
72. x = –1 and y = 0
73. x = 1 and y = 0
74. x = –5 and y = 2
75. x = 3 and y = –1
76. $
10-2

42. 2. Write an equation of a parabola with a vertex
Parabolas
ALGEBRA 2 LESSON 10-2
1. Write an equation for a graph that is the set of all points in the plane equidistant from the point F( – , 0) and the line x = .
2. Write an equation of a parabola with a vertex
at the origin and a focus at (0, –3).
3. Identify the vertex, the focus, and the directrix
of the equation y2 – 4y – 16x = 0.
Then, graph the parabola. 1 8 1 8
x = –2y2
(2, 2), (6, 2), x = –2;
y = – x2 1 12
10-2

43. Find the missing value to complete the square.
Circles
ALGEBRA 2 LESSON 10-3
(For help, go to Lesson 5-7 and Skills Handbook page 855.)
Simplify.
Find the missing value to complete the square.
6. x2 – 2x x x x2 – 6x +
Find the missing value to complete the square.
6. x2 – 2x x x x2 – 6x +
10-3

44. Circles Solutions 1. 16 = 4 2. 49 = 7 3. 20 = 4  5 = 2 5
ALGEBRA 2 LESSON 10-3
Solutions
6. x2 – 2x ; c = = – = (–1)2 = 1
7. x x ; c = = = 22 = 4
8. x2 – 6x ; c = = – = (–3)2 = 9 b 2 4 6
= 4
= 7
=  5 =
=  3 =
=  2 =
10-3

45. Write an equation of a circle with center (3, –2) and radius 3.
Circles
ALGEBRA 2 LESSON 10-3
Write an equation of a circle with center (3, –2) and radius 3.
(x – h)2 + (y – k)2 = r2 Use the standard form of the equation of a circle.
(x – 3)2 + (y – (–2))2 = 32 Substitute 3 for h, –2 for k, and
3 for r.
(x – 3)2 + (y + 2)2 = 9 Simplify.
Check: Solve the equation for y and enter both functions into your graphing calculator.
(x – 3)2 + (y + 2)2 = 9
(y + 2)2 = 9 – (x – 3)2
y = ± – (x – 3)2
y = –2 ± – (x – 3)2
10-3

46. The equation is (x – 2)2 + (y + 1)2 = 16.
Circles
ALGEBRA 2 LESSON 10-3
Write an equation for the translation of x2 + y2 = 16 two units right and one unit down.
(x – h)2 + (y – k)2 = r 2 Use the standard form of the equation of a circle.
(x – 2)2 + (y – (–1))2 = 16 Substitute 2 for h, –1 for k, and 16 for r 2.
(x – 2)2 + (y + 1)2 = 16 Simplify.
The equation is (x – 2)2 + (y + 1)2 = 16.
10-3

47. Circles
ALGEBRA 2 LESSON 10-3
Suppose gear B, from Example 3, rotates 10 times for each rotation of gear A. Assuming all other information from Example 3 remains the same, write the equation for the circle that represents gear A.
Make a table.
Gear (h, k) r Equation
A –5 , x y2 = 25
B (0, 0) x2 + y2 =
( )
( ) 1 2 11 2 2 1 2 1 4
Let the radius of gear B = . 1 2
The radius of gear A must be 10 times the radius of gear B, or 5.
10-3

48. (x – h)2 + (y – k)2 = r 2 Use the standard form.
Circles
ALGEBRA 2 LESSON 10-3
Find the center and radius of the circle with equation (x + 4)2 + (y – 2)2 = 36.
(x – h)2 + (y – k)2 = r 2 Use the standard form.
(x + 4)2 + (y – 2)2 = 36 Write the equation.
(x – (–4))2 + (y – 2)2 = 62 Rewrite the equation in standard form.
h = –4 k = r = 6 Find h, k, and r.
The center of the circle is (–4, 2). The radius is 6.
10-3

49. (x – h)2 + (y – k)2 = r 2 Find the center and radius of the circle.
Circles
ALGEBRA 2 LESSON 10-3
Graph (x – 3)2 + (y + 1)2 = 4.
(x – h)2 + (y – k)2 = r 2 Find the center and radius of the circle.
(x – 3)2 + (y – (–1))2 = 4
h = k = – r 2 = 4, or r = 2
Draw the center (3, –1) and radius 2.
Draw a smooth curve.
10-3

50. Circles pages 552–554 Exercises 12. (x + 1)2 + (y – 3)2 = 81
ALGEBRA 2 LESSON 10-3
pages 552–554  Exercises
1. x2 + y2 = 100
2. (x + 4)2 + (y + 6)2 = 49
3. (x – 2)2 + (y – 3)2 = 20.25
4. (x + 6)2 + (y – 10)2 = 1
5. (x – 1)2 + (y + 3)2 = 100
6. (x + 5)2 + (y + 1)2 = 36
7. (x + 3)2 + y2 = 64
8. (x + 1.5)2 + (y + 3)2 = 4
9. x2 + (y + 1)2 = 9
10. (x + 1)2 + y2 = 1
11. (x – 2)2 + (y + 4)2 = 25
12. (x + 1)2 + (y – 3)2 = 81
13. x2 + (y + 5)2 = 100
14. (x – 3)2 + (y – 2)2 = 49
15. (x + 6)2 + (y – 1)2 = 20
16. (x – 5)2 + y2 = 50
17. (x + 3)2 + (y – 4)2 = 9
18. (x – 2)2 + (y + 6)2 = 16
19. (1, 1), 1
20. (–2, 10), 2
21. (3, –1), 6
22. (–3, 5), 9
23. (0, –3), 5
10-3

51. Circles
ALGEBRA 2 LESSON 10-3
24. (–6, 0), 11
25. (–2, –4), 16
26. (3, 7),
27.
28.
29.
30.
31.
32.
33.
34.
10-3

52. 54. Replacing x with x + 7 and y with y – 7
Circles
ALGEBRA 2 LESSON 10-3
35. x2 + y2 = 16
36. x2 + y2 = 9
37. x2 + y2 = 25
38. x2 + y2 = 3
39. x2 + y2 = 25
40. x2 + y2 = 169
41. x2 + y2 = 169
42. x2 + y2 = 13
43. x2 + y2 = 26
44. x2 + y2 = 52
45. (x + 6)2 + (y – 13)2 = 49
46. (x – 5)2 + (y + 3)2 = 25
47. (x + 2)2 + (y – 7.5)2 = 2.25
48. (x – 1)2 + (y + 2)2 = 10
49. (x – 2)2 + (y – 1)2 = 25
50. (x – 6)2 + (y – 4)2 = 25
51. (x + 1)2 + (y + 7)2 = 36
52. (x – 24)2 + (y – 22)2 = 100
53. Check students’ work.
54. Replacing x with x + 7 and y with y – 7
has the effect of translating the circle
7 units left and 7 units up.
55. (0, 0), 2
56. (0, –1), 5
57. (0, 0),
10-3

53. 67. Let P(x, y) be any point on the circle
Circles
ALGEBRA 2 LESSON 10-3
66. parabola; x = (y + 2)2 + 3;
67. Let P(x, y) be any point on the circle
centered at the origin and having radius r.
If P(x, y) is one of the points (r, 0), (–r, 0),
(0, r), or (0, –r), substitution shows that
x2 + y2 = r 2. If P(x, y) is any other point on
the circle, drop a perpendicular PK from P to
the x-axis (K on the x-axis). OPK is a right
triangle with legs of lengths |x| and |y| and
with hypotenuse of length r. By the
Pythagorean Theorem, |x|2 + |y|2 = r 2.
But |x|2 = x2 and |y|2 = y2. So x2 + y2 = r 2.
58. (0, 4),
59. (–5, 0),
60. (–2, –4),
61. (–3, 5),
62. (–1, 0), 2
63. (3, 1), 6
64. (0, 2),
65. circle; (x – 4)2 + (y – 3)2 = 16;
10-3

54. 68. a. The radius would have length 0. b. point (0, 0) 69. a.
Circles
ALGEBRA 2 LESSON 10-3
68. a. The radius would have length 0.
b. point (0, 0)
69. a.
b. Earth: x2 + y2 = 15,705,369
Mars: x2 + y2 = 4,456,321
Mercury: x2 + y2 = 2,296,740
Pluto: x2 + y2 = 511,225
70. a. (x – 3)2 + (y – 4)2 = 25
b. y = – x x
71. 12
72. 2
74. 5
75. 10
77. x = – y2
78. x = –1
79. x = 2, x = 3
80. no discontinuities
81. 4
82. 2
83. –3
84. 4
85. 4
86.
87. 4
88. –2
89. 1 1 3 10 12 2
10-3

55. 1. Write an equation of a circle with center (–2, –1) and radius 3.
Circles
ALGEBRA 2 LESSON 10-3
1. Write an equation of a circle with center (–2, –1) and radius 3.
Graph the circle.
2. Write and equation for the translation of x2 + y2 = 25 three units left and two units up.
3. Find the center and radius of the circle with equation
(x – 7)2 + (y – 1)2 = 49.
(x + 2)2 + (y + 1)2 = 9;
(x + 3)2 + (y – 2)2 = 25
(7, 1); 7
10-3

56. Evaluate each expression for a = 3 and b = 5.
Ellipses
ALGEBRA 2 LESSON 10-4
(For help, go to Lesson 5-5 and Skills Handbook page 846.)
Solve each equation.
= x x2 = = – x2
Evaluate each expression for a = 3 and b = 5.
4. a2 + b2 5. a2 – b2 6. b2 – 2a2
10-4

57. 6. b2 – 2a2 for a = 3 and b = 5: 52 – 2(3)2 = 25 – 2(9) = 25 – 18 = 7
Ellipses
ALGEBRA 2 LESSON 10-4
= x
16 = x2
x = ± = ±4
2. x2 = 48
x = ±
x = ±
= – x2
–36 = –x2
36 = x2
x = ± = ±6
4. a2 + b2 for a = 3 and b = 5: = = 34
5. a2 – b2 for a = 3 and b = 5: 32 – 52 = 9 – 25 = –16
6. b2 – 2a2 for a = 3 and b = 5: 52 – 2(3)2 = 25 – 2(9) = 25 – 18 = 7
Solutions
10-4

58. + = 1 Substitute 9 for b2 and 16 for a2.
Ellipses
ALGEBRA 2 LESSON 10-4
Write an equation in standard form of an ellipse that has a vertex at (0, –4), a co-vertex at (3, 0), and is centered at the origin.
Since (0, –4) is a vertex of the ellipse, the other vertex is at (0, 4), and the major axis is vertical.
Since (3, 0) is a co-vertex, the other co-vertex is at (–3, 0), and the minor axis is horizontal.
So, a = 4, b = 3, a2 = 16, and b2 = 9.
= 1 Standard form for an equation of an ellipse with a vertical major axis.
x 2 b2
y 2 a2
= 1 Substitute 9 for b2 and 16 for a2. 9 16
An equation of the ellipse is = 1.
x 2 9
y 2 16
10-4

59. Place the co-vertices at (0, ±5).
Ellipses
ALGEBRA 2 LESSON 10-4
Find an equation of an ellipse centered at the origin that is 20 units wide and 10 units high.
Since the largest part of the ellipse is horizontal and the width is 20 units, place the vertices at (±10, 0).
Place the co-vertices at (0, ±5).
So, a = 10, b = 5, a2 = 100, and b2 = 25.
= 1 Standard form for an ellipse with a horizontal major axis.
= 1 Substitute 100 for a2 and 25 for b2.
x 2 a2
y 2 b2
100 25
An equation of the ellipse is = 1.
x 2
100
y 2 25
10-4

60. + = 1 Write in standard form.
Ellipses
ALGEBRA 2 LESSON 10-4
Find the foci of the ellipse with the equation 9x2 + y2 = 36. Graph the ellipse.
9x2 + y2 = 36
Since 36 > 4 and 36 is with y2, the major axis is vertical, a2 = 36, and b2 = 4.
= 1 Write in standard form.
x 2 4
y 2 36
c2 = a2 – b2 Find c.
= 36 – 4 Substitute 4 for a2 and 36 for b2.
= 32
c = ± = ±
The major axis is vertical, so the coordinates of the foci are (0, ±c). The foci are (0, ) and (0, –4 2).
10-4

61. Since the foci have coordinates (0, ±4), the major axis is vertical.
Ellipses
ALGEBRA 2 LESSON 10-4
Write an equation of the ellipse with foci at (0, ±4) and co-vertices at (±2, 0).
Since the foci have coordinates (0, ±4), the major axis is vertical.
Since c = 4 and b = 2, c2 = 16, and b2 = 4.
c2 = a2 – b2 Use the equation to find a2.
16 = a2 – 4 Substitute 16 for c2 and 4 for b2.
a2 = 20 Simplify.
= 1 Substitute 20 for a2 and 4 for b2.
x 2 4
y 2 20
An equation of the ellipse is = 1.
x 2 4
y 2 20
10-4

62. Ellipses pages 559–561 Exercises 10. + = 1 1. + = 1 11. + = 1
ALGEBRA 2 LESSON 10-4
pages 559–561  Exercises
= 1
y2 = 1
y2 = 1
4. x = 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
14. x = 1
= 1
= 1
17. x = 1
x 2 16
y 2 9 4 36 49 25 81
2.25
0.25 64
256
100
12.25
196 y2
900
400
6.25
18. (0, 5), (0, – 5)
19. (0, 4), (0, –4)
10-4

63. Ellipses
ALGEBRA 2 LESSON 10-4
22. (0, 6), (0,–6)
23. (0, 6), (0, – 6)
20. (4 2, 0), (–4 2, 0)
21. (8, 0), (–8, 0)
24. ( , 0), (–2 3, 0)
25. (9, 0), (–9, 0)
10-4

64. Ellipses 33. ( 5, 0), (– 5, 0) 26. (3 15, 0), (–3 15, 0) 41. a. 0.9;
ALGEBRA 2 LESSON 10-4
x 2 64
y 2
100
128 89 4 20 49
245
225
514
33. ( 5, 0), (– 5, 0)
34. (0, ), (0, –2 3)
35. (0, ), (0, –4 2)
36. (0, ), (0, – )
37. (0, ), (0, –2 7)
38. (0, 1), (0, –1)
39. (–3, 8), (–3, 2)
40. (–2, 2), (–2, – 2)
41. a. 0.9;
b. 0.1;
c. The shape is close
to a circle.
d. The shape is close
to a line segment.
26. ( , 0), (– , 0)
= 1
= 1
= 1
= 1
= 1
= 1
10-4

65. 43. a. Yes; since c2 = a2 – b2, if the
Ellipses
ALGEBRA 2 LESSON 10-4
= 1
43. a. Yes; since c2 = a2 – b2, if the
foci are close to 0, then c2 will
be close to 0 and a2 will be close
to b2. This means a will be close
to b and hence the ellipse will be
close to a circle.
b. If F1 and F2 are considered distinct
pts., then a circle is not an ellipse.
If F1 and F2 are the same pt., then
a circle is an ellipse.
= 1
y2 = 1
46. x = 1
x 2 16
y 2 4 9
20.25
= 1
48. The vertices are the points farthest
from the center and the co-vertices
are the points closest to the center.
49. Check students’ work.
= 1
= 1
= 1
= 1
= 1
= 1 3 25 81
121
702.25
210.25
144
169
324
256
10-4

66. 64. a. The vertices are at the points where the curve
Ellipses
ALGEBRA 2 LESSON 10-4
= 1
= 1
= 1
= 1
= 1
= 1
= 1
= 1
x 2 16
y 2 12
72.25
90.25
100
400 25 39 64 36 27 4 9 18 20
64. a. The vertices are at the points where the curve
intersects the line through the holes made by
the tacks. The co-vertices are the points where
the curve intersects the perpendicular bisector
of the segment connecting the vertices.
b. at the points where the tacks are stuck in the paper
c. Check students’ work.
65. When c is close to 0, the values of a and b are almost
the same, and ab is close to a2, that is, close to the
area of a circle of radius a.
66. a. 3  106 mi
b. about 0.016
c = 1
67. area of blue region = 3(area of white region)
or about ft
8.649  1015
 1015
10-4

67. 72. [2] The x-axis is the major axis since 50 > 40. The
Ellipses
ALGEBRA 2 LESSON 10-4
69. D
70. I
71. C
72. [2] The x-axis is the major
axis since 50 > 40. The
equation is of the form
= 1. 2a = 50,
so a = 25, and 2b = 40,
so b = 20. Therefore,
a2 = 625 and b2 = 400.
The equation is = 1.
[1] answer only OR minor error
in calculating a2 and b2
73. (x – 2)2 + (y + 3)2 = 36
74. (x + 4)2 + (y – 7)2 = 121
; x = 0,
; x = 1, –6
; x = –2
78. log 15
79. log3 6
80. log 2
81. log 320
82. log
83. log 1
2 – 3x4 2 3 / 3
x – 6
x – 1 /
x – 5
x2 – 2x + 4 /
x 2 a2
y 2 b2 x2
625 y2
400 x y
5 k 4
10-4

68. 3. Find the foci of an ellipse with the equation 25x2 ± 36y2 = 900.
Ellipses
ALGEBRA 2 LESSON 10-4
1. Write an equation in standard form of an ellipse that has a vertex at (–4, 0), a co-vertex at (0, 1), and is centered at the origin.
2. Find an equation of an ellipse centered at the origin that is 14 units wide and 8 units high.
3. Find the foci of an ellipse with the equation 25x2 ± 36y2 = 900.
Graph the ellipse.
x 2 16
y 2 1 +
= 1
x 2 49
y 2 16 +
= 1
(± 11, 0)
10-4

69. 1. rise –5, run 2, through the origin 2. through (3,1) and (9, 3)
Hyperbolas
ALGEBRA 2 LESSON 10-5
(For help, go to Lesson 2-2 and Skills Handbook page 846.)
Write an equation of a line in slope-intercept form using the given information.
1. rise –5, run 2, through the origin     2.  through (3,1) and (9, 3)
Solve each equation for y.
3. – = – = – = 1
x 2 4
y 2 16
y 2 9
x 2 25
x 2 36
y 2 81
10-5

70. 1. slope = – , point (0, 0) 2. slope = = = , point (3, 1)
Hyperbolas
ALGEBRA 2 LESSON 10-5
1. slope = – , point (0, 0) 2. slope = = = , point (3, 1)
equation: y – 0 = – (x – 0) equation: y – 1 = (x – 3)
y = – x y = x 5 2
3 – 1
9 – 3 6 1 3
Solutions
x 2 4
y 2 16 – 9x 25 9
– = – = 1
16 = (16) = (225)1
4x2 – y2 = y2 – 9x2 = 225
y2 = 4x2 – y2 =
y = ± 4x2 – y = ±
y = ± 2 x2 – 4 = ± x2 + 25
10-5

71. Solutions (continued)
Hyperbolas
ALGEBRA 2 LESSON 10-5
9x2 – 324 4 3 2
= 1
324 = (324)1
9x2 – 4y2 = 324
y2 =
y = ±
= ± x2 – 36
x 2 36
y 2 81 –
Solutions (continued)
10-5

72. – = 1 Rewrite the equation in standard form.
Hyperbolas
ALGEBRA 2 LESSON 10-5
Graph 4x2 – 16y2 = 64.
4x2 – 16y2 = 64
– = 1 Rewrite the equation in standard form.
x 2 16
y 2 4
Since a2 = 16 and b2 = 4, a = 4 and b = 2.
The equation of the form – = 1, so the transverse axis is horizontal.
x 2
a 2
y 2
b 2
Step 1: Graph the vertices. Since the transverse axis is horizontal, the vertices lie on the x-axis. The coordinates are (±a, 0), or (±4, 0).
Step 2: Use the values a and b to draw the central rectangle. The lengths of its sides are 2a and 2b, or 8 and 4.
10-5

73. Hyperbolas
ALGEBRA 2 LESSON 10-5
(continued)
Step 3: Draw the asymptotes. The equations of the asymptotes are y = ± x or y = ± x . The asymptotes contain the diagonals of the central rectangle. b a 1 2
Step 4: Sketch the branches of the hyperbola through the vertices so they approach the asymptotes.
10-5

74. Find the foci of the graph – = 1.
Hyperbolas
ALGEBRA 2 LESSON 10-5
y 2 4
Find the foci of the graph – = 1.
x 2 9
The equation is in the form – = 1, so the transverse axis is horizontal; a2 = 9 and b2 = 4.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
= Substitute 9 for a2 and 4 for b2.
c = Find the square root of each side of the equation.
10-5

75. The asymptotes are the lines y = ± x , or y = ± x.
Hyperbolas
ALGEBRA 2 LESSON 10-5
(continued)
The foci (0, ±c) are approximately (0, –3.6) and (0, 3.6). The vertices (0, ±b) are (0, –2) and (0, 2).
The asymptotes are the lines y = ± x , or y = ± x. b a 2 3
10-5

76. c2 = a2 + b2 Use the Pythagorean Theorem.
Hyperbolas
ALGEBRA 2 LESSON 10-5
As a spacecraft approaches a planet, the gravitational pull of the planet changes the spacecraft’s path to a hyperbola that diverges from its asymptote. Find an equation that models the path of the spacecraft around the planet given that a = 300,765 km and
c = 424,650 km.
Assume that the center of the hyperbola is at the origin and that the transverse axis is horizontal. The equation will be in the form – = 1.
x 2 a2
y 2 b2
c2 = a2 + b2 Use the Pythagorean Theorem.
(424,650)2 = (300,765)2 + b2 Substitute.
1.803  1011 =  b2 Use a calculator.
10-5

77. The path of the spacecraft around the planet can be modeled by
Hyperbolas
ALGEBRA 2 LESSON 10-5
(continued)
b2 =  1011 –  1010 Solve for b2.
=  1010
– = 1 Substitute a2 and b2.
x 2
9.046  1010
y 2
8.987  1010
The path of the spacecraft around the planet can be modeled by
– = 1.
x 2
9.046  1010
y 2
8.987  1010
10-5

78. Hyperbolas pages 566–568 Exercises 1. 2. 3. 4. 5. 6. 10-5
ALGEBRA 2 LESSON 10-5
pages 566–568  Exercises 1. 2. 3. 4. 5. 6.
10-5

79. Hyperbolas 7. 8. 9. 10. (0, 97), (0, – 97) 11. (0, 113), (0, – 113)
ALGEBRA 2 LESSON 10-5 7. 8. 9.
10. (0, ), (0, – 97)
11. (0, ), (0, – )
12. ( , 0), (– , 0)
10-5

80. Hyperbolas
ALGEBRA 2 LESSON 10-5
13. (10, 0), (–10, 0)
14. (0, ), (0, )
15. ( , 0), (– , 0)
16. (0, ), (0, – 29)
17. ( , 0), (– , 0)
18. (0, ), (0, –4 3)
10-5

81. Hyperbolas 19. – = 1 27. 20. – = 1 30. – = 1 21. – = 1 22. – = 1
ALGEBRA 2 LESSON 10-5
19. – = 1
20. – = 1
– = 1
– = 1
23. – = 1
24. – = 1
25. y2 – = 1
26. – y2 = 1
x 2 9
y 2 16
69,169
96,480
144
192,432,384
170,203,465
240,000
10,000
1.865  1012
5.270  1011 25 3 4
27.
28.
29.
– = 1
31. – x2 = 1
20.25
10-5

82. Hyperbolas 32. – = 1 33. y = ± 2x2 – 8; (2, 0), (–2, 0)
ALGEBRA 2 LESSON 10-5
32. – = 1
33. y = ± x2 – 8; (2, 0), (–2, 0)
x 2 32
y 2 64 1 2
34. y = ± x2 – 1; (1, 0), (–1, 0)
35. y = ± 3x2 – 2; (–0.816, 0), (0.816, 0)
10-5

83. 37. Answers may vary. Sample: Similarities—
Hyperbolas
ALGEBRA 2 LESSON 10-5
36. Check students’ work.
37. Answers may vary. Sample: Similarities—
Both have two axes of symmetry that intersect
at the center of the figure. Both have two foci
that lie on the same line as the two “principal”
vertices. Differences—An ellipse consists of
points whose distances from the foci have a
constant sum, but a hyperbola consists of
constant difference. An ellipse is a closed
curve, but a hyperbola is not and has two
separate branches that do not touch.
Hyperbolas have asymptotes, but ellipses
do not. An ellipse intersects both its axes of
symmetry, but a hyperbola intersects only
one of its axes of symmetry.
38. Answers may vary. Sample:
axes of symmetry, vertices,
asymptotes
39. (0, ±1), y = ±x
40. (±1, 0), y = ±
41. (0, ±8), y = ±2x
42. (±5, 0), y = ±
43. (0, ±4), y = ±2x
44. (±7, 0), y = ±
45. Replace x with x – 3 and y
with y + 5 and rewrite
to obtain
– = 1 1 3 4 5 7
(x – 3)2 9
(y + 5)2
10-5

84. the branch that contains the vertex closest to your airport
Hyperbolas
ALGEBRA 2 LESSON 10-5
46. a. your airport
b. 30 km
c. – = 1 d.
the branch that contains the
vertex closest to your airport
y 2
351
x 2
225
47. a. For the x-values in those rows,
the value of x2 – 9 is negative and
so x2 – 9 is not a real number.
b. As x increases, y increases,
but the difference between x
and y gets closer to zero.
c. No; for positive values of x
greater than 3, x = x2 and
x2 = x2 – 9.
d. y = x, y = –x; /
10-5

85. 48. a. If the hyperbola – = 1 crossed either y = x or y = – x,
Hyperbolas
ALGEBRA 2 LESSON 10-5
48. a. If the hyperbola – = 1 crossed either y = x or y = – x,
then there would have to be a value of x and a value of y for which
x2 + b2 = x or x2 + b2 = – x . That would require that
x2 + b2 = x or x2 + b2 = – x. But squaring these equations
gives x2 + b2 = x2. This would mean that b2 = 0 and hence b = 0.
But for any hyperbola, b > 0. So the hyperbola never intersects
its asymptotes.
b. Yes; multiply both sides by to obtain – = 1. This equation is clearly
an equation of a hyperbola.Yes; multiply both sides by –1 to obtain
– = 1. This equation is clearly an equation of a hyperbola.
x 2 a2 a b
y 2 64 36 9 16 1 4
y 2 b2 a b
10-5

86. divide both sides by 1225 and then simplify, which leaves you
Hyperbolas
ALGEBRA 2 LESSON 10-5
49. D
50. H
51. D
52. [2] – = 1 or – = 1;
divide both sides by 1225 and
then simplify, which leaves you
with – = 1.
[1] answer correct, but no explanation
53. a – = 1
b – = 1
x 2 a2
y 2 b2 49 25
1.19  1012
8.86  1011
2.76  1010
2.05  1010
54. (± 34, 0), (0, ±5)
55. (0, ± 3), (± 2, 0)
56. (0, ±10), (±8, 0)
57.
58.
59.
60.
62. 6 3
10x 1
x + 3
x – 3
2x 2 – x – 7
x3 + x2 – 9x – 9
10-5

87. 2. Find the foci of the graph – = 1. Draw the graph.
Hyperbolas
ALGEBRA 2 LESSON 10-5
1. Graph 16y2 – 9x2 = 144.
2. Find the foci of the graph – = 1. Draw the graph.
x 2 4
y 2 36
(± , 0)
3. Find the equation of a hyperbola that has one focus located at (4, 0) and one vertex located at (–2, 0). Assume that the center of the hyperbola is at the origin.
x 2 4
y 2 12 –
= 1
10-5

88. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(For help, go to Lesson 5-3.)
Name the parent function for the equations in Exercises 1–4. Describe each equation as a translation of the parent function.
1. y = x y = (x – 3)2 – 2
3. y – 1 = x2 4. y = (x + 5)2 + 6
Rewrite each equation in vertex form.
5. y = x2 – 6x y = x2 + 10x – 7
7. y = 2x2 + 8x y = 4x2 – 12x + 3
10-6

89. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
1. y = x y = (x – 3)2 – 2
parent function: y = x2; parent function: y = x2; translated
translated 4 units up 3 units right and 2 units down
3. y – 1 = x2, or y = x y = (x + 5)2 + 6
parent function: y = x2; parent function: y = x2;
translated 1 unit up translated 5 units left and
6 units up
5. y = x2 – 6x + 1; c = – y = x2 + 10x – 7; c =
= (–3)2 = 9 = 52 = 25
y = (x2 – 6x + 9) + 1 – 9 y = (x2 + 10x + 25) – 7 – 25
y = (x – 3)2 – 8 y = (x + 5)2 – 32 6 2 10
Solutions
10-6

90. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Solutions (continued)
7. y = 2x2 + 8x y = 4x2 – 12x + 3
y = 2(x2 + 4x) + 5; c = y = 4(x2 – 3x) + 3; c =
= 22 = 4 =
y = 2(x2 + 4x + 4) + 5 – 2(4) y = 4 x2 – 3x – 4
y = 2(x + 2)2 + 5 – 8 y = – 9
y = 2(x + 2)2 – 3 y = – 6 4 2 3 – 9
x –
10-6

91. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Write an equation of an ellipse with center (–2, 4), a vertical major axis of length 10, and minor axis of length 8.
The length of the major axis is 2a. So 2a = 10 and a = 5.
The length of the minor axis is 2b. So 2b = 8 and b = 4.
Since the center is (–2, 4), h = –2 and k = 4.
The major axis is vertical, so the equation has the form
(x – h)2 b2
(y – k)2 a2 +
= 1.
(x – (–2))2 42
(y – 4)2 52 +
= 1.
Substitute –2 for h and 4 for k.
The equation of the ellipse is +
= 1.
(x + 2)2 16
(y – 4)2 25
10-6

92. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
(y – 4)2 = (400 – 25(x + 2)2) 1 16
Check: Solve the equation for y and graph both equations. +
= 1.
(x + 2)2
(y – 4)2 25
25(x + 2)2 + 16(y – 4)2 = 400
16(y – 4)2 = 400 – 25(x + 2)2
y – 4 = ± (400 – 25(x + 2)2)
y = 4 ± 400 – 25(x + 2)2 4
10-6

93. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Write an equation of a hyperbola with vertices (–1, 2) and (3, 2), and foci (–3, 2) and (5, 2).
Draw a sketch. The center is the midpoint of the line joining the vertices. Its coordinates are (1, 2).
The distance between the vertices is 2a, and the distance between the foci is 2c. 2a = 4, so a = 2; 2c = 8, so c = 4.
Find b2 using the Pythagorean Theorem.
c2 = a2 + b2
16 = 4 + b2
b2 = 12
10-6

94. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
The transverse axis is horizontal, so the equation has the form
(x – h)2 a2
(y – k)2 b2 –
= 1.
The equation of the hyperbola is
(x – 1)2 4
(y – 2)2 12 –
= 1.
10-6

95. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Use the information from Example 3. Find the equation of the hyperbola if the transmitters are 80 mi apart located at (0, 0) and (80, 0), and all points on the hyperbola are 30 mi closer to one transmitter than the other.
Since 2c = 80, c = 40. The center of the hyperbola is at (40, 0).
Find a by calculating the difference in the distances from the vertex at (a + 40, 0) to the two foci.
30 = (a + 40) – (80 – (a + 40))
= 2a
15 = a
10-6

96. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
Find b2.
c2 = a2 + b2
(40)2 = (15)2 + b2
1600 = b2
b2 = 1375
The equation of the hyperbola is
(x – 40)2
152
y 2
1375 –
= 1 or
225
= 1.
10-6

97. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Identify the conic section with equation 9x2 – 4y2 + 18x = 27. If it is a parabola, give the vertex. If it is a circle, give the center and the radius. If it is an ellipse or a hyperbola, give the center and foci. Sketch the graph.
Complete the square for the x- and y-terms to write the equation in standard form.
9x2 – 4y2 + 18x = 27
9x2 + 18x – 4y2 = 27 Group the x- and y- terms.
9(x2 + 2x + ) – 4y2 = 27 Complete the square.
9(x2 + 2x + 1) – 4y2 = (12) Add (9)(12) to each side.
9(x2 + 2x + 1) – 4y2 = Simplify
9(x + 1)2 – 4y2 = 36 Write the trinomials as binomials squared.
10-6

98. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
9(x + 1)2 36
4y 2 –
= 1
Divide each side by 36.
(x + 1)2 4
y 2 9
Simplify.
The equation represents a hyperbola. The center is (–1, 0). The transverse axis is horizontal.
Since a2 = 4, a = 2, b2 = 9, so b = 3.
c2 = a2 + b2
= 4 + 9
= 13
c = 13
10-6

99. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
The distance from the center of the hyperbola to the foci is Since the hyperbola is centered at (–1, 0), and the transverse axis is horizontal, the foci are located to the left and right of the center. The foci are at (– , 0) and (–1 – 13, 0).
10-6

100. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
pages 573–576  Exercises
1. + = 1
2. + = 1
= 1
4. + = 1
5. – = 1
6. – = 1
7. – = 1
8. – = 1
9. – = 1
(x + 2)2 9
(y – 1)2 4
(x – 5)2 16
(y – 3)2 36
(x – 3)2
(y + 6)2 49
x 2
(y + 4)2 25
(x + 3)2
(y + 3)2
(x – 4)2 32
(x + 1)2
(y – 2)2 40
(y + 1)2 56
10. – = 1
11. – = 1
12. y = (x – 4)2 + 3; parabola, vertex (4, 3)
13. (x + 6)2 + y2 = 81; circle, center (–6, 0), radius 9
(x – 150)2
1296
y 2
21,204
(x – 175)2
1936
28,689
10-6

101. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
= 1; ellipse,
center (–1, 3), foci (–1, 3 ± )
15. (x – 1)2 + (y + 3)2 = 13; circle,
center (1, –3), radius
(x + 1)2 3
(y – 3)2 9
16. (y – 2)2 – (x – 3)2 = 1; hyperbola,
center (3, 2), foci (3, 2 ± 2)
17. – (y + 1)2 = 1; hyperbola,
center (1, –1), foci (1 ± , –1)
(x – 1)2 4
10-6

102. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
18. x2 + (y + 7)2 = 36; circle,
center (0, –7), radius 6
19. x – 3 = (y – 2)2; parabola,
vertex (3, 2) 1 2
= 1; ellipse,
center (–2, 3), foci (–2 ± 5, 3)
21. (x + 3)2 – (y – 5)2 = 1; hyperbola,
center (–3, 5), foci (–3 ± 2, 5)
(x + 2)2 9
(y – 3)2 4
10-6

103. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
= 1; ellipse,
center (1, 0), foci (1 ± , 0)
– = 1; hyperbola,
center (0, –3), foci (± 13, –3) x2 4
(y + 3)2 9
24. Translate the equation – = 1,
a hyperbola centered at (0, 0), 3 units left.
25. a. hyperbola
b. line
26. a. h is added to each x-coordinate,
and k is added to each y-coordinate.
b. The lengths of the major and minor axes
are unchanged; the x-coordinates of the
vertices are increased (or decreased) by
the same amount, and the same is true
for the y-coordinates. A similar remark
holds for the co-vertices. 16
(x – 1)2 16
y 2 4
y 2 8
10-6

104. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
27. The hyperbola originally had
center (0, 0) and a horizontal
or vertical transverse axis. If the
new center is (h, k), then the
equations of the new asymptotes
are obtained by replacing x with
x – h and y with y – k in the equations
of the original asymptotes.
28. Multiplying both sides of the equation
by 16 gives x2 + y2 = 16. This is an
equation of the circle with center (0, 0)
and radius 4.
29. (x + 6)2 + (y – 9)2 = 81
= 1
(x – 3)2 36
(y – 2)2 9
10-6

105. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
31. y = (x – 2)2 – 3
32. – = 1
(y + 3)2 4
(x – 6)2 5 1 20
(x – 1)2 9
(y + 1)2 16
(x – 3)2 36
(y + 2)2
15.84 64
(y – 2)2
(y – 5)2
(x – 5)2
(y – 8)2 25
(y – 9)2
(x – 2)2
(y – 7)2 x2
(x – 4)2
(y – 10)2 y2
= 1
34. – = 1
35. (x + 4)2 + (y + 4)2 = 25
36. x – 1 = (y + 3)2
37. (x – 8)2 + (y – 2)2 = 4
= 1
39. y – 5 = 4(x – 3)2
40. – = 1
41. – = 1
= 1
= 1
= 1
45. – (y – 9)2 = 1
46. (x – 8)2 = 12(y – 11)
= 1
48. x2 – (y – 10)2 = 1
49.
ellipse, = 1
10-6

106. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
50.
hyperbola, – = 1
51.
52.
circle, x2 + y2 = 36 x2 9 y2 36
53.
ellipse, = 1
54.
parabola, x = y2 + 4
55. Check students’ work.
56. Check students’ work.
10-6

107. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
58. a. Earth: = 1
Mars: = 1
Mercury: = 1
b. Earth: is closest to 1 for Earth.
59. C
60. A
61. A
62. F
57. a. A = B; A = B and A and B have
the same sign; A and B have
opposite signs; A = 0 or B = 0,
but not both A and B are zero.
b. No; answers may vary. Sample:
If C = D = E = 0, then the graph
will be the single point (0, 0).
c. two intersecting lines
(y = x and y = –x) / x2
(149.60)2 y2
(149.58)2
(227.9)2
(226.9)2
(57.9)2
(56.6)2 a b
10-6

108. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
63. [4] First group the x- and y-terms so the
equation is x2 – 14x + y2 + 10y = 26.
Complete the square for both the x
and y groups of terms and factor;
(x – 7)2 + (y + 5)2 = 100 is the
equation of a circle with center
(7, –5) and radius 10.
[3] does not add 49 and 25 to 26 when
completing the square
[2] errors in completing the square and
factoring leading to wrong equation
[1] answer only, without explanation
64. ( 85, 0), ( – 85, 0)
65. (0, ), (0, – 21)
66. (0, ), (0, – )
10-6

109. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
67. –1, 4
68. 5
69. about –0.4315,
70. 1
71. 2
72. 3
73. 8
10-6

110. Translating Conic Sections
ALGEBRA 2 LESSON 10-6
1. Write an equation of an ellipse with center (–5, –2), horizontal major axis of length 6, and minor axis of length 4.
2. Find an equation of a hyperbola with vertices (0, –1) and (4, –1), and foci (–1, –1) and (5, –1).
3. Identify the conic section with equation x2 – 4x + y2 + 2y = –1. If it is a parabola, give the vertex. If it is a circle, give the center and radius. If it is an ellipse or a hyperbola, give the center and foci. Sketch the graph.
(x + 5)2 9
(y + 2)2 4 +
= 1
(x – 2)2 4
(y + 1)2 5 –
= 1
circle; center: (2, –1), radius: 2
10-6

111. 1. center (0, 0), intercepts (±2, 0), (0, ±3);
Quadratic Relations
ALGEBRA 2 CHAPTER 10
8. focus – , 0 , directrix x =
9. focus 0, , directrix y = –
10. y = – x2
11. x = y2
12. x = – y2
13. y = x2
14. center (2, 3), radius 6;
1. center (0, 0), intercepts (±2, 0), (0, ±3);
domain: –2 x 2, range: –3 y 3
2. center (0, 0), intercepts (±2, 0);
domain: x –2 or x 2,
range: all real numbers
3. center (0, 0), intercepts (±2, 0), (0, ±1);
domain: –2 x 2, range: –1 y 1
4. center (0, 0), intercepts (0, ±2);
domain: all real numbers,
range: y –2 or y 2
5. Write the equation in standard form.
6. focus 0, , directrix y = –
7. focus – , 0 , directrix x = 1 12 8 20 18 28 4
Page 582
> –
<
10-A

112. Quadratic Relations 15. center (–5, –8), radius 10;
ALGEBRA 2 CHAPTER 10
15. center (–5, –8), radius 10;
16. center (1, –7), radius 9;
17. center (–4, 10), radius 11;
= 1
y2 = 1
= 1 x2 64 y2 25 36
6.25
20.25
10-A

113. Quadratic Relations 21. foci (±3 5, 0); 22. foci (0, ±4 6)
ALGEBRA 2 CHAPTER 10
21. foci (±3 5, 0);
22. foci (0, ±4 6)
23. foci (0, ±4 3)
24. foci (0, ± 3)
25. circle
26. foci (± , 0)
27. foci (0, ± )
10-A

114. 38. circle; center (2, 3), radius 3 28. foci (±2 17, 0)
Quadratic Relations
ALGEBRA 2 CHAPTER 10
– = 1
36. (x + 4)2 – = 1
37. parabola; vertex (2, 1)
38. circle; center (2, 3), radius 3
28. foci (± , 0)
29. foci (0, ± )
30. Check students’ work.
= 1
= 1
= 1
– = 1 x2 16 y2 9
(x + 2)2
(y – 7)2
(x – 3)2 25
(y + 2)2 36
(x – 2)2 24
(y + 1)2 15
10-A

115. 39. ellipse; center (2, 3), foci (2, 3 ± 2 3)
Quadratic Relations
ALGEBRA 2 CHAPTER 10
39. ellipse; center (2, 3), foci (2, 3 ± )
40. hyperbola; center (2, 3), foci (2 ± , 3)
10-A