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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 + 6
2. 2y = –x – 3
3. 3x – 4y = –12
4. y = x2 – 4
5. y = (x – 3)2
6. y = –4x2 + 1
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Solutions
1. y = 3x + 6
x-intercept (let y = 0):
0 = 3x + 6
–3x = 6
x = –2
y-intercept (let x = 0):
y = 3(0) + 6
y = 0 + 6
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 = – 3 , or –1.5
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 =
– 4
x-intercept (let y = 0):
0 = x2 – 4
4 = x2
x = ±2
y-intercept (let x = 0):
y = 02 – 4
y = –4
x2
10-1
2
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 = –4x2 + 1
x-intercept (let y = 0):
0 = –4x2 + 1
4x2 = 1
x2 = 1
4
x =
±
1
4
= ± 1
y-intercept (let x = 0):
y = –4(0)2 + 1
y = 0 + 1
y = 1
10-1
2
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.
Make a table of values.
x
y
–4
0
–3 ± 7
± 2.6
–2 ± 2 3
± 3.5
–1 ± 15 ± 3.9
0
±4
1 ± 15 ± 3.9
2 ±2 3
± 3.5
3 ± 7
± 2.6
4
0
Plot the points and connect
them with a smooth curve.
10-1
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| –4 <
– x <
– 4}. The range
is {y| –4 <
– y <
– 4}.
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
Graph the equation 9x2 + 4y2 = 36. Describe the graph and
the lines of symmetry. Then find the domain and range.
Make a table of values.
x
–2
–1
0
1
2
Plot the points and connect
them with smooth curves.
y
0
± 2.6
±3
± 2.6
0
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| –2 <
– x <
– 2}.
The range is {y| –3 <
– y <
– 3}.
10-1
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.
Make a table of values.
x
–5
–4
–3
–2
–1
0
1
2
3
4
5
Plot the points and connect
them with smooth curves.
y
± 4.6
± 3.5
± 2.2
0
—
—
—
0
± 2.2
± 3.5
± 4.6
10-1
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 >
– 2}. The range is all real numbers.
10-1
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| –5 <
– x <
– 5}. The range is {y| –4 <
– y <
– 4}.
10-1
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, 25x2 + 4y2 = 100, or x2 + y2 = 16.
a.
b.
c.
a. The equation 25x2 + 4y2 = 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
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
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
pages 538–541 Exercises
2.
1.
Hyperbola: center (0, 0),
y-intercepts at ± 5 3 , no
3
x-intercepts, the lines of
symmetry are the x- and
y-axes; domain: all real
Ellipse: center (0, 0), x-intercepts at ±3 2,
y-intercepts at ±6, the lines of symmetry
are the x- and y-axes; domain:
–3 2<
–x <
– 3 2, range –6 <
–y <
– 6.
5 3
numbers, range: y >
–
5 3.
or y <
–
3
3
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
3.
5.
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.
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.
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 <
– 3 or y >
– 3.
10-1
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.
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
7.
9.
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.
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.
10.
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.
10-1
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.
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
11.
13.
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.
Ellipse: center (0, 0), x-intercepts at ±1,
y-intercepts at ± 1 , the lines of symmetry
3
are the x- and y-axes; domain:
1 y 1.
–1 <
–x <
– 1, range: 3 <
– <
–3
Circle: center (0, 0), radius 5, xand y-intercepts at ± 5, there are
infinitely many lines of symmetry;
domain: – 5 <
–x <
– 5,
range: – 5 <
–y <
– 5.
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
14.
16.
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.
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
Hyperbola: center (0, 0), y-intercepts at ± 1 ,
2
the lines of symmetry are the x- and y-axes;
1
1
domain: all real numbers, range: y <
– – or y >
– .
2
10-1
2
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
23. 19
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
28. 22
10-1
24. 17
25. 18
26. 20
27. 21
29.
Hyperbola: center (0, 0), x-intercepts
±4, the lines of symmetry are the xand y-axes; domain: x <
– –4 or x >
– 4,
range: all real numbers.
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
30.
32.
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.
Ellipse: center (0, 0), x-intercepts at
± 8 5 , y-intercepts at ±2
5
5, the
lines of symmetry are the x- and
8 5,
y-axes; domain: – 8 5 <
x<
–
–
5
5
8 5.
range: – 8 5 <
y
<
– –
5
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.
10-1
5
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
33. a. All lines in the plane that pass through
35–40.
the center of a circle are axes of
symmetry of the circle.
35.
b. The axes of symmetry of an ellipse
intersect at the center of the ellipse.
36.
The same is true for a hyperbola.
37.
This can be confirmed using, for example,
4x2 + 9y2 = 36 and 4x2 – 9y2 = 36.
38.
34. a. Let the lamp sit in a normal, upright position,
39.
but close enough to the wall for the bottom rim
40.
of the shade to almost touch the wall.
b. Hold the lamp so that the shade contacts
41.
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.
10-1
Answers may vary.
Samples are given.
(2, 4)
(
2, 1)
(–2, 2
2)
(2, 0)
(–3,
51)
(0, –
7)
x2 + y2 = 36
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
45. Check students’ work.
42.
46. a.
x2 + y2 = 1
4
43.
b. hyperbola
c. No intercepts, but y = x
and y = –x are lines of
symmetry.
x2
+
y2
= 16
44.
x2 + y2 = 1.5625
d. yes; ƒ(x) = 16
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.
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
48. a.
53. I
54. B
b.
49. a parabola, or part of a hyperbola
50. D
51. F
52. D
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 <
– 11.
[1] includes either description OR domain
and range, but not both
56. 1
2
57. 1
2
58. 1
2
10-1
Exploring Conic Sections
ALGEBRA 2 LESSON 10-1
59. 1
69. y = 1 (2)x
60. z = –5xy; 30
70. x3 – 3x2y + 3xy2 – y3
61. z = –4xy; 24
71. p6 + 6p5q + 15p4q2+ 20p3q3 + 15p2q4+ 6pq5 + q6
4
2
62. z = 40xy; –240
63. z = – 1 xy; 3
72. x4 – 8x3 + 24x2 – 32x + 16
73. 243 – 405x + 270x2 – 90x3 + 15x4 – x5
2
64. y = 4(2)x
65. y = 1 (4)x
2
66. y = 3(2)x
67. y = 1 (3)x
3
68. y = 2(3)x
10-1
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 2 <
– y < }
2. 16x2 + 25y2 = 400
3. 2x2 + 2y2 = 50
ellipse, x-int: (±5, 0),
y-int: (0, ±4);
domain: {x| –5 <
– x <
– 5},
range: {y| –4 <
– y <
– 4}
circle, x-int: (±5, 0),
y-int: (0, ±5);
domain: {x| –5 <
– x <
– 5},
range: {y| –5 <
– y <
– 5}
10-1
Parabolas
ALGEBRA 2 LESSON 10-2
(For help, go to Skills Handbook pages 844 and 856.)
Solve for c.
1. 1 = 1
8
c
2. 1 = 1
8
2c
3. 1 = 1
12
4c
4. 2 = 1
4c
Find the distance between the given points.
5. (2, 3) and (4, 1)
6. (4, 6) and (3, –2)
10-2
7. (–1, 5) and (2, –3)
Parabolas
ALGEBRA 2 LESSON 10-2
Solutions
1. 1 = 1 ; c = 8
8
c
2. 1 = 1 ; 2c = 8; c = 4
8
3.
2c
1 = 1 ; 4c = 12; c = 3
12
4c
4. 2(4c) = 1; 8c = 1; c = 1
8
5. (2, 3) and (4, 1); distance =
(–2)2 + 22 =
4 + 4
(2 – 4)2 + (3 – 1)2 =
=
10-2
8
=
4 • 2
= 2
2
Parabolas
ALGEBRA 2 LESSON 10-2
Solutions (continued)
6. (4, 6) and (3, –2); distance =
12 + 82 =
1 + 64
(4 – 3)2 + (6 – (–2))2 =
=
65
7. (–1, 5) and (2, –3); distance =
(–3)2 + 82
=
9 + 64
(–1 – 2)2 + (5 – (–3))2 =
=
10-2
73
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
Parabolas
ALGEBRA 2 LESSON 10-2
(continued)
FP = PQ
(x – 0)2 + (y – 1)2
x2 + (y – 1)2
x2 + y2 – 2y + 1
x2
=
(x – x)2 + (y – (–1))2
= 02 + (y + 1)2
= y2 + 2y + 1
= 4y
y = 1 x2
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 = 1 x2.
4
10-2
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
Parabolas
ALGEBRA 2 LESSON 10-2
(continued)
1
| a | = 4c
1
= 4(7)
Since the focus is a distance of
7 units from the vertex, c = 7.
1
= 28
Since the parabola opens downward, a is negative.
So a = – 1 .
28
An equation for the parabola is y = – 1 x2.
28
10-2
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)
=
1
16
Since the parabola opens upward, a is positive.
The equation of the parabola is y = 1 x2.
16
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
Identify the focus and directrix of the graph of the
equation x = – 1 y2.
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
| – 1 | =
8
4c
4c = 8
c = 2
The focus is at (–2, 0). The equation of the directrix is x = 2.
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
Identify the vertex, the focus, and the directrix of the graph of
the equation x2 + 4x + 8y – 4 = 0. Then graph the parabola.
x2 + 4x + 8y – 4 = 0
8y = –x2 – 4x + 4
Solve for y, since y is the only term.
8y = –(x2 + 4x + 4) + 4 + 4
Complete the square in x.
y = – 1 (x + 2)2 + 1
8
vertex form
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
Parabolas
ALGEBRA 2 LESSON 10-2
(continued)
1
| a | = 4c
1
| –1 | =
8
4c
4c = 8
Substitute – 1 for a.
8
Solve for c.
c = 2
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
Parabolas
ALGEBRA 2 LESSON 10-2
pages 546–548 Exercises 10. x = – 1 y2
4
1. y
2. y
3. x
4. y
5. y
6. x
7. x
8. y
9. y
= 1 x2
8
= – 1 x2
4
= – 1 y2
12
= – 1 x2
32
= 1 x2 + 2
8
= 1 y2
2
= 1 y2
24
= – 1 x2
16
= 1 x2
28
17. 0, 1 , y = – 1
4
4
11. x = 1 y2
8
12. y = – 1 x2
20
13. y = 1 x2
6
18. (0, –2), y = 2
14. y = 2x2
21. (9, 0), x = –9
15. a. Answers may vary.
22. – 9 , 0 , x = 9
19.
20.
1, 0 , x = –
2
0, 1 , y = –
2
2
1
2
1
2
2
Sample: y = x2
1
1
b. The light produced by the 23. 0, – 8 , y = 8
bulb will reflect off the
parabolic mirror and will
be emitted in parallel rays.
16. (0, 1), y = –1
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
24. (0, 0), (6, 0), x = –6
26. (0, 0), (3, 0), x = –3
28. (0, 0), (0, 1), y = –1
25. (0, 0), (0, –1), y = 1
27. (0, 0), 25 , 0 , x = – 25
29. (0, 0), (0, –1), y = 1
4
10-2
4
Parabolas
ALGEBRA 2 LESSON 10-2
30. (2, 0), (2, 1), y = –1
32. (–2, 4), –2, 17 , y = 15
34. (4, 0), (4, –6), y = 6
31. (0, 0), (–2, 0), x = 2
33. (–3, 0), – 3 , 0 , x = – 9
35. (3, –1), (6, –1), x = 0
4
2
10-2
4
2
Parabolas
ALGEBRA 2 LESSON 10-2
36. x = 1 y2
43. x = – 1 y2
37. y
44. y = 1 x2
38. y
39. x
40. x
41. y
42.
12
= 1 x2
400
= – 1 x2
20
= – 1 y2
28
= – 1 y2
36
= – 5 x2
56
8
48.
4
45. x = y2
46.
49.
47.
50.
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
55. Check students’ work.
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).
56. Answers may vary. Sample:
58.
51.
52. y = 1 (x – 1)2 + 1
6
53. x = – 1 (y – 1)2 + 1
2
54. y = – 1 (x – 1)2 + 1
4
Write the equation in the form
x= 1
4 1
y2. The distance from
8
the focus to the directrix is 2 1 , or 1 .
8
4
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
65. [2] |a| = 1
59. a. the bottom half of the
4c
parabola y2 = x
|a| = 1
b. domain: all non-negative
4.4
real numbers, range: all
|a| = 1
16
non-positive real numbers
so since the focus is at (0, 4) and
60. a. The vertex moves up, and
the directrix is y = –4, a is positive.
the parabola widens.
The equation is y = 1 x2.
b. The vertex moves down, and
16
[1] answer correct, without work shown
the parabola narrows.
c. The parabola would degenerate 66. [2] |a| = 1
4c
into the ray with endpoint (0, –2)
1
= 1
that passes through the origin.
36
4c
36 = 4c
61. B
c =9
62. F
so the focus is at (0, 9) since a > 0
and the equation of the directrix
63. A
is y = –9.
64. H
[1] answer correct, without work shown
10-2
Parabolas
ALGEBRA 2 LESSON 10-2
67.
69.
70. the x- and y-axes
71. x = 0 and y = 4
68.
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. $5254.06
10-2
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( – 1 , 0) and the line x = 1 .
x =
–2y2
8
8
2. Write an equation of a parabola with a vertex
at the origin and a focus at (0, –3).
y = – 1 x2
12
3. Identify the vertex, the focus, and the directrix
of the equation y2 – 4y – 16x + 36 = 0.
Then, graph the parabola.
(2, 2), (6, 2), x = –2;
10-2
Circles
ALGEBRA 2 LESSON 10-3
(For help, go to Lesson 5-7 and Skills Handbook page 855.)
Simplify.
1.
16
2.
49
3.
20
4.
48
5.
72
Find the missing value to complete the square.
6. x2 – 2x +
7. x2 + 4x +
10-3
8. x2 – 6x +
Circles
ALGEBRA 2 LESSON 10-3
Solutions
1.
16 = 4
2.
49 = 7
3.
20 =
4  5 = 2
5
4.
48 =
16  3 = 4
3
5.
72 =
36  2 = 6
2
6. x2 – 2x +
; c =
b
2
2
7. x2 + 4x +
; c =
b
2
8. x2 – 6x +
; c =
b
2
=
–2
2
=
4
2
2
=
–6
2
2
2
2
10-3
= (–1)2 = 1
= 22 = 4
2
= (–3)2 = 9
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
(x – 3)2 + (y – (–2))2 = 32
(x – 3)2 + (y + 2)2 = 9
Use the standard form of the
equation of a circle.
Substitute 3 for h, –2 for k, and
3 for r.
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 + 2 = ±
9 – (x – 3)2
y = –2 ±
9 – (x – 3)2
10-3
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
(x – 2)2 + (y – (–1))2 = 16
(x – 2)2 + (y + 1)2 = 16
Use the standard form of
the equation of a circle.
Substitute 2 for h, –1 for k,
and 16 for r 2.
Simplify.
The equation is (x – 2)2 + (y + 1)2 = 16.
10-3
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
A
B
(h, k)
(
r
)
–5 1 , 0
5
(0, 0)
1
2
2
Equation
(
)
2
x + 11 + y2 = 25
2
x2 + y2 = 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
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
Find h, k, and r.
k = 2
r = 6
The center of the circle is (–4, 2). The radius is 6.
10-3
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 = 3
k = –1
r 2 = 4, or r = 2
Draw the center (3, –1) and radius 2.
Draw a smooth curve.
10-3
Circles
ALGEBRA 2 LESSON 10-3
12. (x + 1)2 + (y – 3)2 = 81
pages 552–554 Exercises
1. x2 + y2 = 100
13. x2 + (y + 5)2 = 100
2. (x + 4)2 + (y + 6)2 = 49
14. (x – 3)2 + (y – 2)2 = 49
3. (x – 2)2 + (y – 3)2 = 20.25
15. (x + 6)2 + (y – 1)2 = 20
4. (x + 6)2 + (y – 10)2 = 1
16. (x – 5)2 + y2 = 50
5. (x – 1)2 + (y + 3)2 = 100
17. (x + 3)2 + (y – 4)2 = 9
6. (x + 5)2 + (y + 1)2 = 36
18. (x – 2)2 + (y + 6)2 = 16
7. (x + 3)2 + y2 = 64
19. (1, 1), 1
8. (x + 1.5)2 + (y + 3)2 = 4
20. (–2, 10), 2
9. x2 + (y + 1)2 = 9
21. (3, –1), 6
10. (x + 1)2 + y2 = 1
22. (–3, 5), 9
11. (x – 2)2 + (y + 4)2 = 25
23. (0, –3), 5
10-3
Circles
ALGEBRA 2 LESSON 10-3
24. (–6, 0), 11
29.
32.
30.
33.
25. (–2, –4), 16
26. (3, 7), 4
6
27.
28.
31.
34.
10-3
Circles
ALGEBRA 2 LESSON 10-3
35. x2 + y2 = 16
47. (x + 2)2 + (y – 7.5)2 = 2.25
36. x2 + y2 = 9
48. (x – 1)2 + (y + 2)2 = 10
37. x2 + y2 = 25
49. (x – 2)2 + (y – 1)2 = 25
38. x2 + y2 = 3
50. (x – 6)2 + (y – 4)2 = 25
39. x2 + y2 = 25
51. (x + 1)2 + (y + 7)2 = 36
40. x2 + y2 = 169
52. (x – 24)2 + (y – 22)2 = 100
41. x2 + y2 = 169
53. Check students’ work.
42. x2 + y2 = 13
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.
43. x2 + y2 = 26
44. x2 + y2 = 52
55. (0, 0),
45. (x + 6)2 + (y – 13)2 = 49
56. (0, –1),
5
46. (x – 5)2 + (y + 3)2 = 25
57. (0, 0),
14
10-3
2
Circles
ALGEBRA 2 LESSON 10-3
58. (0, 4),
66. parabola; x = (y + 2)2 + 3;
11
59. (–5, 0), 3
2
60. (–2, –4), 5
61. (–3, 5),
2
38
62. (–1, 0), 2
67. Let P(x, y) be any point on the circle
centered at the origin and having radius r.
64. (0, 2), 2 5
If P(x, y) is one of the points (r, 0), (–r, 0),
(0, r), or (0, –r), substitution shows that
65. circle; (x – 4)2 + (y – 3)2 = 16;
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.
63. (3, 1),
6
10-3
Circles
ALGEBRA 2 LESSON 10-3
68. a. The radius would have length 0. 70. a. (x – 3)2 + (y – 4)2 = 25
b. y = – 1 x2 + 10 x
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
3
3
81. 4
82. 2
71. 12
83. –3
72. 2
84. 4
73. 3.25
85. 4
74. 5
86. 1
75. 10
87. 4
76. 3.14
88. –2
77. x = – 1 y2
89. 1
2
12
78. x = –1
79. x = 2, x = 3
80. no discontinuities
10-3
Circles
ALGEBRA 2 LESSON 10-3
1. Write an equation of a circle with center (–2, –1) and radius 3.
Graph the circle.
(x + 2)2 + (y + 1)2 = 9;
2. Write and equation for the translation of x2 + y2 = 25 three units left
and two units up.
(x + 3)2 + (y – 2)2 = 25
3. Find the center and radius of the circle with equation
(x – 7)2 + (y – 1)2 = 49.
(7, 1); 7
10-3
Ellipses
ALGEBRA 2 LESSON 10-4
(For help, go to Lesson 5-5 and Skills Handbook page 846.)
Solve each equation.
1. 27 = x2 + 11
2. x2 = 48
3. 84 = 120 – x2
Evaluate each expression for a = 3 and b = 5.
4. a2 + b2
5. a2 – b2
10-4
6. b2 – 2a2
Ellipses
ALGEBRA 2 LESSON 10-4
Solutions
1. 27 = x2 + 11
16 = x2
x = ± 16 = ±4
2. x2 = 48
x = ±
x =
3. 84
–36
36
48
±4 3
= 120 – x2
= –x2
= x2
x = ± 36 = ±6
4. a2 + b2 for a = 3 and b = 5: 32 + 52 = 9 + 25 = 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
10-4
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.
x2
b2 +
y2
a2 = 1
x2
9 +
y2
16
= 1
An equation of the ellipse is
Standard form for an equation of an
ellipse with a vertical major axis.
Substitute 9 for b2 and 16 for a2.
x2
y2
9 + 16 = 1.
10-4
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.
x2
a2 +
y2
b2 = 1
x2
y2
100 + 25
= 1
Standard form for an ellipse with a
horizontal major axis.
Substitute 100 for a2 and 25 for b2.
x2
y2
An equation of the ellipse is 100 + 25 = 1.
10-4
Ellipses
ALGEBRA 2 LESSON 10-4
Find the foci of the ellipse with the equation 9x2 + y2 = 36.
Graph the ellipse.
9x2 + y2 = 36
y2
x2
+
=1
4
36
Write in standard form.
Since 36 > 4 and 36 is with y2, the major axis is vertical, a2 = 36, and b2 = 4.
c2 = a2 – b2
= 36 – 4
Find c.
Substitute 4 for a2 and 36 for b2.
= 32
c=±
32 = ± 4
2
The major axis is vertical, so the coordinates of the foci are (0, ±c).
The foci are (0, 4 2 ) and (0, –4 2).
10-4
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.
x2 y2
+
=1
4
20
Substitute 20 for a2 and 4 for b2.
y2
x2
An equation of the ellipse is
+
= 1.
4
20
10-4
Ellipses
ALGEBRA 2 LESSON 10-4
pages 559–561 Exercises
2
1. x +
y2
=1
9
16
2
2. x + y2 = 1
4
2
3. x + y2 = 1
9
y2
2
4. x +
=1
36
2
2
5. x + y = 1
49
16
2
2
6. x + y = 1
25
36
2
2
7. x + y = 1
4
81
2
2
8. x + y = 1
25
9
2
2
9. x + y = 1
2.25 0.25
2
2
10. x + y = 1
11.
12.
13.
14.
15.
16.
17.
256
64
x2 + y2 = 1
100
36
x2 + y2 = 1
25
12.25
x2 + y2 = 1
49
196
2
x2 + y = 1
16
x 2 + y2
=1
256 12.25
x2 + y2 = 1
400
900
2
x2 + y = 1
6.25
10-4
18. (0,
5), (0, –
19. (0, 4), (0, –4)
5)
Ellipses
ALGEBRA 2 LESSON 10-4
20. (4
2, 0), (–4
21. (8, 0), (–8, 0)
2, 0)
22. (0, 6), (0,–6)
23. (0,
6), (0, –
10-4
24. (2
6)
3, 0), (–2
25. (9, 0), (–9, 0)
3, 0)
Ellipses
ALGEBRA 2 LESSON 10-4
26. (3
15, 0), (–3
15, 0) 33. (
5, 0), (–
34. (0, 2
3), (0, –2
3)
35. (0, 4
2), (0, –4
2)
36. (0,
37. (0, 2
27.
28.
29.
30.
31.
32.
x2
100
x2
64
x2
89
x2
4
x2
245
x2
514
+
+
+
+
+
+
y2
=1
64
y2 = 1
128
y2 = 1
64
y2
=1
20
y2
=1
49
y2
=1
225
41. a. 0.9;
5, 0)
21), (0, –
7), (0, –2
21)
7)
b. 0.1;
38. (0, 1), (0, –1)
39. (–3, 8), (–3, 2)
40. (–2,
2), (–2, –
2)
c. The shape is close
to a circle.
d. The shape is close
to a line segment.
10-4
Ellipses
ALGEBRA 2 LESSON 10-4
42.
x2 + y2 = 1
20.25
4
47.
x2 + y2 = 1
16
4
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.
48. The vertices are the points farthest
from the center and the co-vertices
are the points closest to the center.
x2 + y2 = 1
4
9
2
45. x + y2 = 1
16
2
46. x2 + y = 1
9
53.
44.
49. Check students’ work.
50.
51.
52.
54.
55.
10-4
x2 + y2
3
4
x2 + y2
4
25
x2 + y2
81
121
x2
+
702.25
x2 + y2
169 144
x2 + y2
256 324
=1
=1
=1
y2
=1
210.25
=1
=1
Ellipses
ALGEBRA 2 LESSON 10-4
56.
57.
58.
59.
60.
61.
62.
63.
x2 + y2 = 1
72.25 90.25
x2 + y2 = 1
400 100
x2 + y2 = 1
12
16
x2 + y2 = 1
25
16
x2 + y2 = 1
64
39
x2 + y2 = 1
27
36
x2 + y2 = 1
9
4
x2 + y2 = 1
20
18
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.
y2
x2
+
=1
15
15
8.64675  10
8.649  10
67. area of blue region = 3(area of white region)
68. 10
799 or about 282.7 ft
10-4
Ellipses
ALGEBRA 2 LESSON 10-4
69. D
73. (x – 2)2 + (y + 3)2 = 36
70. I
74. (x + 4)2 + (y – 7)2 = 121
71. C
72. [2] The x-axis is the major
axis since 50 > 40. The
equation is of the form
x 2 + y 2 = 1. 2a = 50,
b2
a2
so a = 25, and 2b = 40,
78. log 15
so b = 20. Therefore,
79. log3 6
a2 = 625 and b2 = 400.
80. log 2
2
2
The equation is x + y = 1.
625
400
[1] answer only OR minor error
in calculating a2 and b2
3
1
; x =/ 0, 2
4
2 – 3x
3
76. x – 6 ; x =/ 1, –6
x–1
77. 2 x – 5
; x =/ –2
x – 2x + 4
75.
81. log 320
82. log x
y
k
83. log 5
4
10-4
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.
x2
y2
+
=1
16
1
2. Find an equation of an ellipse centered at the origin that is 14 units
wide and 8 units high.
x2
y2
+
=1
49
16
3. Find the foci of an ellipse with the equation 25x2 ± 36y2 = 900.
Graph the ellipse.
(±
11, 0)
10-4
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.
x2
y2
3.
–
=1
4
16
y2 x2
4.
–
=1
9
25
10-5
x2
y2
5.
–
=1
36 81
Hyperbolas
ALGEBRA 2 LESSON 10-5
Solutions
1. slope = – 5 , point (0, 0)
2
2. slope = 3 – 1 = 2 = 1 , point (3, 1)
9–3
equation: y – 1 =
2
y=–5x
2
y2
4.
–
9
2
225 y –
9
4x2 – y2 = 16
y2
=
4x2
y =±
y =±2
3
1
(x – 3)
3
y= 1x
3
equation: y – 0 = – 5 (x – 0)
x2 y2
3.
–
=1
4
16
2
2
16 x – y = (16)1
4
16
6
x2
25
x2
25
=1
= (225)1
25y2 – 9x2 = 225
9x2 + 255
=
25
2
y = ± 9x + 255
25
= ± 3 x2 + 25
5
– 16
y2
4x2 – 16
x2 – 4
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
Solutions (continued)
x2
y2
–
5.
36 81 = 1
x2
y2
324
= (324)1
–
36 81
9x2 – 4y2 = 324
y2
9x2 – 324
=
4
y =±
=± 3
2
9x2 – 324
4
x2 – 36
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
Graph 4x2 – 16y2 = 64.
4x2 – 16y2 = 64
y2
x2
–
=1
16
4
Rewrite the equation in standard form.
y2
x2
The equation of the form 2 – 2 = 1, so the transverse axis is horizontal.
a
b
Since a2 = 16 and b2 = 4, a = 4 and 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
Hyperbolas
ALGEBRA 2 LESSON 10-5
(continued)
Step 3: Draw the asymptotes. The equations of the asymptotes are
y=±
b
x or y = ± 1 x . The asymptotes contain the diagonals of the
a
2
central rectangle.
Step 4: Sketch the branches of the hyperbola through the vertices so
they approach the asymptotes.
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
Find the foci of the graph
2
y2
– x = 1.
9
4
y2
x2
The equation is in the form 2 – 2 = 1, so the transverse axis is horizontal;
a
b
a2 = 9 and b2 = 4.
c2 = a2 + b2
Use the Pythagorean Theorem.
=9+4
Substitute 9 for a2 and 4 for b2.
c=
13
3.6
Find the square root of each side of the equation.
10-5
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).
b
2
The asymptotes are the lines y = ± a x , or y = ± 3 x.
10-5
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
2
y2
transverse axis is horizontal. The equation will be in the form x2 – 2 = 1.
a
c2 = a2 + b2
(424,650)2 = (300,765)2 + b2
b
Use the Pythagorean Theorem.
Substitute.
1.803  1011 = 9.046  1010 + b2
Use a calculator.
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
(continued)
b2 = 1.803  1011 – 9.046  1010
Solve for b2.
= 8.987  1010
x2
9.046  1010
–
y2
8.987  1010
Substitute a2 and b2.
=1
The path of the spacecraft around the planet can be modeled
by
x2
9.046  1010
–
y2
8.987  1010
= 1.
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
pages 566–568 Exercises
1.
3.
5.
4.
6.
2.
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
7.
9.
8.
10. (0,
11. (0,
97), (0, –
113)
97)
12. (
10-5
113), (0, –
265, 0), (–
265, 0)
Hyperbolas
ALGEBRA 2 LESSON 10-5
13. (10, 0), (–10, 0)
14. (0, 5
5), (0, 5
15. (
5)
16. (0,
205, 0), (–
29), (0, –
10-5
205, 0)
29)
17. (2
11, 0), (–2
18. (0, 4
3), (0, –4
11, 0)
3)
Hyperbolas
ALGEBRA 2 LESSON 10-5
2
x2
– y
=1
96,480
69,169
2
x2
20.
– y
=1
10,000
240,000
y2
x2
21.
–
=1
170,203,465
192,432,384
y2
x2
22.
–
=1
11
12
5.270  10
1.865  10
2
2
23. x – y = 1
16
9
2
2
24. y – x = 1
25
144
2
25. y2 – x = 1
3
2
26. x – y2 = 1
4
19.
27.
30.
2
y2
– x =1
20.25
4
28.
2
31. y – x2 = 1
9
29.
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
32.
y2
x2
–
=1
64
32
33. y = ± 1
2
2x2 – 8; (2, 0), (–2, 0)
34. y = ±
x2 – 1; (1, 0), (–1, 0)
35. y = ±
3x2 – 2; (–0.816, 0), (0.816, 0)
10-5
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
points whose distances from the foci have a
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.
10-5
38. Answers may vary. Sample:
axes of symmetry, vertices,
asymptotes
39. (0, ±1), y = ±x
40. (±1, 0), y = ± 1
3
41. (0, ±8), y = ±2x
42. (±5, 0), y = ± 4
5
43. (0, ±4), y = ±2x
44. (±7, 0), y = ± 5
7
45. Replace x with x – 3 and y
with y + 5 and rewrite
to obtain
(x – 3)2 – (y + 5)2 = 1
9
4
Hyperbolas
ALGEBRA 2 LESSON 10-5
46. a. your airport
b. 30 km
c.
x2 – y2 = 1
225 351
d.
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;
the branch that contains the
vertex closest to your airport
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
2
2
48. a. If the hyperbola x 2 – y 2 = 1 crossed either y = a x or y = – a x,
b
a
b
b
then there would have to be a value of x and a value of y for which
a
b
x2 + b2 = a x or a
b
x2 + b2 = x or
b
x2 + b2 = – a x . That would require that
b
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.
2
2
b. Yes; multiply both sides by 1 to obtain y – x = 1. This equation is clearly
4
64
36
an equation of a hyperbola.Yes; multiply both sides by –1 to obtain
x 2 – y 2 = 1. This equation is clearly an equation of a hyperbola.
16
9
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
54. (±
49. D
55. (0, ±
50. H
3), (±
2, 0)
56. (0, ±10), (±8, 0)
51. D
2
2
2
y2
y
x
52. [2] 2 – 2 = 1 or 2 – x 2 = 1;
a
b
b
a
then simplify, which leaves you
2
2
with x – y = 1.
49
25
[1] answer correct, but no explanation
x2
–
y2
=1
8.86 
1.19 
y2
x2
b.
–
=1
10
10
2.05

10
2.76  10
1012
1011
3
10x
2x 2 – x – 7
58.
x3 + x2 – 9x – 9
59. x + 3
x–3
60. 1
3
57.
divide both sides by 1225 and
53. a.
34, 0), (0, ±5)
61. 125
62. 6
10-5
Hyperbolas
ALGEBRA 2 LESSON 10-5
1. Graph 16y2 – 9x2 = 144.
x2 y2
2. Find the foci of the graph 4 – 36 = 1. Draw the graph.
(±2
10, 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.
x2 y2
– 12 = 1
4
10-5
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 = x2 + 4
2. 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 + 1
6. y = x2 + 10x – 7
7. y = 2x2 + 8x + 5
8. y = 4x2 – 12x + 3
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Solutions
1. y = x2 + 4
parent function: y = x2;
translated 4 units up
2. y = (x – 3)2 – 2
parent function: y = x2; translated
3 units right and 2 units down
3. y – 1 = x2, or y = x2 + 1
parent function: y = x2;
translated 1 unit up
4. y = (x + 5)2 + 6
parent function: y = x2;
translated 5 units left and
6 units up
5. y = x2 – 6x + 1; c = – 6
2
2
6. y = x2 + 10x – 7; c = 10
2
2
= (–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
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
Solutions (continued)
8. y = 4x2 – 12x + 3
7. y = 2x2 + 8x + 5
y = 2(x2 + 4x) + 5; c = 4
= 22 = 4
y = 4(x2 – 3x) + 3; c = – 3
2
= 9
y = 2(x2 + 4x + 4) + 5 – 2(4)
y = 4 x2 – 3x + 9 + 3 – 4 9
y = 2(x + 2)2 + 5 – 8
y=4 x–
2
3
2
+3–9
y = 2(x + 2)2 – 3
y=4 x–3
2
2
–6
2
2
2
4
4
10-6
4
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
(y – k)2
+
= 1.
b2
a2
(x – (–2))2
(y – 4)2
+
= 1.
42
52
Substitute –2 for h and 4 for k.
The equation of the ellipse is
(x + 2)2
(y – 4)2
+ 25 = 1.
16
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
Check: Solve the equation for y and graph both equations.
(x + 2)2
(y – 4)2
+
= 1.
16
25
25(x + 2)2 + 16(y – 4)2 = 400
16(y – 4)2 = 400 – 25(x + 2)2
1
(y – 4)2 = 16 (400 – 25(x + 2)2)
y–4=±
1
2)
(400
–
25(x
+
2)
16
1
y=4± 4
400 – 25(x + 2)2
10-6
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
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
The transverse axis is horizontal, so the equation has the form
(x – h)2
(y – k)2
–
= 1.
a2
b2
The equation of the hyperbola is
(x – 1)2
(y – 2)2
– 12
= 1.
4
10-6
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
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
Find b2.
c2 = a2 + b2
(40)2 = (15)2 + b2
1600 = 225 + b2
b2 = 1375
The equation of the hyperbola is
or
(x – 40)2
y2
– 1375 = 1
152
(x – 40)2
y2
–
= 1.
225
1375
10-6
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
9(x2 + 2x +
) – 4y2 = 27
Group the x- and y- terms.
Complete the square.
9(x2 + 2x + 1) – 4y2 = 27 + 9(12)
Add (9)(12) to each side.
9(x2 + 2x + 1) – 4y2 = 27 + 9
Simplify
9(x + 1)2 – 4y2 = 36
Write the trinomials as binomials squared.
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
9(x + 1)2
4y 2
– 36 = 1
36
Divide each side by 36.
(x + 1)2
y2
– 9 =1
4
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
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
(continued)
The distance from the center of the hyperbola to the foci is 13. Since the
hyperbola is centered at (–1, 0), and the transverse axis is horizontal, the
foci are located 13 to the left and right of the center. The foci are at
(–1 + 13, 0) and (–1 – 13, 0).
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
pages 573–576 Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
(y – 1)2
(x + 2)2
+
4
9
(y – 3)2
(x – 5)2
+
36
16
(y + 4)2
x2
+
=1
25
36
(y + 6)2
(x – 3)2
+
49
9
(y + 3)2
(x + 3)2
–
9
16
(x – 4)2
(y + 3)2
–
32
4
(x + 1)2 – (y – 2)2
40
9
(x + 1)2
(y + 1)2
–
56
25
(y – 1)2 – x 2 = 1
25
16
=1
=1
y2
(x – 150)2
10.
– 21,204 = 1
1296
y2
(x – 175)2
11.
– 28,689 = 1
1936
12. y = (x – 4)2 + 3; parabola, vertex (4, 3)
=1
=1
=1
13. (x + 6)2 + y2 = 81; circle, center (–6, 0), radius 9
=1
=1
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
14.
(y – 3)2
(x + 1)2
+
= 1; ellipse,
9
3
center (–1, 3), foci (–1, 3 ±
15. (x –
+ (y +
= 13; circle,
center (1, –3), radius 13
1)2
3)2
6 )
16. (y – 2)2 – (x – 3)2 = 1; hyperbola,
center (3, 2), foci (3, 2 ± 2)
17.
(x – 1)2
– (y + 1)2 = 1; hyperbola,
4
center (1, –1), foci (1 ±
10-6
5 , –1)
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
18.
x2
7)2
+ (y +
= 36; circle,
center (0, –7), radius 6
19. x – 3 = 1 (y – 2)2; parabola,
2
vertex (3, 2)
20.
(y – 3)2
(x + 2)2
+
= 1; ellipse,
4
9
center (–2, 3), foci (–2 ±
5, 3)
21. (x + 3)2 – (y – 5)2 = 1; hyperbola,
center (–3, 5), foci (–3 ± 2, 5)
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
y2
(x – 1)2
22.
+ 4 = 1; ellipse,
16
center (1, 0), foci (1 ± 2
3, 0)
y2
x2
24. Translate the equation
– 8 = 1,
16
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
x2
(y + 3)2
vertices are increased (or decreased) by
23.
–
= 1; hyperbola,
4
9
the same amount, and the same is true
center (0, –3), foci (± 13, –3)
for the y-coordinates. A similar remark
holds for the co-vertices.
10-6
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.
29. (x + 6)2 + (y – 9)2 = 81
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.
30.
10-6
(y – 2)2
(x – 3)2
+
=1
9
36
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
31. y = 1 (x – 2)2 – 3
20
(y + 1)2
(x – 1)2
33.
+
=1
16
9
2
(y + 2)2
34. (x – 3) –
=1
15.84
36
35. (x + 4)2 + (y + 4)2 = 25
(x – 6)2
(y + 3)2
32.
–
=1
5
4
2
(y – 7)2
(x
–
2)
43.
+
=1
9
16
x2
(y – 5)2
44.
+
=1
16
4
2
45. (x – 4) – (y – 9)2 = 1
16
36. x – 1 = (y + 3)2
46. (x – 8)2 = 12(y – 11)
37. (x – 8)2 + (y – 2)2 = 4
47.
(y – 2)2
(x – 6)2
38.
+ 36 = 1
64
48. x2 – (y – 10)2 = 1
39. y – 5 = 4(x – 3)2
49.
2
(y – 5)2
(x
–
3)
40.
–
=1
9
16
2
2
41. (x – 5) – (y – 8) = 1
25
36
(y – 9)2
(x – 6)2
42.
+
=1
9
4
10-6
x2
(y – 10)2
+
=1
16
25
y2
x2
ellipse,
+ 36 = 1
9
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
50.
53.
2
y2
hyperbola, x –
=1
9
2
2
ellipse, x + y = 1
36
36
51.
9
54.
2
2
hyperbola, y – x = 1
36
parabola, x = y2 + 4
9
55. Check students’ work.
52.
56. Check students’ work.
circle, x2 + y2 = 36
10-6
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
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).
y2
x2
58. a. Earth:
+
=1
(149.58)2
(149.60)2
y2
x2
Mars:
+
=1
(226.9)2
(227.9)2
2
y2
x
Mercury:
+
=1
(56.6)2
(57.9)2
c. two intersecting lines
(y = x and y = –x)
b. Earth:
59. C
60. A
61. A
62. F
10-6
a
is closest to 1 for Earth.
b
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)
10-6
65. (0,
66. (0, 2
21), (0, –
26), (0, –2
21)
26)
Translating Conic Sections
ALGEBRA 2 LESSON 10-6
67. –1, 4
68. 5
69. about –0.4315, 1.9315
70. 1
71. 2
72. 3
73. 8
10-6
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.
(x + 5)2
(y + 2)2
+
=1
9
4
2. Find an equation of a hyperbola with vertices (0, –1) and (4, –1), and
foci (–1, –1) and (5, –1).
(x – 2)2
(y + 1)2
–
=1
4
5
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.
circle; center: (2, –1), radius: 2
10-6
Quadratic Relations
ALGEBRA 2 CHAPTER 10
Page 582
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
6. focus 0, 1 , directrix y = – 1
12
7. focus – 1 , 0 , directrix x = 1
8
20
20
9. focus 0, 1 , directrix y = – 1
18
18
10. y = – 1 x2
8
11. x = 1 y2
12
12. x = – 1 y2
28
13. y = 1 x2
4
14. center (2, 3), radius 6;
5. Write the equation in standard form.
12
8. focus – 1 , 0 , directrix x = 1
8
10-A
Quadratic Relations
ALGEBRA 2 CHAPTER 10
15. center (–5, –8), radius 10;
17. center (–4, 10), radius 11;
16. center (1, –7), radius 9;
2
y2
18. x +
=1
25
64
2
19. x + y2 = 1
36
2
2
20. x + y
=1
20.25
6.25
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Quadratic Relations
ALGEBRA 2 CHAPTER 10
21. foci (±3
5, 0);
22. foci (0, ±4
6)
23. foci (0, ±4
24. foci (0, ±
25. circle
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3)
3)
26. foci (±2
61, 0)
27. foci (0, ±
569)
Quadratic Relations
ALGEBRA 2 CHAPTER 10
28. foci (±2
17, 0)
2
2
35. y – (x – 2) = 1
24
(y + 1)2
2
36. (x + 4) –
=1
15
25
37. parabola; vertex (2, 1)
29. foci (0, ±
226)
30. Check students’ work.
31.
x2
+
y2
38. circle; center (2, 3), radius 3
=1
9
16
(y – 7)2
(x + 2)2
32.
+
=1
9
16
(y + 2)2
(x – 3)2
33.
+
=1
36
25
2
(y – 7)2
34. x –
=1
16
9
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Quadratic Relations
ALGEBRA 2 CHAPTER 10
39. ellipse; center (2, 3), foci (2, 3 ± 2
40. hyperbola; center (2, 3), foci (2 ± 2
3)
5, 3)
10-A