Section 13.2 The Ellipse
46.
222
222
50 30
cab
c
47. – 53. Answers will vary.
54. does not make sense; Explanations will vary.
Sample explanation: The foci of an ellipse are
always on the major axis.
59. false; Changes to make the statement true will vary.
A sample change is: Ellipses are not functions.
They do not pass the vertical line test.
60. false; Changes to make the statement true will vary.
A sample change is: The only ordered pair that
Using a and the given point, we can solve for b.
22
22
36
36 4 16 36
xy
b
bb
22
2
144 16 36
144 20
bb
b
5
63.
22
22
22
92536501640
9 36 25 50 164
9 4 25 2 164
xyxy
xx yy
xx yy
Chapter 13 Conic Sections and Systems of Nonlinear Equations
22
944252116494251
xx yy
64.
22
22
4 9 32 36 64 0
4 32 9 36 64
xy xy
xx yy
22
22
22
4 8 16 9 4 4 64 4 16 9 4
4 4 9 2 64 64 36
449236
xx yy
xy
xy
Section 13.2 The Ellipse
65. a. The perigee is 5000 – 16 – 4000 = 984 miles above the Earth.
67.
32 2
2
2
248 242
24
222
22
xxx xx x
xx
xxx
xx
1
x
The solution set is
1.
70.
22
22
49 36
49 36
36 36 36
xy
xy
Chapter 13 Conic Sections and Systems of Nonlinear Equations
71.
22
1
16 9
xy
a. Substitute 0 for y.
b.
22
2
2
01
16 9
1
9
9
y
y
y
The equation
2
9y
has no real solutions.
72.
22
1
916
yx
a. Substitute 0 for x.
b.
22
2
2
01
916
1
16
16
x
x
x
The equation
2
16x
has no real solutions.
13.3 Check Points
1. a. Since the
2
termx
is positive, the transverse
axis lies along the x–axis. Also, since
2.
22
1
36 9
xy
2
3. First write the equation in standard form.
22
22
4 and 1ab
. We know the transverse axis
lies on the y-axis and the vertices are
0, 2 and 0, 2 .
Because
22
4 and 1,ab
2 and 1.ab
Construct a rectangle using –2
and 2 on the y–axis, and –1 and 1 on the x–axis.
Draw extended diagonals to obtain the asymptotes.
Draw the two branches of the hyperbola by starting
at each vertex and approaching the asymptotes.
Section 13.3 The Hyperbola
13.3 Concept and Vocabulary Check
1. hyperbola; vertices; transverse
2. (5,0); (5,0)
13.3 Exercise Set
1. Since the
2
termx
is positive, the transverse axis
lies along the x–axis. Also, since
2
4 and 2,aa
3. Since the
2
termy
is positive, the transverse axis
lies along the y–axis. Also, since
2
4 and 2,aa
5.
22
1
925
xy
The equation is in the form
22
22
1
xy
ab
with
6.
22
1
16 25
xy
The transverse axis lies on the x-axis and the
Chapter 13 Conic Sections and Systems of Nonlinear Equations
and 8 on the y–axis. Draw extended diagonals to
obtain the asymptotes. Graph the hyperbola.
8.
22
1
144 81
xy
9.
22
1
16 36
xy
The equation is in the form
22
22
1
xy
ab
with
22
16, and 36ab
. We know the transverse
axis lies on the x-axis and the vertices are
10.
22
1
25 64
yx
The transverse axis lies on the y-axis and the
11.
1
36 25
yx
The equation is in the form
22
22
1
yx
with
Draw extended diagonals to obtain the asymptotes.
Graph the hyperbola.
Section 13.3 The Hyperbola
12.
22
1
100 49
yx
The transverse axis lies on the y-axis and the
13.
22
22
22
94 36
94 36
36 36 36
1
49
xy
xy
xy
14.
22
22
22
425100
425100
100 100 100
xy
xy
xy
15.
22
22
925225
925225
225 225 225
yx
yx
Draw extended diagonals to obtain the asymptotes.
Graph the hyperbola.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
16.
22
22
22
169144
16 9 144
144 144 144
1
916
yx
yx
yx
17.
22
22
22
22
44
44
44
444
1
14
xy
xy
xy
xy
The equation is in the form
22
22
1
xy
ab
with
18.
22
22
22
22
25 225 9
25 9 225
25 9 225
225 225 225
yx
yx
yx
yx
19. The graph shows that the transverse axis lies along
the x–axis and the vertices are
3, 0
and
3, 0 .
This means that a = 3. We also see that b = 5.
22
22
22
22
1
1
35
xy
ab
xy
Section 13.3 The Hyperbola
21. The graph shows that the transverse axis lies along
the y–axis and the vertices are
0, 2
and
0, 2 .
22.
22
22
1
yx
ab
23.
22
1
916
xy
The equation is for a hyperbola in standard form
with the transverse axis on the x-axis. We have
2
9a
and
2
16b
, so
3a
and
4b
.
24.
22
1
xy
From the graph we determine the following:
Domain:
,5 5,
Range:
,
are
0, a
or
0, 4
. The endpoints of the minor
axis are
,0b
or
3, 0
.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
26.
22
1
25 4
xy
The equation is for an ellipse in standard form with
major axis along the x-axis. We have
2
25a
and
2
4b
, so
5a
and
2b
. Therefore, the vertices
27.
22
1
16 9
yx
The equation is in standard form with the transverse
axis on the y-axis. We have
2
16a
and
2
9b
, so
28.
22
1
425
yx
The equation is in standard form with the transverse
axis on the y-axis. We have
2
4a
and
2
25b
, so
2a
and
5b
. Therefore, the vertices are at
Range:
,2 2,
29.
22
22
4
4
xy
xy
Check
2, 0
:
22
204
22
204
Section 13.3 The Hyperbola
30.
22
22
9
9
xy
xy
5
y
(3, 0)
31.
22
22
99
99
xy
yx
or
22
22
1
19
1
91
xy
yx
5
y
(0, 3)
32.
22
22
44
44
xy
yx
or
22
22
1
14
1
41
xy
yx
5
y
44true
44true
The solution set is
0, 2 , 0, 2
.
33.
22
22
22
625 400 250,000
625 400 250,000
250,000 250,000 250,000
1
yx
yx
yx
Chapter 13 Conic Sections and Systems of Nonlinear Equations
34. The graph shows that the transverse axis lies along
the x–axis and the vertices are
3,0 and 3,0 .
This means that
3.a
11 3
3.
yx
35. – 40. Answers will vary.
41.
22
0
49
xy
Solve the equation for y.
22
22
49
94
xy
xy
42. Answers will vary.
43. does not make sense; Explanations will vary.
Sample explanation: This would change the ellipse
to a hyperbola.
46. makes sense
47. false; Changes to make the statement true will vary.
A sample change is: If a hyperbola has a transverse
axis along the x–axis and one of the branches is
49
94
same asymptotes.
51.
22
23
1
16 9
xy
This is the equation of a hyperbola with center
2,3 .
The transverse axis is horizontal and the
vertices lie 4 units to the right and left of
2,3
at
24,3 2,3
and
24,3
6, 3 .
This is the graph of a hyperbola with center
2,1 .
The equation is in the form
22
22
1
xh yk
ab
Section 13.3 The Hyperbola
23,1 1,1.
Because
22
9 and 25,ab
3 and 5.ab
Construct two sides of a rectangle
using –5 and 1 (the x–coordinates of the vertices)
on the x–axis. The remaining two sides of the
53.
22
22
22
34 3 4
343 4
444
33
1
41
xy
xy
xy
This is the equation of a hyperbola with center
3, 3 .
The transverse axis is horizontal and the
54.
22
2440xy xy
Rearrange and complete the squares.
Complete the squares.
22
2
211
22
b
22
axis is horizontal and the vertices lie 1 unit to the
right and left of
()
1, 2
− at
()()
11,2 0,2
−− = − and
()()
11,2 2,2.
+− = − Because
22
1 and 1,ab
==
1 and 1.ab==
Construct two sides of a rectangle
using 0 and 2 (the x–coordinates of the vertices) on
the x–axis. The remaining two sides of the
rectangle are constructed 1 unit above and 1 unit
below the center,
()
1, 2 ,−
at
21 3 and−−=−
This means that
24.b
22
1
xy
Chapter 13 Conic Sections and Systems of Nonlinear Equations
56. Since the vertices are
0,7 and
0, 7 ,
we know that the transverse axis lies along the y–axis and a = 7. Use the
equation of the asymptote, 5,yx to find b. We need to find the x–coordinate that corresponds with y = 7.
This means that
7.
5
b
Using a and b, write the equation of the hyperbola.
57.
245yx x
The vertex is at (−2, 9). The x–intercepts are –5 and 1. The y–intercept is 5.
3
Section 13.3 The Hyperbola
Interval Test Value Substitution Conclusion
1
,3
−1
2
31 111 40
10 0, true
1
,3
belongs in the solution set
3
59.
4
3
log 3 1 3
314
x
x
60.
245yx x
Since
1a
is positive, the parabola opens upward. The x-coordinate of the vertex is
42.
22(1)
b
xa
The
y-coordinate of the vertex is
2
(2) 4(2) 5 9.y
61.
2
312yx
Since
3a
is negative, the parabola opens downward. The vertex of the parabola is
,1,2hk
.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
62. Find the y-intercepts.
2
2
3( 1) 2
03(1)2
xy
y
Mid-Chapter Check Point – Chapter 13
1.
22
9xy
Center:
0, 0
Radius: 93r
2.
22
3225xy
Center:
3, 2
Radius: 25 5r
3.
2
2
14xy
Center:
0,1
4.
22
4240xy xy
Complete the square in both x and y to get the
equation in standard form.
Mid-Chapter Check Point
5.
22
1
25 4
xy
Center:
0, 0
Because the denominator of the
2termx
is
6.
22
94 36xy
Divide both sides by 36 to get the standard form:
22
1
xy
7.
22
21
1
16 25
xy
Center:
2, 1
2
8.
22
21
1
xy
Chapter 13 Conic Sections and Systems of Nonlinear Equations
9.
2
2
1
9
xy
The equation is for a hyperbola in standard form
with the transverse axis on the x-axis. We have
2
9a
and
2
1b
, so
3a
and
1b
. Therefore,
10.
2
2
1
9
yx
The equation is in the form
22
22
1
yx
ab
with
22
9, and 1
ab
. We know the transverse axis
lies on the y-axis and the vertices are
0, 3 and 0, 3 .
Because
22
9 and 1,
ab
11.
22
22
416
1
16 4
yx
yx
The equation is in the form
22
1
yx
with
12.
22
22
4 49 196
1
49 4
xy
xy
The equation is for a hyperbola in standard form
with the transverse axis on the x-axis. We have
249a
and
24b
, so
7a
and
2b
.
Therefore, the vertices are at
,0a
or
7, 0
.
Mid-Chapter Check Point
13.
22
4xy
This is the equation of a circle centered at the origin
with radius
42r
.
14.
4
4
xy
yx
This is the equation of a line with slope
1m
and
a y-intercept of 4. We can plot the point
0, 4
, use
15.
22
22
4
1
44
xy
xy
The equation is for a hyperbola in standard form
with the transverse axis on the x-axis. We have
Graph the hyperbola.
major axis is horizontal. We have
and
21b
, so
2a
and
1b
. The vertices lie 2 units
to the left and right of the center. The endpoints of
the minor axis lie 1 unit above and below the center.
Vertices:
2, 0
and
2,0
Center:
1,1
Radius:
42r
We plot the points that are 2 units to the left, right,
above and below the center.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
18.
2
2
2
2
414
11
41
xy
y
x
13.4 Check Points
1.
2
21xy
This is a parabola of the form
2
xayk h
.
Since
1a
is negative, the parabola opens to the
Replace x with 0 to find the y–intercepts.
2
2
21
021
xy
y
The y–coordinate of the vertex is
88
4.
2212
b
ya
The x–coordinate of the vertex is
22
87(4)8(4)7 9.xy y
The vertex of the parabola is
9, 4 . The axis of
symmetry is 4.y
70 or 10
71
yy
yy