Section 13.4 The Parabola: Identifying Conic Sections
3. a. Collect
22
and termsxy on the same side
of the equation.
22
22
416
416
xy
xy
c. Collect
22
and termsxy
on the same side
of the equation.
22
22
22
4164
44 16
or
4
xy
xy
xy
13.4 Concept and Vocabulary Check
1. parabola; focus
2.
0
9. to the right
10. to the left
11. downward
c. circle
d. parabola
13.4 Exercise Set
The equation’s graph is f.
4. Since a = 1, the parabola opens to the right.
The vertex of the parabola is
1, 2 .
The equation’s graph is e.
5. Since a = −1, the parabola opens to the left.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
7.
2
2
2
200
xy
xy
The vertex is the point
0, 0 .
11.
2
421xy
The vertex is the point
1, 2 .
12.
2
251xy
The vertex is the point
1, 5 .
16.
2
68xy y
The y–coordinate of the vertex is
66
3.
2212
b
a
31 6 7 3 6 7 4.
The vertex is the point
4, 1 .
18.
2
246xyy
The y–coordinate of the vertex is
44
1.
222 4
b
a
Section 13.4 The Parabola: Identifying Conic Sections
Apply the zero product principle.
20.
2
34xy
The parabola opens to the right. The vertex of the
parabola is
4,3 .
The axis of symmetry is y = 3.
21.
2
35xy
This is a parabola of the form
2
xayk h
.
Since
1a
is positive, the parabola opens to the
The x–intercept is 0. Replace x with 0 to find the
y-intercepts.
2
035
y
2
4
bb ac
xa
23 and 23.
23.
54xy
This is a parabola of the form
2
xayk h
.
Since
1a
is negative, the parabola opens to the
Chapter 13 Conic Sections and Systems of Nonlinear Equations
The x–intercept is 0. Replace x with 0 to find the
y-intercepts.
2
054
y
The y–intercepts are 3 and 7 .
24.
34xy
The vertex of the parabola is
4,3 . The axis of
symmetry is y = 3. The x–intercept is –5. The y–
intercepts are 1 and 5 .
25.
2
41xy
This is a parabola of the form
2
xayk h.
17
The x–intercept is 0. Replace x with 0 to find the y–
intercepts.
2
2
041
08161
y
yy
884117
21
86468
2
84
2
82
i
Section 13.4 The Parabola: Identifying Conic Sections
26.
2
23xy
The vertex of the parabola is
3, 2 .
The axis of
27.
2
353xy
This is a parabola of the form
2
xayk h
.
Since
3a
is negative, the parabola opens to the
The x–intercept is 0. Replace x with 0 to find the y–
intercepts.
2
2
03 5 3
03 10253
y
yy
28.
2
262xy
The vertex of the parabola is
2, 6 .
The axis of
29.
2
231xy
This is a parabola of the form
2
xayk h
.
Since
2a
is negative, the parabola opens to the
left. The vertex of the parabola is
1, 3 .
The
The x–intercept is 0. Replace x with 0 to find the y–
intercepts.
2
2
2
02 31
02 691
y
yx
Chapter 13 Conic Sections and Systems of Nonlinear Equations
Since the solutions will be complex, there are no
30.
2
312xy
The vertex of the parabola is
2, 1 . The axis of
31.
2
121
2
xy
This is a parabola of the form
2
xayk h.
Since 1
2
a is positive, the parabola opens to the
2
1
021
y
Solve using the quadratic formula.
2
44416
y
Section 13.4 The Parabola: Identifying Conic Sections
33.
2
23xy y
This is a parabola of the form
2
.xay byc
Since
1a
is positive, the parabola opens to the
right. The y–coordinate of the vertex is
The x–intercept is –3. Replace x with 0 to find the
y–intercepts.
2
023
031
yy
yy
34.
2
68xy y
The parabola opens to the right. The y–coordinate of
the vertex is
66
3.
2212
b
a
The
x-coordinate of the vertex is
35.
2
45xy y
This is a parabola of the form
2
.xay byc
Since
1a
is negative, the parabola opens to the
left. The y–coordinate of the vertex is
2
04050055x
The x–intercept is 5. Replace x with 0 to find the
y-intercepts.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
36.
2
67xy y
The parabola opens to the left. The y–coordinate of
the vertex is
66
3.
221 2
b
a
37.
2
6xy y
This is a parabola of the form
2
.xay byc
Since
1a
is positive, the parabola opens to the
right. The y–coordinate of the vertex is
66
3.
2212
b
a
The x–coordinate of the vertex is
2
3639189.x
The y–intercepts are –6 and 0.
39.
2
24xyy
This is a parabola of the form
2
.xay byc
Section 13.4 The Parabola: Identifying Conic Sections
2
2
02 4
02
02
yy
yy
yy
40.
2
36xyy
The parabola opens to the left. The y–coordinate of
the vertex is
66
1.
b
The
41.
2
241xyy
This is a parabola of the form
2
.xay byc
2
20 40 1 20 0 1
0011
x
The x–intercept is 1. Replace x with 0 to find the
y-intercepts.
The y–intercepts are
26
2
.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
42.
2
243xyy
The parabola opens to the left. The y–coordinate of
43. a. Since the squared term is y, the parabola is
horizontal.
44. a. The parabola is horizontal.
45. a. Since the squared term is x, the parabola is
vertical.
47. a. Since the squared term is x, the parabola is
vertical.
48. a. The parabola is vertical.
b. The parabola opens down.
50. a. The parabola is horizontal.
b. The parabola opens to the left.
c. The vertex is the point
4, 1 .
51. a. Since the squared term is x, the parabola is
vertical.
b. Since a = 1 is positive, the parabola opens up.
c. The x–coordinate of the vertex is
2
22421
f
52. a. The parabola is vertical.
b. The parabola opens up.
53. a. Since the squared term is y, the parabola is
horizontal.
Section 13.4 The Parabola: Identifying Conic Sections
c. The y–coordinate of the vertex is
54. a. The parabola is horizontal.
b. The parabola opens to the left.
c. The y–coordinate of the vertex is
55.
2
78xyy
Since only one variable is squared, the graph of the
equation is a parabola.
56.
2
34 6xyy
coefficients, the equation’s graph is an ellipse.
58.
22
22
436
436
xy
xy
60.
22
36 4
xy
Because
and xy
have the same positive
coefficient, the equation’s graph is a circle.
62.
22
22
3273
33 27
xy
xy
22
64.
22
22
3273
33 27
xy
xy
Because
22
and xy
have opposite signs, the
equation’s graph is a hyperbola.
22
1
16 4
xy
22
Chapter 13 Conic Sections and Systems of Nonlinear Equations
Graph the hyperbola.
66.
22
77 28xy
Because
22
and xy
have opposite signs, the
equation’s graph is a hyperbola.
22
77 28
xy
67.
22
44 16xy
Because
22
and xy
have the same positive
coefficient, the equation’s graph is a circle.
22
44 16
xy
68.
22
77 28xy
The center is
0, 0
and the radius is 2 units.
of the minor axis are
0, 2 and 0, 2 .
Section 13.4 The Parabola: Identifying Conic Sections
70.
22
416xy
Because
22
and xy
have different positive
coefficients, the equation’s graph is an ellipse.
71.
2
14xy
Since only one variable is squared, the graph of the
equation is a parabola.
This is a parabola of the form
2
xayk h
.
Since
1a
is positive, the parabola opens to the
right. The vertex of the parabola is
4,1 .
The
axis of symmetry is y = 1. Replace y with 0 to find
the x–intercept.
72.
2
41xy
Since only one variable is squared, the graph of the
equation is a parabola. The parabola opens to the
coefficient, the equation’s graph is a circle.
The center is
2, 1
and the radius is 4 units.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
75. The y-coordinate of the vertex is
63
221
b
ya
76. The y–coordinate of the vertex is
21
221
b
ya
The x-coordinate of the vertex is
2
12156x
77. The x-coordinate of the vertex is
42
221
b
xa
The y-coordinate of the vertex is
78. The x–coordinate of the vertex is
42
221
b
xa
79. The equation is in the form
2
xayk h
From the equation, we can see that the vertex is
3,1
.
Since the squared term is y and
0a
, the graph
opens to the left.
Domain:
,3
Domain:
,2
Range:
,
The relation is not a function.
Section 13.4 The Parabola: Identifying Conic Sections
81.
2
24
1
2
xy
yx
Check
0, 0
:
2
002 4
1
00
82.
2
32
5
xy
xy
Check
3, 2 :
2
323 2
32 5
83.
2
2
3
3
xy
xy y
84.
2
22
5
25
xy
xy
Check
5, 0:
2
22
Check
4,3 :
2
43 5
495
22
4325
16 9 25
Chapter 13 Conic Sections and Systems of Nonlinear Equations
85.
2
22
21
221
xy
xy
86.
2
22
245
121
xy y
xy
87. a.
2
2
316 1750
yax
a
b. To find the height of the cable 1000 feet from
the tower, find y when x = 1750 – 1000 = 750.
88. a. We want to use the model
2
yax
and are
given the point
640,140 on the graph. Using
the point we can solve for a in our model.
2
140 640
140 0.000342
a
89. a.
2
2
26
236
2
yax
a
a
b.
4
11
18 4
ap
p
90. a.
2
2
22
yax
a
Section 13.4 The Parabola: Identifying Conic Sections
b. 1
4
ap
91. a. ellipse
b.
22
44xy
92. a. hyperbola
101.
2
2
26130
26130
1 2 6 13
yyx
yy x
ab cx
102.
2
2
10 25 0
10 25 0
yyx
yyx
10 4
2
10 2
2
5
x
x
x
A sample change is: Because a = −1, the parabola
will open to the left.
109. true
110. false; Changes to make the statement true will vary.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
111. false; Changes to make the statement true will vary.
A sample change is:
xayk h is not a
parabola. There is no squared variable.
114.
1
2x
fx
x f (x)
−2 8
−1 4
0 2
115.
15
3
15
3
fx x
yx
Interchange x and y and solve for y.
116.
22
22
2
(1)(3) 4
21 694
xx
xx xx
Substitute to find y.
27
2(2.5) 7 2
yx
y
The solution is (2.5, 2).
The solution set is
(2.5, 2) .
212 4
28
4
x
x
x
The solution is (4, 3).
Section 13.5 Systems of Nonlinear Equations in Two Variables
119.
2
2
2(3 9) 10
61810
xx
xx
13.5 Check Points
1.
2
1
41
xy
xy
=−
−=−
Solve the first equation for y.
2
1yx=+
2.
22
20
(1)(1) 5
xy
xy
+=
−+−=
Solve the first equation for x.
x = –2y
530or 10
3or 1
yy
yy
−= +=
==−
3.
22
32 35
43 48
xy
xy
+=
+=
Eliminate the
2
y
-term by multiplying the first
equation by –3 and the second equation by 2. Add
the resulting equations.
22
22
96 105
86 96
xy
xy
−− =−
+=
4.
2
22
5
25
yx
xy
=+
+=
Arrange the first equation so that variable terms
appear on the left, and constants appear on the right.
2
Chapter 13 Conic Sections and Systems of Nonlinear Equations
5. 22 20
21
xy
xy
+=
=
Solve the second equation for x.
21
xy
=
13.5 Concept and Vocabulary Check
1. nonlinear
2.
{
}
(4,3),(0,1)−−
()()
320
xx
+−=
30 or 20
32
xx
xx
+= −=
=− =
Substitute –3 and 2 for x in the second equation to
find y.
find y.
22
01
or
01 11
111
2
xx
yy
yy
y
==
=+ =+
==+
=