39)
36y2–4x2=144
39)
A)
Asymptotes: y = ± 1
3x
B)
Asymptotes: y = ± 1
3x
C)
Asymptotes: y = ± 3x
D)
Asymptotes: y = ± 3x
Graph the parabola with the given equation.
21
40)
(y – 2)2= – 5(x + 1)
40)
A)
B)
C)
D)
22
Find the solution set for the system by graphing both of the system’s equations in the same rectangular coordinate system
and finding points of intersection.
41)
x2–y2=64
x2+y2=64
41)
A)
{(0, 8)}
B)
{(8, 0)}
C)
{(0, 8), (0, –8)}
D)
{(8, 0), (–8, 0)}
Find the standard form of the equation of the ellipse satisfying the given conditions.
42)
Foci: (0, –3), (0, 3); y–intercepts: –8 and 8
42)
A)
x2
9+y2
55 = 1
B)
x2
9+y2
64 = 1
C)
x2
55 +y2
64 = 1
D)
x2
64 +y2
55 = 1
Find the vertices and locate the foci for the hyperbola whose equation is given.
43)
x2
121 –y2
81 = 1
43)
A)
vertices: (–9, 0), (9, 0)
foci: (–202, 0), ( 202, 0)
B)
vertices: (0, –11), (0, 11)
foci: (–202, 0), ( 202, 0)
C)
vertices: (–11, 0), (11, 0)
foci: (–9, 0), (9, 0)
D)
vertices: (–11, 0), (11, 0)
foci: (–202, 0), ( 202, 0)
Find the focus and directrix of the parabola with the given equation.
44)
y2=24x
44)
A)
focus: (0, –6)
directrix: y = – 6
B)
focus: (0, 6)
directrix: y = – 6
C)
focus: (6, 0)
directrix: x =6
D)
focus: (6, 0)
directrix: x = – 6
Find the vertex, focus, and directrix of the parabola with the given equation.
45)
(y + 1)2= – 20(x – 2)
45)
A)
vertex: (–1, 2)
focus: (–6, 2)
directrix: x =4
B)
vertex: (–2, 1)
focus: (–7, 1)
directrix: x =3
C)
vertex: (2, –1)
focus: (7, –1)
directrix: x = – 3
D)
vertex: (2, –1)
focus: (–3, –1)
directrix: x =7
Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes.
46)
x2
4–y2
36 = 1
46)
24
A)
Asymptotes: y = ± 1
3x
B)
Asymptotes: y = ± 3x
C)
Asymptotes: y = ± 1
3x
D)
Asymptotes: y = ± 3x
Convert the equation to the standard form for an ellipse by completing the square on x and y.
47)
36x2+4y2– 72x + 16y – 92 = 0
47)
A)
(x – 1)2
4+(y + 2)2
36 = 1
B)
(x + 2)2
4+(y – 1)2
36 = 1
C)
(x + 1)2
4+(y – 2)2
36 = 1
D)
(x – 1)2
36 +(y + 2)2
4= 1
25
Find the standard form of the equation of the ellipse and give the location of its foci.
48)
48)
A)
x2
81 +y2
4= 1
foci at (–77, 0) and ( 77, 0)
B)
x2
81 –y2
4= 1
foci at (–77, 0) and ( 77, 0)
C)
x2
4+y2
81 = 1
foci at (–77, 0) and ( 77, 0)
D)
x2
81 +y2
4= 1
foci at (–9, 0) and (9, 0)
Graph the ellipse and locate the foci.
49)
x2
49 +y2
40 = 1
49)
26
A)
foci at (210, 0) and (–210, 0)
B)
foci at (0, 7) and (0, –7)
C)
foci at (3, 0) and (–3, 0)
D)
foci at (0, 3) and (0, –3)
Find the standard form of the equation of the parabola using the information given.
50)
Focus: (–4, 4); Directrix: x = – 2
50)
A)
(x + 3)2= – 4(y – 4)
B)
(y – 4)2= – 4(x + 3)
C)
(y + 3)2= – 4(x – 4)
D)
(x – 4)2= – 4(y + 3)
Find the solution set for the system by graphing both of the system’s equations in the same rectangular coordinate system
and finding points of intersection.
51)
x2+y2=16
16x2+4y2=64
51)
A)
{(0, 4), (0, –4)}
B)
{(4, 0), (–4, 0)}
C)
{(2, 0), (–2, 0)}
D)
{(0, 2), (0, –2)}
52)
(y –5)2= x +25
y = – 1
5x
52)
A)
{(–25, 5)}
B)
{(–25, 0), (5, 0)}
C)
{(25, –5), (0, 0)}
D)
{(–25, 5), (0, 0)}
28
53)
x =(y +4)2– 1
(x –4)2+(y +4)2= 1
53)
A)
{(–1, –4), (4, –4)}
B)
{(4, –4)}
C)
{(–1, –4)}
D)
Convert the equation to the standard form for a hyperbola by completing the square on x and y.
54)
9y2–4x2– 18y – 16x – 43 = 0
54)
A)
(y + 1)2
4–(x – 2)2
9= 1
B)
(y – 1)2
9–(x + 2)2
4= 1
C)
(y – 1)2
4–(x + 2)2
9= 1
D)
(x – 2)2
9–(y + 1)2
4= 1
C
D)
Graph the ellipse and locate the foci.
55)
9x2+4y2=36
55)
29
D
D)
A)
foci at (0, 5) and (0, –5)
B)
foci at ( 5, 0) and (–5, 0)
C)
foci at (2 3, 0) and (–2 3, 0)
D)
foci at ( 13, 0) and (–13, 0)
30
Find the standard form of the equation of the ellipse and give the location of its foci.
56)
Center at (–1, 1)
56)
A)
(x – 1)2
25 +(y + 1)2
36 = 1
foci at (–11, 1) and ( 11, 1)
B)
(x + 1)2
36 +(y – 1)2
25 = 1
foci at (–1+11, 1) and (–1–11, 1)
C)
(x + 1)2
25 +(y – 1)2
36 = 1
foci at (1+11, –1) and (1–11, –1)
D)
(x – 1)2
36 +(y + 1)2
25 = 1
foci at (–1+11, –1) and (–1–11, –1)
Solve the problem.
57)
A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 156 feet and a
maximum height of 25 feet. Find the height of the arch at 15 feet from its center.
57)
A)
3.7 ft
B)
24.1 ft
C)
0.2 ft
D)
28.1 ft
Graph the parabola.
58)
y2= – 5x
58)
31
A)
B)
C)
D)
Is the relation a function?
59)
y =x2+ 8x + 18
59)
A)
Yes
B)
No
32
Find the standard form of the equation of the hyperbola.
60)
60)
A)
y2
25 –x2
4= 1
B)
x2
25 –y2
4= 1
C)
y2
4–x2
25 = 1
D)
x2
4–y2
25 = 1
Find the vertex, focus, and directrix of the parabola with the given equation.
61)
(x – 1)2= – 8(y – 2)
61)
A)
vertex: (1, 2)
focus: (1, 0)
directrix: y =4
B)
vertex: (2, 1)
focus: (2, –1)
directrix: y =3
C)
vertex: (–1, –2)
focus: (–1, –4)
directrix: y =0
D)
vertex: (1, 2)
focus: (1, 4)
directrix: x =0
Find the standard form of the equation of the ellipse satisfying the given conditions.
62)
Foci: (–7, 0), (7, 0); x–intercepts: –8 and 8
62)
A)
x2
49 +y2
64 = 1
B)
x2
49 +y2
15 = 1
C)
x2
64 +y2
15 = 1
D)
x2
15 +y2
64 = 1
Determine the direction in which the parabola opens, and the vertex.
63)
y2+ 2y – x + 3 = 0
63)
A)
Opens downward; (–1, – 2)
B)
Opens to the right; (2, –1)
C)
Opens downward; (1, – 2)
D)
Opens to the left; (– 2, 1)
Convert the equation to the standard form for a hyperbola by completing the square on x and y.
64)
y2–25x2– 2y – 100x – 124 = 0
64)
A)
(x + 2)2–(y – 1)2
25 = 1
B)
(y – 2)2
25 –(x + 4)2= 1
C)
(y – 1)2
25 –(x + 2)2= 1
D)
(x – 1)2
25 –(y + 2)2= 1
C
Find the solution set for the system by graphing both of the system’s equations in the same rectangular coordinate system
and finding points of intersection.
65)
25x2+y2=25
y2–25x2=25
65)
A)
{(0, –5), (0, 5)}
B)
{(5, 0), (5, 0)}
C)
{(0, 25)}
D)
{(0, –5)}
A
34
B
Solve the problem.
66)
A reflecting telescope has a parabolic mirror for which the distance from the vertex to the focus is
35 feet. If the distance across the top of the mirror is 76 inches, how deep is the mirror in the center?
66)
A)
361
420 in.
B)
361
5040 in.
C)
1225
152 in.
D)
361
35 in.
Find the standard form of the equation of the hyperbola.
67)
67)
A)
x2
9–y2
4= 1
B)
y2
9–x2
4= 1
C)
x2
4–y2
9= 1
D)
y2
4–x2
9= 1
68)
68)
A)
(y + 1)2
25 –(x – 1)2
4= 1
B)
(x – 1)2
25 –(y + 1)2
4= 1
C)
(x – 1)2
4–(y + 1)2
25 = 1
D)
(y + 1)2
4–(x – 1)2
25 = 1
Find the foci of the ellipse whose equation is given.
69)
(x – 3)2
25 +(y + 3)2
36 = 1
69)
A)
foci at (–3, 3–11) and (–3, 3+11)
B)
foci at (4, –3–11) and (4, –3+11)
C)
foci at (–3, –3–11) and (–3, –3+11)
D)
foci at (3, –3–11) and (3, –3+11)
Solve the problem.
70)
A satellite dish is in the shape of a parabolic surface. Signals coming from a satellite strike the
surface of the dish and are reflected to the focus, where the receiver is located. The satellite dish
shown has a diameter of 10 feet and a depth of 3 feet. The parabola is positioned in a rectangular
coordinate system with its vertex at the origin. The receiver should be placed at the focus (0, p). The
value of p is given by the equation a =1
4p . How far from the base of the dish should the receiver be
placed?
(5, 3)
3 feet
70)
A)
21
12 feet from the base
B)
3
25 feet from the base
C)
12
25 feet from the base
D)
81
3 feet from the base
Find the standard form of the equation of the hyperbola satisfying the given conditions.
71)
Endpoints of transverse axis: (–7, 0), (7, 0); foci: (–8, 0), (–8, 0)
71)
A)
x2
64 –y2
49 = 1
B)
x2
49 –y2
15 = 1
C)
x2
49 –y2
64 = 1
D)
x2
15 –y2
49 = 1
Graph the parabola with the given equation.
37
72)
(x – 1)2=5(y – 1)
72)
A)
B)
C)
D)
Solve the problem.
73)
The arch beneath a bridge is semi–elliptical, a one–way roadway passes under the arch. The width
of the roadway is 30 feet and the height of the arch over the center of the roadway is 13 feet. Two
trucks plan to use this road. They are both 10 feet wide. Truck 1 has an overall height of 13 feet and
Truck 2 has an overall height of 12 feet. Draw a rough sketch of the situation and determine which
of the trucks can pass under the bridge.
73)
A)
Both Truck 1 and Truck 2 can pass under the bridge.
B)
Truck 2 can pass under the bridge, but Truck 1 cannot.
C)
Neither Truck 1 nor Truck 2 can pass under the bridge.
D)
Truck 1 can pass under the bridge, but Truck 2 cannot.
Graph the ellipse and locate the foci.
74)
4x2=64 –16y2
74)
A)
foci at (0, 2 3) and (0, –2 3)
B)
foci at (2 5, 0) and (–2 5, 0)
39
C)
foci at ( 21, 0) and (–21, 0)
D)
foci at (2 3, 0) and (–2 3, 0)
Solve the problem.
75)
An experimental model for a suspension bridge is built. In one section, cable runs from the top of
one tower down to the roadway, just touching it there, and up again to the top of a second tower.
The towers stand 80 inches apart. At a point between the towers and 20 inches along the road from
the base of one tower, the cable is 4 inches above the roadway. Find the height of the towers.
75)
A)
18 in.
B)
16.5 in.
C)
15.5 in.
D)
16 in.
Graph the parabola.
76)
x2=6y
76)