Find the focus and directrix of the parabola with the given equation.
72)
y2=24x
72)
A)
focus: (6, 0)
directrix: x =6
B)
focus: (6, 0)
directrix: x = – 6
C)
focus: (0, 6)
directrix: y = – 6
D)
focus: (0, –6)
directrix: y = – 6
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.
73)
x2
16 +y2
9= 1
y =3
73)
A)
{(0, 3), (0, –3)}
B)
{(3, 0)}
C)
{(3, 3)}
D)
{(0, 3)}
Find the vertices and locate the foci for the hyperbola whose equation is given.
74)
81y2–4x2=324
74)
A)
vertices: (–9, 0), (9, 0)
foci: (–77, 0), ( 77, 0)
B)
vertices: (0, –9), (0, 9)
foci: (0, –85), (0, 85)
C)
vertices: (–2, 0), (2, 0)
foci: (–85, 0), ( 85, 0)
D)
vertices: (0, –2), (0, 2)
foci: (0, –85), (0, 85)
Graph the ellipse.
41
75)
4(x – 1)2+9(y + 1)2=36
75)
A)
B)
C)
D)
Is the relation a function?
76)
x = – (y + 6)2+ 1
76)
A)
Yes
B)
No
Graph the parabola.
42
77)
y2+16x = 0
77)
A)
B)
C)
D)
Find the standard form of the equation of the hyperbola satisfying the given conditions.
78)
Endpoints of transverse axis: (–7, 0), (7, 0); foci: (–8, 0), (–8, 0)
78)
A)
x2
49 –y2
64 = 1
B)
x2
15 –y2
49 = 1
C)
x2
64 –y2
49 = 1
D)
x2
49 –y2
15 = 1
Graph the parabola.
79)
x2= – 20y
79)
A)
B)
C)
D)
Convert the equation to the standard form for a hyperbola by completing the square on x and y.
80)
y2–25x2– 2y – 100x – 124 = 0
80)
A)
(y – 2)2
25 –(x + 4)2= 1
B)
(y – 1)2
25 –(x + 2)2= 1
C)
(x – 1)2
25 –(y + 2)2= 1
D)
(x + 2)2–(y – 1)2
25 = 1
Determine the direction in which the parabola opens, and the vertex.
81)
y2+ 2y – x + 3 = 0
81)
A)
Opens to the left; (– 2, 1)
B)
Opens downward; (–1, – 2)
C)
Opens downward; (1, – 2)
D)
Opens to the right; (2, –1)
Use the vertex and the direction in which the parabola opens to determine the relation’s domain and range.
82)
y2– 4y – x + 3 = 0
82)
A)
Domain: (–, 1)
Range: (–, )
B)
Domain: (–, )
Range: (–, )
C)
Domain: (–1, ]
Range: (–, )
D)
Domain: (–, )
Range: (–, 1]
83)
y =x2– 8x + 14
83)
A)
Domain: (–, )
Range: (–2, )
B)
Domain: [–2, )
Range: (–, )
C)
Domain: (–, )
Range: [–2, )
D)
Domain: (–2, )
Range: (–, )
Graph the ellipse and locate the foci.
45
84)
9x2+4y2=36
84)
A)
foci at ( 5, 0) and (–5, 0)
B)
foci at ( 13, 0) and (–13, 0)
C)
foci at (0, 5) and (0, –5)
D)
foci at (2 3, 0) and (–2 3, 0)
46
Solve the problem.
85)
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 are both 16 inches tall and stand 80 inches apart. At some point along the road from the
lowest point of the cable, the cable is 1.44 inches above the roadway. Find the distance between that
point and the base of the nearest tower.
85)
A)
28.2 in.
B)
28 in.
C)
12.2 in.
D)
11.8 in.
Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes.
86)
y2
9–x2
25 = 1
86)
A)
Asymptotes: y = ± 3
5x
B)
Asymptotes: y = ± 3
5x
47
C)
Asymptotes: y = ± 5
3x
D)
Asymptotes: y = ± 5
3x
Convert the equation to the standard form for an ellipse by completing the square on x and y.
87)
4x2+36y2– 16x + 72y – 92 = 0
87)
A)
(x – 2)2
4+(y + 1)2
36 = 1
B)
(x + 2)2
36 +(y – 1)2
4= 1
C)
(x – 2)2
36 +(y + 1)2
4= 1
D)
(x + 1)2
36 +(y – 2)2
4= 1
Graph the ellipse and locate the foci.
88)
x2
9+y2
25 = 1
88)
48
A)
foci at (4, 0) and (–4, 0)
B)
foci at (0, 4) and (0, –4)
C)
foci at (0, 3 3) and (0, –3 3)
D)
foci at (3 3, 0) and (–3 3, 0)
Find the vertices and locate the foci for the hyperbola whose equation is given.
89)
y = ± x2–7
89)
A)
vertices: (–7, 0), (7, 0)
foci: (–14, 0), ( 14, 0)
B)
vertices: (0, –7), (0, 7)
foci: (0, –14), (0, 14)
C)
vertices: (–7, 0), ( 7, 0)
foci: (–14, 0), ( 14, 0)
D)
vertices: (–7, 0), (7, 0)
foci: (–7, 0), ( 7, 0)
Find the standard form of the equation of the hyperbola.
90)
90)
A)
x2
4–y2
25 = 1
B)
x2
25 –y2
4= 1
C)
y2
4–x2
25 = 1
D)
y2
25 –x2
4= 1
Convert the equation to the standard form for a hyperbola by completing the square on x and y.
91)
9y2–4x2– 18y – 16x – 43 = 0
91)
A)
(y – 1)2
4–(x + 2)2
9= 1
B)
(x – 2)2
9–(y + 1)2
4= 1
C)
(y – 1)2
9–(x + 2)2
4= 1
D)
(y + 1)2
4–(x – 2)2
9= 1
Find the standard form of the equation of the parabola using the information given.
92)
Focus: (–4, 4); Directrix: x = – 2
92)
A)
(x + 3)2= – 4(y – 4)
B)
(y – 4)2= – 4(x + 3)
C)
(x – 4)2= – 4(y + 3)
D)
(y + 3)2= – 4(x – 4)
Graph the ellipse.
50
93)
(x + 1)2
9+(y – 1)2
16 = 1
93)
A)
B)
C)
D)
Use the vertex and the direction in which the parabola opens to determine the relation’s domain and range.
94)
x = – (y + 6)2– 10
94)
A)
Domain: (–, )
Range: (–, –10]
B)
Domain: (–, –10)
Range: (–, )
C)
Domain: (–, –10]
Range: (–, )
D)
Domain: (–, )
Range: (–, )
Find the vertex, focus, and directrix of the parabola with the given equation.
95)
(y + 1)2= – 20(x – 2)
95)
A)
vertex: (2, –1)
focus: (7, –1)
directrix: x = – 3
B)
vertex: (–2, 1)
focus: (–7, 1)
directrix: x =3
C)
vertex: (2, –1)
focus: (–3, –1)
directrix: x =7
D)
vertex: (–1, 2)
focus: (–6, 2)
directrix: x =4
Convert the equation to the standard form for a parabola by completing the square on x or y as appropriate.
96)
y2– 4y + 4x + 0 = 0
96)
A)
(y + 2)2= – 4(x – 1)
B)
(y + 2)2=4(x – 1)
C)
(y – 2)2= – 4(x + 1)
D)
(y – 2)2= – 4(x – 1)
Graph the semi–ellipse.
97)
y = – 16 –9x2
97)
52
A)
B)
C)
D)
Graph the parabola.
98)
x2=6y
98)
53
A)
B)
C)
D)
Find the foci of the ellipse whose equation is given.
99)
25(x + 1)2+36(y + 3)2=900
99)
A)
foci at (–11, –3) and ( 11, –3)
B)
foci at (–1+11, –3) and (–1–11, –3)
C)
foci at (–3+11, –1) and (–3–11, –1)
D)
foci at (–1+11, –1) and (–1–11, –1)
Graph the ellipse and locate the foci.
54
100)
x2
5
2
+y2
9
2
= 1
Round to the nearest tenth if necessary.
100)
A)
foci (0, 1.4) and (0, –1.4)
B)
foci (0, 1.4) and (0, –1.4)
C)
foci (1.4, 0) and (0, –1.4)
D)
foci (1.5, 0) and (0, –1.5)
55
Find the location of the center, vertices, and foci for the hyperbola described by the equation.
101)
(y – 2)2–64(x – 3)2=64
101)
A)
Center: (–3, –2); Vertices: (–3, –10) and (–3, 6); Foci: (–3, –2–65) and (–3, –2+65)
B)
Center: (3, 2); Vertices: (3, –6) and (3, 10); Foci: (3, 2–65) and (3, 2+65)
C)
Center: (3, 2); Vertices: (–3, –8) and (3, 8); Foci: (3, –65) and (3, 65)
D)
Center: (3, 2); Vertices: (4, –5) and (4, 11); Foci: (4, 3–65) and (4, 3+65)
Find the standard form of the equation of the ellipse satisfying the given conditions.
102)
Foci: (–7, 0), (7, 0); x–intercepts: –8 and 8
102)
A)
x2
49 +y2
15 = 1
B)
x2
49 +y2
64 = 1
C)
x2
15 +y2
64 = 1
D)
x2
64 +y2
15 = 1
Find the foci of the ellipse whose equation is given.
103)
(x – 1)2
36 +(y + 3)2
25 = 1
103)
A)
foci at (1+11, –3) and (1–11, –3)
B)
foci at (1+11, 1) and (1–11, 1)
C)
foci at (–3+11, 1) and (–3–11, 1)
D)
foci at (–11, –3) and ( 11, –3)
Use the center, vertices, and asymptotes to graph the hyperbola.
104)
(x + 1)2
4–(y – 1)2
16 = 1
104)