120)
(x +5)2+(y +4)2=25
120)
A)
center (–5, 4), r =5
B)
center (–5, –4), r =5
C)
center (5, 4), r =5
D)
center (5, –4), r =5
121)
The arch beneath a bridge is semi–elliptical, a one–way roadway passes under the arch. The width
of the roadway is 32 feet and the height of the arch over the center of the roadway is 11 feet. Two
trucks plan to use this road. They are both 10 feet wide. Truck 1 has an overall height of 10 feet and
Truck 2 has an overall height of 11 feet. Draw a rough sketch of the situation and determine which
of the trucks can pass under the bridge.
121)
A)
Truck 1 can pass under the bridge, but Truck 2 cannot.
B)
Truck 2 can pass under the bridge, but Truck 1 cannot.
C)
Both Truck 1 and Truck 2 can pass under the bridge.
D)
Neither Truck 1 nor Truck 2 can pass under the bridge.
122)
2x2+ y2= 17
3x2– 2y2= – 6
122)
A)
{(1, 3), (1, –3), (–1, 3), (–1, –3)}
B)
{(1, 3), (–1, –3)}
C)
{(2, –3), (–2, 3)}
D)
{(2, 3), (2, –3), (–2, 3), (–2, –3)}
123)
(x –5)2+(y –3)2=4
123)
A)
center (5, 3), r =2
B)
center (–5, 3), r =2
C)
center (5, –3), r =2
D)
center (–5, –3), r =2
124)
x2
36 +y2
36 = 1
124)
A)
ellipse
B)
circle
C)
hyperbola
D)
parabola
125)
Find the distance between (–5, 6) and (3, –3). Give an exact answer.
125)
A)
72 units
B)
145 units
C)
17 units
D)
145 units
126)
x2+y2– 4x – 2y – 11 = 0
x2–y2– 4x + 2y – 13 = 0
126)
A)
{(2, –5), (2, 3)}
B)
{(6, 1), (2, 1)}
C)
{(2, 5), (2, –3)}
D)
{(6, 1), (–2, 1)}
127)
x – 9 + 6y =3y2
127)
A)
circle
B)
ellipse
C)
hyperbola
D)
parabola
128)
128)
A)
x2+y2
10 = 1
B)
x2
10 +y2= 1
C)
x2
100 +y2= 1
D)
x2+y2
100 = 1
129)
9x2=36 +y2
129)
A)
parabola
B)
ellipse
C)
circle
D)
hyperbola
130)
x =y2– 12y + 40
130)
A)
Axis of symmetry is x =6.
B)
Axis of symmetry is y = – 6.
C)
Axis of symmetry is y =6.
D)
Axis of symmetry is x = – 6.
131)
(–2, –2) and (4, –4)
131)
A)
32 2 units
B)
8 units
C)
32 units
D)
210 units
132)
36y2–4x2=144
132)
A)
B)
C)
D)
A
D
D)
133)
The sum of the squares of two numbers is 41. The difference of the two numbers is 1. Find the two
numbers.
133)
A)
4 and 5 or –5 and –4
B)
4 and 5
C)
–4 and 5 or –5 and 4
D)
–5 and –4
134)
x2+y2– 8x – 8y + 23 = 0
134)
A)
(x – 4)2+(y + 4)2=9
center (4, –4), r =3
B)
(x – 4)2+(y – 4)2=9
center (4, 4), r =3
D)
C)
(x + 4)2+(y + 4)2=9
center (–4, –4), r =3
D)
(x + 4)2+(y – 4)2=9
center (–4, 4), r =3
135)
A rectangular coordinate system with coordinates in miles is placed with the origin at the center of
Los Angeles. A university is located 34.9 miles east and 50.9 miles north of central Los Angeles. A
seismograph on the campus shows a small earthquake occurred. The quake’s epicenter is 26.9 miles
from the university. Write the standard form of the equation for the set of points that could be the
epicenter of the quake.
135)
A)
(x –34.9)2+(y –50.9)2=26.9
B)
(x +34.9)2+(y +50.9)2=723.61
C)
(x –34.9)2+(y –50.9)2=723.61
D)
(x +34.9)2+(y +50.9)2=26.9
136)
x = – 6y2– 96y – 379
136)
A)
(0, –8)
B)
(5, 8)
C)
(5, –8)
D)
(8, 5)
137)
137)
A)
x2
8+y2
6= 1
B)
x2
36 +y2
64 = 1
C)
x2
64 +y2
36 = 1
D)
x2
6+y2
8= 1
138)
x2=384 –6(y +6)2
138)
A)
circle
B)
hyperbola
C)
parabola
D)
ellipse
139)
x2
25 +y2
16 = 1
139)
A)
B)
C)
D)
140)
x2+y2+ 6x + 8y + 21 = 0
140)
A)
(x + 3)2+(y + 4)2=4
center (–3, –4), r =2
B)
(x + 3)2+(y – 4)2=4
center (–3, 4), r =2
C)
(x – 3)2+(y – 4)2=4
center (3, 4), r =2
D)
(x – 3)2+(y + 4)2=4
center (3, –4), r =2
141)
x = – 2(y + 5)2– 8
141)
A)
(–5, –8)
B)
(–8, 5)
C)
(–8, –5)
D)
(5, –8)
142)
y =(x + 5)2+ 1
2x – y + 10 = 0
142)
A)
{(–5, 0)}
B)
{(–4, 2)}
C)
{(–4, 2), (4, 18)}
D)
{(0, 10), (0, 26)}
143)
x2+y2– 16x + 63 = 0
143)
A)
x2+(y – 8)2=1
center (0, 8), r =1
B)
(x + 8)2+y2=1
center (–8, 0), r =1
C)
(x – 8)2+y2=1
center (8, 0), r =1
D)
x2+(y + 8)2=1
center (0, –8), r =1
144)
x2+y2=64
x2–y2=64
144)
A)
{(8, 0)}
B)
{(0, 8), (0, –8)}
C)
{(8, 0), (–8, 0)}
D)
{(0, 8)}
145)
25x2+9y2=225
145)
A)
ellipse
B)
circle
C)
parabola
D)
hyperbola
146)
x = – 2(y – 2)2+ 4
146)
A)
Axis of symmetry is x = – 2.
B)
Axis of symmetry is x =2.
C)
Axis of symmetry is y = – 2.
D)
Axis of symmetry is y =2.
147)
Center (6, 0), r =5
147)
A)
x2+ (y + 6)2=5
B)
x2+ (y – 6)2=5
C)
(x + 6)2+ y2=25
D)
(x – 6)2+ y2=25
148)
x =9y2
148)
A)
(0, 0)
B)
(9, 9)
C)
(0, 9)
D)
(9, 0)
A
149)
A system for tracking ships indicated that a ship lies on a hyperbolic path described by
6x2–y2=135. The process is repeated and the ship is found to lie on a hyperbolic path described
by y2–2x2=9. If it is known that the ship is located in the first quadrant of the coordinate system,
determine its exact location.
149)
A)
(9, 6)
B)
(–6, –9)
C)
(6, 9)
D)
(–9, –6)
C
150)
(–1, –3) and (–9, –9)
150)
A)
10 units
B)
20 units
C)
11 units
D)
100 units
A
151)
x = – y2– 2y + 4
151)
A)
horizontal; opens to the left; (5, 1)
B)
vertical; opens downward; (1, 5)
C)
vertical; opens downward; (–1, 5)
D)
horizontal; opens to the left; (5, –1)
D
D
152)
x2+y2=9
9x2+4y2=36
152)
A)
{(2, 0), (–2, 0)}
B)
{(3, 0), (–3, 0)}
C)
{(0, 3), (0, –3)}
D)
{(0, 2), (0, –2)}
153)
(–5, –3) and (3, –5)
153)
A)
60 15 units
B)
217 units
C)
10 units
D)
60 units
B
154)
x2–9y2=36
154)
A)
B)
C
C)
D)
155)
2x2– 2y2= – 14
3x2+ 3y2=75
155)
A)
{(3, 4), (–3, 4), (3, –4), (–3, –4)}
B)
{(–3, –4), (–4, –3)}
C)
{(3, –4), (3, 4)}
D)
{(3, 4), (4, 3), (–3, –4), (–4, –3)}
156)
y2
16 –x2
9= 1
156)
A)
B)
C)
D)
157)
x2+y2– 14y + 48 = 0
157)