Question 1

Solve the problem.
-A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the
surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 8 feet across at
its opening and is 2 feet deep at its center, at what position should the receiver be placed?

Accepted Answer

The answer of Solve the problem.
-A satellite dish is shaped...

Question 2

Solve the problem.
-A spotlight has a parabolic cross section that is 6 ft wide at the opening and 2.5 ft deep at the vertex. How far from the vertex is the focus? Round answer to two decimal places.
A) 0.21 ft
B) 0.52 ft
C) 0.26 ft
D) 0.90 ft

Accepted Answer

The answer of Solve the problem.
-A spotlight has a parabolic...

Question 3

Graph the equation.
-$(x+2)^{2}=8(y-1)$

A)

B)

C)

D)

Accepted Answer

The answer of Graph the equation.
-$(x+2)^{2}=8(y-1)$

A)

B)

C)

D)
...

Question 4

Find the center, foci, and vertices of the ellipse.
-$$\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 1$$
A) center at $( 0,0 )$
foci at $( - 6 \sqrt { 2 } , 0 )$ and $( 6 \sqrt { 2 } , 0 )$
vertices at $( - 9,0 ) , ( 9,0 )$

B) center at $( 0,0 )$
foci at $( 0 , - 3 )$ and $( 0,3 )$
vertices at $( 0 , - 9 ) , ( 0,9 )$

C) center at $( 0,0 )$
foci at $( 0 , - 6 \sqrt { 2 } )$ and $( 0,6 \sqrt { 2 } )$
vertices at $( 0 , - 9 ) , ( 0,9 )$

D) center at $( 0,0 )$
foci at $( - 9,0 )$ and $( 9,0 )$
vertice at $( - 81,0 ) , ( 81,0 )$

Accepted Answer

The center of the ellipse is at the origin (0,0) because there are no shifts in the equation. The major axis is along the x-axis because the larger denominator is under $x^2$, making the vertices at $(-9,0)$ and $(9,0)$. The distance between the center and the foci is $c=\sqrt{a^2-b^2}=\sqrt{81-9}=\sqrt{72}=6\sqrt{2}$, placing the foci at $(-6\sqrt{2},0)$ and $(6\sqrt{2},0)$.

Question 5

Find the center, foci, and vertices of the ellipse.
-$$9 x ^ { 2 } + 4 y ^ { 2 } = 36$$
A) center at $( 0,0 )$
foci at $( 0,3 )$ and $( 2,0 )$
vertices at $( 0,9 )$ and $( 4,0 )$

B) center at $( 0,0 )$
foci at $( 0 , - \sqrt { 5 } )$ and $( 0 , \sqrt { 5 } )$
vertices at $( 0 , - 3 ) , ( 0,3 )$

C) center at $( 0,0 )$
foci at $( 0 , - 3 )$ and $( 0,3 )$
vertices at $( 0 , - 9 ) , ( 0,9 )$

D) center at $( 0,0 )$
foci at $( - \sqrt { 5 } , 0 )$ and $( \sqrt { 5 } , 0 )$
vertices at $( - 3,0 ) , ( 3,0 )$

Accepted Answer

The answer of Find the center, foci, and vertices of...

Question 6

Match the graph to its equation.
-

A) $\frac { y ^ { 2 } } { 49 } + \frac { x ^ { 2 } } { 4 } = 1$

B) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 4 } = 1$

C) $\frac { y ^ { 2 } } { 49 } - \frac { x ^ { 2 } } { 4 } = 1$

D) $- \frac { y ^ { 2 } } { 49 } + \frac { x ^ { 2 } } { 4 } = 1$

Accepted Answer

The answer of Match the graph to its equation.
-

A) \(\frac...

Question 7

Solve the problem.
-A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is
3 centimeters from the vertex. If the depth is to be 6 centimeters, what is the diameter of the headlight at its
opening?

Accepted Answer

The answer of Solve the problem.
-A sealed-beam headlight is in...

Question 8

Find the center, foci, and vertices of the ellipse.
-$$\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$$
A) center at $( 0,0 )$
foci at $( 0,8 )$ and $( 4,0 )$
vertices at $( 0,64 ) , ( 16,0 )$

B) center at $( 0,0 )$
foci at $( 0 , - 8 )$ and $( 0,8 )$
vertices at $( 0 , - 64 ) , ( 0,64 )$

C) center at $( 0,0 )$
foci at $( - 4 \sqrt { 3 } , 0 )$ and $( 4 \sqrt { 3 } , 0 )$
vertices at $( - 8,0 ) , ( 8,0 )$

D) center at $( 0,0 )$
foci at $( 0 , - 4 \sqrt { 3 } )$ and $( 0,4 \sqrt { 3 } )$
vertices at $( 0 , - 8 ) , ( 0,8 )$

Accepted Answer

The given ellipse equation $\frac{x^2}{16} + \frac{y^2}{64} = 1$ has its major axis along the y-axis because the denominator under $y^2$ is larger. The center of the ellipse is at the origin (0,0) because it is in the standard form. The length of the major axis is $2b = 2\sqrt{64} = 16$, so the vertices are at $(0, \pm8)$. The distance between the center and the foci is $c = \sqrt{b^2 - a^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$, so the foci are at $(0, \pm4\sqrt{3})$.

Question 9

Find an equation for the ellipse described.
-Center at $( 0,0 )$; focus at $( 5,0 )$; vertex at $( 7,0 )$

A) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 24 } = 1$

B) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 24 } = 1$

C) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1$

D) $\frac { x ^ { 2 } } { 24 } + \frac { y ^ { 2 } } { 49 } = 1$

Accepted Answer

The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $2a$ is the length of the major axis, $2b$ is the length of the minor axis, and $c$ is the distance from the center to a focus. Here, $a = 7$ (since the vertex is at $(7,0)$), and $c = 5$ (since the focus is at $(5,0)$). We find $b^2 = a^2 - c^2 = 49 - 25 = 24$, so the equation is $\frac{x^2}{49} + \frac{y^2}{24} = 1$.

Question 10

Solve the problem.
-A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches across at its opening and is 4 feet deep, where will the light be concentrated?
A) 0.1 in. from the vertex
B) 10.1 in. from the vertex
C) 0.2 in. from the vertex
D) 0.8 in. from the vertex

Accepted Answer

The answer of Solve the problem.
-A reflecting telescope contains a...

Question 11

Match the graph to its equation.
-
A) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 9 } = 1$

B) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 16 } = 1$

C) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1$

D) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$

Accepted Answer

The answer of Match the graph to its equation.
-
A) \(\frac...

Question 12

Find the center, foci, and vertices of the ellipse.
-$$4 x ^ { 2 } + 64 y ^ { 2 } = 256$$
A) center at $( 0,0 )$
foci at $( 0 , - 2 )$ and $( 0,2 )$
vertices at $( 0 , - 4 ) , ( 0,4 )$

B) center at $( 0,0 )$
foci at $( 0 , - 2 \sqrt { 15 } )$ and $( 0,2 \sqrt { 15 } )$
vertices at $( 0 , - 8 ) , ( 0,8 )$

C) center at $( 0,0 )$
foci at $( - 2 \sqrt { 15 } , 0 )$ and $( 2 \sqrt { 15 } , 0 )$
vertices at $( - 8,0 ) , ( 8,0 )$

D) center at $( 0,0 )$
foci at $( - 8,0 )$ and $( 8,0 )$
vertices at $( - 64,0 ) , ( 64,0 )$

Accepted Answer

The given equation of the ellipse is $4x^2 + 64y^2 = 256$. Dividing the entire equation by 256 gives $\frac{x^2}{64} + \frac{y^2}{4} = 1$, which is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 64$ and $b^2 = 4$. Since $a^2 > b^2$, the ellipse is oriented along the x-axis. The center is at (0,0), the vertices are at $(\pm a, 0) = (\pm 8, 0)$, and the foci are at $(\pm \sqrt{a^2 - b^2}, 0) = (\pm \sqrt{64 - 4}, 0) = (\pm \sqrt{60}, 0) = (\pm 2\sqrt{15}, 0)$.

Question 13

Solve the problem.
-A searchlight is shaped like a paraboloid of revolution. If the light source is located 5 feet from the base along the axis of symmetry and the opening is 8 feet across, how deep should the searchlight be?
A) 4 ft
B) 0.8 ft
C) 1.6 ft
D) 3.2 ft

Accepted Answer

The answer of Solve the problem.
-A searchlight is shaped like...

Question 14

Graph the equation.
-$(y-1)^{2}=-7(x+2)$

A)

B)

C)

D)

Accepted Answer

The answer of Graph the equation.
-$(y-1)^{2}=-7(x+2)$

A)

B)

C)

D)

...

Question 15

Solve the problem.
-An experimental model for a suspension bridge is built in the shape of a parabolic arch. 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 50 inches apart. At a point between the towers and 15 inches along the road from the
Base of one tower, the cable is 1 inches above the roadway. Find the height of the towers.
A) 6.75 in.
B) 5.75 in.
C) 6.25 in.
D) 8.25 in.

Accepted Answer

Since the cable follows the shape of a parabolic arch, we know that it can be modeled by a quadratic equation of the form y = ax² + bx + c, where y is the height of the cable above the roadway at a given distance x along the road from one tower.

We know that the cable just touches the roadway at two points: at the base of each tower. This means that the x-coordinates of these points are equidistant from the midpoint between the towers, which is 25 inches. We are also given one point on the parabola: (15, 1), which means that we can substitute these values into our equation to get an equation in terms of a, b, and c:

1 = a(15)² + b(15) + c
0 = a(0)² + b(0) + c
0 = a(50)² + b(50) + c

Simplifying each equation and solving the resulting system of equations, we get:

a = -0.001
b = 0.3
c = 1

The height of each tower is given by the vertex of the parabola, which is at x = 25. Plugging x = 25 into our equation, we get:

y = -0.001(25)² + 0.3(25) + 1
y = 6.25

Therefore, the height of each tower is 6.25 inches, and the best choice is C.

Question 16

Solve the problem.
-An experimental model for a suspension bridge is built in the shape of a parabolic arch. 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 12.25 inches tall and stand 70 inches apart. Find the vertical distance from the
Roadway to the cable at a point on the road 14 inches from the lowest point of the cable.
A) 2.16 in.
B) 1.76 in.
C) 7.84 in.
D) 1.96 in.

Accepted Answer

The answer of Solve the problem.
-An experimental model for a...

Question 17

Solve the problem.
-A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 174 feet and a maximum height of 30 feet. Find the height of the arch at 15 feet from its center.
A) 3.6 ft
B) 21.8 ft
C) 0.2 ft
D) 29.1 ft

Accepted Answer

The answer of Solve the problem.
-A bridge is built in...

Question 18

Solve the problem.
-A reflecting telescope has a mirror shaped like a paraboloid of revolution. If the distance of the vertex to the focus is 31 feet and the distance across the top of the mirror is 66 inches, how deep is the mirror in the center? 
A) $\frac { 121 } { 1984 }$ in.

B) $\frac { 363 } { 496 }$ in.

C) $\frac { 961 } { 132 }$ in.

D) $\frac { 1089 } { 124 } \mathrm { in }$.

Accepted Answer

The depth of the mirror can be found using the formula for a paraboloid of revolution, $y = \frac{1}{4f}(x^2)$, where $f$ is the focal length and $x$ is the radius of the mirror. The focal length is 31 feet, which is 372 inches. The diameter of the mirror is 66 inches, so the radius is 33 inches. Plugging these values into the formula gives $y = \frac{1}{4(372)}(33)^2 = \frac{1089}{1488} = \frac{363}{496}$ inches.

Question 19

Graph the equation.
-$(x+1)^{2}=-8(y-2)$

A)

B)

C)

D)

Accepted Answer

The answer of Graph the equation.
-$(x+1)^{2}=-8(y-2)$

A)

B)

C)

D)

...

Question 20

Solve the problem.
-An experimental model for a suspension bridge is built in the shape of a parabolic arch. 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 12.25 inches tall and stand 70 inches apart. At some point along the road from the
Lowest point of the cable, the cable is 1.96 inches above the roadway. Find the distance between that point and
The base of the nearest tower.
A) 21 in.
B) 13.8 in.
C) 14.2 in.
D) 21.2 in.

Accepted Answer

The answer of Solve the problem.
-An experimental model for a...

Quiz 9: Analytic Geometry

Solve the problem. -A spotlight has a parabolic cross section that is 6 ft wide at the opening and 2.5 ft deep at the vertex. How far from the vertex is the focus? Round answer to two decimal places.

Graph the equation. - $(x+2) ^{2}=8(y-1)$ $Graph the equation. - (x+2) ^{2}=8(y-1) A) B) C) D)$

Find the center, foci, and vertices of the ellipse. - $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 1$

Find the center, foci, and vertices of the ellipse. - $9 x ^ { 2 } + 4 y ^ { 2 } = 36$

Solve the problem. -A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 3 centimeters from the vertex. If the depth is to be 6 centimeters, what is the diameter of the headlight at its opening?

Find the center, foci, and vertices of the ellipse. - $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$

Find an equation for the ellipse described. -Center at $( 0,0 )$ ; focus at $( 5,0 )$ ; vertex at $( 7,0 )$

Solve the problem. -A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches across at its opening and is 4 feet deep, where will the light be concentrated?

Find the center, foci, and vertices of the ellipse. - $4 x ^ { 2 } + 64 y ^ { 2 } = 256$

Solve the problem. -A searchlight is shaped like a paraboloid of revolution. If the light source is located 5 feet from the base along the axis of symmetry and the opening is 8 feet across, how deep should the searchlight be?

Graph the equation. - $(y-1) ^{2}=-7(x+2)$ $Graph the equation. - (y-1) ^{2}=-7(x+2) A) B) C) D)$

Solve the problem. -A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 174 feet and a maximum height of 30 feet. Find the height of the arch at 15 feet from its center.

Solve the problem. -A reflecting telescope has a mirror shaped like a paraboloid of revolution. If the distance of the vertex to the focus is 31 feet and the distance across the top of the mirror is 66 inches, how deep is the mirror in the center?

Graph the equation. - $(x+1) ^{2}=-8(y-2)$ $Graph the equation. - (x+1) ^{2}=-8(y-2) A) B) C) D)$

Quiz 9: Analytic Geometry

Solve the problem. -A spotlight has a parabolic cross section that is 6 ft wide at the opening and 2.5 ft deep at the vertex. How far from the vertex is the focus? Round answer to two decimal places.

Graph the equation. - (x+2) 2=8(y−1) (x+2) ^{2}=8(y-1) (x+2) 2=8(y−1)

Find the center, foci, and vertices of the ellipse. - x281+y29=1\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 } = 181x2​+9y2​=1

Find the center, foci, and vertices of the ellipse. - 9x2+4y2=369 x ^ { 2 } + 4 y ^ { 2 } = 369x2+4y2=36

Match the graph to its equation. -