Question 1

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 formula for the focal length (f) of a paraboloid is f = d^2 / (16c), where d is the diameter and c is the depth. Here, d = 24 inches and c = 48 inches (4 feet). Plugging in these values gives f = 24^2 / (16*48) = 0.8 inches.

Question 2

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 is in the form of an ellipse centered at the origin $(0,0)$. To find the foci and vertices, we first need to bring the equation to its standard form by dividing both sides by 256, resulting in $\frac{x^2}{64} + \frac{y^2}{4} = 1$. This shows that $a^2 = 64$ and $b^2 = 4$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. Since $a^2 > b^2$, the major axis is along the x-axis. The length of the semi-major axis $a = 8$, so the vertices are at $(-8,0)$ and $(8,0)$. The foci are found using $c^2 = a^2 - b^2$, which gives $c^2 = 64 - 4 = 60$, so $c = \sqrt{60} = 2\sqrt{15}$. Therefore, the foci are at $(-2\sqrt{15}, 0)$ and $(2\sqrt{15}, 0)$.

Question 3

Write the word or phrase that best completes each statement or answers the question.
-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 Write the word or phrase that best...

Question 4

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 5

Choose the one alternative that best completes the statement or answers the question.
-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 Choose the one alternative that best completes...

Question 6

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) $
vertices at $ (-81,0),(81,0) $

Accepted Answer

The given equation of the ellipse is in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 81$ and $b^2 = 9$, indicating that $a = 9$ and $b = 3$. The center of the ellipse is at the origin $(0,0)$ because there are no terms that shift the ellipse horizontally or vertically. The major axis is along the x-axis because $a > b$, and the vertices are at $(-a,0)$ and $(a,0)$, or $(-9,0)$ and $(9,0)$. The foci are located along the major axis at a distance of $c$ from the center, where $c = \sqrt{a^2 - b^2} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2}$, so the foci are at $(-6\sqrt{2}, 0)$ and $(6\sqrt{2}, 0)$.

Question 7

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 distance from the vertex. Given the focal length is 31 feet (or 372 inches, converting feet to inches) and the diameter of the mirror is 66 inches (thus the radius, $x$, is 33 inches), we substitute these values into the formula to find the depth, $y$. This gives $y = \frac{1}{4 \times 372}(33)^2 = \frac{1089}{1488} = \frac{363}{496}$ inches.

Question 8

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 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, and $2b$ is the length of the minor axis. The distance from the center to a focus is $c$, and the distance from the center to a vertex is $a$. Given a vertex at $(7,0)$, $a = 7$, so $a^2 = 49$. The focus is at $(5,0)$, so $c = 5$, and $c^2 = 25$. Using the relationship $c^2 = a^2 - b^2$, we find $b^2 = a^2 - c^2 = 49 - 25 = 24$. Therefore, the correct equation is $\frac{x^2}{49} + \frac{y^2}{24} = 1$.

Question 10

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 11

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

The height of the towers can be found by using the vertex form of a parabolic equation, $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. Since the towers are 50 inches apart and the cable touches the roadway exactly between them, the vertex is at the midpoint, 25 inches from each tower, and at the roadway level. The point given, 15 inches from one tower and 1 inch above the roadway, provides a way to solve for $a$. Substituting $x = 15$, $y = 1$, and knowing the vertex is at $(25, 0)$ (shifting the coordinate system for simplicity), we get $1 = a(15 - 25)^2$, leading to $a = 1/100$. The height of the towers is the value of $y$ when $x = 0$ or $x = 50$, which gives $y = a(25)^2 = (1/100)(625) = 6.25$.

Question 12

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 13

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)...

Question 14

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 equation of the parabola can be derived using the vertex form and the given dimensions. Solving for the vertical distance at 14 inches from the lowest point gives 1.96 inches.

Question 15

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 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. 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 correct answer is A because the distance between the point where the cable is 1.96 inches above the roadway and the base of the nearest tower is 21 inches.

Question 17

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 18

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 19

Write the word or phrase that best completes each statement or answers the question.
-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 Write the word or phrase that best...

Question 20

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, indicated by the larger denominator under $y^2$. The center of the ellipse is at the origin (0,0) because it is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ without any shifts. The vertices are located at a distance of $b = \sqrt{64} = 8$ units along the y-axis from the center, hence at (0,-8) and (0,8). The foci are found using the formula $c = \sqrt{b^2 - a^2}$, where $a^2 = 16$ and $b^2 = 64$, so $c = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$. Therefore, the foci are at (0, -4\sqrt{3}) and (0, 4\sqrt{3}).

Quiz 9: Analytic Geometry

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. - $4 x ^ { 2 } + 64 y ^ { 2 } = 256$

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

Choose the one alternative that best completes the statement or answers the question. -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.

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

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?

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.

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

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

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

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

Quiz 9: Analytic Geometry

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. - 4x2+64y2=2564 x ^ { 2 } + 64 y ^ { 2 } = 2564x2+64y2=256

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

Choose the one alternative that best completes the statement or answers the question. -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.

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

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?

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.

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

Match the graph to its equation. -

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