Question 1

Graph the equation.
-$$( y - 1 ) ^ { 2 } = - 5 ( x - 1 )$$

Accepted Answer

The answer of Graph the equation.
-\[( y - 1 )...

Question 2

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

Accepted Answer

The answer of Graph the equation.
-\[( y + 1 )...

Question 3

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

Accepted Answer

The answer of Graph the equation.
-\[( x - 2 )...

Question 4

Graph the ellipse and locate the foci.
-$16 x^{2}+4 y^{2}=64$

Accepted Answer

The answer of Graph the ellipse and locate the foci.
-\(16...

Question 5

Graph the ellipse and locate the foci.
-$$\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1$$

Accepted Answer

The answer of Graph the ellipse and locate the foci.
-\[\frac...

Question 6

Find the center, foci, and vertices of the ellipse.
-$$\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$$

Accepted Answer

The center of the ellipse is given by the point $(h, k)$ in the general equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, which is $(3, -3)$ in this case. The lengths of the semi-major and semi-minor axes are $\sqrt{36} = 6$ and $\sqrt{9} = 3$, respectively. Since $a^2 = 36$ and $b^2 = 9$, and the ellipse is wider than it is tall, it is horizontally oriented. The foci are calculated using the formula $c = \sqrt{a^2 - b^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}$. Therefore, the foci are at $(3 \pm 3\sqrt{3}, -3)$. The vertices are at the ends of the major axis, which are $(3 \pm 6, -3)$, simplifying to $(-3, -3)$ and $(9, -3)$.

Question 7

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 8

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 14 feet across, how deep should the searchlight be?

Accepted Answer

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

Question 9

Graph the equation.
-$$( x - 2 ) ^ { 2 } = - 5 ( y + 1 )$$

Accepted Answer

The answer of Graph the equation.
-\[( x - 2 )...

Question 10

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 60 inches apart. At a point between the towers and 18 inches along the road from the
Base of one tower, the cable is 1.44 inches above the roadway. Find the height of the towers.

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 60 inches apart and the cable touches the roadway exactly between them, the vertex is at the midpoint, 30 inches from each tower, and at the roadway level, making $k = 0$. The point (18, 1.44) gives us a way to solve for $a$ because it is 18 inches from one tower and 1.44 inches above the roadway. Plugging these values into the equation gives $1.44 = a(18 - 30)^2$, which simplifies to $1.44 = a(12)^2$, or $1.44 = 144a$. Solving for $a$ gives $a = 0.01$. To find the height of the towers, we use the fact that the towers are at the ends of the parabola, 30 inches from the vertex on either side. Plugging this into the equation with $a = 0.01$ and $x = 30$ (since the towers are 30 inches from the midpoint), we get $y = 0.01(30)^2 = 0.01(900) = 9$. Therefore, the towers are 9 inches tall.

Question 11

Solve the problem.
-A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 142 feet and a maximum height of 25 feet. Find the height of the arch at 25 feet from its center.

Accepted Answer

The equation of a parabolic arch can be represented as $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. Given the maximum height of the arch is 25 feet, this means $k = 25$. The span of the bridge is 142 feet, so the vertex is at the center, making $h = 0$. The distance from the center to either end is 71 feet, so when $x = 71$, $y = 0$. We can use this information to find $a$: $0 = a(71)^2 + 25$, solving for $a$ gives $a \approx -0.00496$. To find the height at 25 feet from the center, substitute $x = 25$ into the equation: $y = -0.00496(25)^2 + 25$, which gives $y \approx 21.9$.

Question 12

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.

Accepted Answer

The focus of a parabola is located at a distance of p/4 from the vertex, where p is the distance between the focus and the directrix. In this case, the distance between the focus and the vertex is 2.5 ft, so the focus is located at a distance of 2.5/4 = 0.625 ft from the vertex.

Question 13

Solve the problem.
-A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 20 inches across at its opening and is 5 feet deep, where will the light be concentrated?

Accepted Answer

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

Question 14

Write an equation for the graph.
-

Accepted Answer

The answer of Write an equation for the graph.
- 
...

Question 15

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 16

Graph the ellipse and locate the foci.
-$4 x^{2}+16 y^{2}=64$

Accepted Answer

The answer of Graph the ellipse and locate the foci.
-\(4...

Question 17

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 30 feet and the distance across the top of the mirror is 50 inches, how deep is the mirror in the center?

Accepted Answer

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

Question 18

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.

Accepted Answer

The equation of a parabola in vertex form is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola. Since the lowest point of the cable is at the middle of the two towers and touches the roadway, we can assume the vertex of the parabola is at $(0,0)$ for simplicity. The towers are 70 inches apart, so one tower is at $x = 35$ inches and the other at $x = -35$ inches, both with a height of $12.25$ inches ($y = 12.25$). We can use one of these points to find $a$, the coefficient that determines the width and direction of the parabola.Using the point $(35, 12.25)$, we substitute into the vertex form equation to solve for $a$:$$12.25 = a(35)^2$$$$12.25 = 1225a$$$$a = \frac{12.25}{1225}$$$$a = 0.01$$So, the equation of the parabola is $y = 0.01x^2$.To find the vertical distance from the roadway to the cable at a point 14 inches from the lowest point of the cable, we substitute $x = 14$ into the equation:$$y = 0.01(14)^2$$$$y = 0.01(196)$$$$y = 1.96$$Therefore, the vertical distance from the roadway to the cable at that point is 1.96 inches.

Question 19

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 9 inches tall and stand 60 inches apart. At some point along the road from the lowest
Point of the cable, the cable is 0.81 inches above the roadway. Find the distance between that point and the base
Of the nearest tower.

Accepted Answer

The parabola can be modeled with the equation $ y = a(x - h)^2 + k $, where the vertex is at the lowest point of the cable. Given the towers are 9 inches tall and 60 inches apart, the vertex is at (30, 0). Solving for when the cable is 0.81 inches above the roadway, we find the distance to be approximately 21 inches from the nearest tower.

Question 20

Graph the ellipse and locate the foci.
-$$\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$$

Accepted Answer

The answer of Graph the ellipse and locate the foci.
-\[\frac...

Quiz 5: Conics

Graph the equation. - $( y - 1 ) ^ { 2 } = - 5 ( x - 1 )$ $Graph the equation. - ( y - 1 ) ^ { 2 } = - 5 ( x - 1 ) A) B) C) D)$

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

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. - $\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 1$

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 14 feet across, how deep should the searchlight be?

Graph the equation. - $( x - 2 ) ^ { 2 } = - 5 ( y + 1 )$ $Graph the equation. - ( x - 2 ) ^ { 2 } = - 5 ( y + 1 ) A) B) C) D)$

Solve the problem. -A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 142 feet and a maximum height of 25 feet. Find the height of the arch at 25 feet from its center.

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.

Solve the problem. -A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 20 inches across at its opening and is 5 feet deep, where will the light be concentrated?

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?

Graph the ellipse and locate the foci. - $4 x^{2}+16 y^{2}=64$

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 30 feet and the distance across the top of the mirror is 50 inches, how deep is the mirror in the center?

Quiz 5: Conics

Graph the equation. - (y−1) 2=−5(x−1) ( y - 1 ) ^ { 2 } = - 5 ( x - 1 ) (y−1) 2=−5(x−1)

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

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

Graph the ellipse and locate the foci. - 16x2+4y2=6416 x^{2}+4 y^{2}=6416x2+4y2=64

Graph the ellipse and locate the foci. - x29+y24=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 19x2​+4y2​=1

Find the center, foci, and vertices of the ellipse. - (x−3) 236+(y+3) 29=1\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y + 3 ) ^ { 2 } } { 9 } = 136(x−3) 2​+9(y+3) 2​=1