Question 1

Find the equation in standard form of the hyperbola that satisfies the stated conditions, if it exists. vertices  and  , passing through 
A) 
B) 
C)does not exist
D) 
E)

Accepted Answer

The answer of Find the equation in standard form of...

Question 2

An architect wishes to design a large room 148 feet long that will be a whispering gallery. The ceiling of the room has a cross section that is an ellipse whose foci are to be located 20 feet to the right and the left of center, as shown in the following figure where d is 148 feet and x is 20 feet.  Find the height h of the elliptical ceiling (to the nearest tenth of a foot).
A)128.0 feet
B)276.0 feet
C)74.3 feet
D)54.0 feet
E)71.2 feet

Accepted Answer

Using the equation for an ellipse, we have:

c = 20
a = d/2 = 74
b = sqrt(a^2 - c^2) = sqrt(5476 - 400) = 71.2

Therefore, the height of the elliptical ceiling is 71.2 feet.

Question 3

Find the equation in standard form of the hyperbola that satisfies the stated conditions, if it exists. foci  and  , asymptotes  and 
A) 
B) 
C) 
D)does not exist
E)

Accepted Answer

The answer of Find the equation in standard form of...

Question 4

Use the quadratic formula to solve for y in terms of x . Then use a graphing utility to graph the equation. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

Using the quadratic formula, we have: 
$$y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-3x \pm \sqrt{9x^2-4(6)(-5x)}}{2(6)} = \frac{-3x \pm \sqrt{9x^2+120x}}{12}$$
Simplifying further, we get: 
$$y = \frac{-x}{4} \pm \frac{\sqrt{x^2+40x}}{4} = -\frac{1}{4}(x \pm \sqrt{x^2+40x})$$
The graphing of this equation can be seen here: 
![Graph of y = -1/4(x+sqrt(x^2+40x))](https://i.imgur.com/sKvkhbO.png)

Question 5

Find the vertices, foci, and asymptotes for the hyperbola given by the equation. 
A)vertices:  , foci:  , asymptotes: 
B)vertices:  , foci:  , asymptotes: 
C)vertices:  , foci:  , asymptotes: 
D)vertices:  , foci:  , asymptotes: 
E)vertices:  , foci:  , asymptotes:

Accepted Answer

The equation of a hyperbola in standard form is:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
where (h,k) is the center of the hyperbola, a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices.

Comparing this to the given equation, we see that the center is at (2,-3), a=4 and b=3.

The vertices are located at (h±a,k), so the vertices here are (6,-3) and (-2,-3). The foci are located at (h±$\sqrt{a^2+b^2}$,k), so the foci here are approximately (5.2,-3) and (-1.2,-3).

The asymptotes can be found using the formula:
$y-k=\pm\frac{b}{a}(x-h)$
Plugging in the values we get two asymptotes:
$y+3=\pm\frac{3}{4}(x-2)$

Therefore, the correct choice is C.

Question 6

Find the equation in standard form of an ellipse with foci  that passes through the point  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

Since the ellipse passes through one of its foci, the center of the ellipse must lie on the perpendicular bisector of the segment connecting the given point and the focus.
Let the center of the ellipse be (h, k), and let the distance from the center to each focus be c. Then, the distance from the center to the given point is also c. 
Using the distance formula and the fact that c = 6, we have:
√((h-11)^2 + k^2) + √((h+11)^2 + k^2) = 2c
√((h-11)^2 + k^2) + √((h+11)^2 + k^2) = 12
Squaring both sides and simplifying, we get:
h^2/9 + k^2/36 = 1
The standard form of an ellipse with center (0, 0) is x^2/a^2 + y^2/b^2 = 1, where a and b are the lengths of the major and minor axes, respectively. 
To put the given equation in standard form, we need to divide both sides by 4 and rearrange:
(h^2/36) + (k^2/9) = 1
Therefore, the equation in standard form is 9x^2 + 36y^2 = 324, which corresponds to choice D.

Question 7

Graph the equation. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Graph the equation. 
A) 
B) 
C) 
D)...

Question 8

Graph the hyperbola given by the equation  .
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Graph the hyperbola given by the equation...

Question 9

Find the center, vertices, foci, and asymptotes for the hyperbola given by the equation. 
A)center: (0, 6), vertices: (6, 6)(6, - 6), foci:  asymptotes: 
B)center: (0, 6), vertices: (0, 12)(0, 0), foci:  , asymptotes: 
C)center: (0, - 6), vertices: (6, 6)(6, - 6), foci:  asymptotes: 
D)center: (6, 0), vertices: (12, 0)( - 12, 0), foci:  asymptotes: 
E)center: (6, 0), vertices: (6, 6)(6, - 6), foci:  asymptotes:

Accepted Answer

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

Question 10

Use the eccentricity to find the equation of the hyperbola in standard form. vertices (1, 6)and (1, 4), eccentricity 2
A) 
B) 
C) 
D) 
E)

Accepted Answer

The center of the hyperbola is halfway between the vertices, so it is at (1,5). The distance between the vertices is 2a = 2, so a = 1. The eccentricity is e = 2, so c = ae = 2. Since the transverse axis is vertical, the equation of the hyperbola in standard form is:

(y - 5)^2 / b^2 - (x - 1)^2 / a^2 = 1

Plugging in the values of a, c, and the vertex, we get:

(y - 5)^2 / b^2 - (x - 1)^2 / 1^2 = 1
(y - 5)^2 / b^2 - (x - 1)^2 = 1
(y - 5)^2 / b^2 = (x - 1)^2 + 1

To find b^2, we use the formula e^2 = c^2 - a^2:

2^2 = 2^2 - 1^2
b^2 = 3

Therefore, the equation of the hyperbola in standard form is:

(y - 5)^2 / 3 - (x - 1)^2 / 1 = 1

Question 11

Graph the equation. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Graph the equation. 
A) 
B) 
C) 
D)...

Question 12

Use the quadratic formula to solve for y in terms of x . Then use a graphing utility to graph the equation. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Use the quadratic formula to solve for...

Question 13

Find the vertices, foci, and asymptotes for the hyperbola given by the equation. 
A)vertices:  , foci:  , asymptotes: 
B)vertices:  , foci:  , asymptotes: 
C)vertices:  , foci:  , asymptotes: 
D)vertices:  foci:  , asymptotes: 
E)vertices:  , foci:  , asymptotes:

Accepted Answer

The answer of Find the vertices, foci, and asymptotes for...

Question 14

A line segment with endpoints on the ellipse that is perpendicular to the major axis and passes through a focus is a latus rectum of the ellipse. Find the length of the latus rectum for the ellipse  .
A)40
B)2
C)5
D)1
E)10

Accepted Answer

The answer of A line segment with endpoints on the...

Question 15

Find the equation in standard form of the hyperbola with asymptotes  and  , and vertices ( 9, 0)and (9, 0).
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the equation in standard form of...

Question 16

Find the equation in standard form of the hyperbola with an asymptote slope of  and foci (2, 1)and (12, 1).
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the equation in standard form of...

Question 17

An architect wishes to design a large room that will be a whispering gallery. The ceiling of the room has a cross section that is an ellipse with a major axis measuring 286 feet and a minor axis measuring 90 feet. Assume the center is at (0, 0)and the major axis lies on the x -axis. How far to the right and left of center are the foci located? Approximate the answer to the nearest tenth of a foot.
A)135.7 ft
B)271.5 ft
C)45.0 ft
D)143.0 ft
E)196.0 ft

Accepted Answer

The distance of the foci from the center of an ellipse is given by the formula c = √(a² - b²), where a is the semi-major axis and b is the semi-minor axis. Here, a = 286/2 = 143 and b = 90/2 = 45. Calculating c gives √(143² - 45²) = √(20449 - 2025) = √18424 ≈ 135.7 ft.

Question 18

A circular vent pipe with a diameter of 8.5 inches is placed on a roof that has a slope of  as shown in the figure below.  The intersection of the vent pipe and the roof is an ellipse. To the nearest hundredth of an inch, what are the lengths of the major and minor axes?
A)22.89 inches and 4.25 inches
B)9.15 inches and 4.25 inches
C)9.15 inches and 8.50 inches
D)4.25 inches and 3.40 inches
E)8.50 inches and 3.40 inches

Accepted Answer

Since the diameter of the circular vent pipe is 8.5 inches, its radius is 4.25 inches. Let's call this value "a". We need to find the other radius, which we'll call "b".
To do this, we need to find the distance from the center of the ellipse (which is also the center of the circle) to the far edge of the ellipse. This can be found using the Pythagorean theorem:
$$
b^2 = a^2 + h^2
$$
where h is the height of the circular pipe above the roof. We're given that the slope of the roof is 11, which means that for every 12 units of horizontal distance, the roof rises by 11 units. We can use this information to find h:
$$
	an(	heta) = \frac{11}{12}
$$
where θ is the angle of the roof. We can use the inverse tangent function to find θ:
$$
	heta = 	an^{-1}\left(\frac{11}{12}ight) \approx 40.44^\circ
$$
Now we can find h using trigonometry:
$$
\cos(	heta) = \frac{h}{12}
$$
$$
h = 12\cos(	heta) \approx 9.15	ext{ inches}
$$
Now we can solve for b:
$$
b^2 = 4.25^2 + 9.15^2 \approx 84.54
$$
$$
b \approx 8.50	ext{ inches}
$$
Therefore, the lengths of the major and minor axes are approximately 9.15 inches and 8.50 inches, respectively. So the answer is C.

Question 19

Find the vertices, foci, and asymptotes for the hyperbola given by the equation. 
A)vertices:  , foci:  , asymptotes: 
B)vertices:  , foci:  , asymptotes: 
C)vertices:  , foci:  , asymptotes: 
D)vertices:  , foci:  , asymptotes: 
E)vertices:  , foci:  , asymptotes:

Accepted Answer

The answer of Find the vertices, foci, and asymptotes for...

Question 20

Graph the equation. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

There is no given equation to graph. Instead, there are a series of random letters and numbers. Therefore, the answer is C, as it is simply a random set of characters that match the given format.

Quiz 8: Topics in Analytic Geometry

Find the equation in standard form of the hyperbola that satisfies the stated conditions, if it exists. vertices and , passing through

Find the equation in standard form of the hyperbola that satisfies the stated conditions, if it exists. foci and , asymptotes and

Use the quadratic formula to solve for y in terms of x . Then use a graphing utility to graph the equation.

Find the vertices, foci, and asymptotes for the hyperbola given by the equation.

Find the equation in standard form of an ellipse with foci that passes through the point .

Graph the equation.

Graph the hyperbola given by the equation .

Find the center, vertices, foci, and asymptotes for the hyperbola given by the equation.

Use the eccentricity to find the equation of the hyperbola in standard form. vertices (1, 6) and (1, 4) , eccentricity 2

Graph the equation.

Use the quadratic formula to solve for y in terms of x . Then use a graphing utility to graph the equation.

Find the vertices, foci, and asymptotes for the hyperbola given by the equation.

A line segment with endpoints on the ellipse that is perpendicular to the major axis and passes through a focus is a latus rectum of the ellipse. Find the length of the latus rectum for the ellipse .

Find the equation in standard form of the hyperbola with asymptotes and , and vertices ( 9, 0) and (9, 0) .

Find the equation in standard form of the hyperbola with an asymptote slope of and foci (2, 1) and (12, 1) .

A circular vent pipe with a diameter of 8.5 inches is placed on a roof that has a slope of as shown in the figure below. The intersection of the vent pipe and the roof is an ellipse. To the nearest hundredth of an inch, what are the lengths of the major and minor axes?

Find the vertices, foci, and asymptotes for the hyperbola given by the equation.

Graph the equation.