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Precalculus : Understand features of hyperbolas and ellipses

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Example Question #21 : Understand Features Of Hyperbolas And Ellipses

The equation of an ellipse is given by

$\frac{(x-4)^2}{9}+\frac{(y+3)^2}{25}=1$

Find the center and foci of the ellipse.

Possible Answers:

Center (4, -3), Foci (4, 1) and (4, -7)

Center (4, -3), Foci (-1, -3) and (9, -3)

Center (-4, 3), Foci (0, -3) and (8, -3)

Center (4, -3), Foci (4, 2) and (4, -8)

Center (4, -3), Foci (0, -3) and (8, -3)

Correct answer:

Center (4, -3), Foci (4, 1) and (4, -7)

Explanation:

Begin by noticing the equation is already in standard form

$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$

First, the center is given by (h,k). In this problem, h=4, y=-3 so the center is (4, -3).

To find the foci, we use the equation a^2=b^2+c^2 , where a^2 is the larger denominator, $b^{2}$ is the smaller denominator, and c is the distance from the center to the foci. Plugging in the values, we have

25=9+c^2

16=c^2

4=c

Therefore, the foci are 4 units from the center. Because the larger denominator is under the y term, the foci are 4 units in either vertical direction. Thus, the foci are at

(4, -3+4)=(4,1) AND (4, -3-4)=(4,-7)

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Example Question #22 : Understand Features Of Hyperbolas And Ellipses

Find the foci for the ellipse with the following equation:

$\frac{x^2}{25}+\frac{y^2}{34}=1$

Possible Answers:

(0, 3), (3, 3)

(3, 3), (0, 0)

(3, 0), (-3, 0)

(0, 3), (0, -3)

Correct answer:

(0, 3), (0, -3)

Explanation:

Recall that the standard form of the equation of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where (h, k) is the center for the ellipse.

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

For the given equation, the center is at (0, 0) . Since b^2>a^2 , the major-axis is vertical.

Plug in the values to solve for .

$c=\sqrt{34-25}=\sqrt9=3$

The foci are then at the points (0, 3) and (0, -3) .

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Example Question #21 : Hyperbolas And Ellipses

Find the eccentricity for the ellipse with the following equation:

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

Possible Answers:

$\frac{7}{24}$

$\frac{24}{7}$

$\frac{2\sqrt6}{7}$

$\frac{49}{25}$

Correct answer:

$\frac{2\sqrt6}{7}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

Next, find the distance from the center to the focus of the ellipse, . Recall that when a>b , the major axis will lie along the -axis and be horizontal and that when b>a , the major axis will lie along the -axis and be vertical.

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question, a>b , so

$c=\sqrt{49-25}=\sqrt{24}=2\sqrt6$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because a>b , the major axis for this ellipse is horizontal. will be the distance from the center to the vertices.

For this ellipse, a=7 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{2\sqrt6}{7}$

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Example Question #22 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

$\frac{x^2}{36}+\frac{y^2}{9}=1$

Possible Answers:

$\frac{1}{9}$

$3\sqrt3$

$\frac{9}{36}$

$\frac{\sqrt3}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question, a>b , so

$c=\sqrt{36-9}=\sqrt{27}=3\sqrt3$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because a>b , the major axis for this ellipse is horizontal. will be the distance from the center to the vertices.

For this ellipse, a=6 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2}$

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Example Question #21 : Hyperbolas And Ellipses

Find the eccentricity of an ellipse with the following equation:

$\frac{(x-4)^2}{27}+\frac{(y-3)^2}{11}=1$

Possible Answers:

$\frac{\sqrt3}{4}$

$\frac{11}{27}$

$\frac{4\sqrt3}{3}$

$\frac{4\sqrt3}{9}$

Correct answer:

$\frac{4\sqrt3}{9}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{27-11}=\sqrt{16}=4$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because a>b , the major axis for this ellipse is horizontal. will be the distance from the center to the vertices.

For this ellipse, $a=\sqrt{27}=3\sqrt3$ .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{4}{3\sqrt3}=\frac{4\sqrt3}{9}$

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Example Question #23 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

$\frac{(x-6)^2}{100}-\frac{(y-9)^2}{400}=1$

Possible Answers:

$\frac{2\sqrt3}{3}$

$\frac{\sqrt3}{2}$

$\frac{1}{4}$

$\frac{\sqrt3}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{400-100}=\sqrt{300}=10\sqrt3$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because b>a , the major axis for this ellipse is vertical. will be the distance from the center to the vertices.

For this ellipse, b=20 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{10\sqrt3}{20}=\frac{\sqrt3}{2}$

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Example Question #111 : Conic Sections

Find the eccentricity of the ellipse with the following equation:

$\frac{(x-10)^2}{24}+\frac{(y+2)^2}{25}=1$

Possible Answers:

$\frac{2\sqrt6}{9}$

$\frac{2\sqrt6}{25}$

$\frac{1}{5}$

$\frac{24}{25}$

Correct answer:

$\frac{1}{5}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{25-24}=\sqrt{1}=1$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because b>a , the major axis for this ellipse is vertical. will be the distance from the center to the vertices.

For this ellipse, b=5 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{1}{5}$

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Example Question #28 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of an ellipse with the following equation:

$\frac{(x-2)^2}{15}+\frac{(y-1)^2}{6}=1$

Possible Answers:

$\frac{\sqrt{15}}{4}$

$\frac{3\sqrt{15}}{5}$

$\frac{2}{5}$

$\frac{\sqrt{15}}{5}$

Correct answer:

$\frac{\sqrt{15}}{5}$

Explanation:

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{15-6}=\sqrt{9}=3$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because a>b , the major axis for this ellipse is horizontal. will be the distance from the center to the vertices.

For this ellipse, $a=\sqrt{15}$ .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$3=\frac{3}{\sqrt{15}}=\frac{\sqrt{15}}{5}$

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Example Question #21 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

4x^2+16y^2+40x-64y+100=0

Possible Answers:

$\frac{\sqrt3}{5}$

$\frac{2\sqrt3}{3}$

$\frac{\sqrt3}{3}$

$\frac{\sqrt3}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

Start by putting this equation in the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where (h, k) is the center of the ellipse.

Group the terms together and the terms together.

4x^2+16y^2+40x-64y+100=0

4x^2+40x+16y^2-64y+100=0

Factor out a from the terms and from the terms.

4(x^2+10x)+16(y^2-4y)+100=0

Now, complete the squares. Remember to add the same amount to both sides!

4(x^2+10x+25)+16(y^2-4y+4)+100=164

Subtract 100 from both sides.

4(x^2+10x+25)+16(y^2-4y+4)=64

Divide both sides by .

$\frac{x^2+10x+25}{16}+\frac{y^2-4y+4}{4}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x+5)^2}{16}+\frac{(y-2)^2}{4}=1$

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{16-4}=\sqrt{12}=2\sqrt3$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because a>b , the major axis for this ellipse is horizontal. will be the distance from the center to the vertices.

For this ellipse, a=4 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{2\sqrt3}{4}=\frac{\sqrt3}{2}$

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Example Question #2 : Ellipses

Find the eccentricity of an ellipse with the following equation:

25x^2+4y^2-400x-8y+1504=0

Possible Answers:

$\frac{4}{25}$

$\frac{\sqrt{21}}{6}$

$\frac{\sqrt{5}}{21}$

$\frac{\sqrt{21}}{5}$

Correct answer:

$\frac{\sqrt{21}}{5}$

Explanation:

Start by putting this equation in the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where (h, k) is the center of the ellipse.

Group the terms together and the terms together.

25x^2+4y^2-400x-8y+1504=0

25x^2-400x+4y^2-8y+1504=0

Factor out from the terms and from the terms.

25(x^2-16x)+4(y^2-2y)+1504=0

Now, complete the squares. Remember to add the same amount to both sides!

25(x^2-16x+64)+4(y^2-2y+1)+1504=1604

Subtract 1504 from both sides.

25(x^2-16x+64)+4(y^2-2y+1)=100

Divide both sides by 100 .

$\frac{(x^2-16x+64)}{4}+\frac{y^2-2y+1}{25}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-8)^2}{4}+\frac{(y+1)^2}{25}=1$

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for a>b , or

$c=\sqrt{b^2-a^2}$ for b>a

For the ellipse in question,

$c=\sqrt{25-4}=\sqrt{21}$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because b>a , the major axis for this ellipse is vertical. will be the distance from the center to the vertices.

For this ellipse, b=5 .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{\sqrt{21}}{5}$

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Precalculus : Understand features of hyperbolas and ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Understand Features Of Hyperbolas And Ellipses

Example Question #22 : Understand Features Of Hyperbolas And Ellipses

Example Question #21 : Hyperbolas And Ellipses

Example Question #22 : Hyperbolas And Ellipses

Example Question #21 : Hyperbolas And Ellipses

Example Question #23 : Hyperbolas And Ellipses

Example Question #111 : Conic Sections

Example Question #28 : Understand Features Of Hyperbolas And Ellipses

Example Question #21 : Understand Features Of Hyperbolas And Ellipses

Example Question #2 : Ellipses

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