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Precalculus : Conic Sections

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Example Questions

Example Question #28 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

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

Possible Answers:

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

$\frac{\sqrt3}{5}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{3}$

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, .

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

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,

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

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 #21 : Hyperbolas And Ellipses

Find the eccentricity of an ellipse with the following equation:

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

Possible Answers:

$\frac{4}{25}$

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

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

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

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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Example Question #31 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

25x^2+36y^2+300x-720y+3600=0

Possible Answers:

$e=\frac{3\sqrt{11}}{11}$

$e=\frac{\sqrt{11}}{6}$

$e=\frac{5\sqrt{11}}{6}$

$e=\frac{\sqrt{11}}{3}$

Correct answer:

$e=\frac{\sqrt{11}}{6}$

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+36y^2+300x-720y+3600=0

25x^2+300x+36y^2-720y+3600=0

Factor out from the terms and from the terms.

25(x^2+12x)+36(y^2-20y)+3600=0

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

25(x^2+12x+36)+36(y^2-20y+100)+3600=4500

Subtract 3600 from both sides.

25(x^2+12x+36)+36(y^2-20y+100)=900

Divide both sides by 900 .

$\frac{x^2+12x+36}{36}+\frac{y^2-20y+100}{25}=1$

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

$\frac{(x+6)^2}{36}+\frac{(y-10)^2}{25}=1$

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,

$c=\sqrt{36-25}=\sqrt{11}$

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.

$e=\frac{\sqrt{11}}{6}$

Report an Error

Example Question #124 : Conic Sections

Find the eccentricity of the ellipse with the following equation:

15x^2+16y^2+90x+32y-89=0

Possible Answers:

$\frac{1}{4}$

$\frac{3}{4}$

$\frac{1}{6}$

Correct answer:

$\frac{1}{4}$

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.

15x^2+16y^2+90x+32y-89=0

15x^2+90x+16y^2+32y-89=0

Factor out from the terms and from the terms.

15(x^2+6x)+16(y^2+2y)-89=0

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

15(x^2+6x+9)+16(y^2+2y+1)-89=151

Add to both sides.

15(x^2+6x+9)+16(y^2+2y+1)=240

Divide both sides by 240 .

$\frac{x^2+6x+9}{16}+\frac{y^2+2y+1}{15}=1$

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

$\frac{(x+3)^2}{16}+\frac{(y+1)^2}{15}=1$

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,

$c=\sqrt{16-15}=\sqrt{1}=1$

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 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.

$e=\frac{1}{4}$

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Example Question #32 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse with the following equation:

$\frac{(x-3)^2}{38}+\frac{(y-2)^2}{42}=1$

Possible Answers:

$\frac{\sqrt{42}}{2}$

$\frac{\sqrt{42}}{21}$

$\frac{4\sqrt{42}}{3}$

$\frac{1}{2}$

Correct answer:

$\frac{\sqrt{42}}{21}$

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,

$c=\sqrt{42-38}=\sqrt{4}=2$

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, $b=\sqrt{42}$ .

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.

$e=\frac{2}{\sqrt{42}}=\frac{\sqrt{42}}{21}$

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Example Question #126 : Conic Sections

Find the eccentricity of the ellipse with the following equation:

$\frac{(x+\pi)^2}{15}+\frac{(y-1)^2}{144}=1$

Possible Answers:

$\frac{8}{97}$

$\frac{2\sqrt2}{3}$

$\frac{\sqrt2}{2}$

$\frac{\sqrt2}{4}$

Correct answer:

$\frac{2\sqrt2}{3}$

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,

$c=\sqrt{144-16}=\sqrt{128}=8\sqrt2$

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=12 .

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.

$e=\frac{8\sqrt2}{12}=\frac{4\sqrt 2}{6}= \frac{2\sqrt2}{3}$

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Example Question #33 : Hyperbolas And Ellipses

Find the eccentricity of the ellipse

$\frac{(x-1)^2 }{ 49 } + \frac{(y - 4) ^2 }{ 25 } = 1$ .

Possible Answers:

$\frac{\sqrt{24}}{7}$

$\frac{\sqrt{24}}{49}$

$\frac{\sqrt{24}}{5 }$

$\frac{\sqrt{14}}{7}$

$\frac{24}{7}$

Correct answer:

$\frac{\sqrt{24}}{7}$

Explanation:

To find the eccentrictity, first we need to find c, the distance from the center to the foci. We can use the equation a^2 + b^2 = c^2 where a and b are the lengths of half the minor and major axes, and c is the distance from the center to the foci.

49 - 25 = c^2

24 = c^2

$\sqrt{24 } = c$

The eccentricity is $\frac{c}{a}$ where c is half the length of the major axis. In this case, $a = \sqrt{49 } = 7$ because 49 is greater than 25.

The eccentricity is $\frac {\sqrt {24 }}{ 7 }$ .

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Example Question #34 : Hyperbolas And Ellipses

Find the eccentricity of $\frac{ (x+3)^2 }{ 14 } + \frac{ (y-1)^2 }{4 } = 1$

Possible Answers:

$\sqrt{\frac{5}{7}}$

$\sqrt{\frac{9}{7}}$

$\frac{ \sqrt{10}}{14}$

$\frac{\sqrt{53}}{7}$

$\frac{3 \sqrt5}{7}$

Correct answer:

$\sqrt{\frac{5}{7}}$

Explanation:

To find the eccentricity, first find "c" as we would if we were finding the focus. The relationship between a, the radius of the major axis, b, the radius of the minor axis, and c for an ellipse is: b^2 = a^2 - c^2

4 = 14 - c^2 add c squared to both sides

4 + c^2 = 14 subtract 4 from both sides

c^2 = 10 take the square root

$c = \sqrt{10}$

Since a^2 = 14 , $a = \sqrt{14}$ .

The eccentricity is $e = \frac{c}{a}$ so here $\frac{\sqrt10}{\sqrt{14 } } = \sqrt{\frac{10}{14} } = \sqrt{\frac{5}{7}}$

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Example Question #31 : Hyperbolas And Ellipses

Find the eccentricity of $\frac{(x-1)^2 }{ 2 } + \frac{ (y+4)^2 }{ 3 } = 1$

Possible Answers:

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

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{\sqrt3}$

$\sqrt{\frac{5}{2}}$

Correct answer:

$\frac{1}{\sqrt3}$

Explanation:

First find "c" by using the relationship b^2 = a^2 - c^2 where a is the radius of the major axis and b is the radius of the minor axis.

2 = 3 - c ^2 add c squared to both sides

2 + c^2 = 3 subtract 2 from both sides

c^2 = 1 take the square root of both sides

c =1

The eccentricity is $e = \frac{c }{a}$ . Since a^2 = 3 , $a = \sqrt{3}$

$e = \frac{1}{\sqrt{3}}$

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Example Question #130 : Conic Sections

The equation of an ellipse is given by

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

Find the eccentricity of the ellipse.

Possible Answers:

$\frac{5}{3}$

$\frac{3}{5}$

$\frac{4}{3}$

$\frac{3}{4}$

$\frac{4}{5}$

Correct answer:

$\frac{4}{5}$

Explanation:

The equation for the eccentricity of an ellipse is given by

$e=\frac{c}a{}$

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

25=9+c^2

16=c^2

4=c

Plugging the values into the equation gives

$e=\frac{4}5{}$

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Precalculus : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #28 : Hyperbolas And Ellipses

Example Question #21 : Hyperbolas And Ellipses

Example Question #31 : Hyperbolas And Ellipses

Example Question #124 : Conic Sections

Example Question #32 : Hyperbolas And Ellipses

Example Question #126 : Conic Sections

Example Question #33 : Hyperbolas And Ellipses

Example Question #34 : Hyperbolas And Ellipses

Example Question #31 : Hyperbolas And Ellipses

Example Question #130 : Conic Sections

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