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

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

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 4x^2+16y^2+40x-64y+100=0\)

\(\displaystyle 4x^2+40x+16y^2-64y+100=0\)

Factor out a \(\displaystyle 4\) from the \(\displaystyle x\) terms and \(\displaystyle 16\) from the \(\displaystyle y\) terms.

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

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

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

Subtract \(\displaystyle 100\) from both sides.

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

Divide both sides by \(\displaystyle 64\).

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

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

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

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{16-4}=\sqrt{12}=2\sqrt3\)

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

For this ellipse, \(\displaystyle a=4\).

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.

\(\displaystyle e=\frac{2\sqrt3}{4}=\frac{\sqrt3}{2}\)

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

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

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 25x^2+4y^2-400x-8y+1504=0\)

\(\displaystyle 25x^2-400x+4y^2-8y+1504=0\)

Factor out \(\displaystyle 25\) from the \(\displaystyle x\) terms and \(\displaystyle 4\) from the \(\displaystyle y\) terms.

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

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

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

Subtract \(\displaystyle 1504\) from both sides.

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

Divide both sides by \(\displaystyle 100\).

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

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

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

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{25-4}=\sqrt{21}\)

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

For this ellipse, \(\displaystyle b=5\).

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.

\(\displaystyle e=\frac{\sqrt{21}}{5}\)

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

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

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 25x^2+36y^2+300x-720y+3600=0\)

\(\displaystyle 25x^2+300x+36y^2-720y+3600=0\)

Factor out \(\displaystyle 25\) from the \(\displaystyle x\) terms and \(\displaystyle 36\) from the \(\displaystyle y\) terms.

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

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

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

Subtract \(\displaystyle 3600\) from both sides.

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

Divide both sides by \(\displaystyle 900\).

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

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

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

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{36-25}=\sqrt{11}\)

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

For this ellipse, \(\displaystyle a=6\).

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.

\(\displaystyle e=\frac{\sqrt{11}}{6}\)

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

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

Group the \(\displaystyle x\) terms together and the \(\displaystyle y\) terms together.

\(\displaystyle 15x^2+16y^2+90x+32y-89=0\)

\(\displaystyle 15x^2+90x+16y^2+32y-89=0\)

Factor out \(\displaystyle 15\) from the \(\displaystyle x\) terms and \(\displaystyle 16\) from the \(\displaystyle y\) terms.

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

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

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

Add \(\displaystyle 89\) to both sides.

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

Divide both sides by \(\displaystyle 240\).

\(\displaystyle \frac{x^2+6x+9}{16}+\frac{y^2+2y+1}{15}=1\)

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

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

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{16-15}=\sqrt{1}=1\)

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

For this ellipse, \(\displaystyle a=4\).

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.

\(\displaystyle e=\frac{1}{4}\)

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{42-38}=\sqrt{4}=2\)

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

For this ellipse, \(\displaystyle b=\sqrt{42}\).

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.

\(\displaystyle e=\frac{2}{\sqrt{42}}=\frac{\sqrt{42}}{21}\)

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, \(\displaystyle e\).

\(\displaystyle e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}\)

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

\(\displaystyle c\) is calculated using the following formula:

\(\displaystyle c=\sqrt{a^2-b^2}\) for \(\displaystyle a>b\), or

\(\displaystyle c=\sqrt{b^2-a^2}\) for \(\displaystyle b>a\)

For the ellipse in question,

\(\displaystyle c=\sqrt{144-16}=\sqrt{128}=8\sqrt2\)

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

For this ellipse, \(\displaystyle b=12\).

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.

\(\displaystyle e=\frac{8\sqrt2}{12}=\frac{4\sqrt 2}{6}= \frac{2\sqrt2}{3}\)

To find the eccentrictity, first we need to find c, the distance from the center to the foci. We can use the equation \(\displaystyle a^2 + b^2 = c^2\) 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.

\(\displaystyle 49 - 25 = c^2\)

\(\displaystyle 24 = c^2\)

\(\displaystyle \sqrt{24 } = c\)

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

The eccentricity is \(\displaystyle \frac {\sqrt {24 }}{ 7 }\).

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: \(\displaystyle b^2 = a^2 - c^2\)

\(\displaystyle 4 = 14 - c^2\) add c squared to both sides

\(\displaystyle 4 + c^2 = 14\) subtract 4 from both sides

\(\displaystyle c^2 = 10\) take the square root

\(\displaystyle c = \sqrt{10}\)

Since \(\displaystyle a^2 = 14\), \(\displaystyle a = \sqrt{14}\).

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

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

\(\displaystyle 2 = 3 - c ^2\) add c squared to both sides

\(\displaystyle 2 + c^2 = 3\) subtract 2 from both sides

\(\displaystyle c^2 = 1\) take the square root of both sides

\(\displaystyle c =1\)

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

\(\displaystyle e = \frac{1}{\sqrt{3}}\)

The equation for the eccentricity of an ellipse is given by 

\(\displaystyle e=\frac{c}a{}\)

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 \(\displaystyle a^2=b^2+c^2\), where \(\displaystyle a^2\) is the larger denominator and \(\displaystyle b^{2}\) is the smaller denominator.  Plugging in the values, we have 

\(\displaystyle 25=9+c^2\)

\(\displaystyle 16=c^2\)

\(\displaystyle 4=c\)

Plugging the values into the equation gives

\(\displaystyle e=\frac{4}5{}\)