# How do you find the center, vertices, foci and eccentricity of (X^2/25)+(y^2/64)=1?

Oct 28, 2016

The center is $\left(0 , 0\right)$
The vertices are $\left(5 , 0\right)$, $\left(- 5 , 0\right)$, $\left(0 , 8\right)$ and $\left(0 , - 8\right)$
The foci are $\left(\sqrt{0} , \sqrt{39}\right)$ and $\left(0 , - \sqrt{39}\right)$
The eccentricity is $= \frac{\sqrt{39}}{8}$

#### Explanation:

This is the equation of an ellipse with a vertical major axis

${\left(x - {x}_{0}\right)}^{2} / {a}^{2} + {\left(y - {y}_{0}\right)}^{2} / {b}^{2} = 1$

Here $\left({x}_{0} , {y}_{0}\right) = \left(0 , 0\right)$
So the center of the ellipse is $\left(0 , 0\right)$
$a = 5$ and $b = 8$ are the axes

The vertices are $\left(5 , 0\right)$, $\left(- 5 , 0\right)$, $\left(0 , 8\right)$ and $\left(0 , - 8\right)$

The distance of the center to the focus is $\sqrt{{b}^{2} - {a}^{2}} = \sqrt{64 - 25} = \sqrt{39}$

So the foci are $\left(\sqrt{0} , \sqrt{39}\right)$ and $\left(0 , - \sqrt{39}\right)$

The eccentricity is $= \frac{\sqrt{39}}{8}$