# How do you find the center, vertices, foci and eccentricity of 9x^2 + 25y^2 = 225?

##### 1 Answer
Nov 3, 2016

The center is $= \left(0 , 0\right)$
The vertices are $\left(5 , 0\right) \left(- 5 , 0\right) \left(0 , 3\right) \left(0 , - 3\right)$
The foci are F$= \left(4 , 0\right)$ and F'$= \left(- 4 , 0\right)$
The eccentricity $e = \frac{4}{5}$

#### Explanation:

Let's start by rewriting the equation
$9 {x}^{2} + 25 {y}^{2} = 225$
By dividing by 225
${x}^{2} / \left(\frac{225}{9}\right) + {y}^{2} / \left(\frac{225}{25}\right) = 1$
${x}^{2} / 25 + {y}^{2} / 9 = 1$
this is the equation of an ellipse
${\left(x - {x}_{0}\right)}^{2} / {a}^{2} + {\left(y - {y}_{0}\right)}^{2} / {b}^{2} = 1$
Here ${x}_{0} = 0$ and ${y}_{0} = 0$
the vertices are $\left(5 , 0\right) \left(- 5 , 0\right) \left(0 , 3\right) \left(0 , - 3\right)$
${c}^{2} = {a}^{2} - {b}^{2}$
$c = \sqrt{25 - 9} = \sqrt{16} = 4$
So the foci are $\left({x}_{0} + c , {y}_{0}\right)$ and $\left({x}_{0} - c , {y}_{0}\right)$
so F$= \left(4 , 0\right)$ and F'$= \left(- 4 , 0\right)$
The eccentricity $e = \frac{c}{a} = \frac{4}{5}$
graph{x^2/25+y^2/9=1 [-10, 10, -5, 5]}