# How do you find the center and vertices of the ellipse x^2/(25/9)+y^2/(16/9)=1?

Jan 14, 2017

The center is $\left(0 , 0\right)$
The vertices are A$= \left(h + a , k\right) = \left(\frac{5}{3} , 0\right)$
A'$= \left(h - a , k\right) = \left(- \frac{5}{3} , 0\right)$
B$= \left(h , k + b\right) = \left(0 , \frac{4}{3}\right)$
B'$= \left(h , k - b\right) = \left(0 , - \frac{4}{3}\right)$

#### Explanation:

The equation is ${x}^{2} / \left(\frac{25}{9}\right) + {y}^{2} / \left(\frac{16}{9}\right) = 1$

We compare this to ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is O $= \left(h , k\right) = \left(0 , 0\right)$

The vertices are :

A$= \left(h + a , k\right) = \left(\frac{5}{3} , 0\right)$

A'$= \left(h - a , k\right) = \left(- \frac{5}{3} , 0\right)$

B$= \left(h , k + b\right) = \left(0 , \frac{4}{3}\right)$

B'$= \left(h , k - b\right) = \left(0 , - \frac{4}{3}\right)$

graph{x^2/(25/9)+y^2/(16/9)=1 [-3.898, 3.897, -1.947, 1.95]}