# How do you find the center and vertices of the ellipse 4x^2+9y^2=36?

Aug 3, 2017

The center is $= \left(0 , 0\right)$.
The vertices are $A ' = \left(- 3 , 0\right)$,$A = \left(3 , 0\right)$, $B ' = \left(- 2 , 0\right)$ and $B = \left(2 , 0\right)$

#### Explanation:

The equation is

$4 {x}^{2} + 9 {y}^{2} = 36$

Divide throughout by $36$

$\frac{4}{36} {x}^{2} + \frac{9}{36} {y}^{2} = \frac{36}{36}$

${x}^{2} / 9 + {y}^{2} / 4 = 1$

${x}^{2} / {3}^{2} + {y}^{2} / {2}^{2} = 1$

This is the standard equation of an ellipse center $= \left(0 , 0\right)$

When $y = 0$

${x}^{2} / {3}^{2} = 1$, $\implies$, ${x}^{2} = 9$, $\implies$, $x = \pm 3$

The vertices are $A ' = \left(- 3 , 0\right)$ and $A = \left(3 , 0\right)$

When $x = 0$

${y}^{2} / {2}^{2} = 1$, $\implies$, ${y}^{2} = 4$, $\implies$, $x = \pm 2$

The vertices are $B ' = \left(- 2 , 0\right)$ and $B = \left(2 , 0\right)$

graph{(x^2/9)+(y^2/4)=1 [-8.89, 8.89, -4.444, 4.445]}