# How do you find the center, foci and vertices of 9x^2+4y^2-36x+8y+31=0?

Jun 17, 2018

#### Explanation:

The equation is

$9 {x}^{2} + 4 {y}^{2} - 36 x + 8 y + 31 = 0$

$\iff$, $9 {x}^{2} - 36 x + 4 {y}^{2} + 8 y + 31 = 0$

$\iff$, $9 \left({x}^{2} - 4 x\right) + 4 \left({y}^{2} + 2 y\right) = - 31$

Complete the squares

$\iff$, $9 \left({x}^{2} - 4 x + 4\right) + 4 \left({y}^{2} + 2 y + 1\right) = - 31 + 36 + 4$

$\iff$, $9 {\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 9$

Dividing by $9$

$\iff$, ${\left(x - 2\right)}^{2} + {\left(y + 1\right)}^{2} / \left(\frac{9}{4}\right) = 1$

This is the equation of an ellipse, center $\left(2 , - 1\right)$

The vertices are $A = \left(2 , 0.5\right)$, $A ' = \left(2 , - 2.5\right)$, $B = \left(3 , - 1\right)$ and $B ' = \left(1 , - 1\right)$

The semi axes are $a = 1$ and $b = \frac{3}{2}$

$c = \sqrt{{b}^{2} - {a}^{2}} = \sqrt{\frac{9}{4} - 1} = \frac{\sqrt{5}}{2}$

The foci are $F = \left(2 , \frac{\sqrt{5}}{2}\right)$ and $F ' = \left(2 , - \frac{\sqrt{5}}{2}\right)$

graph{9x^2+4y^2-36x+8y+31=0 [-3.05, 5.718, -3.592, 0.79]}