# How do you find the standard form of 2y^2+9x^2=9-x and what kind of a conic is it?

Nov 22, 2016

This is an ellipse, ${\left(x + \frac{1}{18}\right)}^{2} / \left(\frac{2917}{2916}\right) + {y}^{2} / \left(\frac{2917}{648}\right) = 1$

#### Explanation:

Let's rewrite the equation

$9 {x}^{2} + 2 {y}^{2} + x - 9 = 0$

Compare this to the general equation of the conics

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

Let's calculate the dscriminant,

$\Delta = {B}^{2} - 4 A C = 0 - 4 \cdot 9 \cdot 2 = - 72$

As, #Delta<0, we expect an ellipse

Completing the squares

$9 \left({x}^{2} + \frac{x}{9}\right) + 2 {y}^{2} = 9$

$9 \left({x}^{2} + \frac{x}{9} + {\left(\frac{1}{18}\right)}^{2}\right) + 2 {y}^{2} = 9 + {\left(\frac{1}{18}\right)}^{2}$

$9 {\left(x + \frac{1}{18}\right)}^{2} + 2 {y}^{2} = \frac{2917}{324}$

$\frac{9 {\left(x + \frac{1}{18}\right)}^{2}}{\frac{2917}{324}} + \frac{2 {y}^{2}}{\frac{2917}{324}} = 1$

${\left(x + \frac{1}{18}\right)}^{2} / \left(\frac{2917}{2916}\right) + {y}^{2} / \left(\frac{2917}{648}\right) = 1$

This is the standard equation of the ellipse

the center is $\left(- \frac{1}{18} , 0\right)$

graph{9x^2+x+2y^2-9=0 [-4.382, 4.386, -2.19, 2.192]}