# How do you find the standard form given 9x^2+4y^2-36x+24y+36=0?

Feb 15, 2017

The answer is ${\left(x - 2\right)}^{2} / 4 + {\left(y + 3\right)}^{2} / 9 = 1$

#### Explanation:

Let's do some rearrangement by completing the squares

$9 {x}^{2} + 4 {y}^{2} - 36 x + 24 y + 36 = 0$

$9 {x}^{2} - 36 x + 4 {y}^{2} + 24 y = - 36$

$9 \left({x}^{2} - 4 x\right) + 4 \left({y}^{2} + 6 y\right) = - 36$

$9 \left({x}^{2} - 4 x + 4\right) + 4 \left({y}^{2} + 6 y + 9\right) = - 36 + 36 + 36$

$9 {\left(x - 2\right)}^{2} + 4 {\left(y + 3\right)}^{2} = 36$

${\left(x - 2\right)}^{2} / 4 + {\left(y + 3\right)}^{2} / 9 = 1$

This is the equation of an ellipse.

graph{9x^2-36x+4y^2+24y+36=0 [-11.41, 11.09, -11.1, 0.15]}