# How do you find the standard form of 5x^2 + 8y^2 + 16y - 32 = 0 and what kind of a conic is it?

Jan 18, 2016

First, complete the square ...

#### Explanation:

$5 {x}^{2} + 8 {y}^{2} + 16 y - 32 = 0$

$5 {x}^{2} + 8 \left({y}^{2} + 2 y\right) = 32$

Now, complete the square ...

$5 {x}^{2} + 8 \left({y}^{2} + 2 y + 1\right) = 32 + 8$

$5 {x}^{2} + 8 {\left(y + 1\right)}^{2} = 40$

Now, divide both sides by 40 ...

${x}^{2} / 8 + {\left(y + 1\right)}^{2} / 5 = 1$

This is an ellipse since the coefficients on x and y are different.

The center $= \left(0 , - 1\right)$.

The major axis is horizontal .

Hope that helped