# How do you find the standard form of y^2 - 8x + 4y + 12 = 0?

Nov 10, 2016

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

#### Explanation:

Rewrite it as ${y}^{2} + 4 y = 8 x - 12$. Now complete the square on the left side as,

#y^2 +4y +4 = 8x +16 (by adding 4 on both sides)

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

This represents a horizontal parabola with its vertex at (-2, -2) with axis of symmetry being y=-2