How do you find the standard form given y^2+8y-4x+16=0?

1 Answer
Feb 13, 2017

$x = \frac{1}{4} {y}^{2} + 2 y + 4$

Explanation:

This is the equation of a parabola, whose standard form is

$x = a {y}^{2} + b y + c$

Here we have ${y}^{2} + 8 y - 4 x + 16 = 0$

Therefore $4 x = {y}^{2} + 8 y + 16$

or $x = \frac{1}{4} {y}^{2} + 2 y + 4$
graph{y^2+8y-4x+16=0 [-7.05, 32.95, -13.84, 6.16]}