# How do you find the equation in standard form of the parabola whose focus is at (-2, 4) and whose directrix is at x = 4?

Mar 3, 2017

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

#### Explanation:

Parabola is the locus of a point which moves whose distance from a given line called directrix and a point called focus is always constant.

Here we have directrix as $x = 4$ and focus is $\left(- 2 , 4\right)$

Distance of $\left(x , y\right)$ from $x = 4$ is $| x - 4 |$

and distance from focus $\left(- 2 , 4\right)$ is $\sqrt{{\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2}}$. Hence

$\sqrt{{\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2}} = | x - 4 |$

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

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

or ${y}^{2} - 8 y + 4 x + 4 = 0$

and for standard form it is $4 x = - {y}^{2} + 8 y - 4$

or $x = - \frac{1}{4} {y}^{2} + 2 y - 1$
graph{x=-1/4y^2+2y-1 [-12.965, 7.035, -1, 9]}