# What is the equation in standard form of the parabola with a focus at (12,-5) and a directrix of y= -6?

Mar 19, 2017

Because the directrix is a horizontal line, then the vertex form is $y = \frac{1}{4 f} {\left(x - h\right)}^{2} + k$ where the vertex is $\left(h , k\right)$ and f is the signed vertical distance from the vertex to the focus.

#### Explanation:

The the focal distance, f, is half of the vertical distance from the focus to the directrix:

$f = \frac{1}{2} \left(- 6 - - 5\right)$

$f = - \frac{1}{2}$

$k = {y}_{\text{focus}} + f$

$k = - 5 - \frac{1}{2}$

$k = - 5.5$

h is the same as the x coordinate of the focus

$h = {x}_{\text{focus}}$

$h = 12$

The vertex form of the equation is:

$y = \frac{1}{4 \left(- \frac{1}{2}\right)} {\left(x - 12\right)}^{2} - 5.5$

$y = \frac{1}{-} 2 {\left(x - 12\right)}^{2} - 5.5$

Expand the square:

$y = \frac{1}{-} 2 \left({x}^{2} - 24 x + 144\right) - 5.5$

Use the distributive property:

$y = - {x}^{2} / 2 + 12 x - 72 - 5.5$

Standard form:

$y = - \frac{1}{2} {x}^{2} + 12 x - 77.5$