# What is the equation in standard form of the parabola with a focus at (12,-3) and a directrix of y= 12?

Sep 3, 2017

$y - \frac{9}{2} = - \frac{1}{30} {\left(x - 12\right)}^{2.}$

#### Explanation:

We will use the following Focus-Directrix Property of Parabola (FDP) :

FDP : Any point on a Parabola is equidistant from the

Focus F and the Directrix d.

Let $P \left(x , y\right)$ be any arbitrary point on the Parabola. Then.

The Distance $P F ,$ btwn. #P(x,y), and, F(12,-3), is,

$P F = \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right) .$

The Dist. btwn $P \left(x , y\right) , \mathmr{and} , d : y - 12 = 0 , \text{ is, } | y - 12 | \ldots \left(2\right) .$

$\therefore \left(1\right) , \left(2\right) ,$ and, FDP,

$\Rightarrow \sqrt{{\left(x - 12\right)}^{2} + \left(y - {\left(- 3\right)}^{2}\right)} = | y - 12 | .$

$\therefore {\left(x - 12\right)}^{2} = | y - 12 {|}^{2} - {\left(y + 3\right)}^{2.}$

$\therefore {\left(x - 12\right)}^{2} = {y}^{2} - 24 y + 144 - {y}^{2} - 6 y - 9 , i . e . ,$

${\left(x - 12\right)}^{2} = - 30 y + 135 = - 30 \left(y - \frac{9}{2}\right) .$

$\therefore y - \frac{9}{2} = - \frac{1}{30} {\left(x - 12\right)}^{2} ,$ is the desired eqn. of the Parabola.