# How do you find an equation of a parabola given focus (10,1) and directrix x=5?

Feb 26, 2017

${\left(y - 1\right)}^{2} = {\left(x - 5\right)}^{2} - {\left(x - 10\right)}^{2} = 5 \left(2 x - 15\right) , \mathmr{and} ,$

${y}^{2} - 2 y - 10 x + 76 = 0.$

#### Explanation:

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

Parabola to solve this Problem :

FDP : Let a pt. $S$ and a line $d ,$ be the focus & directrix of a

parabola, resp. If $P$ is any pt. on the parabola, then, $P$ is

equidistant from $S$ and $d .$

Focus is $S = S \left(10 , 1\right) \text{ and the eqn. of dir. is } d : x - 5 = 0.$

Let $P \left(x , y\right)$ be any pt. on the parabola.

$\therefore \text{ the Dist. SP=} \sqrt{{\left(x - 10\right)}^{2} + {\left(y - 1\right)}^{2}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right) .$

The $\bot - \mathrm{di} s t . \text{ from "P" to "d" is =} | x - 5 \frac{|}{\sqrt{{1}^{2} + {0}^{2}}} = | x - 5 | \ldots \left(2\right)$

FDP$\Rightarrow \sqrt{{\left(x - 10\right)}^{2} + {\left(y - 1\right)}^{2}} = | x - 5 | .$

$\Rightarrow {\left(y - 1\right)}^{2} = {\left(x - 5\right)}^{2} - {\left(x - 10\right)}^{2} = 5 \left(2 x - 15\right) , \mathmr{and} ,$

${y}^{2} - 2 y - 10 x + 76 = 0.$

Enjoy Maths.!