# How do you find the standard form of the equation of the parabola with focus (7, 5) & directrix x=1?

Jul 6, 2016

${\left(y - 5\right)}^{2} = 12 \left(x - 4\right)$ is the reqd. eqn.

#### Explanation:

Let $S \left(7 , 5\right)$ be the Focus & the eqn. of Dir. $d : x - 1 = 0.$

If $P \left(x , y\right)$ be any point on the reqd. Parabola , then by the Focus-Directrix Property of Parabola, we have,

Dist. $S P$ = Perp. Dist. btwn. $\left(P , d\right) .$

$\therefore \sqrt{{\left(x - 7\right)}^{2} + {\left(y - 5\right)}^{2}} = | x - 1 | \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

$\therefore {\left(x - 7\right)}^{2} + {\left(y - 5\right)}^{2} = | x - 1 {|}^{2}$

$\therefore {\left(y - 5\right)}^{2} = {x}^{2} - 2 x + 1 - {x}^{2} + 14 x - 49 = 12 x - 48 ,$ i.e.,

${\left(y - 5\right)}^{2} = 12 \left(x - 4\right)$ is the reqd. eqn.

The Left Member of $\left(1\right)$ is obtained using the Distance Formula, whereas, the Right one uses the Formula for the Perp. Dist. of a Pt. to a Line.

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