# How do you write the equation of a parabola in vertex form that has a focus at (6,5) and a directrix of y=-1?

Sep 1, 2016

${\left(x - 6\right)}^{2} = 12 \left(y - 2\right) , \text{ is the reqd. eqn. of Parabola, with, Vertex } \left(6 , 2\right)$. graph{(x-6)^2=12(y-2) [-10, 10, -5, 5]}

#### Explanation:

Let the Focus be $S \left(6 , 5\right)$ and, the Directrix $d : y + 1 = 0$.

Let $P \left(x , y\right)$ is any pt. on the reqd. Parabola, then, by the Focus-

Directrix Property of Parabola, we have,

$\text{Dist. "SP="the" bot-dist. "btwn. P & d}$.

$\therefore \sqrt{{\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2}} = | y + 1 |$.

$\therefore {\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(y + 1\right)}^{2}$.

$\therefore {\left(x - 6\right)}^{2} = 2 y + 1 + 10 y - 25 = 12 y - 24$, i.e.,

${\left(x - 6\right)}^{2} = 12 \left(y - 2\right) ,$ is the reqd. eqn. of Parabola [ Vertex $\left(6 , 2\right)$].

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