# What is the equation of the parabola with a focus at (3,6) and a directrix of y= 8?

Feb 14, 2016

$y = \left(- \frac{1}{4}\right) {x}^{2} + \left(\frac{6}{4}\right) x + \left(\frac{19}{4}\right)$

#### Explanation:

If the focus of a parabola is (3,6) and the directrix is y = 8, find the equation of the parabola.

Let ( x0 , y0 ) be any point on the parabola. First of all, finding the distance between (x0 , y0) and the focus. Then finding the distance between (x0 , y0) and directrix. Equating these two distance equations and the simplified equation in x0 and y0 is equation of the parabola.

The distance between (x0 , y0) and (3,6) is
sqrt((x0-2)^2+(y0-5)^2

The distance between (x0 , y0) and the directrix, y = 8 is | y0– 8|.

Equating the two distance expressions and square on both sides.
sqrt((x0-3)^2+(y0-6)^2 = | y0– 8|.

${\left(x 0 - 3\right)}^{2} + {\left(y 0 - 6\right)}^{2}$ =${\left(y 0 - 8\right)}^{2}$

Simplifying and bringing all terms to one side:

$x {0}^{2} - 6 x 0 + 4 y 0 - 19 = 0$

Write the equation with y0 on one side:
$y 0 = \left(- \frac{1}{4}\right) x {0}^{2} + \left(\frac{6}{4}\right) x 0 + \left(\frac{19}{4}\right)$

This equation in (x0 , y0) is true for all other values on the parabola and hence we can rewrite with (x , y).

So, the equation of the parabola with focus (3,6) and directrix is y = 8 is
$y = \left(- \frac{1}{4}\right) {x}^{2} + \left(\frac{6}{4}\right) x + \left(\frac{19}{4}\right)$