# What is the vertex form of the equation of the parabola with a focus at (1,20) and a directrix of y=23 ?

Oct 4, 2017

$y = {x}^{2} / - 6 + \frac{x}{3} + \frac{64}{3}$

#### Explanation:

Given -

Focus $\left(1 , 20\right)$
directrix $y = 23$

The vertex of the parabola is in the first quadrant. Its directrix is above the vertex. Hence the parabola opens downward.

The general form of the equation is -

${\left(x - h\right)}^{2} = - - 4 \times a \times \left(y - k\right)$

Where -

$h = 1$ [ X-coordinate of the vertex]
$k = 21.5$ [ Y-coordinate of the vertex]

Then -

${\left(x - 1\right)}^{2} = - 4 \times 1.5 \times \left(y - 21.5\right)$

${x}^{2} - 2 x + 1 = - 6 y + 129$
$- 6 y + 129 = {x}^{2} - 2 x + 1$
$- 6 y = {x}^{2} - 2 x + 1 - 129$
$y = {x}^{2} / - 6 + \frac{x}{3} + \frac{128}{6}$
$y = {x}^{2} / - 6 + \frac{x}{3} + \frac{64}{3}$