# What is the vertex form of the equation of the parabola with a focus at (6,5) and a directrix of y=19 ?

May 22, 2016

Vertex form of the equation of the parabola is
$y = - \frac{1}{28} {\left(x - 6\right)}^{2} + 12$

#### Explanation:

Here the directrix is a horizontal line $y = 19$.

Since this line is perpendicular to the axis of symmetry, this is a regular parabola, where the $x$ part is squared.

Now the distance of a point on parabola from focus at $\left(6 , 5\right)$ is always equal to its distance from the directrix.

Let this point be $\left(x , y\right)$.

Its distance from focus is $\sqrt{{\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2}}$ and from directrix will be $| y - 19 |$

Hence, ${\left(x - 6\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(y - 19\right)}^{2}$

or ${x}^{2} - 12 x + 36 + {y}^{2} - 10 y + 25 = {y}^{2} - 38 y + 361$

or ${x}^{2} - 12 x + 28 y - 300 = 0$

As vertex form of equation is $y = a {\left(x - h\right)}^{2} + k$

the above is $28 y = - {x}^{2} + 12 x + 300$ or $28 y = - \left({x}^{2} - 12 x + 36 - 336\right)$

or $28 y = - {\left(x - 6\right)}^{2} + 336$ or

$y = - \frac{1}{28} {\left(x - 6\right)}^{2} + \frac{336}{28}$ or

$y = - \frac{1}{28} {\left(x - 6\right)}^{2} + 12$

graph{-1/28(x-6)^2+12 [-12.83, 27.17, -2.08, 17.92]}