# How do you find the equation of a parabola when given Focus (-2, 6) Vertex (-2, 9)?

Aug 8, 2015

$y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$

#### Explanation:

[There may be a more direct route to this solution, but the following uses the tools I remember.]

A parabola in vertex form has the general form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$
with a vertex at $\left(a , b\right)$

So a parabola with a vertex at $\left(- 2 , 9\right)$
has $a = - 2$ and $b = 9$ in the general form

The focus of a parabola given the general vertex form (above) is at
$\textcolor{w h i t e}{\text{XXXX}}$$\left(a , b + \frac{1}{4 m}\right)$

With the given focus at $\left(- 2 , 6\right)$, this gives us
$\textcolor{w h i t e}{\text{XXXX}}$$b + 1 \left(/ 4 m\right) = 6$
and since $b = 9$
$\textcolor{w h i t e}{\text{XXXX}}$$9 + \frac{1}{4 m} = 6$

$\textcolor{w h i t e}{\text{XXXX}}$$\frac{1}{4 m} = - 3$

$\textcolor{w h i t e}{\text{XXXX}}$$- 3 m = \frac{1}{4}$

$\textcolor{w h i t e}{\text{XXXX}}$$m = - \frac{1}{12}$

Substituting the derived values for $m , a ,$ and $b$ gives
$\textcolor{w h i t e}{\text{XXXX}}$$y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$