# What is the equation of the parabola with a focus at (-2, 6) and a vertex at (-2, 9)?

Jun 27, 2017

y - 9 = 1/12 ( x + 2 )^2

#### Explanation:

Generic Equation is
y - k = 1/4p ( x - h)^2
p is distance vertex to focus = 3
(h,k) = vertex location = (-2, 9)

Jun 27, 2017

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

#### Explanation:

When talking about the focus and vertex of a parabola, the easiest way to write the equation is in vertex form. Luckily, you already have most of your information.

$y = a {\left(x + 2\right)}^{2} + 9$

However, we do not have the value of $a$.

$a = \frac{1}{4 c}$

$c$ is the distance between the focus and the vertex.

$c = - 3$

We know this because the only difference between the two coordinates is the $y$ part. The reason it is negative is because the vertex is above the focus; this means that the parabola opens downwards.

$\frac{1}{4 c}$

$\frac{1}{\left(4\right) \left(- 3\right)}$

$\frac{1}{-} 12$

$- \frac{1}{12}$

Now that you have your value for $a$, you can plug this in and finalize your equation.

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

Jun 27, 2017

$y = - {x}^{2} / 12 - \frac{x}{3} + \frac{26}{3}$

#### Explanation:

Given -

Vertex $\left(- 2 , 9\right)$
Focus $\left(- 2 , 6\right)$

The focus of the parabola lies below the vertex. Hence, it opens down.

The formula for downward opening parabola having origin as its vertex is -

${x}^{2} = - 4 a y$

The vertex of the given parabola is not at the vertex. it is in the 2nd quarter.

The formula is -

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

$h = - 2$ x-coordinate of the vertex
$k = 9$ y-coordinate of the vertex
$a = 3$Distance between vertex and focus
Substitute the values in the formula
${\left(x + 2\right)}^{2} = - 4 \times 3 \times \left(y - 9\right)$
${x}^{2} + 4 x + 4 = - 12 y + 108$
$- 12 y + 108 = {x}^{2} + 4 x + 4$
$- 12 y = {x}^{2} + 4 x + 4 - 108$
$y = - {x}^{2} / 12 - \frac{4}{12} x + \frac{108}{12}$
$y = - {x}^{2} / 12 - \frac{x}{3} + \frac{26}{3}$