# How do you write the equation of the parabola in vertex form given vertex (3,3) and focus: (-2,3)?

Oct 22, 2017

The vertex form for the equation of a parabola whose focus is shifted horizontally a signed distance, $f$, from its vertex is:

$x = \frac{1}{4 f} {\left(y - k\right)}^{2} + h \text{ }$

where $\left(h , k\right)$ is the vertex.

#### Explanation:

Substitute the vertex $\left(3 , 3\right)$ into equation :

$x = \frac{1}{4 f} {\left(y - 3\right)}^{2} + 3 \text{ }$

Compute the signed distance from the vertex to the focus:

$f = - 2 - 3$

$f = - 5$

Substitute $f = - 5$ into equation :

$x = \frac{1}{4 \left(- 5\right)} {\left(y - 3\right)}^{2} + 3$

Simplify the denominator:

$x = - \frac{1}{20} {\left(y - 3\right)}^{2} + 3$

Oct 22, 2017

$x = - \frac{1}{20} {\left(y - 3\right)}^{2} + 3$

#### Explanation:

First, let's figure out which way we need to draw the parabola.

The vertex is (3,3) and the focus is (-2,3).

These points have the same y-value, so they form a horizontal line. This means that the parabola will be horizontal.

It must be of the form $x = a {\left(y - k\right)}^{2} + h$

Additionally, the focus is to the LEFT of the vertex, so the parabola will point to the LEFT (meaning $a$ is negative).

Let's call the distance between the focus and the vertex $c$.

We know that the value of $a$ is equal to $\pm \frac{1}{4 c}$.

In this case, $c$ is 5, since the vertex and focus are $5$ units apart.

$a = \pm \frac{1}{4 \left(5\right)} = \pm \frac{1}{20} = - \frac{1}{20}$ since we already know $a$ is negative.

This gives us everything we need to write our parabola's equation! The vertex $\left(h , k\right)$ is (3,3), the stretch factor $a$ is -1/20, and the direction is horizontal.

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

$x = - \frac{1}{20} {\left(y - 3\right)}^{2} + 3$