# What is the equation in standard form of the parabola with a focus at (51,14) and a directrix of y= 16?

Apr 22, 2017

$y = \frac{1}{4} {x}^{2} - \frac{51}{2} x + \frac{2661}{4}$

#### Explanation:

Because the directrix is a horizontal line, we know that is x coordinate of the vertex is the same as the x coordinate of the focus, $h = 51$.

The y coordinate of the vertex is halfway from the directrix to the focus, $k = 15$

The signed distance from the vertex to the focus is, $f = 1$

Using the vertex form of a parabola of this type:

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

Substitute the values for h and k:

$y = a {\left(x - 51\right)}^{2} + 15$

Use the formula $a = \frac{1}{4 f}$

$a = \frac{1}{4 \left(1\right)}$

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

Substitute this into the equation:

$y = \frac{1}{4} {\left(x - 51\right)}^{2} + 15$

Expand the square:

$y = \frac{1}{4} \left({x}^{2} - 102 x + 2601\right) + 15$

Distribute the $\frac{1}{4}$:

$y = \frac{1}{4} {x}^{2} - \frac{51}{2} x + \frac{2601}{4} + 15$

Combine the constant terms:

$y = \frac{1}{4} {x}^{2} - \frac{51}{2} x + \frac{2661}{4} \text{ } \leftarrow$ This is the standard form.