# What is the focus of the parabola (x – 1)^2 + 32 = 8y?

Jan 9, 2016

$\left(1 , 6\right)$

#### Explanation:

First, get the equation of the parabola into vertex form:

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

To do this, multiply the entire equation by $\frac{1}{8}$.

$\frac{1}{8} \left({\left(x - 1\right)}^{2} + 32\right) = \frac{1}{8} \left(8 y\right)$

This simplifies to

$y = \frac{1}{8} {\left(x - 1\right)}^{2} + 4$

Thus, this parabola has:

$a = \frac{1}{8}$
$h = 1$
$k = 4$

The focus of a parabola can be found through:

$\left(h , k + \frac{1}{4 a}\right)$

Thus, the focus is

$\left(1 , 4 + \frac{1}{4 \cdot \frac{1}{8}}\right) = \left(1 , 4 + \frac{1}{\frac{1}{2}}\right) = \left(1 , 4 + 2\right) = \left(1 , 6\right)$

Graphed are the focus, parabola (and directrix):

graph{(y-1/8(x-1)^2-4)((x-1)^2+(y-6)^2-.03)(y-2)=0 [-10.97, 14.34, 0.06, 12.74]}