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

Question

5104 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-09T16:49:07

$(1,6)$

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

$y=a(x-h)^2+k$

To do this, multiply the entire equation by $1/8$ .

$1/8((x-1)^2+32)=1/8(8y)$

This simplifies to

$y=1/8(x-1)^2+4$

Thus, this parabola has:

$a=1/8$
$h=1$
$k=4$

The focus of a parabola can be found through:

$(h,k+1/(4a))$

Thus, the focus is

$(1,4+1/(4*1/8))=(1,4+1/(1/2))=(1,4+2)=(1,6)$

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]}

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