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

##(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]}