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QUESTION

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]}