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

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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]}