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

##(16)##
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:
##(hk+1/(4a))##
Thus the focus is
##(14+1/(4*1/8))=(14+1/(1/2))=(14+2)=(16)##
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]}