SOLUTION: What is the focus of the parabola with the equation (x - 1)^2 + 32= 8y?

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Question 145225: What is the focus of the parabola with the equation (x - 1)^2 + 32= 8y?
Answer by solver91311(24713) About Me  (Show Source):
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The equation of a parabola with axis parallel to the y-axis, vertex at (h,k), and focus at (h,k+p) is:

%28x-h%29%5E2=4p%28y-k%29

So, rearrange the given equation:

%28x-1%29%5E2%2B32=8y

Add -32 to both sides:
%28x-1%29%5E2=8y-32

Factor the 8 out of the right hand expression:
%28x-1%29%5E2=8%28y-4%29

That puts the equation into the given form and you can see by inspection that the vertex is at (1,4)

Since 4p=8, p=2 and the focus is at (1,6)