Question 170128
y^2-8y-8x+64=0 
and I'm supposed to write this in standard form
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Complete the square of the y terms, as follows:
y^2 - 8y + 16 = 8x -64 + 16
(y-4)^2 = 8x - 48
(y-4)^2 = 8(x-6)
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This form shows you the vertex is at (6,4) and the 
focus of the parabola is 8/4 = 2 units above the vertex.
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Cheers,
Stan H.