Question

The equation of a parabola is given.y=1/8x^2+4x+20 What are the coordinates of the focus of the parabola?

Answer 1

He equation of a parabola is x = -4(y-1)^2. What is the equation of the directrix? 
You may write the equation as 
(y-1)^2 = (1) (x+4) 
(y-k)^2 = 4p(x-h), where (h,k) is the vertex 
4p=1 
p=1/4 
k=1 
h=-4 

The directrix is a vertical line x= h-p 
x = -4-1/4 
x=-17/4 

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What is the focal length of the parabola with equation y - 4 = 1/8x^2 
(x-0)^2 = 8(y-4) 
The vertex is (0,4) 
4p=8 
p=2 (focal length) -- distance between vertex and the focus 
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(y-0)^2 = (4/3) (x-7) 
vertex = (7,0) 
4p=4/3 
p=1/3 
focus : (h+p,k) 
(7+1/3, 0)

Answer 2

Convert it to  vertex form we get 

y = 1/8(x + 16)^2 - 12     

so the vertex is at (-16,-12) and  the x coordinate of the focus will be -16)

The focus  will be at (-16, k + p)    Here k = -12 from the vertex equation and  p  = 1 / 4a  where a = 1/8    so p = 1 / 1/2 = 2  So k + p = -12 + 2 = -10

So the coordinates of the focus are (-16, -10) answer