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