How do you find the focus and directrix of the graph of the equation:
x=1/16(y-2)^2-4????
The answer is: Focus=(0,2), Directix is x=-8
How do you figure this out?
2006-06-15 14:05:58 · 8 answers · asked by Bouncingnoodles 1 in Science & Mathematics ➔ Mathematics
x = a(y - k)^2 + h
x = (1/16)(y - 2)^2 - 4
a = (1/16)
k = 2
h = -4
a = 1/(4p)
(1/16) = 1/(4p)
4p = 16
p = 4
Vertex : (-4,2)
p = distance from vertex to focus/directrix
Focus = (h + p,k)
Directrix = h - p
Focus = (-4 + 4,2)
Directrix = -4 - 4
Focus = (0,2)
Directrix = -8
2006-06-15 14:57:03 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
Wikipedia really explains this the best, better than I could begin to, so here's the link. The examples they use are concave up, but this one will be to the right (rotated) so you just need to replace x with y and vice versa.
2006-06-15 14:13:09 · answer #2 · answered by drewK 3 · 0⤊ 0⤋
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2016-12-08 09:33:49 · answer #3 · answered by ? 3 · 0⤊ 0⤋
Easiest way - drop math, take more English courses.
2006-06-15 14:08:22 · answer #4 · answered by blairs_smirking_revenge 3 · 0⤊ 0⤋
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2006-06-15 14:08:24 · answer #5 · answered by Anonymous · 0⤊ 0⤋
notes
2006-06-15 14:07:26 · answer #6 · answered by ballin 2 · 0⤊ 0⤋
sherman spoiled the fun!
2006-06-16 04:50:13 · answer #7 · answered by know it all guy 1 · 0⤊ 0⤋
I have ni idea GOOD LUCK
2006-06-15 14:07:29 · answer #8 · answered by bartenderkaty 1 · 0⤊ 0⤋