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anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = 2.
anonymous
 one year ago
Find the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = 2.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0least favorite problems lol, mind if I help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0standard form: y = a(xh)^2 + k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0BTW did my help earlier help you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes quite a bit actually, sorry i didn't help so much I was a bit confused. but I will make it up to you now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440285805656:dw lol kind of failed but you get the point. so the vertex is halfway between these points. so the vertex is actually at (0,0) in the standard from (h,k) is the vertex following so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh no its fine I understand!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y = a(xh)^2 + k y = a(x0)^2 + 0 y = a(x)^2 a = 1/4c c is 2, the difference between the focus and the vertex so a = 1/(4*2) a = 1/8 y = 1/8 (x)^2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Focus (0,2) hence p=2, right? So just plug it in x^2=4py

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Loser66 hey my final answer is y = 1/8(x)^2 Is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome so do you get that @PHUNISH

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay thanks guys1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1But, to focus, direct fix oinformat6, you should apply the formula
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