anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.
Mathematics
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anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
anonymous
  • anonymous
least favorite problems lol, mind if I help?
anonymous
  • anonymous
Sure!

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anonymous
  • anonymous
standard form: y = a(x-h)^2 + k
anonymous
  • anonymous
BTW did my help earlier help you?
anonymous
  • anonymous
yes 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
  • anonymous
|dw: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
  • anonymous
Oh no its fine I understand!
anonymous
  • anonymous
y = a(x-h)^2 + k y = a(x-0)^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
  • Loser66
Focus (0,-2) hence p=-2, right? So just plug it in x^2=4py
anonymous
  • anonymous
@Loser66 hey my final answer is y = 1/8(x)^2 Is that right?
Loser66
  • Loser66
Same
anonymous
  • anonymous
awesome so do you get that @PHUNISH
anonymous
  • anonymous
Oh okay thanks guys1
anonymous
  • anonymous
no problem
Loser66
  • Loser66
But, to focus, direct fix oinformat6, you should apply the formula

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