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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.

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  1. anonymous
    • one year ago
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    @ganeshie8 @Loser66

  2. anonymous
    • one year ago
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    least favorite problems lol, mind if I help?

  3. anonymous
    • one year ago
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    Sure!

  4. anonymous
    • one year ago
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    standard form: y = a(x-h)^2 + k

  5. anonymous
    • one year ago
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    BTW did my help earlier help you?

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    |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?

  8. anonymous
    • one year ago
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    Oh no its fine I understand!

  9. anonymous
    • one year ago
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    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

  10. Loser66
    • one year ago
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    Focus (0,-2) hence p=-2, right? So just plug it in x^2=4py

  11. anonymous
    • one year ago
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    @Loser66 hey my final answer is y = 1/8(x)^2 Is that right?

  12. Loser66
    • one year ago
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    Same

  13. anonymous
    • one year ago
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    awesome so do you get that @PHUNISH

  14. anonymous
    • one year ago
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    Oh okay thanks guys1

  15. anonymous
    • one year ago
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    no problem

  16. Loser66
    • one year ago
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    But, to focus, direct fix oinformat6, you should apply the formula

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