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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at \((0,−9)\) and a directrix \(y=9.\)

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  1. welshfella
    • one year ago
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    this parabola opens upwards and standard form is x^2 = 4ay where a is the y coordinate of the focus and y = -a is the directrix so can you work this one out?

  2. anonymous
    • one year ago
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    Wouldn't the parabola open downwards?

  3. anonymous
    • one year ago
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    It would. The form for a parabola is \[y = 1/4p(x-h)^2+k\] From here you can plug 9 into p since p = distance between vertex and focus. The h and k values are both 0 since the vertex is at (0,0). Can you get it from here?

  4. anonymous
    • one year ago
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    I lied, plug -9 in for p.

  5. anonymous
    • one year ago
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    |dw:1439345400610:dw|

  6. anonymous
    • one year ago
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    P is (x,y)

  7. anonymous
    • one year ago
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    Hang on, I'm trying to solve it. :)

  8. anonymous
    • one year ago
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    Okay, post your answer here so I can check it for you :)

  9. anonymous
    • one year ago
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    So is it \(y=-\frac{ 1 }{ 36 }x^2?\)

  10. anonymous
    • one year ago
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    It sure is!

  11. anonymous
    • one year ago
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    Wow thanks! I was working with this, but I got the vertex wrong. :( How would you find the vertex?

  12. anonymous
    • one year ago
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    No problem! Okay so the vertex is equidistant from both the focus and the directrix. Since they both are -9 away from the origin (as the x value for both is 0) you can tell that the vertex must be at the origin in order for it to be equidistant from both!

  13. anonymous
    • one year ago
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    Awesome! Thanks again. You're great! :)

  14. welshfella
    • one year ago
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    yes surit is correct - i misread the problem

  15. welshfella
    • one year ago
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    - and migillope also

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