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Math Help - Standard form of parabola

  1. #1
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    Standard form of parabola



    For some reason I don't get this at all. I understand ellipses and hyperbolas, but not a simple parabola.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dataspot View Post


    For some reason I don't get this at all. I understand ellipses and hyperbolas, but not a simple parabola.
    Is it the y^2 screwing with you? If so then just find all the info for

    y=\frac{x^2}{-6}

    and then be sure to remember to adjust so that it is facing the proper direciton.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Is it the y^2 screwing with you? If so then just find all the info for

    y=\frac{x^2}{-6}

    and then be sure to remember to adjust so that it is facing the proper direciton.
    The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be (y-k)^2=4p(x-h). But how is that -6x going to 4(\frac{-3}{2})x?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dataspot View Post
    The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be (y-k)^2=4p(x-h). But how is that -6x going to 4(\frac{-3}{2})x?
    because 4\cdot\frac{-3}{2}=-6? Which is neccesary.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dataspot View Post
    The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be (y-k)^2=4p(x-h). But how is that -6x going to 4(\frac{-3}{2})x?
    The parabola takes on the form y^2=4px

    Since our equation is y^2=-6x, comparing it to the standard form we see that 4p=-6, solving for p, we get p=-\frac{3}{2}. Thus, that's how they get -6x=4\left(-\frac{3}{2}\right)x

    Does that make sense?

    --Chris
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  6. #6
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    They're flowing pretty easy now. Thanks!
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