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Thread: Parabola problem

  1. #1
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    Smile Parabola problem

    I'm trying to figure out what the heck I'm supposed to do with this:

    x^2=8y.

    I'm supposed to solve for vertex, focus, and directx... how do I put this into the y=ax^2+bc+c form??
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  2. #2
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    Quote Originally Posted by Kevin Green View Post
    I'm trying to figure out what the heck I'm supposed to do with this:

    x^2=8y.

    I'm supposed to solve for vertex, focus, and directx... how do I put this into the y=ax^2+bc+c form?? ... divide both sides by 8 ; a = 1/8 , b = 0, c = 0
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  3. #3
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    Thank you.... where do I go from here? The math book I have does a horrible job describing this whole matter.
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  4. #4
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  5. #5
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    Quote Originally Posted by Kevin Green View Post
    I'm trying to figure out what the heck I'm supposed to do with this:

    x^2=8y.

    I'm supposed to solve for vertex, focus, and directx... how do I put this into the y=ax^2+bc+c form??
    Parabolas can also be expressed in this form:
    (x - h)^2 = 4p(y - k) or (y - k)^2 = 4p(x - h).
    The first is a parabola that opens upward or downward, and the second is a parabola that opens left or right. Sometimes this form is preferred because you can see what the vertex is.

    Looking at your equation:
    x^2=8y
    h = k = 0, so the vertex is (0, 0). The x is squared, and the y coefficient is positive, so it opens upward. Solve 4p = 8 to get the focal length, which is 2.

    Since the parabola opens upward, the focus would be defined by (h, k + p), or (0, 2). The directrix is an equation of a line, in our case, defined by y = k - p, or y = -2.


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