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Homework Help: General form of a parabola

  1. Oct 12, 2009 #1
    For a parabola whose Directrix is given by the equation x=p and whose Focus is (h,k).

    Is this by any chance the correct general form of the parabola?

    x=1/2(h-p) [y^2 - 2yk + h^2+k^2-p^2]
  2. jcsd
  3. Oct 12, 2009 #2


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    The vertex of a parabola is halfway between the focus and directrix. Here, that is at ((h+p)/2, k) so the focal length is (h+p)/2- p= (h-p)/2. Since the "standard" parabola, with horizontal axis, is [itex]4d(x- x_0)= (y- y_0)^2[/itex], here that would be [4(h-p)/2](x- (h+p)/2)= (y- k)^2 which can be written as [itex]x= \frac{1}{2(h-p)}(y- k)^2+ \frac{h+p}{2}[/itex]. If you take that "(h+p)/2" inside the parentheses you get exactly what you have. Well done!
  4. Oct 12, 2009 #3
    Alright perfect! thanks a lot for your time.
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