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Simple parabola problem

  1. Feb 19, 2008 #1
    First off this is not a homework or a test problem, I just need some help understanding the problem.

    "Find the standard form of the equation of the parabola with vertex (2,1) and focus (5,1)."

    I think you use Y=-1/4x^2 and F(0,p) to find the ax^2 part but I'm confused about how to make the adjustments so that the vertex is (2,1)

    Any help?

  2. jcsd
  3. Feb 20, 2008 #2


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    Knowing the line for the directrix may help. Do you see how the parabola is oriented if the focus and the vertex have the same y value? The axis of the parabola is horizontal. If you could begin to draw a picture, you will see that the directrix is the line x=-1. (make a crude sketch so you see this).

    Be aware that standard form of a parabola in this oreintation is x = (y-k)^2 +c;
    and the graph has been shifted to have a vertex at (k, c) instead of being in standard position.

    Since you have only one point "given" on the parabola, you may only find a variablized result, but that is probably all you need according to your exercise. IF you have another second point on the parabola, then you can fix the contants of k and c. You probably do not need to resort to the distance formula; just use a little bit of simultaneous equations (two of them, actually; one for each point on the parabola).
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