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Conics- Parabola Question

  1. Jun 12, 2006 #1
    If I am given (y-2)2= 4 (x-3), how would I find the vertex, equation of the axis of symmetry, and the direction of the opening? I'm guessing that I have to use PF=PD, but it's confusing because it looks different from other things that I have done involving PF=PD.

  2. jcsd
  3. Jun 12, 2006 #2
    the right hand side of your equation is positive right?, what is the minimum value of x for which this is posible? and the corresponding value of y? this is the vertex. Now rebember that p=1 is the distance from the vertex to the focus and that you have and horizontal parabola.
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