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Find an equation of the parabola with vertex at the origin.

Passes through the point (4,6); horizontal axis.

I know it has to be y^2=4px. But I don't know what to do with (4,6). Please show step by step. Thanks.

  • Algebra2/Trig -

    Horizontal axis means of form:
    (y-k)^2 = 4p(x-h)
    if the vertex is at the origin (h,k) = (0,0)
    y^2 = 4 p x
    36 = 4 p (4)
    p = (9/4)
    y^2 = 9 x

  • Algebra2/Trig -

    Oh, I see. Solve for P first then substitute it in. I guess I forgot that concept. Thanks!

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