Algebra2/Trig

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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)
so
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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