Algebra2/Trig
posted by
Vincent
.
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 
Damon
Horizontal axis means of form:
(yk)^2 = 4p(xh)
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 
Vincent
Oh, I see. Solve for P first then substitute it in. I guess I forgot that concept. Thanks!