P(2ap, ap^2) is a any point on the parabola x^2=4ay other than its vertex. The normal at P meets the parabola again at Q. Find the coordinates of Q.
the tangent at P has slope x/(2a)
So, the normal line is
y-ap^2 = p/(2a) (x-2ap)
y = p/(2a)(x-2ap)+ap^2
So, where does that line intersect y = x^2/(4a) ?
p/(2a)(x-2ap)+ap^2 = x^2/4a
(1/4a)x^2 - p/(2a)x + p^2(1-a) = 0
Now just solve that for x and figure y.
One solution better be x=2ap (P itself)!