I did this question a different way...
As you can see from the graph, a=4, so one point is drawn from off the directrix.
First thing I did was find the value of p and q, by substituting the points into the tangent formula.
y=px - ap^2
-4= 0p -4p^2
p = 1
Therefore the points of P(2ap, ap^2) is going to be (8, 4)
For q:
y=qx - aq^2
-2=2q - 4q^2 (this simplifies into a simple quadratic)
2(2q + 1)(q - 1) = 0
Obviously, if you use q=1, you will get exactly the same point, so you have to use q=-1/2
so Q = (-4, 1)
Using these two points, and the fact that the gradients of the normals are -1/p, -1/q (-1 and 2 respectively), you can get the equations of the normals and prove the intersect at x=0.
I have been working on this question for a quite a while now, and I while I think my working up until here is correct, when I find the equations of the normals, I don't get the intersecting on the axis. I get the intersecting and (1, 11)
I think my methodology is correct, but I must have made a mistake someone. Can anyone see it?
