Question 1998
simple relationship between t and u (your t1 and t2..i cannot do subscripts :-( )... PSQ is a straight line, so find its gradient on PS ans SQ...

for PS:
m = {{{(y2-y1)/(x2-x1)}}}
m = {{{(2at - 0)/(at^2 - a)}}}
m = (2t)/(t^2-1)

for SQ:
m = {{{(y2-y1)/(x2-x1)}}}
m = {{{(2au - 0)/(au^2 - a)}}}
m = (2u)/(u^2-1)

these 2 gradients are the same, so

{{{(u)/(u^2-1) = (t)/(t^2-1)}}}

To find the tangents at P and Q...we need the gradient of the parabola, so differentiate {{{y^2 = 4ac}}}

2y(dy/dx) = 4a
so dy/dx = {{{(4a)/(2y)}}}

as y=2aT (generic value), then (dy/dx) = {{{(4a)/(4aT)}}}
dy/dx = 1/T

at P, gradient = 1/t, so tangent equation is y=mx+c..sub it

at Q, gradient = 1/u, so tangent equation is y=mx+c..sub it

Got 2 equations then {{{y = (x/u) + au}}} and {{{y = (x/t) + at}}}

equate these for y to find the coordinates of the point where the 2 tangents meet... x=atu, y=a(t+u).

I think this is it..it is late here and i am tired, so please check my answers with your working.


jon