If the tangents to parabola $y^2=4ax$ at $(at^2,2at)$ and $(as^2,2as)$ meet at $(p,q)$, then $a^2(t-s)^2=q^2-4ap$

Question

If the tangents to the parabola $y^2 = 4ax$ at the points $(at^2, 2at)$ and $(as^2, 2as)$ meet at the point $(p, q)$. Show that
$$a^2(t – s)^2 = q^2 – 4ap$$

My work so far: Using $yy = 2a(x + x)$ for point $(at^2, 2at)$, I got the equation to be
$$yt = x + at^2$$ For point $(as^2, 2as)$, I got
$$ys = x + as^2$$

Then I put in $(p, q)$ for $x$ and $y$ and got
$$qt = p + at^2 \quad\text{and}\quad qs = p + as^2$$ This is where I'm stuck, because if I equate them, one of the variables will cancel out but all the variables are involved.

Answer 1

calculate the point of intersection, after writing the equation of tangent using T=0.