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.

Answer 2

You have $$a t^2 - q t + p = 0 \qquad a s^2 - q s + p = 0$$ Therefore, $s$ and $t$ are the two roots of the quadratic equation $$am^2 - qm + p = 0$$ By Vieta's Formulas, $$s + t = \frac{q}{a} \qquad s t = \frac{p}{a}$$ so that $$a^2(s-t)^2 = a^2\left(s^2 - 2 s t + t^2\right) = a^2\left((s+t)^2 - 4 s t\right) = a^2\left(\frac{q^2}{a^2}-\frac{4p}{a}\right) = q^2-4ap$$ as desired. $\square$