Prove that $QG^2 - PG^2 = constant$(Parabola)

Question

The normal at a point P to the parabola $y^2 = 4ax$ meets its axis at G. Q is another point on the parabola such that QG is perpendicular to the axis of the parabola. Prove that $QG^2 - PG^2 = constant$

Answer 1

Let $\;P=\left(\frac{t^2}{4a}\,,\,\,t\right)\;,\;\;t\neq0\;$ (why?) . The normal to the parabola at this point is

$$y-t=\frac t{2a}\left(x-\frac{t^2}{4a}\right)\implies y=\frac t{2a}x-\frac{t^3}{8a^2}+t$$

The parabola's axis is the $\;x\,$ axis, $\,y=0\;$ , and the above line crosses this line when

$$\frac x{2a}=\frac{t^2}{8a^2}-1\implies x=\frac{t^2}{4a}-2a\implies G=\left(\frac{t^2}{4a}-2a\,,\,\,0\right)$$

The perpendicular to the axis (the $\;x\,-$ axis, that is) at $\;G\;$ is the vertical line $\; x=\frac{t^2}{4a}-2a\;$ , and this line intersects the parabola at the points

$$Q:=\left( \frac{t^2}{4a}-2a\,,\,\,\pm\sqrt{4a\left(\frac{t^2}{4a}-2a\right)}\right)=\left( \frac{t^2}{4a}-2a\,,\,\,\pm\sqrt{t^2-8a^2}\right)\implies$$

$$\begin{cases} \left\|QG\right\|^2=t^2-8a^2\\{}\\ \left\|PG\right\|^2=4a^2+t^2\end{cases}\;\;\implies\left\|QG\right\|^2-\left\|PG\right\|^2=-12a^2 =\text{ a constant}$$