Show that if a tangent to a parabola at P is parallel to a normal at Q, then PQ passes through the focus

Question

I study maths purely as a hobby. I am struggling with the final part of this question in a textbook I am working through.

Find the equations of the tangent and the normal to the parabola $y^2=4ax$ at the point P with parameter p. (i) Show that, if the tangent at P meets the directrix at L then $PL=a(p^2+1)^\frac{3}{2}\div p$

(ii) Show that, if the tangent at P is parallel to the normal at a point Q, then PQ passes through the focus of the parabola.

I have said, for the tangent, $y-2ap=\frac{1}{p}(x-ap^2) \rightarrow py=x+ap^2$

For the equation of the normal at P: $y-2ap = -p(x-ap^2) \rightarrow y+px=ap^3 +2ap$

So I find the equations of the tangent and normal at P to be $y=\frac{x+ap^2}{p}$ and $y= ap^3+2ap-px$ respectively.

I have also verified that $PL=a(p^2+1)^\frac{3}{2}\div p$

But I cannot do the last part which is to show that PQ passes through the focus.

Answer 1

Point $P$ is given by $(ap^2, 2ap)$

Slope of tangent to the parabola at any point is given by,

$y' = \cfrac{2a}{y} = \cfrac{1}{p}$

But as this is also the slope of the normal at $Q$, the slope of tangent at $Q$ must be $ \ - p$.

So at point $Q$, we have, $y' = \cfrac{2a}{y} = - p \implies y = - \cfrac{2a}{p}$

Hence $x = \cfrac{y^2}{4a} = \cfrac{a}{p^2}$

So point $Q$ is $ \ \left(\cfrac{a}{p^2}, -\cfrac{2a}{p}\right)$

Equation of line passing through $PQ$ is,

$y - 2 ap = \cfrac{2ap + 2a/p}{ap^2 - a/p^2} (x-ap^2)$

$y - 2 ap = \cfrac{2p (p^2+1)}{p^4-1} (x - ap^2)$

$y - 2ap = \cfrac{2p}{p^2-1} (x-ap^2)$

Plugging in $y = 0$, we show that $x = a$. So we know the line through $PQ$ passes through focus and as $P$ and $Q$ are on the opposite side of x-axis on which the focus lies, focus must lie in the segment $PQ$.