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

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.

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 $$PQ$$.