I am doing the end of chapter questions on parabolas in a pure maths book out of interest. I am struggling with this:
A fixed point $P(ap^2,2ap)$ is taken on the parabola $y^2=4ax$. Two points Q and R are chosen on the parabola such that the lines PQ and PR are perpendicular. Prove that the line QR passes through a fixed point, F, independent of Q and R, and that PF is normal to the parabola at P.
I started by working out the slopes of PQ,QR and PR:
$(\frac{2}{p+q},\frac{2}{r+q},\frac{-(p+q)}{2})$ and then I worked out the equation of QR, which I found to be:
$y-2aq=\frac{2}{r+q}(x-aq^2) \rightarrow y=\frac{2x+2aqr}{r+q}$
The question wants an answer independent of Q and R so I reasoned I need to find an expression in p only.
I did find such an expression but even then I couldn't find where the fixed point should be. But even then, the question only gives the fixed point F after we are supposed to have found that such a point exists.
Best Answer
Point $P$ on the parabola is $(ap^2, 2ap)$. Say, coordinates of $Q$ and $R$ are $(aq^2, 2aq)$ and $(ar^2, 2ar)$.
Equation of line passing through $QR$,
$y - 2aq = \cfrac{2aq - 2ar}{aq^2-ar^2} (x - a q^2)$
Simplifying,
$ (q + r) y = 2x + 2 a q r \tag1$
As $PQ$ and $PR$ are perpendicular,
$\cfrac{2ap - 2aq}{ap^2 - aq^2} \times \cfrac{2ap - 2ar}{ap^2 - ar^2} = - 1$
$\implies \cfrac{4}{(p+q)(p+r)} = - 1$
Simplifying, $p^2 + (q+r) p + 4 + q r = 0$
Multiply by $- 2a$ and rearrange to bring it in the same form as $(1)$.
$ - 2 a p (q+r) = 2 a (p^2 + 4) + 2 a q r \tag2$
From $(1)$ and $(2)$, we can see that $x = a (p^2 + 4), y = - 2 a p$ satisfies the equation of the line $QR$. So, regardless of the value of $q$ and $r$, point $F \big(a (p^2 + 4), - 2 a p \big)$ lies on $QR$.
Equation of normal through point $P$ is,
$y - 2 ap = - p (x - ap^2)$
and you can check that $F$ lies on the normal through $P$.