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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.

enter image description here

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    $\begingroup$ That $PF$ is the normal at $P$ can be proved independently: in the limit $Q\to P$ line $PQ$ is tangent and chord $QR=PR$ is by construction perpendicular to it. $\endgroup$ Aug 20 at 19:32
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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 $P$.

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    $\begingroup$ For anyone who's interested, I worked on a Desmos graph when I tried to solve this question a while ago. Thanks to MathLover's answer and something I completely missed, the graph is now complete, and you can visualise the fixed point if you wish (playing with the graph at first I was convinced a fixed point didn't exist, but it all checks out): here $\endgroup$
    – FShrike
    Aug 20 at 18:04
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    $\begingroup$ @FShrike thank you. that's a good interactive graph. On the contrary, when the question said that the fixed point lied on the normal, I had an intuition it would be circumcenter of the right angled triangle $ \triangle PQR$ but it became difficult to use that information in the working. I finally had to abandon that approach :) $\endgroup$
    – Math Lover
    Aug 20 at 18:20
  • $\begingroup$ and it took me a while to see that my intuition was wrong. $\endgroup$
    – Math Lover
    Aug 20 at 18:34
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    $\begingroup$ You're welcome. I found the quadratic equation right off, and noticed that it looked very similar to the line equation, but missed the multiplication entirely. Good job $\endgroup$
    – FShrike
    Aug 20 at 18:43
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    $\begingroup$ It was a very good job indeed. I too explored the fact that PQ is the diameter of a circle but got nowhere. I certainly would never had done it without the help of MathLover. $\endgroup$
    – Steblo
    Aug 20 at 19:33

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