# Prove a chord of a parabola passes through a fixed point

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.

• 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. Aug 20 at 19:32

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 $$P$$.

• 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 Aug 20 at 18:04
• @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 :) Aug 20 at 18:20
• and it took me a while to see that my intuition was wrong. Aug 20 at 18:34
• 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 Aug 20 at 18:43
• 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. Aug 20 at 19:33