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The line passing through the point $P$ ($p^2,2p$) on the curve $y^2=4x$ and the point $Q$(2,0) intersects the curve once again at point $R$, find the coordinates of point $R$ in terms of $p$.

I am able to solve this part of the question and found that the coordinates of $R$ in terms of $p$ is $R\Big(\displaystyle\frac{4}{p^2},-\frac{4}{p}\Big)$.

Working for the first part of the question: Let coordinates of R be ($x,y$).

$m_{PQ}=m_{QR}$

$\displaystyle\frac{0-2p}{2-p^2}=\frac{y-0}{x-2}$

$-2px+4p=2y-p^2y \tag{1}$

Substitute $\displaystyle\frac{y^2}{4}$ into $(1)$

$-2p\Big(\displaystyle\frac{y^2}{4}\Big)+4p=y(2-p^2)$

$py^2+(4-2p^2)y-8p=0$

$y=\displaystyle\frac{-(4-2p^2)\pm\sqrt{(4-2p^2)^2-4p(-8p)}}{2p}$

$y=\displaystyle\frac{-4+2p^2\pm(2p^2+4)}{2p}$

$y=2p$ or $\displaystyle\frac{-4}{p}$

When $y=\displaystyle\frac{-4}{p},x=\frac{4}{p^2}$

Therefore, coordinates of R = $\Big(\displaystyle\frac{4}{p^2},\frac{-4}{p}\Big)$

However, I am not sure how to prove the second part of the question, that is, to show that when $p$ varies, the midpoint of $PR$ lies on the curve $y^2=2x-4$.

gc3941d
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1 Answers1

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The midpoint of $PR$ in terms of $p$ is:

$$\left(\frac{p^2+4/p^2}{2}, \frac{2p-4/p}{2} \right)$$

and you can surely show that $\left(\frac{2p-4/p}{2} \right)^2 = 2 \left( \frac{p^2+4/p^2}{2} \right)-4$, right?

Toby Mak
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  • What does the question mean when it says 'when $p$ varies'? – gc3941d Apr 01 '20 at 07:46
  • $p$ can be any real number. The notation $(p^2, 2p)$ describes a parametric equation that can be described with only one variable ($p$ in this case). This means you need to prove the equation holds for any point on the curve $y^2=4x$, where the curve is described by $(p^2, 2p)$. – Toby Mak Apr 01 '20 at 07:49
  • So I don't have to write anything about $p$ varying in the working, right? – gc3941d Apr 01 '20 at 07:50
  • No. Using algebra, you have shown this can hold for any value of $p$, not just for a particular choice of $p$. Everything is fine in your working. – Toby Mak Apr 01 '20 at 07:53
  • Got it, thank you for your helpful explanation. – gc3941d Apr 01 '20 at 07:54