Show that when $p$ varies, the midpoint of $PR$ lies on the curve $y^2=2x-4$. 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 $y^2=2x-4$.

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?
