Question:
Using $y^2=4ax$ $y+px=2ap+ap^3$
Show that $y=-2a\left( \frac{2+p^2}{p} \right)$
Working: $Substitute \rightarrow x=\frac{y^2}{4a}$ $\begin{align} y+p\left(\frac{y^2}{4a}\right)&=2ap+ap^3\\ 4ay+py^2&=8a^2p +4a^2p^3\\ y(4a+py)&=8a^2p +4a^2p^3\\ 0&=py^2 + 4ay −8a^2p − 4a^2 p^3 \\ \end{align}$
And I am stuck there. Any suggestions on how to move on?