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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?

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up vote 1 down vote accepted

Hint: $$0=py^2 + 4ay −8a^2p − 4a^2 p^3= (2ap^2+4a+py)(2ap-y)$$

Well the basic approach is to use the quadratic formaula: treat the above as a quadratic in $y$. so $$\begin{align*} y &= \frac{-4a \pm \sqrt {16a^2 + 4p(8a^2+4a^2p^4)}}{2p}\\ & =\ldots \end{align*} $$

Can you take it from here?

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I know how to solve from here, the thing is I can't reach this point without looking at the answer. Mind sharing your method on how you found this? – εν οίδα ότι ουδέν οίδα May 28 '12 at 20:40
@Panayiotis: I added to my post. see if you okay. if not let me know – Nana May 28 '12 at 20:55

You are trying to solve for y, aren't you? Look at the last equation you wrote and think of it as an equation where y is the variable. Does anything suggest itself in terms of solving for y?

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Okay I know that is a quadratic, but I have no idea how to factorize it – εν οίδα ότι ουδέν οίδα May 28 '12 at 20:43

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