I am looking for a way to understand the last steps found at this site: http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=196&bodyId=203
The page finishes with showing in the modern style of Apollonius ellipse equation as
$$ y^2=x \left( p-\frac{p}{2a}x \right) $$
together with the comment: "Can this be written in the standard form of an ellipse? Complete the square and see."
I try to get this to the standard form as in: $$ \frac{y^2}{a^2}+\frac{x^2}{b^2}=1 $$
But the closest my algebra gets me is: $$ \frac{y^2}{xp}+\frac{x}{2a}=1 $$
Can anyone show me how they are connected?
$$ y^2=x \left( p-\frac{p}{2a}x \right)$$
$$2a\cdot y^2=p(2ax-x^2)=-p\{(x-a)^2-a^2\}$$
$$2a\cdot y^2+p(x-a)^2=p\cdot a^2$$
$$\frac{(x-a)^2}{a^2}+\frac{y^2}{\frac{pa}2}=1$$
$$\left(\frac{x-a}a\right)^2+\left(\frac y{\sqrt\frac{pa}2}\right)^2=1$$