# Apollonius ellipse equation $y^2=x \left( p-\frac{p}{2a}x \right)$ to standard form.

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?

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You need the standard form $(y-y_0)^2/a^2+(x-x_0)^2/b^2=1$ instead. This is the equation of an (axis-aligned) ellipse centered at $(x_0,y_0)$ instead of at the origin. – Rahul Nov 1 '12 at 22:02

$$y^2=x \left( p-\frac{p}{2a}x \right)$$

$$2a\cdot y^2=p(2ax-x^2)=-p\{(x-a)^2-a^2\}$$

$$2a\cdot y^2+p(x-a)^2=p\cdot a^2$$

$$\frac{(x-a)^2}{a^2}+\frac{y^2}{\frac{pa}2}=1$$

$$\left(\frac{x-a}a\right)^2+\left(\frac y{\sqrt\frac{pa}2}\right)^2=1$$

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Thank you! I thought there was a error in your calculations first therefore i made an edit but i was wrong. (Is there any way for me to cancel my edit?) – user1532054 Nov 4 '12 at 12:43