# Find the eclipse focal point

A conic with equation $$a x^2 + b y^2 = c$$ has two focus points, where $a=4$, $b=24$ and $c=65$. One of those focus points has a positive x-coordinate. To two decimal places, what is the value of that positive x-coordinate?

I got the answer as 14.83, seems a bit too big is that right? The above answer is according to this in my textbook

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title:"Just convert this to radians?" the question seems to be asking for a coordinate, not an angle... – Navin Oct 3 '12 at 4:38
@Navin Tks for pointing that out my old question was in my browsers cache, changed the title – JackyBoi Oct 3 '12 at 4:43

Put the ellipse in standard form. Start from $4x^2+24y^2=65$. Divide through by $65$. We get $\frac{4}{65}x^2+\frac{24}{65}y^2=1$.
This is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=\frac{65}{4}$ and $b^2=\frac{65}{24}$. Finally, you want $\sqrt{a^2-b^2}$.
Your intuition is right: this is substantially smaller than your answer. To two decimal places, I get $3.68$.
I found the $x$-coordinate of the focal point with positive $x$-coordinate, as the question asks. The eccentricity is not $a^2/b^2$, it is $\frac{\sqrt{a^2-b^2}}{a}$. If your book/notes do not have the relevant formulas, I am sure you can find them in the wikipedia article on the ellipse, there must be one! – André Nicolas Oct 3 '12 at 5:00
That is equivalent to what I wrote. But I guess if you follow the book, you will first find $e$, which is $\sqrt{1-\frac{b^2}{a^2}}$ (mine was different but equivalent) and then you will multiply by $a$. You will get $\sqrt{a^2-b^2}$. Anyway, same answer. Eccentricity is about $0.91287$. – André Nicolas Oct 3 '12 at 5:16