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I got this solution, is this right?

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up vote 2 down vote accepted

here equation of ellipse is $$\frac{x^2}{11}+\frac{y^2}{(\frac{55}{27})}=1$$

So, $a^2=11$ and $b^2=\frac{55}{27}\implies e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5}{27}}=\frac{1}{3}\sqrt{\frac{22}{3}}$

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Just realised that eccentric cant be more than 0<e<1 – JackyBoi Oct 3 '12 at 8:19
For ellipse $0\lt e\lt 1$ and one more thing $\sqrt{1-5}\neq 2$ but $2i$ which is certainly can't be eccentricity of any conic – Aang Oct 3 '12 at 8:22

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