Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

I got this solution, is this right?

enter image description here

share|improve this question
add comment

1 Answer

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.