Conic Sections Question 362
Question: If the foci of an ellipse are $ (\pm \sqrt{5},0) $ and its eccentricity is $ \frac{\sqrt{5}}{3} $ , then the equation of the ellipse is
[J & K 2005]
Options:
A) $ 9x^{2}+4y^{2}=36 $
B) $ 4x^{2}+9y^{2}=36 $
C) $ 36x^{2}+9y^{2}=4 $
D) $ 9x^{2}+36y^{2}=4 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \because ae=\pm \sqrt{5} $
therefore $ a=\pm \sqrt{5}( \frac{3}{\sqrt{5}} )=\pm 3 $
therefore $ a^{2}=9 $
$ b^{2}=a^{2}(1-e^{2})=9( 1-\frac{5}{9} )=4 $
Hence, equation of ellipse $ \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 $
therefore $ 4x^{2}+9y^{2}=36 $ .
BETA
