Question

If the foci of an ellipse are $ (\pm \sqrt{5},0) $ and its eccentricity is $ \sqrt{5}/3, $ then the equation of the ellipse is

• A. $ 9{{x}^{2}}+4{{y}^{2}}=36 $
• B. $ 4{{x}^{2}}+9{{y}^{2}}=36 $
• C. $ 36{{x}^{2}}+9{{y}^{2}}=4 $
• D. $ 9{{x}^{2}}+36{{y}^{2}}=4 $

Answer 1

Answer: (B)

Given, eccentricity $ e=\frac{\sqrt{5}}{3} $
and foci $ =(\pm \sqrt{5},\,0) $
$ \Rightarrow $ $ ae=\sqrt{5} $
$ \Rightarrow $ $ a=\frac{\sqrt{5}\times 3}{\sqrt{5}}=3 $
Now, $ {{b}^{2}}={{a}^{2}}(1-{{e}^{2}})=9\left( 1-\frac{5}{9} \right) $
$ \Rightarrow $ $ {{b}^{2}}=4 $
$ \therefore $ The equation of ellipse is
$ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $
$ \Rightarrow $ $ \frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1 $
$ \Rightarrow $ $ 4{{x}^{2}}+9{{y}^{2}}=36 $