Question

The eccentricity of an ellipse whose pair of a conjugate diameter are $ y=x $ and $ 3y=-2x $ is

• A. $ \frac{2}{3} $
• B. $ \frac{1}{3} $
• C. $ \frac{1}{\sqrt{3}} $
• D. None of these

Answer 1

Answer: (C)

Let the ellipse be $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $
. Since, $ y=x $ and $ 3y=-2x $ is a pair of conjugate diameters. Therefore, $ {{m}_{1}}{{m}_{2}}=-\frac{{{b}^{2}}}{{{a}^{2}}} $
$ \Rightarrow $ $ (1)\left( -\frac{2}{3} \right)=-\frac{{{b}^{2}}}{{{a}^{2}}} $
$ \Rightarrow $ $ 2{{a}^{2}}=3{{b}^{2}} $
$ \Rightarrow $ $ 2{{a}^{2}}=3{{a}^{2}}(1-{{e}^{2}}) $
$ \Rightarrow $ $ 2=3(1-{{e}^{2}}) $
$ \Rightarrow $ $ {{e}^{2}}=\frac{1}{3} $
$ \Rightarrow $ $ e=\frac{1}{\sqrt{3}} $