Question

If $y=x$ and $3y+2x=0$ are the equations of a pair of conjugate diameter of an ellipse, then the eccentricity of the ellipse is:

• A. $\sqrt{\frac{2}{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}$

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Slope of the given diameter are $m_1=1,m_2=-\frac{2}{3}$.

$\Rightarrow m_1{m}_2=-\frac{2}{3}=-\frac{b^2}{a^2}$

{Using the condition of conjugacy of two diameter}

$\Rightarrow 3b^2=2a^2\Rightarrow 3a^2(1-e^2)=2a^2$

$\Rightarrow 1-e^2=\frac{2}{3}\Rightarrow e^2=\frac{1}{3}\Rightarrow e=\frac{1}{\sqrt{3}}$