Question

If $2y=x$ and $3y+4x=0$ are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is

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

Answer 1

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

The equation of a pair of conjugate diameters of an ellipse are
$2y=x$ and $3y+4x=0$

Slope of $2y=x$ is $\frac{1}{2}(m_1,say)$

and slope of $3y+4x=0$ is $-\frac{4}{3}(m_2,say)$

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

$\Rightarrow \left(\frac{1}{2}\right)(-\frac{4}{3})=-\frac{b^2}{a^2}$

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

$\therefore Eccentricity,e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{2}{3}}$ $=\sqrt{\frac{1}{3}}$