Question

If $y=x$ and $3y+2x=0$ are the equations of a pair of conjugate diameters of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then its eccentricity is

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

Answer 1

Answer: (C)

$\frac{1}{\sqrt{3}}$

We know relation of slope of conjugate diameter, $m^{\prime }=-\frac{b^2}{a^{2}m}$
$\Rightarrow -\frac{2}{3}=-\frac{b^2}{a^2\left(1\right)}\Rightarrow \frac{b^2}{a^2}=\frac{2}{3}$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt{3}}$