Question

If $3x^2+xy-4y^2=0$ is the equation of a pair of conjugate diameters of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then its eccentricity is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\sqrt{2}$

Answer 1

The correct option is B $\frac{1}{2}$

For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ equations of conjugate diameters are given by $3x^2+xy-4y^2=0$
$\Rightarrow (x-y)(3x+4y)=0\Rightarrow y=x;3x+4y\Rightarrow =0$
$\Rightarrow m_1=1;m_2=-3/4$
for conjugate diameters $m_1{m}_2=-\frac{b^2}{a^2}=-\frac{3}{4}$
$\Rightarrow \frac{b^2}{a^2}=\frac{3}{4}\therefore 1-\frac{b^2}{a^2}=\frac{1}{4}$
$\Rightarrow e^2=\frac{1}{4}\therefore e=\frac{1}{2}$ rejecting $e=-\frac{1}{2}$