Question

The equation of chord of ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ whose sum and difference of eccentric angles are $\frac{\pi }{3}$ and $\frac{2\pi }{3}$ respectively is

• A. $2\sqrt{3}x-3y=6$
• B. $2\sqrt{3}x+3y=6$
• C. $2\sqrt{3}x+3y=\sqrt{3}$
• D. $2\sqrt{3}x+3y=6\sqrt{3}$

Answer 1

The correct option is B $2\sqrt{3}x+3y=6$

Given : $\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow a^2=9,b^2=4$
Also, $\alpha +\beta =\frac{\pi }{3},|\alpha -\beta |=\frac{2\pi }{3}$
We know, equation of chord joining two points having eccentric angle $\alpha$ and $\beta$ is
$\frac{x}{a}\cos \left(\frac{\alpha +\beta }{2}\right)+\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)$
$\Rightarrow \frac{x}{3}\cos \left(\frac{\pi }{6}\right)+\frac{y}{2}\sin \left(\frac{\pi }{6}\right)=\cos \left(\frac{\pi }{3}\right)$
$\Rightarrow \frac{\sqrt{3}x}{6}+\frac{y}{4}=\frac{1}{2}$
$\Rightarrow 2\sqrt{3}x+3y=6$