Question

If $\frac{\pi }{4}\text{and}\frac{\pi }{3}$ are the eccentric angles of the extremeties of the chord of the standard horizontal ellipse, and passes through the focus $(ae,0)$, then eccentricity of the ellipse is

Answer 1

Given: Chord passing through the focus $(ae,0)$ and eccentric angles of the extremeties of the chord are $\frac{\pi }{4}$ and $\frac{\pi }{3}$.

To Find: Eccentricity of the ellipse

Step-$1:$ Consider the equation of the focal chord

Step-$2:$ Use the coordinates of the foci to find the value of eccentricity.

If $\alpha$ and $\beta$ are the extremeties of the focal chord $d=ae$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{e-1}{e+1}$

$e=\frac{\sin \alpha +\sin \beta }{\sin (\alpha +\beta )}$

$\Rightarrow e=\frac{\sin (\frac{\pi }{4})+\sin (\frac{\pi }{3})}{\sin (\frac{\pi }{4}+\frac{\pi }{3})}$

$\Rightarrow e=\frac{\sin (\frac{\pi }{4})+\sin (\frac{\pi }{3})}{\sin (\frac{7\pi }{12})}$

$\Rightarrow e=\frac{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}+\sqrt{6}}{4}}$

$\Rightarrow e=2\frac{(\sqrt{2}+\sqrt{3})}{(\sqrt{2}+\sqrt{6})}$

$\Rightarrow e=\sqrt{2}\frac{(\sqrt{2}+\sqrt{3})}{(1+\sqrt{3})}$