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If $\alpha $ and $\beta $ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is
$\frac{{\cos \,\alpha \, + \,\cos \,\beta }}{{\cos \,\left( {\alpha \, - \,\beta } \right)}}$
$\frac{{\sin \,\alpha \, - \,\sin \,\beta }}{{\sin \,\left( {\alpha \, - \,\beta } \right)}}$
$\frac{{\cos \,\alpha \, - \,\cos \,\beta }}{{\cos \,\left( {\alpha \, - \,\beta } \right)}}$
$\frac{{\sin \,\alpha \, + \,\sin \,\beta }}{{\sin \,\left( {\alpha \, + \,\beta } \right)}}$
Solution
The equation of a chord joining points having eccentric angles $\alpha$ and $\beta$ is given by
$\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)$
If it passes through $(a e, 0)$ then
$e \cos \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)$
$\Rightarrow e=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} $
$\Rightarrow \frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)} $
$\Rightarrow e=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}$
