Let $\mathrm{a}, \mathrm{b}$ are the length of semi-major axis and semi-minor axis respectively, then $(a \cos \alpha, b \sin \alpha),(a \cos \beta, b \sin \beta),(a e, 0)$ are collinear.
$\frac{b(\sin \alpha-\sin \beta)}{a(\cos \beta-\cos \cos \alpha)}=\frac{b \sin \alpha-0}{a \cos \alpha-a e}$

$\begin{aligned} & (\cos \alpha-e)(\sin \beta-\sin \alpha)=\sin \alpha(\cos \beta-\cos \alpha) \end{aligned}$
$\begin{aligned} & e=\frac{\cos \alpha(\sin \beta-\sin \alpha)-\sin \alpha(\cos \beta-\cos \alpha)}{\sin \beta-\sin \alpha} \\ & e=\frac{\sin \sin (\alpha-\beta)}{\sin \alpha-\sin \beta} \\ & e=\frac{\sin (\alpha-\beta)(\sin \alpha+\sin \beta)}{\sin ^2 \alpha-\sin ^2 \beta}\end{aligned}$
$\therefore e=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}$ $\left[\because \sin \sin (A+B) \cdot \sin \sin (A-B)=\sin ^2 A-\sin ^2 B\right]$