The chord joining the parametric points $\mathrm{\alpha }$and $\mathrm{\beta }$ on $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ cuts the positive major axis at a distance d from the centre, then $\tan \frac{\mathrm{\alpha }}{2}\tan \frac{\mathrm{\beta }}{2}$ is

Let the ellipse be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$($a>b)$

Now equation of chord joining points whose eccentric angles are $\mathrm{\alpha }$ and $\mathrm{\beta }$ is 

$\frac{\mathrm{x}}{\mathrm{a}}\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)+\frac{\mathrm{y}}{\mathrm{b}}\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$ 

It cuts major axis at the point (d, 0) then $\frac{\mathrm{d}}{\mathrm{a}}\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$

$\Rightarrow \frac{\cos \left({\displaystyle \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}}\right)}{\cos \left({\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}\right)}=\frac{\mathrm{a}}{\mathrm{d}}$

By applying Componendo and divedendo rule we get

$$\begin{gathered} \frac{\cos \left({\displaystyle \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}}\right)+\cos \left({\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}\right)}{\cos \left({\displaystyle \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}}\right)-\cos \left({\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}\right)}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}} \\ \Rightarrow \frac{2\cos {\displaystyle \frac{\alpha }{2}}\cos {\displaystyle \frac{\beta }{2}}}{-2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}} \\ \Rightarrow cot\frac{\alpha }{2}cot\frac{\beta }{2}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{d}-\mathrm{a}} \end{gathered}$$

$\Rightarrow \tan \frac{\mathrm{\alpha }}{2} \tan \frac{\mathrm{\beta }}{2}=\frac{d-a}{d+a}$