Question

Let $A,B,C$ and $D$ be four points on the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b,$ such that $AB$ and $CD$ cut its major axis at two distinct equidistant points from centre.
Let $\alpha ,\beta ,\gamma ,\delta$ represent eccentric angles of $A,B,C$ and $D$ respectively. Then
$\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}\tan \frac{\delta }{2}=$

Answer 1

$A(a\cos \alpha ,b\sin \alpha ),B(a\cos \beta ,b\sin \beta )$
Equation of $AB$
$y-b\sin \alpha =\frac{b(\sin \alpha -\sin \beta )}{a(\cos \alpha -\cos \beta )}(x-a\cos \alpha )$
$\frac{x}{a}\cos \left(\frac{\alpha +\beta }{2}\right)+\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)$
Equation of $CD$
$\frac{x}{a}\cos \left(\frac{\gamma +\delta }{2}\right)+\frac{y}{b}\sin \left(\frac{\gamma +\delta }{2}\right)=\cos \left(\frac{\gamma -\delta }{2}\right)$

$\because \text{Magnitude of x-intercepts are equal}$
$\therefore a\frac{\cos \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)}=-a\frac{\cos \left(\frac{\gamma -\delta }{2}\right)}{\cos \left(\frac{\gamma +\delta }{2}\right)}$
$\Rightarrow \frac{\cos \left(\frac{\alpha -\beta }{2}\right)-\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)+\cos \left(\frac{\alpha +\beta }{2}\right)}=\frac{-\cos \left(\frac{\gamma -\delta }{2}\right)-\cos \left(\frac{\gamma +\delta }{2}\right)}{-\cos \left(\frac{\gamma -\delta }{2}\right)+\cos \left(\frac{\gamma +\delta }{2}\right)}$
$\Rightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\cot \frac{\gamma }{2}\cot \frac{\delta }{2}$
$\therefore \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}\tan \frac{\delta }{2}=1$