Question

If a chord joining two points $A,B$ whose eccentric angles are $\alpha ,\beta$ cuts the major axis of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at a distance $1$ from the centre, then $\left|3\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\right|=$

Answer 1

Equation of the chord $AB$ is
$\frac{x}{5}\cos \frac{\alpha +\beta }{2}+\frac{y}{4}\sin \frac{\alpha +\beta }{2}=\cos \frac{\alpha -\beta }{2}$
It also passes through $(\pm 1,0)$
$\therefore \pm \frac{1}{5}=\frac{\cos \frac{\alpha -\beta }{2}}{\cos \frac{\alpha +\beta }{2}}\Rightarrow \pm \left[\frac{\cos \frac{\alpha -\beta }{2}-\cos \frac{\alpha +\beta }{2}}{\cos \frac{\alpha -\beta }{2}+\cos \frac{\alpha +\beta }{2}}\right]=\frac{1-5}{1+5}\Rightarrow \pm \left[\frac{2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}{2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}\right]=-\frac{2}{3}\Rightarrow \left[\frac{2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}{2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}\right]=\pm \frac{2}{3}\left|3\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\right|=2$