Question

If the chord through the points whose eccentric angles are $\theta$ and $\varphi$ on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ Passes through a focus,then the value of tan $\left(\frac{\theta }{2}\right)$tan$\left(\frac{\varphi }{2}\right)$is

Answer 1

Equation of chord

$\cos \left(\frac{\theta +\varphi }{2}\right)\frac{x}{5}+\sin \left(\frac{\theta +\varphi }{2}\right)\frac{y}{3}=\cos \left(\frac{\theta -\varphi }{2}\right)$

If it is focal chord then it will pass through $(4,0)$ is focus.

$\cos \left(\frac{\theta +\varphi }{2}\right)\frac{4}{5}=\cos \left(\frac{\theta -\varphi }{2}\right)$

$\frac{4}{5}=\frac{\cos \frac{\theta }{2}\cos \frac{\alpha }{2}-\sin \frac{\theta }{2}\sin \frac{\alpha }{2}}{\cos \frac{\theta }{2}\cos \frac{\alpha }{2}+\sin \frac{\theta }{2}\sin \frac{\alpha }{2}}$

$\frac{4}{5}=\frac{1+\tan \frac{\theta }{}2}{\tan }\frac{\varphi }{2}1-\tan \frac{\theta }{2}\tan \frac{\varphi }{2}$

$\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{4/5-1}{4/5+1}=\frac{-1/5}{9/5}=\frac{-1}{9}$