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

• A. $\frac{1}{9}$
• B. $-9$
• C. $-\frac{1}{9}$
• D. $9$

Answer 1

Answer: (C), (D)

The correct options are
C $-\frac{1}{9}$
D $9$

The equation of the line joining $\theta$ and $\varphi$ is $\frac{x}{5}\cos \left(\frac{\theta +\varphi }{2}\right)+\frac{y}{3}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$
If it passes through the point $(4,0)$, then
$\frac{4}{5}\cos \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$
$\Rightarrow \frac{4}{5}=\frac{\cos \left(\frac{\theta -\varphi }{2}\right)}{\cos \left(\frac{\theta +\varphi }{2}\right)}$
$\Rightarrow \frac{4+5}{4-5}=\frac{\cos \left(\frac{\theta -\varphi }{2}\right)+\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)-\cos \left(\frac{\theta +\varphi }{2}\right)}$
$=\frac{2\cos \frac{\theta }{2}\cos \frac{\varphi }{2}}{2\sin \frac{\varphi }{2}\sin \frac{\theta }{2}}$
$\Rightarrow \tan \frac{\theta }{2}\tan \frac{\varphi }{2}=-\frac{1}{9}$
If it passes through the point $(-5,0)$
then $\tan \frac{\varphi }{2}\tan \frac{\theta }{2}=9$