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 possible value(s) of $\tan \left(\frac{\theta }{2}\right)\tan \left(\frac{\varphi }{2}\right)$ is/are

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

Answer 1

Answer: (B), (C)

$-\frac{1}{9}$

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 a focus $(ae,0)\equiv (4,0),$ then
$\frac{4}{5}\cos \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$
$\Rightarrow \frac{4}{5}=\frac{\cos \left(\frac{\theta -\varphi }{2}\right)}{\cos \left(\frac{\theta +\varphi }{2}\right)}$

By applying componendo and dividendo rule, we get
$\Rightarrow \frac{4+5}{4-5}=\frac{\cos \left(\frac{\theta -\varphi }{2}\right)+\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)-\cos \left(\frac{\theta +\varphi }{2}\right)}$
$\Rightarrow -\frac{9}{1}=\frac{2\cos \left(\frac{\varphi }{2}\right)\cos \left(\frac{\theta }{2}\right)}{2\sin \left(\frac{\varphi }{2}\right)\sin \left(\frac{\theta }{2}\right)}$
$\Rightarrow \tan \left(\frac{\theta }{2}\right)\tan \left(\frac{\varphi }{2}\right)=-\frac{1}{9}$

If it passes through another focus $(-4,0)$
then (by applying above process again)
$\Rightarrow \tan \left(\frac{\theta }{2}\right)\tan \left(\frac{\varphi }{2}\right)=-9$