Question

If $(a\sec \theta ,b\tan \theta )$ and $(a\sec \varphi ,b\tan \varphi )$ be the coordinates of the ends of a focal chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=$

• A. $\frac{1-e}{1+e}$
• B. $\frac{1+e}{1-e}$
• C. $\frac{e-1}{e+1}$
• D. None of these

Answer 1

The correct option is A $\frac{1-e}{1+e}$

Equation of the chord joining $(a\sec \theta ,b\tan \theta )$ and $(a\sec \varphi ,b\tan \varphi )$ is

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

If it passes through $(ae,0)$, then

$\frac{ae}{a}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{0}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)$

$e\cos \left(\frac{\theta -\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)\Rightarrow e=\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}=\frac{1-\tan \frac{\theta }{2}\tan \frac{\varphi }{2}}{1+\tan \frac{\theta }{2}\tan \frac{\varphi }{2}}$

$\Rightarrow \tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{1-e}{1+e}$