Question

If $(a\sec \theta ,b\tan \theta )$ and $(a\sec \varphi ,b\tan \varphi )$ are the ends of a focal chord of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the value of $\tan \frac{\theta }{2}\cdot \tan \frac{\varphi }{2}$ can be equal to :

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

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1-e}{1+e}$
C $\frac{1+e}{1-e}$

If $(a\sec \theta ,b\tan \theta )$ and $(a\sec \varphi ,b\tan \varphi )$ are the ends of a focal chord
Then, equation of chord $AB$ is
$\frac{x}{a}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)$
Put $(\pm ae,0)$ in the above equation
$\Rightarrow \frac{\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta -\varphi }{2}}=\frac{e}{1}$ or $\frac{-e}{1}$
Applying componendo and dividendo,
$\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{1-e}{1+e}$ or $\frac{1+e}{1-e}$