Question

If $A\left(\theta \right)$ and $B\left(\varphi \right)$ are the parametric ends of a chord of hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ which passes through $(4,0)$, then the value of $\tan \frac{\theta }{2}\cdot \tan \frac{\varphi }{2}$is

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $0$

Given: $\frac{x^2}{16}-\frac{y^2}{9}=1\Rightarrow a=4,b=3$
The endpoints of the chord are
$A=(a\sec \theta ,b\tan \theta ),B=(a\sec \varphi ,b\tan \varphi )$
The equation of chord $AB$ is
$\frac{x}{a}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)\Rightarrow \frac{x}{4}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{3}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)$

It passes through $(4,0)$, so
$\frac{\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta -\varphi }{2}}=1$
Applying componendo and dividendo,
$\Rightarrow \frac{\cos \frac{\theta +\varphi }{2}-\cos \frac{\theta -\varphi }{2}}{\cos \frac{\theta -\varphi }{2}+\cos \frac{\theta +\varphi }{2}}=\frac{1-1}{1+1}\Rightarrow -\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=0\therefore \tan \frac{\theta }{2}\tan \frac{\varphi }{2}=0$