Question

Let $A\left(\theta \right)$ and $B\left(\varphi \right)$ are the parametric ends of a chord of the hyperbola $\frac{x^2}{144}-\frac{y^2}{25}=1$. If the equation of $AB$ is $2x+3y=1$, then

• A. $\tan \left(\frac{\theta +\varphi }{2}\right)=15$
• B. $23\left(\tan \frac{\theta }{2}\tan \frac{\varphi }{2}\right)=25$
• C. $\tan \left(\frac{\theta +\varphi }{2}\right)=-15$
• D. $25\left(\tan \frac{\theta }{2}\tan \frac{\varphi }{2}\right)=23$

Answer 1

Answer: (C), (D)

The correct options are
C $\tan \left(\frac{\theta +\varphi }{2}\right)=-15$
D $25\left(\tan \frac{\theta }{2}\tan \frac{\varphi }{2}\right)=23$

Given: $\frac{x^2}{144}-\frac{y^2}{25}=1\Rightarrow a=12,b=5$
Endpoints of the chord are $A=(a\sec \theta ,b\tan \theta )$ and $B=(a\sec \varphi ,b\tan \varphi )$
Then, the equation of chord $AB$ is
$\frac{x}{a}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)\Rightarrow \frac{x}{12}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{5}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)\Rightarrow 5\cos \left(\frac{\theta -\varphi }{2}\right)x-12\sin \left(\frac{\theta +\varphi }{2}\right)y=60\cos \left(\frac{\theta +\varphi }{2}\right)\cdots \left(1\right)$

Given equation of chord $AB$ is
$2x+3y=1\cdots \left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$, we get
$\frac{5\cos \left(\frac{\theta -\varphi }{2}\right)}{2}=\frac{-12\sin \left(\frac{\theta +\varphi }{2}\right)}{3}=\frac{60\cos \left(\frac{\theta +\varphi }{2}\right)}{1}\Rightarrow \frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta +\varphi }{2}\right)}=-15\Rightarrow \tan \left(\frac{\theta +\varphi }{2}\right)=-15$

Also,
$\Rightarrow \frac{\cos \frac{\theta -\varphi }{2}}{\cos \frac{\theta +\varphi }{2}}=24$
Applying componendo and dividendo,
$\Rightarrow \frac{\cos \frac{\theta -\varphi }{2}-\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta +\varphi }{2}+\cos \frac{\theta -\varphi }{2}}=\frac{24-1}{24+1}\therefore \tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{23}{25}$