Question

If $A\left(\theta \right)$ and $B\left(\varphi \right)$ are the parametric ends of a focal chord of $\frac{x^2}{144}-\frac{y^2}{25}=1$, then the maximum value of $\left|\tan \frac{\theta }{2}\tan \frac{\varphi }{2}\right|$ is

Answer 1

Given: $\frac{x^2}{144}-\frac{y^2}{25}=1\Rightarrow a=12,b=5$
$\Rightarrow e^2=1+\frac{b^2}{a^2}\Rightarrow e^2=1+\frac{25}{144}\Rightarrow e=\frac{13}{12}$

The endpoints of the focal chord are $A=(a\sec \theta ,b\tan \theta )$ and $B=(a\sec \varphi ,b\tan \varphi )$
Then, the equation of chord $AB$ is
$\frac{x}{a}\cos \left(\frac{\theta -\varphi }{2}\right)-\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta +\varphi }{2}\right)$

It passes through $(\pm ae,0)$, so
$\frac{\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta -\varphi }{2}}=\pm e$

When $\frac{\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta -\varphi }{2}}=e$
Applying componendo and dividendo, we get
$-\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{e-1}{e+1}\Rightarrow -\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{\frac{13}{12}-1}{\frac{13}{12}+1}=\frac{1}{25}$
When $\frac{\cos \frac{\theta +\varphi }{2}}{\cos \frac{\theta -\varphi }{2}}=-e$
Applying componendo and dividendo, we get
$-\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{-e-1}{-e+1}\Rightarrow -\tan \frac{\theta }{2}\tan \frac{\varphi }{2}=\frac{-\frac{13}{12}-1}{-\frac{13}{12}+1}=25$
Therefore, the maximum value of $\left|\tan \frac{\theta }{2}\tan \frac{\varphi }{2}\right|$ is $25$