If $PQ$ is a focal chord of hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ , where eccentric angles of $P$ and $Q$ are $\alpha$ and $\beta$ respectively, then value of $3\tan \frac{\alpha }{2}\tan \frac{\beta }{2}$ is ____ .
My approach is as follow
Let $y = mx - 5m$ is the equation of the focal chord as it passes through $(5,0)$ which is one focus of the hyperbola.
$\left( {4\sec \alpha ,3\tan \alpha } \right)\& \left( {4\sec \beta ,3\tan \beta } \right)$ are the end points of the focal chord
$3\tan \alpha = \left( {\frac{{3\left( {\tan \alpha - \tan \beta } \right)}}{{4\left( {\sec \alpha - \sec \beta } \right)}}} \right)\left( {4\sec \alpha - 5} \right)$
$ \Rightarrow 3\tan \alpha = \left( {\frac{{3\left( {\frac{{\sin \alpha }}{{\cos \alpha }} - \frac{{\sin \beta }}{{\cos \beta }}} \right)}}{{4\left( {\frac{1}{{\cos \alpha }} - \frac{1}{{\cos \beta }}} \right)}}} \right)\left( {4\sec \alpha - 5} \right) \Rightarrow 3\tan \alpha = \left( {\frac{{3\left( {\frac{{\sin \alpha \cos \beta - \sin \beta \cos \alpha }}{{\cos \alpha \cos \beta }}} \right)}}{{4\left( {\frac{{\cos \beta - \cos \alpha }}{{\cos \alpha \cos \beta }}} \right)}}} \right)\left( {4\sec \alpha - 5} \right)$
$ \Rightarrow \tan \alpha = \left( {\frac{{\sin \left( {\alpha - \beta } \right)}}{{4\left( {\cos \beta - \cos \alpha } \right)}}} \right)\left( {4\sec \alpha - 5} \right) \Rightarrow \tan \alpha = \left( {\frac{{\sin \left( {\alpha - \beta } \right)}}{{4\left( {2\sin \left( {\frac{{\alpha - \beta }}{2}} \right)\sin \left( {\frac{{\alpha + \beta }}{2}} \right)} \right)}}} \right)\left( {4\sec \alpha - 5} \right)$
$ \Rightarrow \tan \alpha = \left( {\frac{{2\sin \left( {\frac{{\alpha - \beta }}{2}} \right)\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}}{{4\left( {2\sin \left( {\frac{{\alpha - \beta }}{2}} \right)\sin \left( {\frac{{\alpha + \beta }}{2}} \right)} \right)}}} \right)\left( {4\sec \alpha - 5} \right)$
$\Rightarrow \sin \alpha = \frac{{\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}}{{4\sin \left( {\frac{{\alpha + \beta }}{2}} \right)}}\left( {4 - 5\cos \alpha } \right) \Rightarrow \frac{{2\tan \frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}} = \frac{{\cos \left( {\frac{{\alpha - \beta }}{2}} \right)}}{{4\sin \left( {\frac{{\alpha + \beta }}{2}} \right)}}\left( {4 - 5\left( {\frac{{1 - {{\tan }^2}\frac{\alpha }{2}}}{{1 + {{\tan }^2}\frac{\alpha }{2}}}} \right)} \right)$
$ \Rightarrow 2\tan \frac{\alpha }{2} = \frac{{\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2}}}{{4\left( {\sin \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\beta }{2}\cos \frac{\alpha }{2}} \right)}}\left( {9{{\tan }^2}\frac{\alpha }{2} - 1} \right) \Rightarrow 2\tan \frac{\alpha }{2} = \frac{{1 + \tan \frac{\alpha }{2}\tan \frac{\beta }{2}}}{{4\left( {\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}} \right)}}\left( {9{{\tan }^2}\frac{\alpha }{2} - 1} \right)$
$ \Rightarrow 8{\tan ^2}\frac{\alpha }{2} + 8\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = 9{\tan ^2}\frac{\alpha }{2} + 9{\tan ^3}\frac{\alpha }{2}\tan \frac{\beta }{2} - 1 - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}$
$ \Rightarrow {\tan ^2}\frac{\alpha }{2} + 9{\tan ^3}\frac{\alpha }{2}\tan \frac{\beta }{2} - 1 - 9\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = 0 \Rightarrow {\tan ^2}\frac{\alpha }{2}\left( {1 + 9\tan \frac{\alpha }{2}\tan \frac{\beta }{2}} \right) - \left( {1 + 9\tan \frac{\alpha }{2}\tan \frac{\beta }{2}} \right) = 0$
Not able to approach from here , answer of ${3\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}$ is 8