(d) Given $P$ $(a\sec \theta ,\,b\tan \theta )$ and $Q\,(a\sec \phi ,\,b\tan \phi )$

The equation of tangent at point $P$ is $\frac{{x\sec \theta }}{a} – \frac{{y\tan \theta }}{b} = 1$

$m$ of tangent $ = \frac{b}{{\tan \theta }} \times \frac{{\sec \theta }}{a} = \frac{b}{a}.\frac{1}{{\sin \theta }}$

Hence the equation of perpendicular at $P$ is

$y – b\tan \theta = – \frac{{a\sin \theta }}{b}(x – a\sec \theta )$

or $by – {b^2}\tan \theta = – a\sin \theta \,x + {a^2}\tan \theta $

or $a\sin \theta \,x + by = ({a^2} + {b^2})\tan \theta $…..$(i)$

Similarly the equation of perpendicular at $Q$ is

$a\sin \phi \,x + by = ({a^2} + {b^2})\tan \phi $…..$(ii)$

On multiplying $(i)$ by $\sin \phi $ and $(ii)$ by $\sin \theta $

$a\sin \theta \sin \phi \,x + b\sin \phi y = ({a^2} + {b^2})\tan \theta \sin \phi $

$a\sin \phi \sin \theta \,x + b\sin \theta y = ({a^2} + {b^2})\tan \phi \sin \theta $

On subtraction by,

$(\sin \phi – \sin \theta ) = ({a^2} + {b^2})(\tan \theta \sin \phi – \tan \phi \sin \theta )$

$\therefore y = k = \frac{{{a^2} + {b^2}}}{b}.\frac{{\tan \theta \sin \phi – \tan \phi \sin \theta }}{{\sin \phi – \sin \theta }}$

$\because \theta  + \phi  = \frac{\pi }{2}$ ==> $\phi = \frac{\pi }{2} – \theta $

==> $\sin \phi = \cos \theta $ and $\tan \phi = \cot \theta $

$\therefore y = k = \frac{{{a^2} + {b^2}}}{b}.\frac{{\tan \theta \cos \theta – \cot \theta \sin \theta }}{{\cos \theta – \sin \theta }}$

$ = \frac{{{a^2} + {b^2}}}{b}\left( {\frac{{\sin \theta – \cos \theta }}{{\cos \theta – \sin \theta }}} \right) = – \frac{{({a^2} + {b^2})}}{b}$.