Find the slope of the normal to the curve $x = 1 - a\sin \theta ,\,y = b{\cos ^2}\theta$ at $\theta = \cfrac{\pi }{2}$.

We have $x = 1 - a\sin \theta$ and …(i)
$y = b{\cos ^2}\theta$

Differentiating (i) \& (ii) w.r.t $\theta$,we get
$\cfrac{{dx}}{{d\theta }} = 0 - a\cos \theta = - a\cos \theta$ and

$\cfrac{{dy}}{{d\theta }} = 2b\cos \theta ( - \sin \theta ) = - 2b\sin \theta \cos \theta$

So, $\cfrac{{dy}}{{dx}} = \cfrac{{\left( {\cfrac{{dy}}{{d\theta }}} \right)}}{{\left( {\cfrac{{dx}}{{d\theta }}} \right)}} = \cfrac{{ - 2bcos\theta (\sin \theta )}}{{ - a\cos \theta }} = \cfrac{{2b}}{a}\sin \theta$

Therefore the slope of the normal at $\theta = \cfrac{\pi }{2}$ is
$\cfrac{{ - 1}}{{{{\left( {\cfrac{{dy}}{{dx}}} \right)}_{\theta = \pi /2}}}}$

$= \cfrac{{ - 1}}{{\cfrac{{2b}}{a}sin\left( {\cfrac{\pi }{2}} \right)}} = \cfrac{{ - a}}{{2b}}$