Find the slope of the normal to the curve $x = a{\cos ^3}\theta ,y = a{\sin ^3}\theta$at $\theta = \cfrac{\pi }{4}$.

We have $x = a{\cos ^3}\theta$ …(i)
$y = a{\sin ^3}\theta$ …(ii)

Differentiating (i) \& (ii) w.r.t $\theta$,we get
$\cfrac{{dx}}{{d\theta }} = 3a{\cos ^2}\theta ( - \sin \theta ) = - 3a{\cos ^2}\theta \sin \theta$

$\cfrac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \cos \theta$
$\cfrac{{dy}}{{dx}} = \cfrac{{\left( {\cfrac{{dy}}{{d\theta }}} \right)}}{{\left( {\cfrac{{dx}}{{d\theta }}} \right)}} = \cfrac{{3a{{\sin }^2}\theta \cos \theta }}{{ - 3a{{\cos }^2}\theta \sin \theta }} = - \tan \theta$

Therefore the slope of normal at $\theta = \cfrac{\pi }{4}$ is
$\cfrac{{ - 1}}{{{{\left( {\cfrac{{dy}}{{dx}}} \right)}_{\theta = \cfrac{\pi }{4}}}}} = \cfrac{{ - 1}}{{ - \tan (\pi /4)}} = 1$