The equation of the normal to the curve

Given that the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ have a normal at the point $\theta=\frac{\pi}{4}$

Differentiating both w.r.t. $\theta$,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \cos ^{2} \theta \sin \theta, \frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \operatorname{asin}^{2} \theta \cos \theta$

$\Rightarrow \frac{d y}{d x}=-\tan \theta$

For $\theta=\frac{\pi}{4}$

Slope of the tangent $=-1$

$x=\frac{a}{2 \sqrt{2}}, y=\frac{a}{2 \sqrt{2}}$

Equation of normal:

$\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$

$x=y$