Question:
The equation of the normal to the curve $x=\operatorname{acos}^{3} \theta, y=a \sin ^{3} \theta$ at the point $\theta=\frac{\pi}{4}$ is
A. $\mathrm{x}=0$
B. $y=0$
C. $x=y$
D. $x+y=a$
Solution:
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:
$\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$
$x=y$