Question

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

Answer 1

It is given that $x=a{\cos }^3\theta ,y=a{\sin }^3\theta$

$\Rightarrow \frac{dx}{d\theta }=3a{\cos }^2\theta (-\sin \theta )=-3a{\cos }^2\theta \sin \theta$

& $\frac{dy}{d\theta }=3a{\sin }^2\theta \left(\cos \theta \right)$

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

Therefore, the slope of the tangent at $\theta =\frac{\pi }{4}$ is given by,

${\left(\frac{dy}{dx}\right)}_{\theta =\pi /4}={(-\tan \theta )}_{\theta =\pi /4}=-1$

Hence, the slope of the normal at $\theta =\frac{\pi }{4}$ is ,

$=-\frac{1}{\text{slope of the tangent at}\theta =\frac{\pi }{4}}=\frac{-1}{-1}=1$