Question

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

Answer 1

It is given that $x=1-a\sin \theta$ and $y=b{\cos }^2\theta$.

$\therefore \frac{dx}{d\theta }=-a\cos \theta$

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

$\frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{-2b\sin \theta \cos \theta }{-a\cos \theta }=\frac{2b}{a}\sin \theta$

Therefore, the slope of the tangent at $\theta =\frac{\pi }{2}$ is given by,

${\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{4}}=\frac{2b}{a}$

Hence, the slope of the normal at $\theta =\frac{\pi }{2}$ is given by,

$-\frac{1}{\text{slope of the tangent at}\theta =\frac{\pi }{4}}=\frac{-1}{\frac{2b}{a}}=-\frac{a}{2b}$