Question

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

• A. $\frac{a}{2b}$
• B. $\frac{2a}{b}$
• C. $\frac{a}{b}$
• D. $\frac{-a}{2b}$

Answer 1

The correct option is D $\frac{-a}{2b}$

Given, $x=1-a\sin \theta$ and $y=b{\cos }^2\theta$
On differentiating with respect to $\theta$, we get
$\frac{dx}{d\theta }=-a\cos \theta$
and $\frac{dy}{d\theta }=2b\cos \theta (-\sin \theta )$
Then, $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{2b}{a}\sin \theta$
$\therefore$ Slope of normal at the point $\theta =\frac{\pi }{2}$ is
$-\frac{dx}{dy}=-\frac{1}{dy/dx}$
$=-\frac{1}{\frac{2b}{a}\sin \left(\frac{\pi }{2}\right)}=-\frac{a}{2b}$