Question

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

Answer 1

It is given that x=1-a sin $\theta andy=bco{s}^2\theta$

On differentiating x and y both w.r.t. $\theta$, we get

$\frac{dx}{d\theta }=\frac{d}{d\theta }[1-asin\theta ]=-acos\theta$ and

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left[bco{s}^2\theta \right]=2bcos\theta (-sin\theta )=-2bcos\theta sin\theta$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-2bcos\theta sin\theta }{-acos\theta }=\frac{2b}{a}sin\theta$

$\therefore$ Slope of normal at the point $\theta =\frac{\pi }{2},-\frac{dx}{dy}=\frac{-1}{dy/dx}$

$\frac{-1}{{\left(\frac{dy}{dx}\right)}_{(\theta =\pi /2)}}=\frac{-1}{\frac{2b}{a}sin\left(\frac{\pi }{2}\right)}=\frac{-a}{2b}$