Find the slope of the normal to the curve $x=1-a \sin \theta$, $y=b\cos^2 \theta$ at $\theta = \dfrac{\pi}{2}$
$x = 1-a\sin \theta$
$\therefore$ $\;$ $\dfrac{dx}{d\theta} = -a \cos \theta$
$y = b \cos^2 \theta$
$\therefore$ $\;$ $\dfrac{dy}{d\theta}=-2b\cos \theta \sin \theta$
$\therefore$ $\;$ Slope of tangent to the given curve $= \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$
i.e. $\dfrac{dy}{dx} = \dfrac{-2b\cos \theta \sin \theta}{-a\cos \theta}$
i.e. $\dfrac{dy}{dx} = \dfrac{2b}{a} \sin \theta$
$\therefore$ $\;$ $\dfrac{dy}{dx} \bigg|_{\theta = \frac{\pi}{2}} = \dfrac{2b}{a} \sin \left(\dfrac{\pi}{2}\right) = \dfrac{2b}{a}$
Slope of normal to the curve $\left(\text{at }\theta = \dfrac{\pi}{2}\right)$ $= \dfrac{-1}{\text{slope of tangent at }\theta = \frac{\pi}{2}}$
$\therefore$ $\;$ Slope of normal $=-\dfrac{a}{2b}$