Answer:
We have, \[x=1-a\sin \theta \] ?(i) \[y=b{{\cos }^{2}}\theta \] ?(ii) On differentiating Eqs. (i) and (ii) w.r.t.\[\theta ,\] we get \[\frac{dx}{d\theta }=-\,a\cos \theta \] and \[\frac{dy}{d\theta }=-\,2b\cos \theta \,\sin \theta \] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-\,2b\cos \theta \,\sin \theta }{-\,a\cos \theta }=\frac{2b\sin \theta }{a}\] Hence, required slope of normal \[=\frac{-\,1}{{{\left( \frac{dy}{dx} \right)}_{at\,\,\theta \,=\,\frac{\pi }{2}}}}=\frac{-\,1}{\frac{2b}{a}\sin \left( \frac{\pi }{2} \right)}=\frac{-\,a}{2b}\]
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