question_answer

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

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             \[=\frac{-\,1}{{{\left( \frac{dy}{dx} \right)}_{at\,\,\theta \,=\,\frac{\pi }{2}}}}=\frac{-\,1}{\frac{2b}{a}\sin \left( \frac{\pi }{2} \right)}=\frac{-\,a}{2b}\]