Find the slope of normal to the curve x = 1 - \(a sin\theta\), y = \(b cos^2\theta\) at \(\theta\) = \(\pi\over 2\).

Solution :

We have, x = 1 – \(a sin\theta\), y = \(b cos^2\theta\)

\(\implies\) \(dx\over d\theta\) = \(-a cos\theta\) and \(dy\over d\theta\) = \(-2b cos\theta sin\theta\)

\(\therefore\) \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) = \(2b\over a\) \(sin\theta\)

\(\implies\) \(dy\over dx\) at \(\pi\over 2\) = \(2b\over a\)

Hence, Slope of normal at \(\theta\) = \(\pi\over 2\) = \(-a\over 2b\)