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 \(\theta\) = \(\pi\over 2\) = \(-a\over 2b\)


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