Find the slope of the normal to the curve x = \(a cos^3\theta\), y = \(a sin^3\theta\) at \(\theta\) = \(\pi\over 4\).

Solution :

We have, x = \(a cos^3\theta\), y = \(a sin^3\theta\)

\(\implies\) \(dx\over d\theta\) = \(-3 a cos^2\theta sin\theta\), \(dy\over d\theta\) = \(3 a sin^2\theta cos\theta\)

Now, \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\)

\(\implies\) \(dy\over dx\) = -\(tan\theta\)

\(\therefore\) Slope of the normal at any point on the curve = \(-1\over dy/dx\) = \(cot\theta\)

Hence, the slope of the normal at \(\theta\) = \(\pi\over 4\) = \(cot\pi\over 4\) = 1.