Find the equations of the tangent and the normal at the point 't' on the curve x = \(a sin^3 t\), y = \(b cos^3 t\).

Solution :

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

\(\implies\) \(dx\over dt\) = \(3a sin^2t cos t\) and, \(dy\over dt\) = \(-3b cos^2t sin t\)

\(\therefore\) \(dy\over dx\) = \(dy/dt\over dx/dt\) = \(-b\over a\) \(cos t\over sin t\)

So, the equation of the tangent at the point ‘t’ is

y – \(b cos^3 t\) = (\(dy\over dx\))(x – \(a sin^3 t\))

or, y – \(b cos^3 t\) = \(-b\over a\) \(cos t\over sin t\)(x – \(a sin^3 t\))

or, bx cos t + ay sin t = ab sin t cos t

The equation of the normal at the point ‘t’ is

y – \(b cos^3 t\) = \((-1\over ({dy\over dx})\)(x – \(a sin^3 t\))

or, y – \(b cos^3 t\) = \((-1\over ({-b\over a}{cos t\over sin t})\)(x – \(a sin^3 t\))

or, ax sin t – by cos t = \(a^2 sin^4 t – b^2 cos^4 t\)