Find the equation of tangent to the curve given by`x=asin^3t ,y=bcos^3t` ... (1)at a point where `t=pi/2`.

Explanation

Given `x=asin^3t ,y=bcos^3t`


Differentiate given curve w.r.t `t`


`(dx)/(dt)=3asin^2tcost` and `(dy)/(dt)=-3bcos^2tsint`


So, `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`


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