Question

please solve this with step by step

Answer 1

$x = a { \cos }^{3} \alpha \\ y = a {sin}^{3} \alpha \\ (dy \div d \alpha ) = 3a {sin}^{2} \alpha \cos( \alpha )...(1) \\ (dx \div d \alpha ) = - 3a {cos}^{2} \alpha \sin( \alpha ) ...(2) \\ divide(1) \div (2) \\ (dy \div dx) = - \tan( \alpha ) \\ at \: \alpha = (\pi \div 4) \\ dy \div dx = - 1 \\ slope \: of \: normal = - 1 \div ( - ) = 1$