Question

Solve this tricky Math question [Class 12]

Points: 20 ☺

Answer 1

x=3cost-cos³t

dx/dt=-3sint+3cos²tsint


y=3sint-sin³t

dy/dt =3cost-3sin²tcost


now slope of a normal. will be -dx/dy which is tan³t


now equation of normal =(y-y')=m(x-x')

y-3sint+sin³t = tan³t(x-3cost+cos³t)

by expanding we will get

ycos³t-xsin³t=3sintcos³t-3sin³tcost

=3sintcost(cos²t-sin²t)

=3/2(sin2t)(cos2t)

=3/4(sin4t)


4(ycos³t-xsin³t)=3sin4t



I hope this will help u ;)

Answer 2

REFER TO THE ATTACHMENT!!!!+!!!!!!