x=asin2t(1+cos2t)
Differentiating w.r.t t, we get
dxdt=2acos2t(1+cos2t)+asin2t(−2sin2t)
=2acos2t(1+cos2t)−2asin22t
[dxdt]t=π4=2acos(2×π4){1+cos(2×π4)}−2asin2(2×π4)
=0−2a=−2a
y=bcos2t(1−cos2t)
Differentiating w.r.t t, we get,
dydt=−2bsin2t(1−cos2t)+bcos2t(2sin2t)
=−2bsin2t(1−cos2t)+2bcos2tsin2t
[dydt]t=π4=−2bsinπ2(1−cosπ2)+2bcosπ2sinπ2
=−2b
Now, dydx=dy/dtdx/dt=−2b−2a=ba