Answer 1

`x=α sin 2t(1+cos 2t)`

`⇒x=α sin 2t+α/2×2 sin 2tcos 2t`

`⇒x=α sin 2t+α/2 sin 4t`

Differentiating both sides w.r.t. t, we get

`dx/dt=α cos2t xx 2+α/2 cos4txx4`

`⇒dx/dt=2α(cos2t+cos4t)`

`⇒dx/dt=2α(cos2t+2cos^2 2t−1)`

`⇒dx/dt=2α(cos2t+1)(2cos2t−1)`

Now,

`y=β cos2t(1−cos2t)`

`⇒y=β cos2t−β cos^2 2t`

Differentiating both sides w.r.t. t, we get 

`dy/dt=−β sin2t xx 2+β xx 2cos2t xx sin2txx2`

`⇒dy/dt=−2β sin2t+4β cos2t.sin2t`

`⇒dy/dt=2β sin2t(2cos2t−1)`

We know

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

`=(2α(cos2t+1)(2cos2t−1))/(dy/dt=2β sin2t(2cos2t−1))`

`=(βsin2t)/(α(cos2t+1))`

`=(βxx2sintcost)/(αxx2cos^2t)`

`=β/αtant`