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Question
If x=α sin 2t (1 + cos 2t) and y=β cos 2t (1−cos 2t), show that `dy/dx=β/αtan t`
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Solution
`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`
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