If x = a sin 2theta (1 + cos 2theta ) and y = | vedclass

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(A) Given $x = a \sin 2	heta (1 + \cos 2	heta ) = a(2 \sin 	heta \cos 	heta)(2 \cos^2 	heta) = 4a \sin 	heta \cos^3 	heta$. Given $y = b \cos 2

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$\frac{b 	an 	heta}{a}$

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(A) Given $x = a \sin 2	heta (1 + \cos 2	heta ) = a(2 \sin 	heta \cos 	heta)(2 \cos^2 	heta) = 4a \sin 	heta \cos^3 	heta$. Given $y = b \cos 2	heta (1 - \cos 2	heta ) = b(2 \cos^2 	heta - 1)(2 \sin^2 	heta) = 2b \sin^2 	heta (2 \cos^2 	heta - 1)$. Alternatively,using the chain rule: $\frac{dx}{d	heta} = a[2 \cos 2	heta (1 + \cos 2	heta) + \sin 2	heta (-2 \sin 2	heta)] = 2a(\cos 2	heta + \cos^2 2	heta - \sin^2 2	heta) = 2a(\cos 2	heta + \cos 4	heta) = 4a \cos 3	heta \cos 	heta$. $\frac{dy}{d	heta} = b[-2 \sin 2	heta (1 - \cos 2	heta) + \cos 2	heta (2 \sin 2	heta)] = 2b(-\sin 2	heta + \sin 2	heta \cos 2	heta + \sin 2	heta \cos 2	heta) = 2b(\sin 4	heta - \sin 2	heta) = 4b \cos 3	heta \sin 	heta$. Therefore,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{4b \cos 3	heta \sin 	heta}{4a \cos 3	heta \cos 	heta} = \frac{b}{a} 	an 	heta$.

If $x = a \sin 2\theta (1 + \cos 2\theta )$ and $y = b \cos 2\theta (1 - \cos 2\theta )$,then $\frac{dy}{dx} = $

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