If x=frac{1-t^2}{1+t^2} and y=frac{2at}{1+t^2 | vedclass

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(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating wi

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$\frac{a(t^2-1)}{2t}$

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(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.Substitute $t=	an 	heta$,then $x=\cos 2	heta$ and $y=a \sin 2	heta$.Differentiating with respect to $	heta$:$\frac{dx}{d	heta} = -2 \sin 2	heta$ and $\frac{dy}{d	heta} = 2a \cos 2	heta$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2a \cos 2	heta}{-2 \sin 2	heta} = -a \cot 2	heta$.Since $\cot 2	heta = \frac{1}{	an 2	heta} = \frac{1-	an^2 	heta}{2 	an 	heta} = \frac{1-t^2}{2t}$,$\frac{dy}{dx} = -a \left( \frac{1-t^2}{2t} ight) = \frac{a(t^2-1)}{2t}$.

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$,then $\frac{dy}{dx}=$

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