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

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(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.On differentiating $x$ and $y$ with respect to $t$,we get:$\frac{dx}{dt} = \frac{(

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

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(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.On differentiating $x$ and $y$ with respect to $t$,we get:$\frac{dx}{dt} = \frac{(1+t^{2})(-2t) - (1-t^{2})(2t)}{(1+t^{2})^{2}} = \frac{-2t - 2t^{3} - 2t + 2t^{3}}{(1+t^{2})^{2}} = \frac{-4t}{(1+t^{2})^{2}}$.$\frac{dy}{dt} = \frac{(1+t^{2})(2a) - (2at)(2t)}{(1+t^{2})^{2}} = \frac{2a + 2at^{2} - 4at^{2}}{(1+t^{2})^{2}} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}} 	imes \frac{(1+t^{2})^{2}}{-4t}$.$\frac{dy}{dx} = \frac{2a(1-t^{2})}{-4t} = \frac{a(1-t^{2})}{-2t} = \frac{a(t^{2}-1)}{2t}$.

If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$,then $\frac{d y}{d x}$ is equal to

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