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

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(B) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$.Differentiating $x$ with respect to $t$ using the quotient rule:$\frac{dx}{dt} =

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

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(B) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$.Differentiating $x$ with respect to $t$ using the quotient rule:$\frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}$.Differentiating $y$ with respect to $t$ using the quotient rule:$\frac{dy}{dt} = \frac{(1 + t^2)(2a) - (2at)(2t)}{(1 + t^2)^2} = \frac{2a + 2at^2 - 4at^2}{(1 + t^2)^2} = \frac{2a(1 - t^2)}{(1 + t^2)^2}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} = \frac{2a(1 - t^2)}{-4t} = \frac{a(1 - t^2)}{-2t} = \frac{a(t^2 - 1)}{2t}$.

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

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