Area enclosed by the curve y = f(x) that is d | vedclass

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(A) Given the parametric equations $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.Let $t = 	an 	heta$. Then $x = \frac{1 - 	an^2 	heta

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$\pi \ sq. \ units$

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(A) Given the parametric equations $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.Let $t = 	an 	heta$. Then $x = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \cos 2	heta$ and $y = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \sin 2	heta$.Squaring and adding,we get $x^2 + y^2 = \cos^2 2	heta + \sin^2 2	heta = 1$.This represents a circle with radius $r = 1$ centered at the origin $(0, 0)$.The area of a circle with radius $r$ is given by $A = \pi r^2$.Substituting $r = 1$,we get $A = \pi(1)^2 = \pi \ sq. \ units$.

Area enclosed by the curve $y = f(x)$ that is defined parametrically as $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$ (where $t \in R$) is equal to

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