Question

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

• A. $\pi sq.units$
• B. $\left(\pi /2\right)sq.units$
• C. $\left(\pi /4\right)sq.units$
• D. $\left(3\pi /4\right)sq.units$

Answer 1

The correct option is A $\pi sq.units$

$x=\frac{1-t^2}{1+t^2}y=\frac{2t}{1+t^2}$

$t=tan\theta$

$x=\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }y=\frac{2tan\theta }{1+ta{n}^2\theta }$

$x=cos2\theta$ $y=sin2\theta$

$x^2+y^2=1$

Area $=\pi r^2$$=\pi .............\because r=1$