Question

Area enclosed by the y=f(x) defined para metrically as $x=\frac{1-t^2}{1+t^2}.y=\frac{2t}{1+t^2}$is equal to

• A. $\pi sq.units$
• B. $\pi /2sq.units$
• C. $\frac{3\pi }{4}sq.units$
• D. $\frac{3\pi }{2}sq.units$

Answer 1

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

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

Let $t=\tan \theta$

$\Rightarrow$ $x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\Rightarrow$ $x=\cos 2\theta$

Now,

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

Let $t=\tan \theta$

$\Rightarrow$ $y=\frac{2\tan \theta }{1+{\tan }^2\theta }$

$\Rightarrow$ $y=\sin 2\theta$

$\Rightarrow$ $x^2+y^2={\cos }^{2}2\theta +{\sin }^{2}2\theta =1$

$\therefore$ $r=1$

$\Rightarrow$ Required area $=\pi r^2=\pi (1{)}^2=\pi sq.units$