Question

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

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

Answer 1

The correct option is A $\pi \text{sq. units}$

Since, $t\in R$
Let $t=\tan \theta$
$\Rightarrow x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$

$y=\sin 2\theta$

So, curve is $x^2+y^2=1$

Area bounded $=\pi \times (1{)}^2=\pi$