Question

The area bounded by the curve $x=a(\frac{1-t^2}{1+t^2})$ , $y=\frac{2at}{1+t^2}$ in sq. units is

• A. $\frac{\pi a^2}{2}$
• B. $\pi a^2$
• C. $\frac{\pi a^2}{4}$
• D. $\frac{3\pi a^2}{2}$

Answer 1

Answer: (B)

$\pi a^2$

$x=a(\frac{1-t^2}{1+t^2})$ , $y=\frac{2at}{1+t^2}$
$(x/a{)}^2+(+y/a{)}^2=1$ [replace + by $\tan x$ we get ]
$\Rightarrow x^2+y^2=a^2$
If the area of circle with r=a
So, $\pi r^2$
$=\pi a^2$