Question

Area of the region enclosed by the parametric curve $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$, the tangent to it at $t=1$ and the normal with slope $\sqrt{3}$ is

• A. $\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{\pi }{6}\right)$
• B. $\frac{1}{3}\left(\frac{1}{2}-\frac{\pi }{3}\right)$
• C. $\frac{1}{2}\left(\frac{1}{\sqrt{3}}+\frac{\pi }{3}\right)$
• D. $\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{\pi }{2}\right)$

Answer 1

Answer: (A)

Put $t=\tan \theta ,x=\cos 2\theta ,y=\sin 2\theta$
$\therefore$ curve is $x^2+y^2=1$
when $t=1,x=0,y=1$
Tangent at $(0,1)$ is $y=1$
Also, required normal $y=\sqrt{3}x$
$\therefore$ Shaded area $=\frac{1}{2}\times 1\times \frac{1}{\sqrt{3}}-\frac{\pi (30)}{360}=\frac{1}{2\sqrt{3}}-\frac{\pi }{12}$
$A=\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{\pi }{6}\right)$