Question

Area of the region enclosed by the parametric curve $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 t}{1+t^2}$, 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 $t =1, x =0, \quad y =1$
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) $