A line passing through the point $\mathrm{P}(\mathrm{a}, \theta)$ makes an acute angle $\alpha$ with the positive x -axis. Let this line be rotated about the point $P$ through an angle $\frac{\alpha}{2}$ in the clock-wise direction. If in the new position, the slope of the line is $2-\sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$, then the value of $3 a^2 \tan ^2 \alpha-2 \sqrt{3}$ is

• Correct answer
  A

  4
• B

  6
• C

  5
• D

  8

Solution

$\begin{aligned}<br/>& \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \quad \Rightarrow \alpha=30^{\circ}<br/>\end{aligned}$
equation of PR :
$\begin{aligned}<br/>& y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a)<br/>\end{aligned}$
$\perp$ distance from origin $=\frac{1}{\sqrt{2}}$
$\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \\ & |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \\ & a^2=2(2+\sqrt{3}) \\ & 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \\ & 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4\end{aligned}$

Hence, the correct answer is option: 'A'.