JEE MAIN - Mathematics (2025 - 8th April Evening Shift - No. 5)
A line passing through the point P($a$, 0) 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 clockwise 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 $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is :
8
4
5
6
Explanation
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$$\begin{aligned} & \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \end{aligned}$$
$\Rightarrow \alpha=30^{\circ}$
equation of PR:
$$\begin{aligned} & y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a) \end{aligned}$$
$\perp$ distance from origin $=\frac{1}{\sqrt{2}}$
$$\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \\ & |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \\ & a^2=2(2+\sqrt{3}) \\ & 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \\ & 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4 \end{aligned}$$
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