JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Straight Lines

The slope after rotation is $2-\sqrt{3}=\tan 15^{\circ }$, so the rotated line makes $15^{\circ }$ with the positive x-axis. Turning clockwise through $\frac{\alpha }{2}$ lowers the inclination from $\alpha$ to $\alpha -\frac{\alpha }{2}=\frac{\alpha }{2}$, hence $\frac{\alpha }{2}=15^{\circ }$ and $\alpha =30^{\circ }$. The rotated line through $P(a,0)$ is $y=(2-\sqrt{3})(x-a)$, that is $(2-\sqrt{3})x-y-(2-\sqrt{3})a=0$. Its distance from the origin is $\frac{|a|(2-\sqrt{3})}{\sqrt{(2-\sqrt{3}{)}^2+1}}$, and since $(2-\sqrt{3}{)}^2+1=8-4\sqrt{3}=4(2-\sqrt{3})$ this equals $\frac{|a|\sqrt{2-\sqrt{3}}}{2}$. Setting it to $\frac{1}{\sqrt{2}}$ gives $a^2(2-\sqrt{3})=2$, so $a^2=\frac{2}{2-\sqrt{3}}=2(2+\sqrt{3})=4+2\sqrt{3}$. With ${\tan }^2\alpha ={\tan }^{2}30^{\circ }=\frac{1}{3}$, $3a^2{\tan }^2\alpha -2\sqrt{3}=3(4+2\sqrt{3})\cdot \frac{1}{3}-2\sqrt{3}=(4+2\sqrt{3})-2\sqrt{3}=4$.