JEE Main 2025 Apr 8 Shift 2, Mathematics Q12: Points and Straight Lines

The rotated line has slope $2-\sqrt{3}=\tan 15^{\circ }$, and since it is obtained from the original by clockwise rotation through $\alpha /2$, the original inclination satisfies $\alpha -\frac{\alpha }{2}=15^{\circ }$, giving $\frac{\alpha }{2}=15^{\circ }$ and $\alpha =30^{\circ }$.

Equation of the rotated line through $P=(a,0)$: $y=(2-\sqrt{3})(x-a)$, or $(2-\sqrt{3})x-y-(2-\sqrt{3})a=0$.

Distance from origin: $\frac{|(2-\sqrt{3})a|}{\sqrt{(2-\sqrt{3}{)}^2+1}}=\frac{1}{\sqrt{2}}$.

Compute the denominator: $(2-\sqrt{3}{)}^2+1=4-4\sqrt{3}+3+1=8-4\sqrt{3}=4(2-\sqrt{3})$, so $\sqrt{\text{denom}}=2\sqrt{2-\sqrt{3}}$.

Setting up: $\frac{|a|(2-\sqrt{3})}{2\sqrt{2-\sqrt{3}}}=\frac{1}{\sqrt{2}}$, so $|a|\sqrt{2-\sqrt{3}}=\frac{2}{\sqrt{2}}=\sqrt{2}$.

Squaring: $a^2(2-\sqrt{3})=2$, so $a^2=\frac{2}{2-\sqrt{3}}=2(2+\sqrt{3})$ after rationalising.

With $\alpha =30^{\circ }$, ${\tan }^2\alpha =\frac{1}{3}$. Substituting: $3a^2{\tan }^2\alpha -2\sqrt{3}=3\cdot 2(2+\sqrt{3})\cdot \frac{1}{3}-2\sqrt{3}=2(2+\sqrt{3})-2\sqrt{3}=4$, matching option (1).