JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Properties of Triangle

The relation $x-\sqrt{3}|y|=\alpha$ splits into $x-\sqrt{3}y=\alpha$ (for $y\ge 0$) and $x+\sqrt{3}y=\alpha$ (for $y\le 0$), two rays meeting at $P=(\alpha ,0)$ with directions $(\sqrt{3},1)$ and $(\sqrt{3},-1)$; the angle between them has cosine $\frac{3-1}{2\cdot 2}=\frac{1}{2}$, so $\angle APB=60^{\circ }$. Since $PA=PB=\alpha$ and the included angle is $60^{\circ }$, triangle $PAB$ is equilateral of side $\alpha$. By symmetry the bisector is the $x$-axis and $Q$ is the midpoint of $AB$, so $PQ$ is the altitude $\frac{\sqrt{3}}{2}\alpha =\frac{9}{2}$, giving $\alpha =3\sqrt{3}$. The circumradius of an equilateral triangle of side $\alpha$ is $R=\frac{\alpha }{\sqrt{3}}=3$. Therefore $\frac{{\alpha }^2}{R}=\frac{(3\sqrt{3}{)}^2}{3}=\frac{27}{3}=9$.