JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Parabola

Take $A=(t^2,2t)$ on $y^2=4x$; the slope of $OA$ is $\frac{2t}{t^2}=\frac{2}{t}$, and by symmetry $B=(t^2,-2t)$. For the equilateral triangle $OA$ makes $30^{\circ }$ with the axis, so $\frac{2}{t}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, giving $t=2\sqrt{3}$. Thus $A=(12,4\sqrt{3})$ and $B=(12,-4\sqrt{3})$. The circle on $AB$ as diameter has centre $C=(12,0)$ and radius $R=\frac{1}{2}\cdot 8\sqrt{3}=4\sqrt{3}$. As $|OC|=12>R$, the origin lies outside the circle, and the minimum distance from the origin to it is $|OC|-R=12-4\sqrt{3}=4(3-\sqrt{3})$.