Parabola Part 2

Concept:Use parametric coordinates on the parabola $y^2=4x$ (with $a=1$) and the equilateral triangle condition to find points $A$ and $B$.Explanation:Let $A=(t^2,2t)$ and $B=(t^2,-2t)$ (symmetry about x‑axis).Triangle $OAB$ is equilateral, so $\angle AOB = 60^\circ$ and each side equal.Slope of $OA$ is $\frac{2t}{t^2} = \frac{2}{t}$.Since $OA$ makes $30^\circ$ with the x‑axis, slope $= \tan30^\circ = \frac{1}{\sqrt3}$.Thus $\frac{2}{t} = \frac{1}{\sqrt3}$ gives $t = 2\sqrt3$.Coordinates: $A = (12, 4\sqrt3)$, $B = (12, -4\sqrt3)$.Circle with diameter $AB$: center $C = \left(\frac{12+12}{2}, \frac{4\sqrt3-4\sqrt3}{2}\right) = (12,0)$.Radius $r = \frac12\sqrt{(12-12)^2 + (4\sqrt3+4\sqrt3)^2} = \frac12 \cdot 8\sqrt3 = 4\sqrt3$.Minimum distance from origin $O(0,0)$ to the circle $= |OC - r|$ where $OC = 12$.Thus $d = 12 - 4\sqrt3 = 4(3-\sqrt3)$.Answer:Option C: $4(3-\sqrt3)$.