JEE Main202623 January 2026Evening ShiftMathematicsParabolaActual
An equilateral triangle OAB is inscribed in the parabola y²=4 x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having A B as a diameter from the origin is
Options
- A2(8-3 3 )
- B2(3+ 3 )
- C4(6+ 3 )
- D4(3- 3 )
Correct answer
D. 4(3- 3 )
Step-by-step solution
Parabola y^2 = 4x has vertex O = (0,0) . By symmetry, let A = (t^2, 2t) , B = (t^2, -2t) . OA = t t^2+4 , AB = 4t . For equilateral: OA = AB t t^2+4 = 4t t^2 = 12 t = 2 3 . A = (12, 4 3 ) , B = (12, -4 3 ) . Circle with AB as diameter: centre (12, 0) , radius = 4 3 . Minimum distance from origin = 12 - 4 3 = 4(3 - 3 ) .