Question
An equilateral triangle OAB is inscribed in the parabola y^ 2 =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
JEE Main
Mathematics
Parabola
2026
JEE Main 2026 (23 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 2(8-3 3 )
- B. 2(3+ 3 )
- C. 4(6+ 3 )
- D. 4(3- 3 )
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 ).
Related: Mathematics — Parabola · All PYQ Banks