Q27 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
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
- A$2(8-3 \sqrt{3})$
- B$2(3+\sqrt{3})$
- C$4(6+\sqrt{3})$
- Correct answerD$4(3-\sqrt{3})$
Solution
Parabola $y^2 = 4x$ has vertex $O = (0,0)$. By symmetry, let $A = (t^2, 2t)$, $B = (t^2, -2t)$.
$OA = t\sqrt{t^2+4}$, $AB = 4t$.
For equilateral: $OA = AB \Rightarrow t\sqrt{t^2+4} = 4t \Rightarrow t^2 = 12 \Rightarrow t = 2\sqrt{3}$.
$A = (12, 4\sqrt{3})$, $B = (12, -4\sqrt{3})$.
Circle with $AB$ as diameter: centre $(12, 0)$, radius $= 4\sqrt{3}$.
Minimum distance from origin $= 12 - 4\sqrt{3} = 4(3 - \sqrt{3})$.
$OA = t\sqrt{t^2+4}$, $AB = 4t$.
For equilateral: $OA = AB \Rightarrow t\sqrt{t^2+4} = 4t \Rightarrow t^2 = 12 \Rightarrow t = 2\sqrt{3}$.
$A = (12, 4\sqrt{3})$, $B = (12, -4\sqrt{3})$.
Circle with $AB$ as diameter: centre $(12, 0)$, radius $= 4\sqrt{3}$.
Minimum distance from origin $= 12 - 4\sqrt{3} = 4(3 - \sqrt{3})$.
Hence, the correct answer is option: 'D'.