JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

An equilateral triangle $O A B$ 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
$4 (3 - 3)$
B
$2 (8 - 33)$
C
$4 (6 + 3)$
D
$2 (3 + 3)$

Solution

Parametrise $A = (t^{2}, 2 t)$ and $B = (t^{2}, - 2 t)$ on $y^{2} = 4 x$ . By symmetry $A B ⊥$ x-axis.

For $O A B$ equilateral, the angle $∠ A O x = 3 0^{\circ}$ , so slope of $O A = tan 3 0^{\circ}$ :

$\frac{2 t}{t ^{2}} = \frac{2}{t} = \frac{1}{3} \Rightarrow t = 23$ .

So $A = (12, 43)$ , $B = (12, - 43)$ .

Circle with $A B$ as diameter: centre $C = (12, 0)$ , radius $R = 43$ .

Minimum distance from $O$ to the circle $= ∣ O C ∣ - R = 12 - 43 = 4 (3 - 3)$ .

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