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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Parabola

An equilateral triangle OAB is inscribed in the parabola y^ 2 =4 x...

An equilateral triangle OAB is inscribed in the parabola y2=4xy^{2}=4 x with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having ABA B as a diameter from the origin is

Options

  1. A2(8−33)2(8-3 \sqrt{3})
  2. B2(3+3)2(3+\sqrt{3})
  3. C4(6+3)4(6+\sqrt{3})
  4. D4(3−3)4(3-\sqrt{3})

Correct answer

D. 4(3−3)4(3-\sqrt{3})

Step-by-step solution

Parabola y2=4xy^2 = 4x has vertex O=(0,0)O = (0,0). By symmetry, let A=(t2,2t)A = (t^2, 2t), B=(t2,−2t)B = (t^2, -2t).
OA=tt2+4OA = t\sqrt{t^2+4}, AB=4tAB = 4t.
For equilateral: OA=AB⇒tt2+4=4t⇒t2=12⇒t=23OA = AB \Rightarrow t\sqrt{t^2+4} = 4t \Rightarrow t^2 = 12 \Rightarrow t = 2\sqrt{3}.
A=(12,43)A = (12, 4\sqrt{3}), B=(12,−43)B = (12, -4\sqrt{3}).
Circle with ABAB as diameter: centre (12,0)(12, 0), radius =43= 4\sqrt{3}.
Minimum distance from origin =12−43=4(3−3)= 12 - 4\sqrt{3} = 4(3 - \sqrt{3}).
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