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Mathematics · Parabola

JEE Main 2026 — 23 January, Evening Shift — Question 19

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

  1. Option A:

    4(3−3)4(3-\sqrt{3})

    Correct
  2. Option B:

    2(8−33)2(8-3 \sqrt{3})

  3. Option C:

    4(6+3)4(6+\sqrt{3})

  4. Option D:

    2(3+3)2(3+\sqrt{3})

Answer: A

Step-by-step solution

MOA=2t−0t2−0=2f\mathrm{M}_{\mathrm{OA}}=\frac{2 \mathrm{t}-0}{\mathrm{t}^{2}-0}=\frac{2}{\mathrm{f}}

2t=tan⁡30∘\frac{2}{\mathrm{t}}=\tan 30^{\circ}

t=23\mathrm{t}=2 \sqrt{3}

Req. Circle : (x−12)2+y2=(43)2(x-12)^{2}+y^{2}=(4 \sqrt{3})^{2}

Least distance =∣CP−R∣=|\mathrm{CP}-\mathrm{R}|

=∣2−43∣=4(3−3)=\mid 2-4 \sqrt{3}|=4(3-\sqrt{3})

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Mathematics
Chapter
Parabola
Topic
Chords connected with a Parabola
An equilateral triangle OAB is inscribed in the parabola y 2 =4 x… | JEE Main 2026 PYQ with Solution · DhiX AI