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MCQMediumJEE 2026Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2026 Question with Solution

An equilateral triangle OAB is inscribed in the parabola y2=4xy^2 = 4x 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

  • A

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

  • B

    4(3−3)4(3-\sqrt{3})

  • C

    2(8−33)2(8-3\sqrt{3})

  • D

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: An equilateral triangle OABOAB is inscribed in the parabola y2=4xy^2 = 4x with O(0,0)O(0,0) at the vertex.

Find: The minimum distance of the circle having ABAB as diameter from the origin.

Let the vertices be O(0,0)O(0,0) and A(t2,2t)A(t^2, 2t). Due to symmetry, B(t2,−2t)B(t^2, -2t).

In equilateral △OAB\triangle OAB, the line OAOA makes 30∘30^\circ with the x-axis.

mOA=2tt2=2t=tan⁡30∘=13m_{OA} = \frac{2t}{t^2} = \frac{2}{t} = \tan 30^\circ = \frac{1}{\sqrt{3}}

So, t=23t = 2\sqrt{3}.

Therefore, the x-coordinate of AA and BB is

t2=(23)2=12t^2 = (2\sqrt{3})^2 = 12

and the y-coordinate of AA is

2t=432t = 4\sqrt{3}

Circle Geometry Step

The circle has diameter ABAB, so its center CC is the midpoint of ABAB, which is (12,0)(12,0). The radius is half the length of ABAB, hence

r=43r = 4\sqrt{3}

The distance of the center from the origin is

d=12d = 12

So the minimum distance of the circle from the origin is

∣d−r∣=12−43=4(3−3)|d-r| = 12 - 4\sqrt{3} = 4(3-\sqrt{3})

Therefore, the correct option is B.

Common mistakes

  • Taking the slope condition incorrectly as tan⁡60∘\tan 60^\circ instead of tan⁡30∘\tan 30^\circ. The side OAOA makes 30∘30^\circ with the x-axis in the symmetric equilateral configuration. Use 2t=tan⁡30∘\frac{2}{t} = \tan 30^\circ.

  • Using the full length of ABAB as the radius of the circle. Since ABAB is the diameter, the radius is half of it. Here that gives r=43r = 4\sqrt{3}, not 838\sqrt{3}.

  • Finding the distance from the center to the origin and stopping at 1212. The question asks for the minimum distance from the circle to the origin, so subtract the radius: 12−4312 - 4\sqrt{3}.

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