An equilateral triangle OAB is inscribed in the parabola $y^{2} = 4x$ with the vertex O at the verte

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An equilateral triangle OAB is inscribed in the parabola $y^{2} = 4x$ 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

Accepted Answer

$${M_{OA} = \frac{2t - 0}{t^{2} - 0} = \frac{2}{f} }{\frac{2}{t} = tan30^{\circ} }{t = 2\sqrt{3} }$$Req. Circle : $(x - 12)^{2} + y^{2} = (4\sqrt{3})^{2}$

Least distance $= |CP - R|$

$$= \mid 2 - 4\sqrt{3} = 4(3 - \sqrt{3})$$

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