An equilateral triangle OAB 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 AB as a diameter from the origin is [2026]

Question

An equilateral triangle OAB 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 AB as a diameter from the origin is [2026]

Smart Achievers · Accepted Answer

(2)

$M_{O A} = \frac{2 t - 0}{t^{2} - 0} = \frac{2}{t}$

$\frac{2}{t} = \tan 30 °$

$t = 2 \sqrt{3}$

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

$Least distance = | C P - R |$

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

Solution

Q.
An equilateral triangle OAB 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 AB as a diameter from the origin is [2026]

1 $4 (6 + \sqrt{3})$

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

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

4 $2 (8 - 3 \sqrt{3})$

Ans.
(2)

$M_{O A} = \frac{2 t - 0}{t^{2} - 0} = \frac{2}{t}$

$\frac{2}{t} = \tan 30 °$

$t = 2 \sqrt{3}$

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

$Least distance = | C P - R |$

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

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Solution

Q. An equilateral triangle OAB is inscribed in the parabola y2=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 [2026]

1 4(6+3)

2 4(3-3)

3 2(3+3)

4 2(8-33)

Ans. (2) MOA=2t-0t2-0=2t 2t=tan30° t=23 Req. Circle: (x-12)2+y2=(43)2 Least distance =|CP-R| =|2-43|=4(3-3)