JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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:
- A \( 4(3-\sqrt{3}) \)
- B \( 2(8-3\sqrt{3}) \)
- C \( 4(6+\sqrt{3}) \)
- D \( 2(3+\sqrt{3}) \)
Answer & Solution
Correct Answer
(A) \( 4(3-\sqrt{3}) \)
Step-by-step Solution
Detailed explanation
\(M _{ OA }=\frac{2 t -0}{ t ^2-0}=\frac{2}{ f }\) \(\frac{2}{ t }=\tan 30^{\circ}\) \(t =2 \sqrt{3}\) Req. Circle : \((x-12)^2+y^2=(4 \sqrt{3})^2\) Least distance \(=| CP - R |\) \(=2-4 \sqrt{3}=4(3-\sqrt{3})\)
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