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
- 2(8 - 3 3)
- 4(6 + 3)
- 4(3 - 3)
- 2(3 + 3)
Answer
(3) 4(3 - 3)
Solution
By symmetry, A and B are mirror images in the x-axis: A(t2, 2t), B(t2, -2t).
OAB is equilateral, OA makes 30 with the x-axis:
2tt2 = 30 = 1 3 t = 2 3
A(12, 4 3), B(12, -4 3)
Circle on AB as diameter: centre (12, 0), radius = AB2 = 4 3
Origin lies outside the circle (12 > 4 3), so
minimum distance = 12 - 4 3 = 4(3 - 3)
Answer: (3)