Question

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. $4(6+\sqrt{3})$
• B. $4(3-\sqrt{3})$
• C. $2(8-3\sqrt{3})$
• D. $2(3+\sqrt{3})$

Hint:

For an equilateral triangle in $y^2=4ax$ with a vertex at $(0,0)$, the side length is $8a\sqrt{3}$.

Answer 1

The Correct Option is B

Let the vertices be $O(0,0)$ and $A(t^2, 2t)$. Due to symmetry, $B(t^2, -2t)$.
In equilateral $\triangle OAB$, the line OA makes $30^\circ$ with the x-axis.
Slope $m_{OA} = \frac{2t}{t^2} = \frac{2}{t} = \tan 30^\circ = \frac{1}{\sqrt{3}}$.
So, $t = 2\sqrt{3}$.
The x-coordinate of A and B is $t^2 = (2\sqrt{3})^2 = 12$.
The y-coordinate of A is $2t = 4\sqrt{3}$.
The circle has diameter AB, so its center C is the midpoint of AB, which is $(12, 0)$.
The radius $r$ is half the length of AB, which is the y-coordinate of A. So $r = 4\sqrt{3}$.
The distance of the center from the origin is $d = 12$.
The minimum distance of the circle from the origin is $|d - r| = 12 - 4\sqrt{3}$.
Factoring out 4 gives $4(3 - \sqrt{3})$.