Question

An equilateral triangle \( OAB \) is inscribed in the parabola \( y = 4x^2 \) whose vertex is \( O \). Find the least distance of the circle described on \( AB \) as diameter from the origin.

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

Hint:

For symmetric curves like parabolas, assume symmetric points to simplify geometry problems.

Answer 1

The Correct Option is B

Step 1: Assume coordinates of points.
Let the equilateral triangle have vertex at the origin \( O(0,0) \). Let the other two vertices be \[ A(x,4x^2), \quad B(-x,4x^2) \] since the parabola is symmetric about the \( y \)-axis.
Step 2: Use equilateral triangle condition.
Distance \( OA = OB = AB \). Using distance formula, \[ OA^2 = x^2 + 16x^4 \] \[ AB = 2x \] Equating \( OA = AB \), \[ x^2 + 16x^4 = 4x^2 \Rightarrow 16x^4 - 3x^2 = 0 \]
Step 3: Find coordinates.
Solving, \[ x^2 = \frac{3}{16} \Rightarrow x = \frac{\sqrt{3}}{4} \]
Step 4: Find the center of the circle on \( AB \) as diameter.
The midpoint of \( AB \) is \[ \left(0,4x^2\right) = \left(0,\frac{3}{4}\right) \]
Step 5: Find the least distance from origin.
Distance from origin to center of the circle is \[ \frac{3}{4} - \frac{\sqrt{3}}{4} = \frac{3-\sqrt{3}}{4} \]