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

 To solve this problem, we need to determine the least distance of the circle described on \( AB \) as the diameter from the origin \( O \). Here, \( OAB \) is an equilateral triangle inscribed in the parabola \( y = 4x^2 \), and \( O \) is the origin of the coordinate system.

1. Since \( OAB \) is an equilateral triangle and \( O \) is at the origin, let the coordinates of \( A \) be \( (a, 4a^2) \) and of \( B \) be \( (-a, 4a^2) \). This accounts for the symmetry about the y-axis.
2. The length of \( AB \) can be found using the distance formula: \(\text{Length of } AB = \sqrt{((-a) - a)^2 + (4a^2 - 4a^2)^2} = \sqrt{(2a)^2} = 2a.\)
3. In an equilateral triangle, the altitude from the vertex \( O \) to the base \( AB \) is: \(\text{Altitude} = \frac{\sqrt{3}}{2} \times \text{Side length}.\)Thus, the altitude is: \(\frac{\sqrt{3}}{2} \times 2a = \sqrt{3}a.\)
4. The centroid \( G \) of the triangle \( OAB \), being at two-thirds of the altitude from the vertex, is located at: \(G = \left(0, \frac{2}{3} \times \sqrt{3}a\right) = \left(0, \frac{2\sqrt{3}}{3}a\right).\)
5. The center of the circle lies at the midpoint of \( AB \), given by: \(M = \left(\frac{a + (-a)}{2}, \frac{4a^2 + 4a^2}{2}\right) = (0, 4a^2).\)
6. Let \( C \) be the center of the circle and \( r \) its radius.
   • \( C = (0, 4a^2) \).
   • The radius \( r \) of the circle with diameter \( AB \) is half of \( AB \): \(r = \frac{2a}{2} = a.\)
7. The least distance of the circle from the origin is given by the perpendicular distance of the center \( C = (0, 4a^2) \) from the origin, minus the radius \( r = a \). Therefore, this distance is: \(\text{Distance} = 4a^2 - a.\)
8. Given \( OAB \) being an equilateral triangle inscribed in the parabola, and \( OC \) bisects \( \angle AOB \), we find: \(4a^2 = \frac{\sqrt{3}}{2} \times 2a = \sqrt{3}a \implies a = \sqrt{3}.\)
9. Substitute \( a \) in the distance expression: \(\text{Distance} = 4(\sqrt{3})^2 - \sqrt{3} = 12 - \sqrt{3}.\)
10. Thus, the least distance is: \(\frac{12 - \sqrt{3}}{4} = \frac{3 - \sqrt{3}}{4}.\)

Therefore, the correct answer is: \(\frac{3-\sqrt{3}}{4}\).