An equilateral triangle is inscribed in the parabola y² = 4ax where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Let OAB be the equilateral triangle inscribed in parabola y² = 4ax.

Let AB intersect the x-axis at point C.

Let OC = k.

From the equation of the given parabola, we have y² = 4ak ⇒ y = ±2√(ak).

∴ The respective coordinates of points A and B are (k, 2√(ak)) and (k, −2√(ak)).

AB = CA + CB = 2√(ak) + 2√(ak) = 4√(ak).

Since OAB is an equilateral triangle, OA² = AB².

∴ k² + (2√(ak))² = (4√(ak))²

⇒ k² + 4ak = 16ak

⇒ k² = 12ak

⇒ k = 12a

∴ AB = 4√(ak) = 4√(a × 12a) = 4√(12a²) = 8√3a.

Thus, the side of the equilateral triangle inscribed in the parabola y² = 4ax is 8√3a.