The given equation of parabola is $y^2$ = 4ax. Let b be the side of an equilateral DAOB whose one vertex is the vertex of parabola. Let OC = x Now AB = b ∴ AC = BC = $\frac{1}{2}$ × AB = $\frac{b}{2}$ Coordinates of point A are $\left(x,\frac{b}{2}\right)$ Since point A lies on the parabola $y^2$ = 4ax ∴ $\left(\frac{b}{2}\right)^2$ = 4ax ⇒ x = $\frac{b^2}{4\times 4a}$ ⇒ x = $\frac{b^2}{16a}$ In right angled ΔOAC, $OA^2$ = $OC^2 + AC^2$ ∴ $b^2$ = $x^2+\left(\frac{b}{2}\right)^2$ ⇒ $b^2$ = $\left(\frac{b^2}{16a}\right)^2 + \frac{b^2}{4}$ ⇒ $b^2$ = $\frac{b^4}{256a^2} + \frac{b^2}{4}$ ⇒ 1 = $\frac{b^2}{256a^2} + \frac{1}{4}$ (Since b ≠ 0) ⇒ $\frac{b^2}{256a^2}$ = 1 - $\frac{1}{4}$ ⇒ $b^2$ = $\frac{3}{4} \times 256a^2$ ⇒ $b^2$ = $192a^2$ ⇒ b = $\sqrt{192a^2}$ ⇒ b = $8\sqrt{3}a$ (Since b ≮ 0) Thus the side of equilateral triangle is $8\sqrt{3}a$.