The given equation of parabola is y2 = 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 =

1

2

× AB =

b

2

Coordinates of point A are (x,

b

2

)
Since point A lies on the parabola y2 = 4ax
∴ (

b

2

)2 = 4ax ⇒ x =

b2

4×4a

⇒ x =

b2

16a

In right angled ΔOAC, OA2 = OC2+AC2
∴ b2 = x2+(

b

2

)2 ⇒ b2 = (

b2

16a

)2+

b2

4

⇒ b2 =

b4

256a2

+

b2

4

⇒ 1 =

b2

256a2

+

1

4

(Since b ≠ 0)
⇒

b2

256a2

= 1 -

1

4

⇒ b2 =

3

4

×256a2
⇒ b2 = 192a2
⇒ b = √192a2 ⇒ b = 8√3a (Since b ≮ 0)
Thus the side of equilateral triangle is 8√3a.