Step 1 :
Let 0AB be the equilateral triangle inscribed in parabola $y^2=4ax$
AB intersects the x - axis at point C.
Let 0C = k
Substituting in the general equation for x we get,
$y^2=4ak$
$ \Rightarrow y = \pm 2\sqrt{ak}$
Step 2 :
The coordinates of A and B are $(k, 2\sqrt{ak} )$ and $(k, -2\sqrt{ak} )$
$AB = CA + CB$
$ = 2\sqrt{ak}+2\sqrt{ak}$
$ = 4\sqrt{ak}$
0AB is an equilateral triangle.
$0A^2=0B^2$ ( Hint : But $0C^2+AC^2=0A^2$ )
(i.e) : $k^2+ (2 \sqrt{ak} )^2=(4 \sqrt{ak} )^2$
$ \Rightarrow k^2+4ak = 16ak$
$ \Rightarrow k^2=12ak$
$ k = 12a$
Step 3 :
$ \therefore AB = 4\sqrt{ak}$
$ = 4\sqrt{a \times 12a}$
$ 4\sqrt{12a^2}$
$ = 8\sqrt{3a}$
Hence the side of the equilateral triangle inscribed in parabola $y^2=4ax$ is $8\sqrt 3a$ units.