Given equation of parabola, $y^2=16 x$ and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola


Since, given parabola is symmetrical about $X$-axis So, $\angle A O M=30^{\circ}$
In $\triangle A O M, \tan 30^{\circ}=\frac{A M}{O M}=\frac{8 t}{4 t^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \Rightarrow$
$t=2 \sqrt{3}$
$\therefore$ Coordinate of point $A$ lying on parabola is
$\left(4(2 \sqrt{3})^2, 8(2 \sqrt{3})\right) \equiv(48,16 \sqrt{3})$
$\therefore \text { Distance } O A=\sqrt{\left(x_A-x_0\right)^2+\left(y_A-y_0\right)^2}$
$=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}$
$O A=32 \sqrt{3}$
$\therefore$ Length of the side of equilateral triangle $=32 \sqrt{3}$.