1. Write the parabola in standard form

   Given parabola: $y^2=8x$ Comparing with $y^2=4ax$, we get $4a=8 \implies a=2$ So the vertex is at $V=(0,0)$

2. Set up the other two vertices

   One vertex of the equilateral triangle is at the vertex of the parabola, i.e. at $(0,0)$.

   Since the parabola is symmetric about the $x$-axis, let the other two vertices be symmetric points on the parabola: $P=(2t^2,4t), \quad Q=(2t^2,-4t)$ using the parametric form for $y^2=4ax$ with $a=2$: $(at^2,2at)=(2t^2,4t).$

3. Compute the side lengths

   First, length $PQ$:

   Since $P$ and $Q$ have same $x$-coordinate, $PQ=|4t-(-4t)|=8|t|$

   Next, length $VP$: $VP=\sqrt{(2t^2)^2+(4t)^2}$ $=\sqrt{4t^4+16t^2}$ $=2|t|\sqrt{t^2+4}$

   Similarly, $VQ=2|t|\sqrt{t^2+4}$

4. Use the equilateral triangle condition

   For an equilateral triangle, $VP=PQ$ So, $2|t|\sqrt{t^2+4}=8|t|$

   For nonzero $t$ (otherwise triangle collapses), divide by $2|t|$: $\sqrt{t^2+4}=4$ Squaring, $t^2+4=16$ $t^2=12$

5. Find the side length

   $PQ=8|t|=8\sqrt{12}=16\sqrt{3}$

6. Area of the equilateral triangle

   Area of an equilateral triangle of side $s$ is $\frac{\sqrt{3}}{4}s^2$

   Here $s=16\sqrt{3}$, so $\text{Area}=\frac{\sqrt{3}}{4}(16\sqrt{3})^2$ $=\frac{\sqrt{3}}{4}\cdot 256\cdot 3$ $=192\sqrt{3}$

7. Check options

   • A: $256\sqrt3$ ❌
   • B: $64\sqrt3$ ❌
   • C: $128\sqrt3$ ❌
   • D: $192\sqrt3$ ✅

Therefore, the required area is $\boxed{192\sqrt3}$