A parabola has the origin as its focus and the line $$x = 2$$ as the directrix. Then the vertex of the parabola is at

A parabola is the locus of a point $$P(x,y)$$ that remains equidistant from a fixed point (focus) and a fixed straight line (directrix).

Here the focus is the origin $$F(0,0)$$ and the directrix is the vertical line $$x = 2$$.

Distance of $$P(x,y)$$ from the focus:
$$PF = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$$

Perpendicular distance of $$P(x,y)$$ from the directrix $$x = 2$$:
$$PD = |x - 2|$$

Equating the two distances, because every point on a parabola satisfies $$PF = PD$$:
$$\sqrt{x^2 + y^2} = |x - 2|$$

Square both sides (squaring eliminates the absolute value as well):
$$x^2 + y^2 = (x - 2)^2$$

Expand the right-hand side:
$$x^2 + y^2 = x^2 - 4x + 4$$

Simplify (the $$x^2$$ terms cancel):
$$y^2 = -4x + 4 = 4(1 - x)$$

Rewrite to highlight the standard horizontal-axis form $$ (y - k)^2 = 4p(x - h) $$:
$$y^2 = -4(x - 1)$$

Comparing with $$ (y - 0)^2 = 4p(x - 1) $$ gives $$h = 1,\; k = 0$$.
Thus the vertex $$V(h,k)$$ of the parabola is $$V(1,0)$$.

Therefore, the vertex is at (1, 0).

Option B which is: (1, 0)