The locus of the mid-point of the line segment joining the focus of the parabola $$y^2 = 4ax$$ to a moving point of the parabola, is another parabola whose directrix is:

The parabola is $$y^2 = 4ax$$, whose focus is at $$(a, 0)$$. Let a moving point on the parabola be $$(at^2, 2at)$$.

The mid-point of the line segment joining the focus $$(a, 0)$$ and $$(at^2, 2at)$$ is $$(h, k)$$, where $$h = \frac{a + at^2}{2}$$ and $$k = \frac{0 + 2at}{2} = at$$.

From $$k = at$$, we get $$t = \frac{k}{a}$$.

Substituting in the expression for $$h$$: $$h = \frac{a + a \cdot \frac{k^2}{a^2}}{2} = \frac{a^2 + k^2}{2a}$$.

So $$2ah = a^2 + k^2$$, which gives $$k^2 = 2ah - a^2 = 2a\left(h - \frac{a}{2}\right)$$.

Replacing $$h$$ with $$x$$ and $$k$$ with $$y$$, the locus is $$y^2 = 2a\left(x - \frac{a}{2}\right)$$.

This is a parabola with vertex at $$\left(\frac{a}{2}, 0\right)$$ and $$4A = 2a$$, so $$A = \frac{a}{2}$$.

The directrix of this parabola is $$x = \frac{a}{2} - \frac{a}{2} = 0$$.

Hence, the correct answer is Option B.