The locus of a point which divides the line segment joining the point $$(0, -1)$$ and a point on the parabola $$x^2 = 4y$$ internally in the ratio 1 : 2 is:

Let us denote the fixed point by $$A(0,-1)$$ and take an arbitrary point on the parabola $$x^{2}=4y$$ as $$B(X,Y)$$, where the capital letters $$X$$ and $$Y$$ will later be related by the equation of the parabola.

The required point $$P(x,y)$$ divides the segment $$AB$$ internally in the ratio $$1:2$$, that is, $$AP:PB = 1:2$$.

For internal division, we first recall the section-formula: if a point $$P$$ divides the segment joining $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$ in the ratio $$m:n$$ (with $$m$$ attaching to the farther end $$B$$ and $$n$$ to $$A$$), then

$$P\bigl(\,\dfrac{mx_{2}+nx_{1}}{m+n},\; \dfrac{my_{2}+ny_{1}}{m+n}\bigr).$$

Here $$A(0,-1)$$, $$B(X,Y)$$, $$m=1$$, $$n=2$$. Substituting each symbol, we obtain the coordinates of $$P$$:

$$x \;=\; \dfrac{1\cdot X + 2\cdot 0}{1+2} \;=\; \dfrac{X}{3},$$

$$y \;=\; \dfrac{1\cdot Y + 2\cdot (-1)}{1+2} \;=\; \dfrac{Y-2}{3}.$$

Next we make use of the fact that $$B(X,Y)$$ lies on the parabola. From $$x^{2}=4y$$ we have for point $$B$$

$$X^{2}=4Y \;\;\Longrightarrow\;\; Y=\dfrac{X^{2}}{4}.$$

Our aim is to eliminate $$X$$ and $$Y$$ in favour of the point $$P(x,y)$$. From the relations for $$x$$ and $$y$$ found above we solve for $$X$$ and $$Y$$:

From $$x = \dfrac{X}{3} \; \Longrightarrow\; X = 3x.$$

From $$y = \dfrac{Y-2}{3} \; \Longrightarrow\; Y = 3y + 2.$$

Substituting these expressions into $$X^{2}=4Y$$ gives

$$\bigl(3x\bigr)^{2} \;=\; 4\bigl(3y+2\bigr).$$

Carrying out the algebra step by step,

$$9x^{2} \;=\; 12y + 8.$$

Re-arranging, we bring all terms to one side:

$$9x^{2} - 12y = 8.$$

Thus the locus of the point $$P(x,y)$$ is represented by the equation

$$9x^{2} - 12y = 8.$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.