Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $$x^2 = 8y$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio 1 : 3, then the locus of $$P$$ is

Let us denote $$O(0,0)$$ as the vertex and take an arbitrary point on the given parabola $$x^{2}=8y$$ as $$Q(x,y)$$, so that $$Q$$ automatically satisfies the relation $$x^{2}=8y$$.

We are told that the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, that is

$$OP:PQ = 1:3.$$

First, recall the coordinate-division formula. For two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$, a point $$R$$ that divides $$AB$$ internally in the ratio $$m:n$$ (measured from $$A$$ to $$B$$) has coordinates

$$\left(\dfrac{nx_1+mx_2}{m+n},\;\dfrac{ny_1+my_2}{m+n}\right).$$

Here we take $$A=O(0,0)$$ and $$B=Q(x,y)$$, with the ratio $$m:n = 1:3$$, where $$m$$ is attached to $$B(Q)$$ and $$n$$ to $$A(O)$$. Substituting $$x_1=0,\;y_1=0,\;x_2=x,\;y_2=y,\;m=1,\;n=3$$, we get

$$P\Bigl(X,Y\Bigr)=\left(\dfrac{3\cdot 0+1\cdot x}{1+3},\;\dfrac{3\cdot 0+1\cdot y}{1+3}\right)=\left(\dfrac{x}{4},\;\dfrac{y}{4}\right).$$

Thus the coordinates of $$P$$ can be expressed in terms of those of $$Q$$ as

$$X=\dfrac{x}{4},\qquad Y=\dfrac{y}{4}.$$

Rewriting these gives the reverse substitutions

$$x = 4X,\qquad y = 4Y.$$

Because $$Q$$ lies on the parabola $$x^{2}=8y$$, we have

$$x^{2}=8y.$$

Now substitute $$x=4X$$ and $$y=4Y$$ into this relation:

$$\bigl(4X\bigr)^{2}=8\bigl(4Y\bigr).$$

Carrying out the algebra step by step,

$$16X^{2}=32Y.$$

Divide every term by $$16$$:

$$X^{2}=2Y.$$

The symbols $$X$$ and $$Y$$ represent the running coordinates of the point $$P$$, so its locus is given by

$$x^{2}=2y.$$

Hence, the correct answer is Option A.