The area of triangle formed by the lines joining the vertex of the parabola, $$x^2 = 8y$$, to the extremities of its latus rectum is

The given parabola is $$x^2 = 8y$$.
Comparing with the standard form $$x^2 = 4ay$$, we get $$4a = 8 \Rightarrow a = 2$$.

Hence:
Vertex $$V(0,0)$$,
Focus $$S(0,a) = (0,2)$$,
Axis is the $$y$$-axis.

The latus rectum is the line through the focus perpendicular to the axis, i.e. $$y = a = 2$$.
Its extremities are obtained by putting $$y = 2$$ in $$x^2 = 8y$$:

$$x^2 = 8(2) \Rightarrow x^2 = 16 \Rightarrow x = \pm4.$$
Therefore the extremities are $$L_1(4,2)$$ and $$L_2(-4,2)$$.

The required triangle has vertices $$V(0,0),\,L_1(4,2),\,L_2(-4,2)$$.
Using the coordinate‐geometry area formula for a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$:

$$\text{Area} = \frac12 \left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$
Substituting $$V(0,0),\,L_1(4,2),\,L_2(-4,2)$$:

$$\text{Area} = \frac12 \left|0(2-2)+4(2-0)+(-4)(0-2)\right|$$
$$= \frac12 \left|0 + 8 + 8\right|$$
$$= \frac12 (16)$$
$$= 8.$$

Hence, the area of the triangle is $$8$$.

Option B which is: $$8$$