# Find the area of a triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum.

The problem can be represented as follows, where $AB$ is the latus rectum of the parabola $x^2 = 12y$.
Given $x^2 =12y \rightarrow 4a = 12 \rightarrow a = 3$.
Therefore, the coordinates of point $C = (0,3)$
When $y = 3 \rightarrow x^2 = 12 \times 3 = 36 \rightarrow x = \pm 6$
Therefore $A = (-6,3)$ and $B = (6,3)$.
Given the 3 vertices of the triangle, we can calculate the area as follows: $\text{Area} = |\frac12(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$
Now that we know the vertices of $\Delta AOB$, we can calculate the area.
$\qquad = |\frac12( 0 (3-3) + (-6) (3-0) + 6(0-3)|$
$\qquad = |\frac12(-18-18)|$
$\qquad = \large\frac{1}{2}$$\times 36 = 18\; \text{sq. units}$