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(A) The given parabola is $x^{2}=12y$.Comparing this with the standard form $x^{2}=4ay$, we get $4a=12$, which implies $a=3$.The focus of the parabola

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$18$

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(A) The given parabola is $x^{2}=12y$.Comparing this with the standard form $x^{2}=4ay$, we get $4a=12$, which implies $a=3$.The focus of the parabola is $S(0, a) = (0, 3)$.The latus rectum is a line segment passing through the focus perpendicular to the axis of the parabola. For $y=3$, $x^{2}=12(3)=36$, so $x=\pm 6$.The ends of the latus rectum are $A(-6, 3)$ and $B(6, 3)$.The vertex of the parabola is $O(0, 0)$.The area of $\Delta OAB$ with vertices $(0, 0)$, $(-6, 3)$, and $(6, 3)$ is given by:$	ext{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$	ext{Area} = \frac{1}{2} |0(3-3) + (-6)(3-0) + 6(0-3)|$$	ext{Area} = \frac{1}{2} |0 - 18 - 18| = \frac{1}{2} |-36| = 18 	ext{ unit}^{2}$.

Find the area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the ends of its latus rectum. (in $\text{ unit}^{2}$)

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