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(D) The given equation of the parabola is $x^2 = 8y$. Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.The vert

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$8$

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(D) The given equation of the parabola is $x^2 = 8y$. Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.The vertex of the parabola is $V(0, 0)$.The endpoints of the latus rectum for $x^2 = 4ay$ are $(2a, a)$ and $(-2a, a)$.Substituting $a = 2$,the endpoints are $L_1(4, 2)$ and $L_2(-4, 2)$.The triangle is formed by the vertices $V(0, 0)$,$L_1(4, 2)$,and $L_2(-4, 2)$.The area of the triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(2 - 2) + 4(2 - 0) + (-4)(0 - 2)| = \frac{1}{2} |0 + 8 + 8| = \frac{1}{2} |16| = 8$ square units.

What is the area of the triangle formed by the vertex and the endpoints of the latus rectum of the parabola $x^2 = 8y$?

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