What is the area of the triangle formed by th | vedclass

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(C) The given parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.The vertex of the parabola is at $(0, 0)$.The endpo

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$18$

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(C) The given parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.The vertex of the parabola is at $(0, 0)$.The endpoints of the latus rectum are $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$.The area of the triangle formed by the vertex $(0, 0)$ and the points $(6, 3)$ and $(-6, 3)$ is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,the base is the length of the latus rectum,which is $|6 - (-6)| = 12$.The height is the $y$-coordinate of the latus rectum,which is $3$.Therefore,$	ext{Area} = \frac{1}{2} 	imes 12 	imes 3 = 18$ square units.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

What is the area of the triangle formed by the vertex of the parabola $x^2 = 12y$ and the endpoints of its latus rectum (in square units)?

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