The area of the triangle formed by the lines | vedclass

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

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$18 	ext{ sq. units}$

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(A) The given equation of the parabola is $x^{2}=12y$. Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$. The vertex of the parabola is at $(0,0)$. The ends of the latus rectum are at $(2a, a)$ 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 $4a = 12$. The height is the distance from the vertex to the latus rectum,which is $a = 3$. $	ext{Area} = \frac{1}{2} 	imes 12 	imes 3 = 18 	ext{ sq. units}$.

The area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the ends of the latus rectum is

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