The area of the triangle formed by the lines | vedclass

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(B) Given the parabola equation $x^{2}=12y$.Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$.The vertex of the par

Vedclass · Accepted Answer

$18 	ext{ sq. units}$

Vedclass · Answer

(B) Given the parabola equation $x^{2}=12y$.Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$.The vertex of the parabola is at the origin $O(0,0)$.The extremities of the latus rectum are given by $(2a, a)$ and $(-2a, a)$.Substituting $a=3$,the coordinates are $L_{1}(6, 3)$ and $L_{2}(-6, 3)$.The triangle is formed by the vertices $O(0,0)$,$L_{1}(6, 3)$,and $L_{2}(-6, 3)$.The base of the triangle $L_{1}L_{2}$ is the length of the latus rectum,which is $4a = 12$.The height of the triangle from the vertex $O$ to the line $L_{1}L_{2}$ is the $y$-coordinate of the latus rectum,which is $a = 3$.Therefore,the area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 12 	imes 3 = 18 	ext{ sq. units}$.Thus,the correct option is $B$.

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

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