The area of the triangle formed by the lines | vedclass

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(D) The equation of the 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 $(0, 0)$

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(D) The equation of the 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 ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$.The triangle is formed by the vertices $(0, 0)$,$(6, 3)$,and $(-6, 3)$.The base of the triangle is the length of the latus rectum,which is $4a = 12$.The height of the triangle is the distance from the vertex to the latus rectum,which is $a = 3$.The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 12 	imes 3 = 18$ sq. units.

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 is equal to $...$ sq. units.

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