The area of a triangle (in sq. units) formed | vedclass

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(C) The given equation of the parabola is $x^2=16y$. Comparing this with the standard form $x^2=4ay$,we get $4a=16$,which implies $a=4$. The focus of

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$32$

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(C) The given equation of the parabola is $x^2=16y$. Comparing this with the standard form $x^2=4ay$,we get $4a=16$,which implies $a=4$. The focus of the parabola is $F(0, a) = (0, 4)$. The latus rectum is the line $y=4$. Substituting $y=4$ in the parabola equation $x^2=16y$,we get $x^2=16(4)=64$,so $x=\pm 8$. The endpoints of the latus rectum are $P(8, 4)$ and $Q(-8, 4)$. The vertex of the parabola is $O(0, 0)$. The triangle is formed by the vertices $O(0, 0)$,$P(8, 4)$,and $Q(-8, 4)$. The base of the triangle $PQ$ has length $8 - (-8) = 16$ units. The height of the triangle from the vertex $O$ to the line $PQ$ is the $y$-coordinate of the latus rectum,which is $4$ units. Area of $	riangle OPQ = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 16 	imes 4 = 32$ sq. units.

The area of a triangle (in sq. units) formed by the latus rectum of the parabola $x^2=16y$ and the lines joining the vertex of the parabola to the ends of the latus rectum is

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