The area of the triangle formed by the lines | vedclass

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(D) The equation of the parabola is $x^2 = 20y$. Comparing this with $x^2 = 4ay$,we get $4a = 20$,so $a = 5$. The vertex of the parabola is at $(0, 0)

Vedclass · Accepted Answer

$50 	ext{ sq.units}$

Vedclass · Answer

(D) The equation of the parabola is $x^2 = 20y$. Comparing this with $x^2 = 4ay$,we get $4a = 20$,so $a = 5$. The vertex of the parabola is at $(0, 0)$. The coordinates of the ends of the latus rectum are $(2a, a)$ and $(-2a, a)$,which are $(10, 5)$ and $(-10, 5)$. The triangle is formed by the vertices $(0, 0)$,$(10, 5)$,and $(-10, 5)$. The base of the triangle is the length of the latus rectum,which is $4a = 20$. The height of the triangle is the distance from the vertex to the latus rectum,which is $a = 5$. The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 20 	imes 5 = 50 	ext{ sq.units}$.

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 20y$ to the ends of its latus rectum is

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