The area of the triangle formed by the lines | vedclass

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(B) The equation of the parabola is $x^2 = 48y$. Comparing this with $x^2 = 4ay$, we get $4a = 48$, so $a = 12$.The vertex of the parabola is $V(0, 0)

Vedclass · Accepted Answer

$288 	ext{ sq. units}$

Vedclass · Answer

(B) The equation of the parabola is $x^2 = 48y$. Comparing this with $x^2 = 4ay$, we get $4a = 48$, so $a = 12$.The vertex of the parabola is $V(0, 0)$.The ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$, which are $(24, 12)$ and $(-24, 12)$.The area of the triangle formed by vertices $(0, 0)$, $(24, 12)$, and $(-24, 12)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(12 - 12) + 24(12 - 0) + (-24)(0 - 12)|$.Area $= \frac{1}{2} |0 + 288 + 288| = \frac{1}{2} |576| = 288 	ext{ sq. units}$.

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 48y$ to the ends of its latus rectum is .....

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