The area of the triangle formed by the lines | vedclass

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(B) The given equation of the parabola is $x^2 = 8y$.Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.The verte

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$8$

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(B) The given equation of the parabola is $x^2 = 8y$.Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.The vertex of the parabola is $C = (0, 0)$.The extremities of the latus rectum are $A = (-2a, a) = (-4, 2)$ and $B = (2a, a) = (4, 2)$.We need to find the area of the triangle formed by the vertices $A(-4, 2)$,$B(4, 2)$,and $C(0, 0)$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the coordinates: Area $= \frac{1}{2} |-4(2 - 0) + 4(0 - 2) + 0(2 - 2)|$.Area $= \frac{1}{2} |-4(2) + 4(-2) + 0| = \frac{1}{2} |-8 - 8| = \frac{1}{2} |-16| = 8$.Thus,the area of the triangle is $8$ square units.

The area of the triangle formed by the lines joining the vertex of the parabola,$x^2 = 8y$,to the extremities of its latus rectum is

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