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Find 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.

6.  Find 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.

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Given the parabola,

x^2=12y

Comparing this equation with x^2=4ay, we get

a=3

Now, As we know the coordinates of ends of latus rectum are:

(2a,a)\:and\:(-2a,a)

So, the coordinates of latus rectum are,

(2a,a)\:and\:(-2a,a)=(6,3)\:and\:(-6,3)

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Widht of the triangle = 2*6=12

Height of the triangle = 3

So The area = 

\frac{1}{2}\times base\times height=\frac{1}{2}\times12\times3=18

Hence the required area is 18 unit square.

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