# 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.

P Pankaj Sanodiya

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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