Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.
The given parabola is x2 = 12y.
On comparing this equation with x2 = 4ay, we get,
4a = 12
⇒ a = 3
Thus, the coordinates of foci are S(0,a) = S(0,3).
Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At y = 3, x2 = 12(3)
⇒ x2 = 36
⇒ x = 
6
Thus, the coordinates of A are (-6,3), while the coordinates of B are (6,3)
Therefore, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).
Area of ΔOAB = 
unit2
= 
unit2
= 
unit2
= 
unit2
unit2
=18unit2
7