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


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