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.

Question

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.

Philoid · Accepted Answer

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

=18unit²

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.