Question

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.

Answer 1

The given parabola is x² = 12y.
On comparing this equation with x2 = 4ay, we obtain 4a = 12 ⇒ a = 3
∴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
∴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).