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 $x ^2 = 12y.$
On comparing this equation with $x^ 2 = 4ay$, 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 $y = 3, x^2 = 12 (3)$
$⇒ x^2 = 36$
$⇒ x = ±6$

∴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).

Area of$\triangle OAB = \frac{1}{2} [0(3-3) + (-6)(3-0) + 6(0-3)]$ unit2

$= \frac{1}{2} [(-6) (3) + 6 (-3)]$ unit2

$= \frac{1}{2} [-18-18]$ unit2

$= \frac{1}{2}[-36]$ unit2

$= 18$ unit2

Thus, the required area of the triangle is 18 unit2 .