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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.
Solution
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).
Area of Δ OAB = `1/2 |0(3 - 3) + (-6) (3 - 0) + 6 (0 - 3)|"unit"^2`
= `1/2 |(-6) (3) + 6(-3)| "unit"^2`
= `1/2 |-18 - 18| "unit"^2`
= `1/2 |-36| "unit"^2`
= `1/2 xx 36 "unit"^2`
= 18 unit2
Thus, the required area of the triangle is 18 unit2.
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