The given parabola is x2 = 12y.
On comparing this equation with x2 = 4ay, we get,
4a = 12
a =\( \dfrac{12}{4}\)
= 3
The coordinates of foci are S(0,a) = S(0,3).
Now 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
So, the coordinates of A are (-6, 3), while the coordinates of B are (6, 3)
Then, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).
By using the formula,
Area of ΔOAB = \( \dfrac{1}{2}\) [0(3-3) + (-6)(3-0) + 6(0-3)] unit2
= \( \dfrac{1}{2}\) [(-6) (3) + 6 (-3)] unit2
= \( \dfrac{1}{2}\) [-18-18] unit2
= \( \dfrac{1}{2}\) [-36] unit2
= 18 unit2
Area of ΔOAB is 18 unit2
Answered by Aaryan | 9 months agoAn equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
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