An equilateral triangle is inscribed in the parabola y2=4ax such that one vertex of this triangle coincides with the vertex of the parabola. The length of side of this triangle is
A
2a√3
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B
4a√3
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C
6a√3
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D
8a√3
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Solution
The correct option is D8a√3 Here, parabola lies at origin.
Let △OAB be an equilateral triangle. ⇒∠AOB=π3 ∴ perpendicular distance of O to AB =k×sinπ3=k×√32 =x− coordinate of A
Similarly, y−coordinate of A =kcosπ3=k2
Since, point A lies on parabola y2=4ax
so, (k2)2=4a√32k ⇒k=8a√3
Here, the length of the triangle is 2×k2=k ∴ length of the triangle 8a√3