An equilateral triangle is inscribed in the parabola $y^{2} = 4 a x,$ such that one vertex of this triangle coincides with the vertex of the parabola. Side length of this triangle is

$4 a \sqrt{3}$

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$6 a \sqrt{3}$

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$2 a \sqrt{3}$

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$8 a \sqrt{3}$

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Solution

The correct option is D
$8 a \sqrt{3}$

If triangle OAB is equilateral then OA = OB = AB. Thus AB will be a double ordinate of the parabola.
$∠ A O X = ∠ X O B = \frac{π}{6} L e t A = (a t_{1}^{2}, 2 a t_{1}) t h e n B = (a t_{1}^{2}, - 2 a t_{1}) S l o p e o f O A = \frac{2}{t_{1}} = \frac{1}{\sqrt{3}} \Rightarrow t_{1} = 2 \sqrt{3} \Rightarrow A B = 4 a t_{1} = 8 a \sqrt{3} u n i t s$

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