An equilateral triangle $S A B$ is inscribed in the parabola $y^{2} = 4 a x$ having its focus at $S$ . If chord $A B$ lies towards the left of $S$ , then side length of this triangle is

2 a (2 - \sqrt{3})

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4 a (2 - \sqrt{3})

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a (2 - \sqrt{3})

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8 a (2 - \sqrt{3})

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Solution

The correct option is C $4 a (2 - \sqrt{3})$
Let $A (a t_{1}^{2}, 2 a t_{1})$ , $B (a t_{1}^{2}, - 2 a t_{1})$
We have, $m_{A S} = tan (\frac{5 π}{6})$
$\Rightarrow \frac{2 a t_{1}}{a t_{1}^{2} - a} = - \frac{1}{\sqrt{3}}$
$\Rightarrow t_{1}^{2} + 2 \sqrt{3} t_{1} - 1 = 0$ $\Rightarrow t_{1} = - \sqrt{3} \pm 2$
Clearly, $t_{1} = - \sqrt{3} - 2$ is rejected.
Thus, $t_{1} = (2 - \sqrt{3})$ .
Hence, $A B = 4 a t_{1} = 4 a (2 - \sqrt{3})$

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