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Area of triangle formed by pair of line ${(x + 2 a)}^{2} - 3 y^{2} = 0 a n d x = a$ is

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$3 \sqrt{3} a$

$\sqrt{3} a^{2}$

$\sqrt{\frac{3}{4}} a^{2}$

$3 \sqrt{3} a^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} x + \sqrt{3} y + 2 a = 0 \\ x - \sqrt{3} y + 2 a = 0 \\ Δ = \frac{1}{2} (\sqrt{3} a + \sqrt{3} a) (a + 2 a) \\ = (\sqrt{3} a) (3 a) = 3 \sqrt{3} a^{2} \end{array}$

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