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

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

$3 \sqrt{3} a$

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

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

answer is C.

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Detailed Solution

Given Equation is

$\begin{array}{l} x^{2} - 3 y^{2} + 4 a x + 4 a^{2} = 0 \\ c o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} + 2 f x + 2 f y + c = 0 \\ a = 1, h = 0, b = - 3, g = 2 a, f = 0, c = 4 a^{2} \\ P . I = (\frac{- g}{a}, \frac{- f}{b}) \Rightarrow (\frac{- 2 a}{1}, \frac{0}{- 3}) \Rightarrow (- 2 a, 0) = (x_{1}, y_{1}) \\ a n d t h e l i n e x - a = 0 \\ l x_{1} + m y_{1} + n = - 2 a - a = - 3 a \\ A r e a = \frac{{(l x_{1} + m y_{1} + n)}^{2} \sqrt{h^{2} - a b}}{|a m^{2} - 2 h l m + b l^{2}|} \Rightarrow \frac{9 a^{2} \sqrt{0 + 3}}{|1 (0) - 0 + 3 (1)|} \Rightarrow \frac{9 a^{2} \sqrt{3}}{3} \Rightarrow 3 \sqrt{3} a^{2} \end{array}$