The equation of the bisectors of the angels between the lines joining the origin to the points of intersention of the curve $x^{2} + x y + y^{2} + x = 3 y + 1 = 0$ and the straight line x+y+2=0 is

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$x^{2} - 4 x y + y^{2} = 0$

$x^{2} + 4 x y - y^{2} = 0$

$2 x^{2} - 4 x y + y^{2} = 0$

$2 x^{2} + 4 x y + y^{2} = 0$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{2} + x y + y^{2} + x + 3 y + 1 = 0 - - (1) \\ a n d x + y + 2 = 0 \Rightarrow \frac{x + y}{- 2} = 1 - - (2) \\ H o m o g e n i z i n g (1) w i t h (2) \\ \Rightarrow x^{2} + x y + y^{2} + x (\frac{x + y}{- 2}) + 3 y (\frac{x + y}{- 2}) + 1 {(\frac{x + y}{- 2})}^{2} = 0 \\ \Rightarrow x^{2} + x y + y^{2} - (\frac{x^{2} + x y}{- 2}) - (\frac{3 x y + 3 y^{2}}{2}) + \frac{x^{2} + y^{2} + 2 x y}{4} = 0 \\ \Rightarrow 4 x^{2} + 4 x y + 4 y^{2} - 2 x^{2} - 2 x y - 6 x y - 6 y^{2} + x^{2} + y^{2} + 2 x y = 0 \\ \Rightarrow 3 x^{2} - 2 x y - y^{2} = 0 \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 3, 2 h = - 2 \Rightarrow h = - 1, b = - 1 \\ P a i r o f A n g l e b i \sec t o r i s h (x^{2} - y^{2}) = (a - b) x y \\ \Rightarrow - 1 (x^{2} - y^{2}) = (3 + 1) x y \\ \Rightarrow - x^{2} + y^{2} = 4 x y \\ \Rightarrow x^{2} + 4 x y - y^{2} = 0 \end{array}$