The lines joining the origin to the points of intersection of $y = 6 x + 8 w i t h 3 x^{2} + 4 x y - 4 y^{2} - 11 x + 2 y + 6 = 0$ are equally inclined to

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$y = 6 x + 8$

$x y = 0$

$y = x$

$y = - x$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 3 x^{2} + 4 x y - 4 y^{2} - 11 x + 2 y + 6 = 0 - - (1) \\ a n d l i n e y = 6 x + 8 \\ \Rightarrow y - 6 x = 8 \\ \Rightarrow \frac{y - 6 x}{8} = 1 - - (2) \\ H o m o g e n i z i g (1) w i t h (2) \\ 3 x^{2} + 4 x y - 4 y^{2} - 11 x (1) + 2 y (1) + 6 {(1)}^{2} = 0 \\ \Rightarrow 3 x^{2} + 4 x y - 4 y^{2} - 11 x (\frac{y - 6 x}{8}) + 2 y (\frac{y - 6 x}{8}) + 6 {(\frac{y - 6}{8})}^{2} = 0 \\ \Rightarrow 3 x^{2} + 4 x y - 4 y^{2} - 11 x (\frac{y - 6 x}{8}) + 2 y (\frac{y - 6 x}{8}) + 6 (\frac{y^{2} + 36 x^{2} - 12 x y}{64}) = 0 \end{array}$

$\begin{array}{l} \frac{96 x^{2} + 128 x y - 128 y^{2} - 44 x y + 264 x^{2} + 8 y^{2} - 48 x y + 3 y^{2} + 108 x^{2} - 36 x y}{32} = 0 \\ \Rightarrow 468 x^{2} - 117 y^{2} = 0 \\ \Rightarrow 4 x^{2} - y^{2} = 0 \\ c o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 4, h = 0, b = - 1 \\ A n g l e b i \sec t o r i s h (x^{2} - y^{2}) = (a - b) x y \\ \Rightarrow 0 (x^{2} - y^{2}) = 5 x y \\ \Rightarrow x y = 0 \\ T h e E q u a t i o n 4 x^{2} - y^{2} = 0 i s e q u a l l y i n c l i n e d w i t h c o - o r d i n a t e a x i s \end{array}$