When the origin is shifted to the point (2, 3) and then the coordinate axes are rotated through an angle π3 in the counter clockwise,sense then the transformed equation of 3x2 + 2xy + 3y2 – 18x

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When the origin is shifted to the point (2, 3) and then the coordinate axes are rotated through an angle $\frac{π}{3}$ in the counter clockwise,
sense then the transformed equation of $3 x^{2} + 2 x y + 3 y^{2} - 18 x - 22 y + 50 = 0$ is

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$(6 - \sqrt{3}) x^{2} + (6 - \sqrt{3}) y^{2} + 2 x y = 0$

$4 x^{2} + 2 y^{2} - 1 = 0$

$3 x^{2} + 3 x y - 1 = 0$

$(6 + \sqrt{3}) x^{2} - 2 x y + (6 - \sqrt{3}) y^{2} - 2 = 0$

answer is B.

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Detailed Solution

$G i v e n (x_{1}, y_{1}) = (2, 3), θ = 60^{0}$

$G i v e n e q u a t i o n 3 x^{2} + 2 x y + 3 y^{2} - 18 x - 22 y + 50 = 0 w h e n t h e a x e s a r e t r a n s l a t e d t o t h e p o i n t (2, 3) t h e n t r a n s f o r m e d e q u a t i o n i s 3 x^{2} + 2 x y + 3 y^{2} + g x_{1} + f y_{1} + c = 0 \Rightarrow 3 x^{2} + 2 x y + 3 y^{2} - 1 = 0 \to (1) N o w a x e s a r e r o t a e d t h r o u g h a n a n g l e 60^{0} t h e n x = \frac{X - \sqrt{3} Y}{2} a n d y = \frac{\sqrt{3} X + Y}{2} s u b t h e s e v a l u e s i n (1) w e g e t 3 {(\frac{X - \sqrt{3} Y}{2})}^{2} + 2 (\frac{X - \sqrt{3} Y}{2}) (\frac{\sqrt{3} X + Y}{2}) + 3 {(\frac{\sqrt{3} X + Y}{2})}^{2} - 1 = 0 \Rightarrow 6 \sqrt{3} X^{2} - 2 X Y + (6 - 2 \sqrt{3}) Y^{2} - 2 = 0$