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(C) When coordinate axes are rotated through an angle $	heta$ in anti-clockwise direction,the transformation is given by:$x = X \cos 	heta - Y \sin

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) When coordinate axes are rotated through an angle $	heta$ in anti-clockwise direction,the transformation is given by:$x = X \cos 	heta - Y \sin 	heta$$y = X \sin 	heta + Y \cos 	heta$Substituting these into $x^2+y^2+2xy+2x+6y+1=0$:$(X \cos 	heta - Y \sin 	heta)^2 + (X \sin 	heta + Y \cos 	heta)^2 + 2(X \cos 	heta - Y \sin 	heta)(X \sin 	heta + Y \cos 	heta) + 2(X \cos 	heta - Y \sin 	heta) + 6(X \sin 	heta + Y \cos 	heta) + 1 = 0$Simplifying the coefficients:$X^2(1 + \sin 2	heta) + 2XY(\cos 2	heta) + Y^2(1 - \sin 2	heta) + X(2 \cos 	heta + 6 \sin 	heta) + Y(6 \cos 	heta - 2 \sin 	heta) + 1 = 0$Comparing with $(2+\sqrt{3})X^2 + 2XY + (2-\sqrt{3})Y^2 + aX + bY + 2 = 0$:Note: The constant term in the given equation is $2$,so we multiply the derived equation by $2$ to match:$2(1 + \sin 2	heta) = 2 + \sqrt{3}$ $\Rightarrow \sin 2	heta = \frac{\sqrt{3}}{2}$ $\Rightarrow 2	heta = 60^\circ$ $\Rightarrow 	heta = 30^\circ$Then $a = 2(2 \cos 30^\circ + 6 \sin 30^\circ) = 2(2 \cdot \frac{\sqrt{3}}{2} + 6 \cdot \frac{1}{2}) = 2(\sqrt{3} + 3) = 6 + 2\sqrt{3}$$b = 2(6 \cos 30^\circ - 2 \sin 30^\circ) = 2(6 \cdot \frac{\sqrt{3}}{2} - 2 \cdot \frac{1}{2}) = 2(3\sqrt{3} - 1) = 6\sqrt{3} - 2$$3a - b = 3(6 + 2\sqrt{3}) - (6\sqrt{3} - 2) = 18 + 6\sqrt{3} - 6\sqrt{3} + 2 = 20$

When the coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction,if the transformed equation of $x^2+y^2+2xy+2x+6y+1=0$ is $(2+\sqrt{3})X^2+2XY+(2-\sqrt{3})Y^2+aX+bY+2=0$,then $3a-b=$

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