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(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$. The rotation transformation is given by: $x = X \cos \frac{\pi}{6} -

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$\sqrt{3} y^2-x y=0$

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(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$. The rotation transformation is given by: $x = X \cos \frac{\pi}{6} - Y \sin \frac{\pi}{6} = \frac{\sqrt{3}X - Y}{2}$ $y = X \sin \frac{\pi}{6} + Y \cos \frac{\pi}{6} = \frac{X + \sqrt{3}Y}{2}$ Substituting these into the equation $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$: $\sqrt{3} \left( \frac{\sqrt{3}X - Y}{2} \right)^2 - 4 \left( \frac{\sqrt{3}X - Y}{2} \right) \left( \frac{X + \sqrt{3}Y}{2} \right) + \sqrt{3} \left( \frac{X + \sqrt{3}Y}{2} \right)^2 = 0$ $\frac{\sqrt{3}}{4} (3X^2 - 2\sqrt{3}XY + Y^2) - \frac{4}{4} (\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) + \frac{\sqrt{3}}{4} (X^2 + 2\sqrt{3}XY + 3Y^2) = 0$ Multiplying by $4$: $3\sqrt{3}X^2 - 6XY + \sqrt{3}Y^2 - 4\sqrt{3}X^2 - 8XY + 4\sqrt{3}Y^2 + \sqrt{3}X^2 + 6XY + 3\sqrt{3}Y^2 = 0$ Combining like terms: $(3\sqrt{3} - 4\sqrt{3} + \sqrt{3})X^2 + (-6 - 8 + 6)XY + (\sqrt{3} + 4\sqrt{3} + 3\sqrt{3})Y^2 = 0$ $0X^2 - 8XY + 8\sqrt{3}Y^2 = 0$ Dividing by $-8$: $XY - \sqrt{3}Y^2 = 0$ Thus,the transformed equation is $\sqrt{3}Y^2 - XY = 0$.

If the coordinate axes are rotated through an angle $\frac{\pi}{6}$ about the origin,then the transformed equation of $\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$ is

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