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

  • A
    $\sqrt{3} y^2+x y=0$
  • B
    $x^2 - y^2 = 0$
  • C
    $\sqrt{3} y^2-x y=0$
  • D
    $\sqrt{3} y^2- 2x y=0$

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