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

The transformed equation of $3x^2 - 6xy + 8y^2 = 8$ when the axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction is:

If $\theta_1, \theta_2, \theta_3$ are respectively the angles by which the coordinate axes are to be rotated to eliminate the $xy$ term from the following equations,then the descending order of these angles is:
$A_1 = 3x^2 + 5xy + 3y^2 + 2x + 3y + 4 = 0$
$A_2 = 5x^2 + 2\sqrt{3}xy + 3y^2 + 6 = 0$
$A_3 = 4x^2 + \sqrt{3}xy + 5y^2 - 4 = 0$

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 $3a-b=$

When the origin is shifted to the point $(h, k)$ by translating the coordinate axes,the equation $S = 2x^2 - xy + y^2 + 2x + 3y + 1 = 0$ is changed to $S' = ax^2 + 2hxy + by^2 + C' = 0$. If the coordinate axes are then rotated about the new origin through an angle $\theta$ in the positive direction to eliminate the $xy$ term,the equation $S' = 0$ becomes $Ax^2 + By^2 + C = 0$. Find the value of $h + k + \tan 2\theta$.

In order to eliminate the first degree terms from the equation $4x^2+8xy+10y^2-8x-44y+14=0$,the point to which the origin has to be shifted is