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:

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 $h + k + \tan 2\theta$.

The angle by which the coordinate axes are to be rotated about the origin so that the transformed equation of $\sqrt{3} x^2+(\sqrt{3}-1) x y-y^2=0$ is free from the $xy$ term is: (in $^{\circ}$)

If the axes are rotated through an angle $45^{\circ}$,then the coordinates of the point $(4 \sqrt{2}, -6 \sqrt{2})$ in the new system are . . . . . .

By rotating the coordinate axes in the positive direction about the origin by an angle $\alpha$,if the point $(1,2)$ is transformed to $\left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}, \frac{\sqrt{3}+3}{2 \sqrt{2}}\right)$ in the new coordinate system,then $\alpha=$

If the origin of a coordinate system is shifted to $(-\sqrt{2}, \sqrt{2})$ and the coordinate system is rotated anti-clockwise through an angle $45^{\circ}$,then the point $P(1, -1)$ in the original system has new coordinates