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

The coordinates of the point $(3,-7,5)$ in the new system,when the origin is shifted to the point $(-1,-1,-1)$ by the translation of axes,are

If the equation $3x^2 + 4y^2 - xy + k = 0$ is the transformed equation of $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$ after shifting the origin to the point $(\alpha, \beta)$ by the translation of axes,then $\alpha + \beta - k =$

$(a, b)$ is the point to which the origin has to be shifted by translation of axes so as to remove the first-degree terms from the equation $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. If the angle by which the axes are to be rotated in the positive direction about the origin to remove the $xy$-term from the equation $ax^2 + 23abxy + by^2 = 0$ is $\theta$,then $\tan 2\theta =$

If $(a, b)$ are the new coordinates of the point $(2, 3)$ after shifting the origin to the point $(3, 2)$ by translation of axes,and $(c, d)$ are the new coordinates of the point $(a, b)$ after rotating the axes through an angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction,then find the value of $d-c$.

If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$,then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ is: