Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ by an angle $\theta$. If the equation $x^2+2 \sqrt{3} x y-y^2=4 a^2$ is transformed to $X^2-Y^2=2 a^2$ with respect to the $XY$-axes,then $\theta$ is equal to (in $^{\circ}$)

If the origin is shifted to remove the first degree terms from the equation $2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0$,then with respect to this new coordinate system,the transformed equation of $x^2 + y^2 - 3xy + 4y + 3 = 0$ is

By shifting the origin to the point $(2, 3)$ and then rotating the coordinate axes through an angle $\theta$ in the counter-clockwise direction,if the equation $3x^2 + 2xy + 3y^2 - 18x - 22y + 50 = 0$ is transformed to $4X^2 + 2Y^2 - 1 = 0$,then the angle $\theta =$

The origin is translated to $(1,2)$. The point $(7,5)$ in the old system undergoes the following transformations successively.
$I$. Moves to the new point under the given translation of origin.
$II$. Translated through $2$ units along the negative direction of the new $X$-axis.
$III$. Rotated through an angle $\frac{\pi}{4}$ about the origin of the new system in the clockwise direction. The final position of the point $(7,5)$ is

If the point $P(1,3)$ undergoes the following transformations successively:
$(i)$ Reflection with respect to the line $y=x$.
(ii) Translation through $3$ units along the positive direction of the $X$-axis.
(iii) Rotation through an angle of $\frac{\pi}{6}$ about the origin in the clockwise direction.
Then,the final position of the point $P$ is

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: