The point to which the origin should be shifted so that the equation $y^2-6y-4x+13=0$ is transformed to the form $y^2+Ax=0$ 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 $\theta_1, \theta_2, \theta_3$ 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:
$A_1 = 3x^2 + 5xy + 3y^2 + 2x + 3y + 4 = 0$
$A_2 = 5x^2 + 2\sqrt{3}xy + 3y^2 + 6 = 0$
$A_3 = 4x^2 + \sqrt{3}xy + 5y^2 - 4 = 0$

If the coordinates of a point $P$ change to $(2, -6)$ when the coordinate axes are rotated through an angle of $135^{\circ}$,then the coordinates of $P$ in the original system are

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 . . . . . .

If $a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of $4 x^2+\sqrt{3} x y+5 y^2-4=0$ obtained by using $\alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2}$ and $\beta=-\frac{x}{2}+\frac{\sqrt{3}}{2} y$,then $c(a+b+d)=$