When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi }{4}$, if the equation $49x^2+25y^2=1225$ is transformed to $p{x}^2+qxy+r{y}^2=t$ and the G.C.D of $p,q,r,t$ is $1$, then

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

$A_1=3x^2+5xy+3y^2+2x+3y+4=0$
$A_2=5x^2+2\sqrt{3}xy+3y^2+6=0$

$A_3=4x^2+\sqrt{3}xy+5y^2-4=0$

Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree) terms in the equation $x^2+y^2-4x+6y-7=0$ are eliminated. Then the point (h, k) is

The transformed equation of $3{\text{x}}^2+3{\text{y}}^2+2\text{xy}=2$, when the coordinate axes are rotated through an angle of ${45}^{°}$ , is

Reflecting the point (2, -1) about y - axis, coordinate axes are rotated at ${45}^o$ angle in negative direction without shifting the origin. The new coordinates of the point are