The angle through which the coordinate axes are to be rotated to remove the $xy$ term in the equation $x^2+2xy-y^2=0$ is

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 $px^2+qxy+ry^2=t$ and the $G.C.D$ of $p, q, r, t$ is $1$,then:

When the coordinate axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction,the equation $ax^2+2hxy+by^2=c$ is transformed to $25x^2+9y^2=225$,then $(a+2h+b-\sqrt{c})^2=$

Find the transformed equation of $x \cos \theta + y \sin \theta = p$,when the axes are rotated through an angle $\theta$.

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:

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 $\alpha=$