When the origin is shifted to $(2, 3)$,the transformed equation of a curve becomes $x^2+3xy-2y^2+17x-7y-11=0$. Find the original equation of the curve.

The transformed equation of $x^2-y^2+2x+4y=0$ when the origin is shifted to the point $(-1, 2)$ is

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

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 axes about the origin in an anti-clockwise direction by a certain angle,if the equation $x^2+4xy+y^2=1$ is transformed to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,then find the value of $\sqrt{\frac{a^2+b^2}{a^2}}$.

The coordinate axes are rotated about the origin in the counter-clockwise direction through an angle $60^{\circ}$. If $a$ and $b$ are the intercepts made on the new axes by a straight line whose equation referred to the original axes is $x+y=1$,then $\frac{1}{a^2}+\frac{1}{b^2}=$