Let $P$ be the point to which the origin is shifted by the translation of axes so as to remove the first-degree terms from the equation $3x^2+y^2-6x+4y+4=0$. If the origin is shifted to $P$ by the translation of axes,then the transformed equation of $2x^2+3xy-5y^2+2x-23y-24=0$ is

By shifting the origin to the point $(-1, 2)$ through translation of axes,if $ax^2+2hxy+by^2+2gx+2fy+c=0$ is the transformed equation of $2x^2-xy+y^2-3x+4y-5=0$,then $2(f+g+h)=$

After the coordinate axes are rotated through an angle $\frac{\pi}{4}$ in the anti-clockwise direction without shifting the origin,if the equation $x^2+y^2-2x-4y-20=0$ transforms to $ax^2+2hxy+by^2+2gx+2fy+c=0$ in the new coordinate system,then $\left|\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right|=$

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:

When the origin is shifted to the point $(h, k)$ by translating the coordinate axes,the equation $S = 2x^2 - xy + y^2 + 2x + 3y + 1 = 0$ is changed to $S' = ax^2 + 2hxy + by^2 + C' = 0$. If the coordinate axes are then rotated about the new origin through an angle $\theta$ in the positive direction to eliminate the $xy$ term,the equation $S' = 0$ becomes $Ax^2 + By^2 + C = 0$. Find the value of $h + k + \tan 2\theta$.

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