The point to which the origin should be shifted in order to eliminate the $x$ and $y$ terms from the equation $9x^2+4y^2+10x+12y+1=0$ is

The equation of a curve $C$ is transformed to $X^2+Y^2-6X+8Y+21=0$ by the rotation of coordinate axes about the origin through an angle of $\frac{\pi}{4}$ in the positive direction. If $ax^2+by^2+cx+dy+e=0$ is the equation of the curve $C$ before the transformation,then find the value of $(a+b+c^2+d^2-5e)^2$.

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 coordinate axes are rotated through an angle $\frac{\pi}{4}$ in the positive direction,an equation is transformed to $x^2+y^2-6x+8y+21=0$. Then the original equation is

Transforming to parallel axes through a point $(p, q)$, the equation $2x^2 + 3xy + 4y^2 + x + 18y + 25 = 0$ becomes $2x^2 + 3xy + 4y^2 = 1$. Then:

If the equation $3x^2 + 4y^2 - xy + k = 0$ is the transformed equation of $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$ after shifting the origin to the point $(\alpha, \beta)$ by the translation of axes,then $\alpha + \beta - k =$