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 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 $\frac{1}{a^2}+\frac{1}{b^2}=$

If the origin is shifted to the point $\left(\frac{3}{2},-2\right)$ by the translation of axes,then the transformed equation of $2x^2+4xy+y^2+2x-2y+1=0$ 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 $\alpha=$

The origin is shifted to the point $(2,3)$ by translation of axes and then the coordinate axes are rotated about the origin through an angle $\theta$ in the counter-clockwise sense. Due to this,if the equation $3x^2+2xy+3y^2-18x-22y+50=0$ is transformed to $4x^2+2y^2-1=0$,then the angle $\theta=$

$A$ vector $\vec{a}$ has components $3p$ and $1$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\vec{a}$ has components $p+1$ and $\sqrt{10}$,then a value of $p$ is equal to