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:

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 $\theta=$

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 $h + k + \tan 2\theta$.

Point $(-1, 2)$ is changed to $(a, b)$ when the origin is shifted to the point $(2, -1)$ by translation of axes. Point $(a, b)$ is changed to $(c, d)$ when the axes are rotated through an angle of $45^{\circ}$ about the new origin. Point $(c, d)$ is changed to $(e, f)$ when $(c, d)$ is reflected through the line $y = x$. Then $(e, f) =$

If the axes are rotated through an angle $45^{\circ}$ in the positive direction without changing the origin,then the coordinates of the point $(\sqrt{2}, 4)$ in the old system are

Suppose the axes are to be rotated through an angle $\theta$ so as to remove the $xy$ term from the equation $3x^2+2\sqrt{3}xy+y^2=0$. Then in the new coordinate system,the equation $x^2+y^2+2xy=2$ is transformed to: