If the point $(2, -3)$ lies on the curve $kx^2 - 3y^2 + 2x + y - 2 = 0$,then $k$ is equal to:

The point to which the origin is to be shifted to remove the first degree terms from the equation $2x^2+4xy-6y^2+2x+8y+1=0$ is

The equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$,where $k > 1$,represents:

To remove the $xy$ term from the second-degree equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$,the coordinate axes are rotated through an angle $\theta$. Then $\theta$ equals:

In order to eliminate the first degree terms from the equation $2x^2+4xy+5y^2-4x-22y+7=0$,the point to which the origin is to be shifted is:

The equation $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ for $r > 1$ represents: