If the equations $x=t^2+t+1$ and $y=t^2-t+1$ represent a curve $C$ with parameter $t$,then the Cartesian equation of $C$ is

Consider the lines $L_1$ and $L_2$ defined by $L_1: x \sqrt{2} + y - 1 = 0$ and $L_2: x \sqrt{2} - y + 1 = 0$. For a fixed constant $\lambda$,let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is $\lambda^2$. The line $y = 2x + 1$ meets $C$ at two points $R$ and $S$,where the distance between $R$ and $S$ is $\sqrt{270}$. Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R^{\prime}$ and $S^{\prime}$. Let $D$ be the square of the distance between $R^{\prime}$ and $S^{\prime}$.
$(1)$ The value of $\lambda^2$ is
$(2)$ The value of $D$ is

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

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:

The lines $y = -\frac{3}{2} x$ and $y = -\frac{2}{5} x$ intersect the curve $3x^2 + 4xy + 5y^2 - 4 = 0$ at the points $P$ and $Q$ respectively. The tangents drawn to the curve at $P$ and $Q$:

$A$ chord through the point $(1,-2)$ cuts the curve $3x^2-y^2-2x+4y=0$ at $P$ and $Q$. If $PQ$ subtends an angle $\theta$ at the origin,then $\theta$ equals (in $^{\circ}$)