If the distance from a variable point $P$ to the point $(4,3)$ is equal to the perpendicular distance from $P$ to the line $x+2 y-1=0$, then the equation of the locus of the point $P$ is

$(0, k)$ is the point to which the origin is to be shifted by the translation of the axes so as to remove the first degree terms from the equation $a x^2-2 x y+b y^2-2 x+4 y+1=0$ and $\frac{1}{2} \tan ^{-1}(2)$ is the angle through which the coordinate axes are to be rotated about the origin to remove the $x y$-term from the given equation, then $a+b=$

$\beta$ is the angle made by the perpendicular drawn from origin to the line $L \equiv x+y-2=0$ with the positive $X$-axis in the anticlockwise direction. If $a$ is the $X$-intercept of the line $L=0$ and $p$ is the perpendicular distance from the origin to the line $L=0$, then $a \tan \beta+p^2=$

The line $2 x+y-3=0$ divides the line segment joining the points $A(1,2)$ and $B(-2,1)$ in the ratio $a: b$ at the point $C$. If the point $C$ divides the line segment joining the points $P\left(\frac{b}{3 a},-3\right)$ and $Q\left(-3,-\frac{b}{3 a}\right)$ in the ratio $p: q$, then $\frac{p}{q}+\frac{q}{p}=$