If the slopes of the lines represented by the equation $6 x^2+2 h x y+4 y^2=0$ are in the ratio $2: 3$, then the value of $h$ such that both the lines make acute angles with the positive $X$-axis measured in positive direction is

If $2 x^2+x y-6 y^2+k=0$ is the transformed equation of $2 x^2+x y-6 y^2-13 x+9 y+15=0$ when the origin is shifted to the point $(a, b)$ by translation of axes, then $k=$

The line $L \equiv 6 x+3 y+k=0$ divides the line segment joining the points $(3,5)$ and $(4,6)$ in the ratio $-5: 4$. If the point of intersection of the lines $L=0$ and $x-y+1=0$ is $P(g, h)$, then $h=$

A straight line through the point $P(1,2)$ makes an angle $\theta$ with positive X -axis in anticlockwise direction and meets the line $x+\sqrt{3 y}-2 \sqrt{3}=0$ at $Q$. If $P Q=\frac{1}{2}$, then $\theta=$