The shaded region in the following figure is the solution set of the inequations

The diagonals of a parallelogram $A B C D$ are along the lines $x+3 y=4$ and $6 x-2 y=7$. Then ABCD must be a

If $x=\sec \theta-\cos \theta, y=\sec ^{10} \theta-\cos ^{10} \theta$ and $\left(x^2+4\right)\left(\frac{d y}{d x}\right)^2=k\left(y^2+4\right)$, then the value of $k$ is

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+\mathrm{e}^x}{2+y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x=0$ satisfying $y(0)=1$, then a value of $y(\log 13)$ is