The area enclosed between the parabolas ${y^2 = 4x}$ and ${x^2 = 4y}$ is

Let $T$ and $C$ respectively be the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$. Then the area of the region above the parabola $x^2 = y + 4$,below the transverse axis $T$ and on the right of the conjugate axis $C$ is:

If the area of the region $S=\{(x, y): 2y - y^2 \leq x^2 \leq 2y, x \geq y\}$ is equal to $\frac{n+2}{n+1} - \frac{\pi}{n-1}$,then the natural number $n$ is equal to $...............$.

The parabola $y^2=4x+1$ divides the disc $x^2+y^2 \leq 1$ into two regions with areas $A_1$ and $A_2$. Then,$|A_1-A_2|$ equals

The area (in sq. units) of the smaller region lying above the $X$-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is

If the area of the region bounded by the curves $y^2-2y=-x$ and $x+y=0$ is $A$,then $8A$ is equal to