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:

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

The area bounded by the parabolas $y=4x^2$, $y=\frac{x^2}{9}$ and the line $y=2$ is

The following figure shows the graph of a continuous function $y=f(x)$ on the interval $[1,3]$. The points $A, B, C$ have coordinates $(1,1), (3,2), (2,3)$ respectively,and the lines $l_1$ and $l_2$ are parallel,with $l_1$ being tangent to the curve at $C$. If the area under the graph of $y=f(x)$ from $x=1$ to $x=3$ is $4$ sq units,then the area of the shaded region is

If the area of the region ${(x, y) : -2x + 1 \le y \le 4 - x^2, x \ge 0, y \ge 0}$ is $\frac{\alpha}{\beta}$, where $\alpha, \beta \in N$ and $\gcd(\alpha, \beta) = 1$, then the value of $(\alpha + \beta)$ is:

The area (in square units) of the region enclosed by the ellipse $x^2+3y^2=18$ in the first quadrant below the line $y=x$ is: