The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

Let $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_{1}$, then $\alpha$ is equal to:

Let $R$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0\}$ is

The area bounded by the parabola $y^2 = 4x$ and the line $2x - 3y + 4 = 0$,in square units,is

If the area (in $sq. units$) of the region $\{(x,y): y^2 \le 4x, x + y \le 1, x \ge 0, y \ge 0\}$ is $a\sqrt{2} + b$,then $a - b$ is equal to

The area (in sq units) bounded by the curves $y^2=4x$ and $x^2=4y$ is