Question 2

The region $S$ is in the first quadrant. The curves are $P_1: y^2 = 4x$ and $P_2: y^2 = 12 - 2x$. They intersect where $4x = 12 - 2x \Rightarrow 6x = 12 \Rightarrow x = 2, y = 2\sqrt{2}$.

The line is $3y + 2\sqrt{2}x = 10\sqrt{2}$. At $x=2$, $3(2\sqrt{2}) + 2\sqrt{2}(2) = 10\sqrt{2}$, so the line also passes through $(2, 2\sqrt{2})$.

For $y \in [0, 2\sqrt{2}]$, the region is bounded on the left by $x = y^2/4$ and on the right by the line $x = 5 - \frac{3y}{2\sqrt{2}}$ (since for $y < 2\sqrt{2}$, the line is more restrictive than the curve $x = 6 - y^2/2$).

Area $= \int_{0}^{2\sqrt{2}} (x_{right} - x_{left}) dy = \int_{0}^{2\sqrt{2}} \left( 5 - \frac{3y}{2\sqrt{2}} - \frac{y^2}{4} \right) dy$

$Area = \left[ 5y - \frac{3y^2}{4\sqrt{2}} - \frac{y^3}{12} \right]_{0}^{2\sqrt{2}}$

$Area = 5(2\sqrt{2}) - \frac{3(8)}{4\sqrt{2}} - \frac{16\sqrt{2}}{12} = 10\sqrt{2} - 3\sqrt{2} - \frac{4\sqrt{2}}{3} = 7\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{17\sqrt{2}}{3}$

Comparing with $\alpha \sqrt{2}$, we get $\alpha = 17/3$.