Let $S=\left\{(x, y) \in \mathbb{R} imes \mathbb{R}: x \geq 0, y \geq 0, y^2 \leq 4 x, y^2 \leq 12-2 x ight.$ and $\left.3 y+\sqrt{8} x \leq 5 \sqrt{8} ight\}$. If the area of the region $S$ is $\alpha \sqrt{2}$, then $\alpha$ is equal to

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Let $S=\left\{(x, y) \in \mathbb{R} 	imes \mathbb{R}: x \geq 0, y \geq 0, y^2 \leq 4 x, y^2 \leq 12-2 xight.$ and $\left.3 y+\sqrt{8} x \leq 5 \sqrt{8}ight\}$. If the area of the region $S$ is $\alpha \sqrt{2}$, then $\alpha$ is equal to

Aicharya · Accepted Answer

Point of intersection of all curves is $(2,2 \sqrt{2})$ Area $=\mathrm{A}_1+\mathrm{A}_2$ $\alpha \sqrt{2}=\int_0^2 2 \sqrt{\mathrm{x}} \mathrm{dx}+\frac{1}{2} 	imes 3 	imes 2 \sqrt{2}$ $\begin{aligned} & \alpha \sqrt{2}=2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}ight]_0^2+3 \sqrt{2} \ & \alpha \sqrt{2}=\frac{17 \sqrt{2}}{3} \ & \alpha=\frac{17}{3}\end{aligned}$

Let $S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x \geq 0, y \geq 0, y^2 \leq 4 x, y^2 \leq 12-2…

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