Let the area of the region $\{(x,y) : x - 2y + 4 \geq 0\}$, $\{x + 2y^{2} \geq 0, x + 4y^{2} \leq 8, y \geq 0\}$ be $\frac{m}{n}$, where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to ______.

Question

Aicharya · Accepted Answer

Given, $ x - 2y + 4 \geq 0, \quad x + 2y^2 \geq 0, \quad x + 4y^2 \leq 8, \quad y \geq 0 $ Now, finding the point of intersection of $x - 2y + 4 = 0$ and $x + 2y^2 = 0$, we get, $ (x, y) \equiv (-2, 1) $ And point of intersection of $x - 2y + 4 = 0$ and $x + 4y^2 = 8$ will be, $ (x, y) \equiv (-1, \frac{3}{2}) $

Now, plotting the diagramw we get,

Now, from the above diagram, the required area will be,

$A = \int_{- 2}^{- 1} (\frac{x + 4}{2} - \sqrt{\frac{- x}{2}}) + \int_{- 1}^{0} (\frac{\sqrt{8 - x}}{2} - \sqrt{\frac{- x}{2}}) d x + \int_{0}^{8} \frac{\sqrt{8 - x}}{2} d x$

$\Rightarrow A = \frac{1}{2} {[\frac{{(x + 4)}^{2}}{2}]}_{- 2}^{- 1} - \frac{1}{\sqrt{2}} {[{(- x)}^{\frac{3}{2}} (\frac{- 2}{3})]}_{- 2}^{- 1} + \frac{1}{2} {[{(8 - x)}^{\frac{3}{2}} (\frac{- 2}{3})]}_{- 1}^{0} - \frac{1}{\sqrt{2}} {[{(- x)}^{\frac{3}{2}} (\frac{- 2}{3})]}_{- 1}^{0} + \frac{1}{2} {[{(8 - x)}^{\frac{3}{2}} (\frac{- 2}{3})]}_{0}^{8}$

$\Rightarrow A = 9 + \frac{5}{4} - \frac{4}{3}$

$\Rightarrow A = \frac{107}{12}$

$\Rightarrow m + n = 119$

Let the area of the region $\{(x,y) : x - 2y + 4 \geq 0\}$, $\{x + 2y^{2} \geq 0, x + 4y^{2} \leq 8, y \geq…

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