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The area enclosed by the curves $y^{2} + 4 x = 4$ and $y - 2 x = 2$ is

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22/3

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answer is C.

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Detailed Solution

Given curves are y² + 4x = 4 and y - 2x = 2
Curves are Intersecting
$x = \frac{4 - y^{2}}{4} = \frac{y - 2}{2} \Rightarrow \frac{y^{2}}{4} + \frac{y}{2} - 2$
Area $= \frac{Δ^{3 / 2}}{6 a^{2}} = 9$

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