The area bounded by the curve $y^{2} = 4 x$ and the line $2 x - 3 y + 4 = 0$ is

\frac{1}{3}

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\frac{2}{3}

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\frac{4}{3}

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\frac{5}{3}

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Solution

The correct option is B $\frac{1}{3}$
$y^{2} = 4 x \Rightarrow x = \frac{y^{2}}{4}$

$2 x - 3 y + 4 = 0 \Rightarrow x = \frac{3}{2} y - 2$

$\frac{y^{2}}{4} = \frac{3}{2} y - 2$

$y^{2} = 6 y - 8$

$y^{2} - 6 y + 8 = 0$

$(y - 4) (y - 2) = 0$

$y_{1} = 2, y_{2} = 4$

$A = \int_{2}^{4} (\frac{3}{2} y - 2) - (\frac{y^{2}}{4}) d y$

$= {[\frac{3 y^{2}}{4} - 2 y - \frac{y^{3}}{12}]}_{2}^{4}$

$= \frac{36 - 24 - 16 - 9 + 12 + 2}{3}$

$= \frac{1}{3}$ sq. units

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Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is

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