The area of the region bounded by the curve x 2 = 4 y and the straight line x = 4 y - 2 is:

Question

The area of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is:

Smart Achievers · Accepted Answer

(4)

Ans. $\frac{9}{8}$ sq. units

Explanation:

Given equation of curve is $x^{2} = 4 y$ and the straight line is $x = 4 y - 2 .$

For the point of intersection, put $x = 4 y - 2$ in the equation of the curve:

${(4 y - 2)}^{2} = 4 y$

$\Rightarrow$ $16 y^{2} + 4 - 16 y = 4 y$

$\Rightarrow$ $16 y^{2} - 20 y + 4 = 0$

$\Rightarrow$ $4 y^{2} - 5 y + 1 = 0$

$\Rightarrow$ $4 y^{2} - 4 y - y + 1 = 0$

$\Rightarrow$ $4 y (y - 1) - 1 (y - 1) = 0$

$\Rightarrow$ $(4 y - 1) (y - 1) = 0$

$∴$ $y = 1, \frac{1}{4}$

For $y = 1$ , $x = \sqrt{4 \cdot 1} = 2$

For $y = \frac{1}{4}$ , $x = \sqrt{4 \cdot \frac{1}{4}} = - 1$

So, the intersection points are $(2, 1)$ and $(- 1, \frac{1}{4})$

Hence, the area of the shaded region is

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

$= \frac{1}{4} {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{2} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 1}^{2}$

$= \frac{1}{4} [\frac{4}{2} + 4 - \frac{1}{2} + 2] - \frac{1}{4} [\frac{8}{3} + \frac{1}{3}]$

$= \frac{1}{4} \cdot \frac{15}{2} - \frac{1}{4} \cdot \frac{9}{3} = \frac{45 - 18}{24}$

$= \frac{27}{24} = \frac{9}{8} sq. units.$

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