Area Question 17
Question 17
- The area (in sq units) of region described by $(x, y) y^{2} \leq 2 x$ and $y \geq 4 x-1$ is
(2015 JEE Main) (a) $\frac{7}{32}$ (b) $\frac{5}{64}$ (c) $\frac{15}{64}$ (d) $\frac{9}{32}$
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Solution:
- Given region is $\left{(x, y): y^{2} \leq 2 x\right.$ and $\left.y \geq 4 x-1\right}$
$y^{2} \leq 2 x$ repressents a region inside the parabola
$$ y^{2}=2 x $$
and $y \geq 4 x-1$ represents a region to the left of the line
$$ y=4 x-1 $$
The point of intersection of the curves (i) and (ii) is
$$ \begin{aligned} & (4 x-1)^{2}=2 x \Rightarrow 16 x^{2}+1-8 x=2 x \ \Rightarrow \quad & 16 x^{2}-10 x+11=0 \Rightarrow \quad x=\frac{1}{2}, \frac{1}{8} \end{aligned} $$
So, the points where these curves intersect are $\frac{1}{2}, 1$ and $\frac{1}{8}, \frac{1}{2}$.
$\therefore$ Required area $=\int_{-1 / 2}^{1} \frac{y+1}{4}-\frac{y^{2}}{2} d y$
$$ \begin{aligned} & =\frac{1}{4} \frac{y^{2}}{2}+y{ }{-1 / 2}^{-1}-\frac{1}{6}\left(y^{3}\right){-1 / 2}^{1} \ & =\frac{1}{4} \frac{1}{2}+1-\frac{1}{8}-\frac{1}{2}-\frac{1}{6} 1+\frac{1}{8} \ & =\frac{1}{4} \frac{3}{2}+\frac{3}{8}-\frac{1}{6} \frac{9}{8} \ & =\frac{1}{4} \times \frac{15}{8}-\frac{3}{16}=\frac{9}{32} \text { sq units } \end{aligned} $$
