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(A, D) The equation of the parabola is $y^{2}=x$ and the line is $y=mx$.To find the intersection points, substitute $x=y^{2}$ into the line equation:$

Vedclass · Accepted Answer

$2$

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(A, D) The equation of the parabola is $y^{2}=x$ and the line is $y=mx$.To find the intersection points, substitute $x=y^{2}$ into the line equation:$y=m(y^{2}) \Rightarrow my^{2}-y=0 \Rightarrow y(my-1)=0$.Thus, $y=0$ or $y=\frac{1}{m}$.For $y=0$, $x=0$. For $y=\frac{1}{m}$, $x=\frac{1}{m^{2}}$.So, the intersection points are $(0,0)$ and $P\left(\frac{1}{m^{2}}, \frac{1}{m}ight)$.The required area is given by the integral:$	ext{Area} = \int_{0}^{1/m} \left(\frac{y}{m} - y^{2}ight) dy = \left[\frac{y^{2}}{2m} - \frac{y^{3}}{3}ight]_{0}^{1/m} = \left|\frac{1}{2m^{3}} - \frac{1}{3m^{3}}ight| = \left|\frac{1}{6m^{3}}ight|$.Given that the area is $\frac{1}{48}$, we have:$\left|\frac{1}{6m^{3}}ight| = \frac{1}{48} \Rightarrow |m^{3}| = 8$.This implies $m^{3} = 8$ or $m^{3} = -8$.Therefore, $m = 2$ or $m = -2$.

The area of the region enclosed between the parabola $y^{2}=x$ and the line $y=mx$ is $\frac{1}{48}$. Then, the value of $m$ is

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