Question

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

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (A), (D)

Equation of parabola is $y^2=x$ and line $y=mx$ For intersection point of both curves put $x=y^2$, we get
$y=m{y}^2$
$\Rightarrow y(my-1)=0$
$\Rightarrow y=0\text{ or }y=\frac{1}{m}$
Then, $x=0$ or $x=\frac{1}{m^2}$
$\therefore$ Intersection points are $(0,0)$ and $P\left(\frac{1}{m^2},\frac{1}{m}\right)$

$\therefore$ Required area $={\int }_0^{1/m}\left|\left(\frac{y}{m}-y^2\right)\right|dy=\left|{\left[\frac{y^2}{2m}-\frac{y^3}{3}\right]}_0^{1/m}\right|$
$=\left|\frac{1}{2m^3}-\frac{1}{3m^3}\right|=\left|\frac{1}{6m^3}\right|=\frac{1}{48}($ given $)$
$\Rightarrow \frac{1}{6m^3}=\pm \frac{1}{48}$
$\Rightarrow m^3=\pm 8$
Now, if $m^3=8$
$\Rightarrow m^3=(2{)}^3\Rightarrow m=2$
If $m^3=-8$
$\Rightarrow m^3=(-2{)}^3\Rightarrow m=-2$