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

The Correct Option is D

Equation of parabola is $y^{2}=x$ and line $y=m x$ For intersection point of both curves put $x=y^{2}$, we get
$ y =m y^{2}$
$ \Rightarrow y(m y-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 $=\int_\limits{0}^{1 / m}\left|\left(\frac{y}{m}-y^{2}\right)\right| d y=\left|\left[\frac{y^{2}}{2 m}-\frac{y^{3}}{3}\right]_{0}^{1 / m}\right|$
$=\left|\frac{1}{2 m^{3}}-\frac{1}{3 m^{3}}\right|=\left|\frac{1}{6 m^{3}}\right|=\frac{1}{48}$(given)
$\Rightarrow \frac{1}{6 m^{3}}=\pm \frac{1}{48}$
$ \Rightarrow m^{3}=\pm 8$
Now, if $m^{3}=8$
$\Rightarrow m^{3}=(2)^{3} \Rightarrow m=2$
If $\quad m^{3}=-8$
$\Rightarrow m^{3}=(-2)^{3} \Rightarrow m=-2$