Question

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

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (A), (D)

$-2$
$2$

Equation of parabola is $y^2=x$ and line $y=mx$
For intersection point of both curves put $x=y^2$, we get
$y=m{y}^2\Rightarrow$ $y(my-1)=0$
$\Rightarrow$ $y=0$ or $y=\frac{1}{m}$
Then, $x=0$ or $x=\frac{1}{m^2}$
$\therefore$ Intersection points are $(0,0)$ and $P(\frac{1}{m^2},\frac{1}{m})$
$\therefore$ Required area
$={\int }_0^{1/m}\left|(\frac{y}{m}-y^2)\right|dy=\left|{[\frac{y^2}{2m}-\frac{y^3}{3}]}_0^{1/m}\right|$
$=|\frac{1}{2m^3}-\frac{1}{3m^3}|=\left|\frac{1}{6m^3}\right|=\frac{1}{48}$
$\Rightarrow \frac{1}{6m^3}=\pm \frac{1}{48}\Rightarrow m^3=\pm 8$
Now if $m^3=8$
$m^3=8\Rightarrow m^3={\left(2\right)}^3\Rightarrow m=2$
If $m^3=-8$
$\Rightarrow m^3=-8$

$\Rightarrow m^3={-\left(2\right)}^3$

$\Rightarrow m=-2$