The area of the region enclosed between the p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

The area bounded by the lines $y=x$,$x=-1$,$x=2$ and the $x$-axis is

Find the area of the region lying in the first quadrant and bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$.

If the area bounded by the curve $x^2y + y^2x = \alpha xy$ is $2$ units,then the possible value$(s)$ of $\alpha$ is/are:

Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

Find the area of the parabola $y^{2}=4ax$ bounded by its latus rectum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module