If the area between y=m x^{2} and x=m y^{2} ( | vedclass

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

Question

(NONE) The given curves are $y=m x^{2}$ and $x=m y^{2}$ where $m>0$.
To find the intersection points,substitute $y=m x^{2}$ into $x=m y^{2}$:
$x=

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(NONE) The given curves are $y=m x^{2}$ and $x=m y^{2}$ where $m>0$.
To find the intersection points,substitute $y=m x^{2}$ into $x=m y^{2}$:
$x=m(m x^{2})^{2} = m^{3} x^{4}$
$m^{3} x^{4}-x=0 \Rightarrow x(m^{3} x^{3}-1)=0$
The intersection points are $x=0$ and $x=1/m$.
The area $A$ between the curves is given by:
$A = \int_{0}^{1/m} (\sqrt{x/m} - m x^{2}) dx$
$A = [\frac{1}{\sqrt{m}} \cdot \frac{x^{3/2}}{3/2} - m \cdot \frac{x^{3}}{3}]_{0}^{1/m}$
$A = [\frac{2}{3 \sqrt{m}} \cdot (1/m)^{3/2} - \frac{m}{3} \cdot (1/m)^{3}]$
$A = \frac{2}{3 m^{2}} - \frac{1}{3 m^{2}} = \frac{1}{3 m^{2}}$
Given $A = 1/4$,so $\frac{1}{3 m^{2}} = \frac{1}{4}$
$3 m^{2} = 4 \Rightarrow m^{2} = 4/3$
Since $m>0$,$m = \frac{2}{\sqrt{3}}$.
Note: None of the provided options match the calculated result. The correct value is $m = 2/\sqrt{3}$.

If the area between $y=m x^{2}$ and $x=m y^{2}$ $(m>0)$ is $1/4$ sq units,then the value of $m$ is

Similar Questions

The polynomial $f(x)$ satisfies the condition $f(x + 1) = x^2 + 4x$. The area enclosed by $y = f(x - 1)$ and the curve $x^2 + y = 0$ is

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant,is

The area bounded by the curve $y^2=2x+1$ and the line $x-y=1$ is

The positive values of the parameter $a$ for which the area of the figure bounded by the curve $y = \cos ax$,$y = 0$,$x = \frac{\pi}{6a}$,and $x = \frac{5\pi}{6a}$ is greater than $3$ are:

The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^2 - 4x$ and $y = 2x - x^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module