If the area bounded by the curve x^2y + y^2x | vedclass

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

Question

(A) The given equation is $x^2y + y^2x = \alpha xy$. Assuming $x 
eq 0$ and $y 
eq 0$,we can divide by $xy$ to get $x + y = \alpha$. The curve repre

Vedclass · Accepted Answer

$\pm 2$

Vedclass · Answer

(A) The given equation is $x^2y + y^2x = \alpha xy$. Assuming $x 
eq 0$ and $y 
eq 0$,we can divide by $xy$ to get $x + y = \alpha$. The curve represents a triangle formed by the lines $x = 0$,$y = 0$,and $x + y = \alpha$. The vertices of this triangle are $(0, 0)$,$(\alpha, 0)$,and $(0, \alpha)$. The area of a right-angled triangle with base $|\alpha|$ and height $|\alpha|$ is given by $\frac{1}{2} 	imes |\alpha| 	imes |\alpha| = \frac{\alpha^2}{2}$. Given that the area is $2$ units,we have $\frac{\alpha^2}{2} = 2$,which implies $\alpha^2 = 4$. Therefore,$\alpha = \pm 2$.

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:

Similar Questions

The area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and the line $x = 2a$ above the $x$-axis is

If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

The area of the region (in square units) bounded by the curve ${x^2} = 4y$,the line $x = 2$,and the $x$-axis is:

The area of the region bounded by the curve $y = -2\sqrt{x}$ and the lines $x = 0$,$x = 1$,and $y = 0$ is . . . . . . sq. units.

The area of the region bounded by the curve $y = \cos x$,the $X$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module