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

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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:

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