Let a curve y = y x pass through the point 3 , 3 and the area of the region under this curve, above the x…

Given, the area of the region under this curve, above the $x$-axis and between the abscissae $3$ and $x\left(>3\right)$ be ${\left(\frac{y}{x}\right)}^3$

So, ${\int }_3^{x}ydx={\left(\frac{y}{x}\right)}^3$

$\Rightarrow x^3{\int }_3^{x}ydx=y^3$

Now differentiating both side w.r.t $x$, we get

$\Rightarrow x^{3}y+3x^2\frac{y^3}{x^3}=3y^2\frac{dy}{dx}$ $\left\{\text{as given }{\int }_3^{x}ydx={\left(\frac{y}{x}\right)}^3\right\}$

$\Rightarrow x^4+3y^2=3yx\frac{dy}{dx}$

$\Rightarrow 3xy\frac{dy}{dx}=3y^2+x^4$

Now put $y^2=t,y\frac{dy}{dx}=\frac{1}{2}\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}-\frac{2}{x}t=\frac{2}{3}{x}^3$

Now $IF=e^{\int -\frac{2}{x}dx}=\frac{1}{x^2}$

So, solution is given by 

$t\times IF=\int \frac{2}{3}{x}^3\times IF$

$\Rightarrow \frac{t}{x^2}=\frac{x^2}{3}+C$

$$Now given curve passes through  $\left(3,3\right)$ so $C=-2$

$\Rightarrow \frac{y^2}{x^2}=\frac{x^2}{3}-2$

So the equation of curve is $3y^2=x^4-6x^2$

Now curve also passes through $\left(\alpha ,6\sqrt{10}\right)$

So, ${\alpha }^4-6{\alpha }^2=1080$

$\Rightarrow \alpha =6$