Let a curve $y=y(x)$ pass through the point $(3,3)$ and the area of the region under this curve, above the $x$-axis and between the abscissae 3 and $x(>3)$ be $\left(\frac{y}{x}\right)^{3}$. If this curve also passes through the point $(\alpha, 6 \sqrt{10})$ in the first quadrant, then $\alpha$ is equal to ___________.

<p>$\int\limits_3^x {f(x)dx = {{\left( {{{f(x)} \over x}} \right)}^3}}$</p> <p>${x^3}\,.\,\int\limits_3^x {f(x)dx = {f^3}(x)}$</p> <p>Differentiate w.r.t. x</p> <p>${x^3}f(x) + 3{x^2}\,.\,{{{f^3}(x)} \over {{x^3}}} = 3{f^2}(x)f'(x)$</p> <p>$\Rightarrow 3{y^2}{{dy} \over {dx}} = {x^3}y + {{3{y^3}} \over x}$</p> <p>$3xy{{dy} \over {dx}} = {x^4} + 3{y^2}$</p> <p>Let ${y^2} = t$</p> <p>${3 \over 2}{{dt} \over {dx}} = {x^3} + {{3t} \over x}$</p> <p>${{dt} \over {dx}} - {{2t} \over x} = {{2{x^3}} \over 3}$</p> <p>$I.F. = \,.\,{e^{\int { - {2 \over x}dx} }} = {1 \over {{x^2}}}$</p> <p>Solution of differential equation</p> <p>$t\,.\,{1 \over {{x^2}}} = \int {{2 \over 3}x\,dx}$</p> <p>${{{y^2}} \over {{x^2}}} = {{{x^2}} \over 3} + C$</p> <p>${y^2} = {{{x^4}} \over 3} + C{x^2}$</p> <p>Curve passes through $(3,3) \Rightarrow C = - 2$</p> <p>${y^2} = {{{x^4}} \over 3} - 2{x^2}$</p> <p>Which passes through $\left( {\alpha ,6\sqrt {10} } \right)$</p> <p>${{{\alpha ^4} - 6{\alpha ^2}} \over 3} = 360$</p> <p>${\alpha ^4} - 6{\alpha ^2} - 1080 = 0$</p> <p>$\alpha = 6$</p>