Let a curve y = y(x) pass through the point (3, 3) and the area of...

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

So, ${∫}_{3}^{x}y d x = {\left(\dfrac{y}{x}\right)}^{3}$

$⇒ {x}^{3}{∫}_{3}^{x}y d x = {y}^{3}$

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

$⇒ {x}^{3}y + 3{x}^{2}\dfrac{{y}^{3}}{{x}^{3}} = 3{y}^{2}\dfrac{dy}{dx}$ \left\{\text{as given}{∫}_{3}^{x}y d x = {\left(\dfrac{y}{x}\right)}^{3}\right\}$</p><p>$⇒ {x}^{4} + 3{y}^{2} = 3yx\dfrac{dy}{dx}$</p><p>$⇒ 3xy\dfrac{dy}{dx} = 3{y}^{2} + {x}^{4}$</p><p>Now put${y}^{2} = t, y\dfrac{dy}{dx} = \dfrac{1}{2}\dfrac{dt}{dx}$</p><p>$⇒ \dfrac{dt}{dx} - \dfrac{2}{x}t = \dfrac{2}{3}{x}^{3}$</p><p>Now$IF = {e}^{∫ - \dfrac{2}{x}dx} = \dfrac{1}{{x}^{2}}$</p><p>So, solution is given by </p><p>$t \times IF = ∫ \dfrac{2}{3}{x}^{3} \times IF$</p><p>$⇒ \dfrac{t}{{x}^{2}} = \dfrac{{x}^{2}}{3} + C$</p><p>Now given curve passes through \left(3, 3\right)$ so $C = - 2$

$⇒ \dfrac{{y}^{2}}{{x}^{2}} = \dfrac{{x}^{2}}{3} - 2$

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

Now curve also passes through \left(\alpha, 6\sqrt{10}\right)$</p><p>So,${\alpha}^{4} - 6{\alpha}^{2} = 1080$</p><p>$⇒ \alpha = 6