Question

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 ______

Answer 1

$ x^4=3 y x \cdot y^{\prime}-3 y^2$
$ \Rightarrow 3 x y \frac{d y}{d x}=3 y^2+x^4 $
Put $ y^2=t, y \frac{d y}{d x}=\frac{1}{2} \frac{d t}{d x} $
$ \frac{d t}{d x}-\frac{2}{x} t=\frac{2}{3} x^3$
$ \therefore \frac{t}{x^2}=\frac{x^2}{3}+C $
$ \Rightarrow \frac{y^2}{x^2}=\frac{x^2}{3}-2$
Put $(3,3), C=-2$
$ \therefore \frac{y^2}{x^2}=\frac{x^2}{3}-2$
$3 y^2=x^4-6 x^2 $
$ x^4-6 x^2=1080$
$ \therefore x=6$