Through any point $(x, y)$ of a curve which passes through the origin, lines are drawn parallel to the co-ordiante axes. The curve, given that it divides the rectangle 

formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

Let $P(x, y)$ be the point on the curve passing 

through the origin $O(0, 0),$ and let $PN$ and $PM$ be the lines

 parallel to the $x$- and $y$-axes, respectively . If the

 equation of the curve is $y = y\left(x\right),$ the area

$\text{ POM }={\int }_0^x y\mathrm{d}x$ and the area $\text{ PON }=xy-{\int }_0^x y\mathrm{d}x$ Assuming

that $\left(POM\right)=PON$, we therefore have 

$2{\int }_0^x y dx=xy-{\int }_0^x y dx\Rightarrow 3{\int }_0^x y dx=xy.$

Differentiating both sides of this gives

$3y=x \frac{\mathrm{d}y}{\mathrm{d}x}+y\Rightarrow 2y=x \frac{\mathrm{d}y}{\mathrm{d}x}\Rightarrow \frac{\mathrm{d}y}{y}=2 \frac{\mathrm{d}x}{x}$

$\Rightarrow \log \left|y\right|=2 \log \left|x\right|+C$

$\Rightarrow y=C{x}^2$ with $C$being a constant.

This solution represents a parabola. We will get a similar

 result if we had started instead with $2\left(PON\right)=POM$