Let $(a, b)$ be a fixed point, and $(x, y)$ a variable point, on the curve $y = f(x)$, $(x \geq a, f ′ (x) \geq 0).$ The curve divides the rectangle with vertices $(a, b),(a, y),(x, y)$ and ($x, b$) into two portions, the lower of which has always half the area of the upper. Show that the curve is an arc of a parabola with its vertex at $(a, b)$.
I have tried fixing ($a,b$) as ($0,0$) and finding the directrix and focus of the parabola but I can't seem to justify how the information about the area implies that f(x) is a parabola in the first place.
Wlog $a=b=0$. Then, your constraint on areas reads $\int_0^xf=\frac13xf(x)$ so by differentiation $2f(x)=xf'(x)$, hence $f(x)= Cx^2$.