Through a point (x, y) on the curve, the lines parallel to the axes form a rectangle of area xy with the axes. The curve splits it into two parts, and the part under the curve has area \(\int_0^x y\,dx\).

Suppose that part is one-third of the rectangle: \(\int_0^x y\,dx = \frac{1}{3}xy\).

Differentiate with respect to x: \(y = \frac{1}{3}(y + xy') \Rightarrow 2y = xy' \Rightarrow \frac{dy}{y} = \frac{2\,dx}{x}\)

So \(y = cx^2\), a family of parabolas. (The other case, area = \(\frac{2}{3}xy\), gives \(y^2 = cx\), also parabolas.)

Answer: (B) parabolas