The curve $y = f(x)$ is such that the area of the trapezium formed by the coordinate axes ordinate of an arbitrary point and the tangent at this point equals half the square of its abscissa. The curve is:
Step-by-Step Solution
Key Concept: The trapezoid area condition leads to a linear differential equation that requires an integrating factor for solution.
The length of intercept on the $y$-axis by a tangent at $P(x,y)$ is $P(x,y) = OT = y - x\frac{dy}{dx}$. The area of trapezoid $OLPTO = \frac{1}{2}(PL + OT)OL = \frac{1}{2}\left(2y - x\frac{dy}{dx}\right)x$. Setting this equal to $\frac{1}{2}x^2$ gives $\frac{dy}{dx} - \frac{2y}{x} = \pm 1$. The integrating factor is $e^{\int -2/x\,dx} = \frac{1}{x^2}$, yielding $y = cx^2 \pm x$.
Correct Answer: 1