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: Set up the trapezium area using coordinate geometry: the tangent at P(x,y) creates a trapezoid with vertices at origin O, point L on y-axis, point P, and the foot of perpendicular. The area formula involves (y + (y - x·dy/dx))·x/2, which must equal x²/2, leading to the linear differential equation dy/dx - 2y/x = ±1.
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