Question

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 equation of the curve can be

• A. $y=c{x}^2\pm x$
• B. $y=c{x}^2\pm 1$
• C. $y=cx\pm x^2$
• D. $y=c{x}^2\pm x^2\pm 1$

Answer 1

The correct option is A $y=c{x}^2\pm x$

Let P(x, y) be any point on the curve. Length of intercept on y-axis by any tangent at P is

$OT=y-x\frac{dy}{dx}$
$\therefore$ Area of trapezium $OLPTO=\frac{1}{2}(PL+OT)OL$
$=\frac{1}{2}(y+y-x\frac{dy}{dx})x=\frac{1}{2}(2y-x\frac{dy}{dx})x$
According to question
Area of trapezium $OLPTO=\frac{1}{2}{x}^2$
$i.e.,\frac{1}{2}(2y-x\frac{dy}{dx})x=\pm \frac{1}{2}{x}^2\Rightarrow 2y-x\frac{dy}{dx}=\pm xor\frac{dy}{dx}-\frac{2y}{x}=\pm 1$
Which is linear differential equation and $I.F.=e^{-2Inx}=\frac{1}{x^2}$
$\therefore$ The solution is $\frac{y}{x^2}=\int \frac{1}{x^2}dx+c=\pm \frac{1}{x}+c$
$\therefore y=\pm x+c{x}^{2}ory=c{x}^{2}ory=c{x}^2\pm x$, where c is an arbitrary constant