Question

A curve is such that the mid point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x. If the curve passes through (1, 0), then the curve is

• A. $2y=x^2-x$
• B. $y=x^2-x$
• C. $y=x-x^2$
• D. $y=2(x-x^2)$
  

Answer 1

The correct option is C $y=x-x^2$

The point on y-axis is $(0,y-x\frac{dy}{dx})$
According to given condition,
$\frac{x}{2}=y-\frac{x}{2}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=2\frac{y}{x}-1$
Putting $\frac{y}{x}=v$ we get
$x\frac{dv}{dx}=v-1\Rightarrow ln|\frac{y}{x}-1|=ln|x|+c\Rightarrow 1-\frac{y}{x}=x$ (as f(1)=0).