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 the $y-$axis lies on the line $y=x.$ If the curve passes through $(1,0),$ then the curve is

• A. $2y=x^2-x,x>0$
• B. $y=x^2-x,x>0$
• C. $y=x-x^2,x>0$
• D. $y=2(x-x^2),x>0$

Answer 1

The correct option is C $y=x-x^2,x>0$

The point on $y-$ axis where tangent cuts is $(0,y-x\frac{dy}{dx}).$
The given mid point will be $(\frac{x}{2},y-\frac{x}{2}\frac{dy}{dx})$
According to given condition,
$\frac{x}{2}=y-\frac{x}{2}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=2\frac{y}{x}-1$
Putting $y=vx,$ we get
$x\frac{dv}{dx}=v-1\Rightarrow \frac{dv}{v-1}=\frac{dx}{x}$
On integrating both sides we have:
$\Rightarrow \ln |\frac{y}{x}-1|=\ln |x|+c$
$\because y\left(1\right)=0,c=0$
$\therefore 1-\frac{y}{x}=\pm x,x>0\Rightarrow y=x\mp x^2$