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 equation of curve is $y=ax+b{x}^2+c$ and it passes through $(1,0),$ then the value of $a-b-c$ is

Answer 1

Let tangent is drawn at point $(x,y).$ Then the point on $y-$ axis where tangent meets, 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,
$\frac{x}{2}=y-\frac{x}{2}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2y}{x}-1$
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:
$\Rightarrow \ln |\frac{y}{x}-1|=\ln |x|+C$
$\because y\left(1\right)=0,C=0$
$\Rightarrow y=x-x^2$ (only)
$\therefore a=1,b=-1,c=0$