Question

A curve is such that the midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and 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 $y=x-x^2$

Let $P(x,y)$ be a point on the curve then equation of the tangent is $Y-y=\frac{dy}{dx}(X-x)$

Given that tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis

$\Rightarrow x-$coordinate$=0$

$\Rightarrow X=0$

$\Rightarrow Y-y=\frac{dy}{dx}(X-x)$ becomes

$\Rightarrow Y-y=\frac{dy}{dx}(0-x)$

$\Rightarrow Y=y-x\frac{dy}{dx}$

$\therefore A=(0,y-x\frac{dy}{dx})$

Given that midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis, lies on the line $y=x$

$\therefore$Midpoint of the line $AP$ lies on the line $y=x$

Midpoint of the line $AP=(\frac{x+0}{2},\frac{y+y-x\frac{dy}{dx}}{2})$ lies on the line $y=x$

$\therefore x-$coordinate$=y-$coordinate

$\Rightarrow \frac{x+0}{2}=\frac{2y-x\frac{dy}{dx}}{2}$

$\Rightarrow x=2y-x\frac{dy}{dx}$ is a linear differential equation.

$\frac{dy}{dx}-\frac{2}{x}y=-1$

Integrating factor is $=e^{\int pdx}=e^{\int \frac{-2}{x}dx}=e^{-\ln x}=\frac{1}{x^2}$

Now, $\frac{1}{x^2}\times \frac{dy}{dx}-\frac{1}{x^2}\times \frac{2}{x}y=\frac{-1}{x^2}$

$\Rightarrow \frac{1}{x^2}\frac{dy}{dx}-\frac{2}{x^3}y=\frac{-1}{x^2}$

$\Rightarrow \frac{1}{x^2}dy-\frac{2}{x^3}ydx=\frac{-dx}{x^2}$

Integrating both sides,we get

$\Rightarrow d\left(\frac{y}{x^2}\right)=d\left(\frac{1}{x}\right)$

$\Rightarrow d\left(\frac{y}{x^2}\right)=d\left(\frac{1}{x}\right)$

$\Rightarrow \frac{y}{x^2}=\frac{1}{x}+c$ is the required curve.

The curve passes through $(1,0)$

$\Rightarrow 0=1+c$

$\Rightarrow c=-1$

$\Rightarrow \frac{y}{x^2}=\frac{1}{x}-1$

$\Rightarrow \frac{y}{x^2}=\frac{1-x}{x}$

$\Rightarrow y=x-x^2$ is the required equation of the curve passing through $(1,0)$