Question

A curve is such that the mid point of the segment of its normal intercepted between the point from where it is drawn and the point where the normal meets the $x-$axis lies on the parabola $2y^2=x.$ If the curve passes through origin, then the curve is

• A. $y^2=2x+1-e^{2x}$
• B. $y=x+1-e^{2x}$
• C. $y^2=x+1-e^x$
• D. $y=2x+1-e^x$

Answer 1

The correct option is A $y^2=2x+1-e^{2x}$

Equation of normal at $P(x,y)$ is
$Y-y=-\frac{dx}{dy}(X-x)$
This meets the $x-$axis at $A(x+y\frac{dy}{dx},0)$
Mid point of $AP$ is $(x+\frac{1}{2}y\frac{dy}{dx},\frac{y}{2})$, which lies on the parabola $2y^2=x.$
$\therefore 2\times \frac{y^2}{4}=x+\frac{1}{2}y\frac{dy}{dx}$
$\Rightarrow y^2=2x+y\frac{dy}{dx}$
Put $y^2=t\Rightarrow y\frac{dy}{dx}=\frac{1}{2}\frac{dt}{dx}$
We get $\frac{dt}{dx}-2t=-4x$ (which is linear)
$I.F.=e^{-2\int dx}=e^{-2x}$
$\therefore$ Solution is
$t{e}^{-2x}=-4\int x{e}^{-2x}dx+c$
On integrating by parts, we have:
$t{e}^{-2x}=-4[x\frac{e^{-2x}}{-2}-\int -\frac{1}{2}{e}^{-2x}dx]+c$
$\Rightarrow y^2{e}^{-2x}=2x{e}^{-2x}+e^{-2x}+c$
Since the curve passes through origin, the value of $c=-1$
$\therefore y^2=2x+1-e^{2x}$ is the equation of the required curve.