Question

The equation of the curve passing through the origin if the middle point of the segment of its normal form any point of the curve to the x-axis, lies on the parabola $2y^2=x$

• A. $y^2=2x+1-e^{2x}$
• B. $y^2=2x+1+e^{2x}$
• C. $y^2=2x+e^{-2x}$
• D. $y^2=x+1-e^{-4x}$

Answer 1

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

Equation of normal at any point $P(x,y)$ is
$\frac{dy}{dx}(Y-y)+(X-x)=0$
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 $2y^2=x$.
$\therefore 2\times \frac{y^2}{4}=x+\frac{1}{2}y\frac{dy}{dx}\Rightarrow y^2=2x+y\frac{dy}{dx}$
Putting $y^2=t$
we get $\frac{dt}{dx}-2t=-4x$
$I.F.=e^{-2\int dx}=e^{-2x}$
Therefore, the solution is given by
$t{e}^{-2x}=-4\int x{e}^{-2x}dx+c\Rightarrow t{e}^{-2x}=-4[-\frac{1}{2}{e}^{-2}dx]+c\Rightarrow y^2{e}^{-2x}=2x{e}^{-2x}+e^{-2x}+c$
Since, curve passes through $(0,0),c=-1$. Therefore,
$y^2=2x+1-e^{2x}$ is the equation of the required curve.