Question

The mid-point of the segment of the normal of a curve from any point of the curve to the x-axis lies on the parabola $4y^2=x.$ If the curve passes through the origin, then the value of $-4c$, where $c$ is the constant of integration, is

Answer 1

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 $4y^2=x$
$\therefore 4\times \frac{y^2}{4}=x+\frac{1}{2}y\frac{dy}{dx}$
$\Rightarrow y^2=x+\frac{1}{2}y\frac{dy}{dx}$
$\Rightarrow 4y^2=4x+2y\frac{dy}{dx}$

Putting $y^2=t$, so that $2y\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore 4t=4x+\frac{dt}{dx}$
$\Rightarrow \frac{dt}{dx}-4t=-4x$

I.F. $=e^{-4\int dx}=e^{-4x}$
Therefore, the curve is given by
$t.e^{-4x}=-4\int x{e}^{-4x}dx$
$t.e^{-4x}=-4[\frac{-1}{4}x{e}^{-4x}-\int \frac{-1}{4}{e}^{-4x}dx]+c$
$\Rightarrow y^2.e^{-4x}=x{e}^{-4x}+\frac{1}{4}{e}^{-4x}+c$
Curve passes through $(0,0)$
$\Rightarrow 0=0+\frac{1}{4}+c\Rightarrow c=-\frac{1}{4}$
$\Rightarrow -4c=1$