I'm at the beggining of a differential equation's course, and I need to solve the following problem:
Find every curve in the plane $XY$ so that the middle point of the normal's segment between each point $(x,y)$ of the curve and the abscissa axis ($OX$ axis) travels the $y^2=px$ parabola.
I tried to translate the problem the best I can, so I apologise if it's not the best way to express it (You can tell me in the comments if it's not clear). I just want help finding the equation. I don't need help solving the equation, my problem is translating this problem into a differential equation I can solve. I want the $y$ variable to be a function of $x$, to be said, the variables of my equation will be $x$ and $y(x)$ (I'm not sure if it's easier doing it any other way or it just doesn't matter using $y$ and $x(y)$ instead). Any help or hint will be appreciated.
At $(X,Y)$, the normal to the curve is given by$$\frac{y-Y}{x-X}=-\frac 1{y'}$$The normal's $x$ intercept is $(X+Yy',0)$ and the midpoint of the segment between $(X,Y)$ and $(X+Yy',0)$ is $(X+Yy'/2,Y/2)$. This point lies on $y^2=px$, giving you the differential equation$$Y^2/4=p(X+Yy'/2)$$
We get$$y^2=4px+2pyy'\implies yy'-y^2/2p=-2x$$Keep $y^2=v,2yy'=v'$, so$$v'/2-v/2p=-2x\implies v'-v/p=-4x$$which is a first order linear ODE with solution$$\begin{align*}ve^{-x/p}&=-\int4xe^{-x/p}dx\\&=-4p^2\int me^mdm~~[\because m=-x/p]\\&=-4p^2(m-1)e^m+k\end{align*}$$Finally we get $y^2=4p^2+4px+ke^{x/p}$.