Finding the Curve that Satisfies a Geometric ODE

[Telemachus]

Hi there. I have this exercise in my practice for differential equations, and it asks me to find the curve that satisfice for every point (on the xy plane) the distance from (x,y) to the points of intersection for the tangent line and the x axis, and the normal with the x-axis remains constant

So I stated it this way, I don't know if it's right:

I called x1 to the point of intersection for the normal line and the x-axis :
$$\displaystyle\frac{-1}{y'}(x_1-x)+y=0\Rightarrow{x_1=yy'+x}$$

And x2 to the point of intersection for the tangent line and the x axis:
$$y'(x_2-x)+y=0\Rightarrow{x_2=\displaystyle\frac{-y}{y'}+x}$$

Then I stated this differential equation, which is the sum of the distances from the (x,y) point on the curve to the intersections on the normal and the tangent line for that point of the curve and the x axis.
$$\sqrt[ ]{y^2+(yy')^2}+\sqrt[ ]{y^2+\left (\displaystyle\frac{y}{y'} \right )^2}=C$$

I'm not sure if this is right, and I don't know how to solve this diff. equation.

Bye there, and thanks for your help in advance

 

[Telemachus]

I've made a mistake posting this here (I didn't realize in which subforum I was posting, really silly from my part). Would somebody move this to the calculus section please?