I am trying to find points $(x,y)$ which satisfy the following constraint:


Where $A\simeq0.0008$. My idea was to plot $z=1000F_r(x,y)$ and $z=1$, then if I could get a 2D plot of the intersection I would have a curve $y(x)$. This is the plot I'm getting:

enter image description here

How can I get the 2D plot of the intersection?


2 Answers 2


real solutions need -A<y<A and x>0.

ContourPlot[(1000*x*Sqrt[1/y^2 - 1/A ^2])/(1 + x*Sqrt[1/y^2 - 1/A ^2]) == 1/2, {y, -A, A}, {x, 0, 10^-6}, FrameLabel -> {y, x}]

enter image description here


First, you can simply solve the equation and find `y[x]'

A = .0008;

s = Solve[
  1000*x*Sqrt[1/y^2 - 1/A1^2]/(1 + x*Sqrt[1/y^2 - 1/A1^2]) == 1/2, y]
(*{{y -> -((1999 A1 x)/Sqrt[A1^2 + 3996001 x^2])}, {y -> (1999 A1 x)/
   Sqrt[A1^2 + 3996001 x^2]}}*)
{Plot[y /. First[s] /. A1 -> A, {x, -1, 1}], 
 Plot[y /. Last[s] /. A1 -> A, {x, -1, 1}]}



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