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

$$1000F_r(x,y)=\frac{1000x\sqrt{1/y^2-1/A^2}}{1+x\sqrt{1/y^2-1/A^2}}=1/2$$

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?

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2 Answers 2

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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

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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}]}

fig1

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