# Get the 2d plot of the intersection of 3d surfaces

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:

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

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}]


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}]}
`