I assume you know that a lemniscate $r^2 = \cos{(2\phi)}$ (polar coordinates) transforms during the inversion w.r.t. a unit circle into $r^{-2} = \cos{(2\phi)}$.
I wonder what happpens with the two foci of the hyperbola.
Do they become the two foci of the lemniscate?
I wonder also whether the two relations of lemniscate and hyperbola are related.
The lemniscate has the product of the two distances to the two foci constant.
The hyperbel has the difference of two distances to the two foci constant.
Do these formulas transform into each other during the inversion?
$r^2=\cos(2\phi)$ describes a Bernoulli lemniscate $r^2=2a^2\cos(2\phi)$ with parameter $a=\sqrt{1/2}$. Its foci are $(\pm a,0)=(\pm\sqrt{1/2},0)$. The hyperbola $r^{-2}=\cos(2\phi)$ has the canonical form $x^2-y^2=1$ so it is an instance of $(x/a)^2+(y/b)^2=1$ with $a=b=1$ and $e=\sqrt2$. It has foci at $(\pm ae,0)=(\pm\sqrt2,0)$. So yes, the foci of the lemniscate are related to those of the hyperbola by inversion in the unit circle.
Obviously, since the conditions involving the distances from the foci are characteristic for both of these curves, they have to translate to one another. But I haven't come up with an easy argument for this yet. It might be worthwhile to consider the situation in $\mathbb C$ where expressing inversion is particularly easy, but expressing unsquared lengths involves square roots, which makes the difference of lengths a bit ugly.