Linear fractional transformation and ellipses. Let $C$ be an ellpse on the complex plane. Then by the linear fractional transformation $\tau \colon z \mapsto -1/z$ what is the image of $C$?
It is well-known that when $C$ a circle, $\tau(C)$ is also a circle including a line, i.e., its radius being infinite. I wonder what is the case when $C$ being an ellpse. I think in the typical case of $C \colon x^2/a^2 + y^2/b^2 = 1$ with $z = x + iy$, where $z$ is the variable of the complex plane ${\Bbb C}$, its image $\tau(C)$ should be an ellipse. I am checking it.
I would like to know what if $C$ being located in the general position on ${\Bbb C}$.
 A: Here is an example in the case of the (black) ellipse whose equation
$$ax^2+2bxy+cy^2+2dx+2ey+f=0$$
is displayed on the top left line with coefficients' values for $a,b,c,d,e,f$ that can be read on the cursors.
Transformation $z \to \frac{1}{z}$ gives as the image of the ellipse the blue curve looking like a bumped circle (it is rather surprizing that the bottom part of this blue curve is almost a perfect circular arc).

Let us explain how I have obtained it in two steps using Geogebra :
$$\underbrace{z \to Z=\frac{1}{\bar{z}}}_{\text{inversion}} \ \ \text{followed by} \ \   \underbrace{Z \to \bar{Z}}_{\text{symmetry wrt x-axis}}$$
being understood that these two transformations are resp. called (by Geogebra) "reflexion" wrt the unit circle (featured in green) and the $x$-axis.
The important thing one can observe (and that can be rather easily proven) is that the resulting curve is a quartic (algebraic curve with degree 4).
Changing the values of coefficients $a,b,c,d,e,f$ yields a good diversity of curves. Here is another example:

