Express by a single equation a region delimited by confocal ellipse and hyperbola

Question

I am trying to find the equation of a shape formed by a conic (ellipse) that excludes its intersection with another conic (hyperbola). The hyperbola and ellipse are assumed to share common foci. But they are general conics in 2D which means they can be rotated about the z-axis and translated along x and y axes. As it can be seen in the figure, the conics are rotated by an angle theta and not centered at the origin.

In the figure, I want the equation of resulting shape shown in grey shade (formed by the two conics). This new shape can be shown as follows:

Please let me know if it is possible to get an equality representing this new shape in the quadratic form as follows:

$ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Answer 1

Such a curve having the shape of a double-sided axe (https://en.wikipedia.org/wiki/Labrys) with mixed elliptical/hyperbolic boundaries can be given by the general equation

$$x^2+k|y^2-1|=k^2 \tag{1}$$

with a "behavorial change" when variable $y$ crosses values $y=1$ or $y=-1$ as can be seen on the following graphics for different values of $k$:

Fig. 1: Curves with equation (1) for different values of $k$, with $k=0.25$ for the most internal one (in two parts) and $k=2$ for the most external, with steps $0.25$. For our question, only values $k>1$ (blue curves) make sense.

It remains now to tilt the "axe" using the following coordinates change in (1):

$$\begin{cases}x&=& \ \ \ (\cos \theta) X + (\sin \theta) Y \\ y&=&-(\sin \theta) X + (\cos \theta) Y \end{cases}$$

(It looks like a pun: a change of axes inducing a different axe...)

Explanations:

1. I have obtained (1) by imposing common foci to the ellipse and the hyperbola. How can it be done ? If $f$ is the common distance from the origin to the foci, it is known that $f^2=a^2-b^2$ for an ellipse and $f^2=a^2+b^2$ for a hyperbola with canonical equations $\dfrac{x^2}{a^2}\pm\dfrac{y^2}{b^2}=1$ ; one obtains in particular $f=\sqrt{k^2-1}$.

2. As a consequence, the elliptical arcs are orthogonal to the hyperbolic arcs. This wasn't necessary (see the solution by @David K).

3. Case $k=1$ is particular as can be seen on Fig. 1 : the elliptical arcs become circular arcs.

Edit: Here is the Matlab program I have written for the generation of the figure:

   clear all;close all;hold on;axis equal
   for k=0.25:0.25:2;
      c='r';
      if k>1;c='b';end;
      e=ezplot([num2str(k),'*abs(y^2-1)+x^2-',num2str(k^2)]);
      get(e);set(e,'linecolor',c);
   end;
   u=2.5;plot([-u,u],[0,0],'k','linewidth',1.5);
   v=2;plot([0,0],[-v,v],'k','linewidth',1.5);