Transforming an Ellipse into a Hyperbola Could it be possible to transform the equation of
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
into
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
without considering the complex transformation of replacing $b$ by $ib$, $i$ being imaginary? 
 A: I have developed a generalization of the superellipse, called superconics. The general form is given by
$$f(X) = b(1-c^2|X/a|^q)^{1/p}$$
or its canonical form
$$f(x) = (1-c^2|x|^q)^{1/p}$$
Here, $a$ and $b$ scale the $x$ and $y$ axes, resp. and $c^2=\pm1$. $c^2=1$ corresponds to elliptic and parabolic types and $c^2=-1$ corresponds to hyperbolic types. More to the point, $c^2$ can vary smoothly between.
In addition to the above intrinsic equation for the superconics, we can derive the following $parametric$ equation for superconics in the complex plane:
$$z=|\cos^{2/q}(u)|\text{sgn}(\cos(u))+i|(1-c^2\cos(u))^p|\text{sgn}(\sin(u))$$
where $u = [0,\pi]$ for the upper half plane and $u = [0,2\pi]$ for the full plane, e.g., a closed curve.
I've previously demonstrated analytic solutions for the area under the curve, the centroid, moments, and moments of inertia of all the superconics. See here: Superelliptical Trig Functions.
In addition, you can find some illustrations and animations of superconics here Superconics and here A New Twist on Möbius. Here is an animation demonstrating a smooth transition from a sphere to a hyperboloid of one and two sheets using only superconics (with variable $c^2$).

A: EDIT1:
With $a,c$ fixed and parameter $\theta$ made a variable parameter the equation of set of confocal ellipses/hyperbolas 
$$ \theta = (  0,\pi/2 ,\pi)$$
$$\frac{x^2}{a^2}+\frac{y^2}{ (a^2- \cos\theta \cdot c^2)}=1, \pi> \theta> 0 $$
respectively for ellipse, circle radius $a$  and hyperbolas are governed by same equation of confocal conics (as their orthogonal trajectories with same foci)  
ConfocalEll/Hyp
A: I don't know what you want to do, but if you just look for a map who converts an ellipse into a hyperbola, here you have an example (in the quadrant I): 
$$(u,v)=(|x|+|{ay\over b}|,\sqrt{2|{bxy\over a}|})$$
Then,
\begin{align}{u^2\over a^2}-{v^2\over b^2}&={x^2+{a^2y^2\over b^2}+2|{axy\over b}|\over a^2}-{2|{bxy\over a}|\over b^2}\\
&={x^2\over a^2}+{y^2\over b^2}\\
&=1
\end{align}
A: The following bijections work well:
$$
\text{Ellipse without 2 points on y-axis} \to \text{Hyperbola}:(x,y) \mapsto \left(\frac{a^2}{x},\frac{by}{\sqrt{b^2-y^2}}\right),
$$
$$
\text{Hyperbola} \to \text{Ellipse without 2 points on y-axis} :(x,y) \mapsto \left(\frac{a^2}{x},\frac{by}{\sqrt{b^2+y^2}}\right).
$$
