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Gilbert Strang notes a parabola is an intermediary between ellipses and hyperbola. As discussed here, this is easy to see in polar coordinates, but quite hard to see in Cartesian coordinates.

In Cartesian coordinates, an ellipse is $x = a \cos \theta, y = b \sin \theta$, and a hyperbola is $x = a \cosh \theta, y = b \sinh \theta$. This suggests that an "intermediate function" between cos and cosh might produce a parabola.

Therefore: Is there a continuous transformation between $\cos x$ to $\cosh x$? This paper claims there is, but the selection shown doesn't define it.

Since $$ \cos x = \frac {e^{ix} + e^{-ix}} 2 \\ \cosh x = \frac {e^{x} + e^{-x}} 2 \\ \cos x = \cosh ix \\ \cosh x = \cos ix \\ $$

it would follow that $$f(x;\theta) := \cos e^{i\theta}x$$ is exactly what we need, with $\theta = 0$ producing $\cos$ and $\theta = \frac \pi 2$ producing $\cosh$.

Questions:

  1. Does there exist a $\theta$ such that the parametric equations $x = f(x; \theta), y = f(x + \frac {\pi} 2; \theta)$ produces a parabola?
  2. What does $f$ with intermediate values of $\theta$ look like, both analytically and visually?
  3. What do the parametric equations of intermediate values of $\theta$ produce?

Also: Mathematical plotting tools, such as Desmos and Wolfram Alpha, are quite powerful; is there a way I can use them to plot these equations?

SRobertJames
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    The problem (I think) is that $f(x;\theta) = \cos e^{i\theta}x$ produces complex (non-real) values for real $x$, unless $ e^{i\theta}$ is $\pm 1$ or $\pm i$. – Martin R Jan 02 '23 at 18:30
  • The paper you cite is talking about a different kind of transition; namely, the piecewise-defined function $$f(x) = \begin{cases} \cos x & x \leq 0 \ \cosh x & x \geq 0. \end{cases}$$ Taking that function along with its sine version won't trace out a parabola, it will trace out a circle when $x \leq 0$ and then switch to tracing out (part of) a hyperbola. – Ravi Fernando Jan 02 '23 at 18:34
  • @MartinR Won't other values of $\theta$ cancel out the imaginary parts, like they do for $\theta = \frac \pi 2$? – SRobertJames Jan 02 '23 at 18:42
  • Do the Jacobi elliptic functions answer your question? See also my older answer here. – Ng Chung Tak Jan 03 '23 at 06:38
  • For pure geometric point of view and besides polar coordinates, I'd suggest using stereographic projection for smooth transition. – Ng Chung Tak Jan 03 '23 at 08:08

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