We know that the degree 2 equations $x^2 + y^2 =1$ and $x^2 - y^2 =1$ can be parametrized by exponential functions. How come exponential functions show up in this seemingly unrelated area? I think it has to do with the fact that exponential functions satisfy $y'=y$, while the degree 2 equations satisfy $y'=\frac{-x}{y}$ and $y'=\frac{x}{y}$ respectively.

Also, there are two extreme cases of degree two equations- parabola and straight lines. Exponential functions don't show up in those. What exactly happens as we approach those extreme cases which causes the exponential behavior to collapse?

Can degree 3 or higher curves be parametrized by exponentials? $y=\frac{1}{x}$ is also a hyperbola. Can exponentials parametrize that?

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    $\begingroup$ Good question. Partial answer. trigonometric functions (which are exponentials with complex exponents) show up for the circle because $x^2 + y^2 = 1$ expresses the conservation of energy for the second order differential equation $f" + f = 0$ when you think of that equation as modeling an oscillator. $\endgroup$ – Ethan Bolker Mar 3 at 2:37
  • $\begingroup$ Partial answer for your last question: Only singular degree-3 curves can be parametrized (with elementary functions). For instance, the singular elliptic curve $y^2=x^3+2x^2+x$ can be parametrized using polynomial functions. For general cubic curves, you can use the elliptic functions, which are basically trigonometric functions for ellipses. $\endgroup$ – Trebor Mar 3 at 2:58
  • $\begingroup$ Straight lines can be parameterized with exponential functions. $\endgroup$ – amd Mar 3 at 3:52
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    $\begingroup$ If $\{(x,y): x^2-y^2=2\}$ $= \{(f(t),g(t)): t\in T\}$ for some suitable $f,g,T$ then $\{(x,y):xy=1\}=$ $=\{(\, (f(t)-g(t))/\sqrt 2\,,(f(t)+g(t))/\sqrt 2\;):t\in T\}.$ $\endgroup$ – DanielWainfleet Mar 3 at 11:01

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