Ellipses and circles are "really" the same: a circle is an ellipse in which the focal distance is $0$. Or, put another way, from among all the ellipses
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
if you are going to choose a "canonical" one to define some functions, the natural choice is to use $a=b=1$, which just leads you back to the unit circle. That is, trying to do "elliptic trigonometric functions" pretty soon either drops you into either an arbitrary choice of parameters, or the regular circular trigonometric functions.
Now, one way in which we can "unify" the circular and the hyperbolic functions is through the complex exponential: the circular trigonometric functions correspond to certain exponentials with purely imaginary arguments, while the hyperbolic ones correspond to purely real arguments: for $t$ a real number,
$$\begin{align*}
\cos t&= \frac{e^{it}+e^{-it}}{2} &\qquad \sin t &= \frac{e^{it}-e^{-it}}{2}\\
\cosh t &= \frac{{e^t}+e^{-t}}{2} & \sinh t &= \frac{e^t-e^{-t}}{2}
\end{align*}$$
Where do these come from? One place to find them is the differential equation
$$y'' + \lambda y = 0.$$
If $\lambda\gt 0$, the solution set is spanned by $\cos(\sqrt{\lambda}\;t)$ and $\sin(\sqrt{\lambda}\;t)$. You get the standard functions by normalizing with $\lambda=1$. If $\lambda\lt 0$, then the solution set is spanned by $\cosh(\sqrt{|\lambda|t})$ and $\sinh(\sqrt{|\lambda|}t)$, and you get the standard functions by normalizing with $\lambda=-1$.
This would suggest looking for any putative "parabolic trigonometric functions" in the only remaining case: $\lambda=0$ (this corresponds to the fact that if you view conic sections as being given by slicing a cone with a plane, you obtain the parabola in the boundary between ellipses and hyperbolas). But the differential equation $y''=0$ has solution space spanned by $1$ and $t$, so that the natural "parabolic functions" are just $1$ and $t$.
Here is the handout from a talk I gave on deriving the hyperbolic trig functions—this is actually a packet guiding a student through the derivation. More or less, it starts with the circular trig functions, shifts the definition to depend on area rather than arc length, constructs the comparable definition in terms of a unit hyperbola, and then bashes through some calculus to get a simpler formula, which is what you're after.
Best Answer
Here's a simple view of Jacobi's elliptic functions. I want to be honest that I'm not really answering your question here, because this deals with constructing analogues to sine and cosine specifically for ellipses. However, you might try to do something similar with other curves, and you might find the history interesting. Also, I tried to aim this exposition at my former self when I was in high school. I have no idea if this is not enough detail, not the right approach, etc. Please leave questions and I'll try to answer them.
Circles are very symmetric. This makes them very good at doing geometry, which is basically all about symmetries. Classical mathematicians did geometry using circles for thousands of years, and by and large, because their goal was to do constructions of various shapes and figures, they didn't really care too much about various lengths that arose associated to segments. When they did care, it was not too hard to give formulas, such as for the arc length of a segment of a circle. Since the circle is symmetric, every arc length chunk is the same as every other of the same size, so it's just the fraction of the circle that the angle sweeps out, i.e. $s = \theta$ for a unit circle.
Later on, analytic geometry got invented, due to the needs of explorers on the sea, as well as of mathematical interest. This lead to trigonometry. The goal was now to calculate e.g. given an angle on a circle, what are the coordinates of the point where a line making that angle with the $x$ axis intersects the circle? This is the point $(\cos \theta, \sin \theta)$, but how do you calculate those numbers? Turns out it's really hard. You can't even really do it with exact numbers for most values of $\theta$, but for numbers of the form $p \pi/q$, with $p,q \in \mathbb{Z}$, you can do it if you're really really patient. So the strategy was to approximate the values well enough, you would find two nearby numbers where you could calculate it reasonably nicely, and then do some kind of averaging trick or interpolation to get an approximation.
Good enough for sailors, but what about for mathematicians? Well modern mathematics has given us another way to compute the arclength of a circle, using calculus to capture that simple geometric idea. Integration works by adding up infinitely small chunks, more or less, so to measure the arc length of a curve given to you by a formula $f(x,y) = 0$, you parametrize the curve as $f(x(t), y(t))$ and then you write $$s = \int_a^b \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+{\bigg(\frac{dy}{dt}\bigg)^2}} dt$$
Once we understand $\sin$ and $\cos$, we see that they parametrize a circle, and they are each others derivatives, and also $\sin^2 \theta + \cos^2 \theta = 1$ by Pythagoras, so putting all these ingredients together, we can obtain a formula:
$$s = \int_0^\theta \sqrt{\cos^2 t + (-\sin t})^2 dt= \int_0^{\theta}1 dt = \theta $$
and we've given a modern calculation of that ancient fact.
This motivated people to want to do the same thing with other shapes. The first thing to try is an ellipse. Already here, you have some problems. An ellipse is parametrized by e.g. $(a \cos(t), b \sin(t))$, which does not satisfy a Pythagoras identity in any obvious way, and so the problem of calculating the arc length of the ellipse was not easily solved. Sine and cosine were already really hard to calculate except for special angles. These guys are downright impossible in comparison. In fact, in a sense we now understand using modern "algebraic geometry", it is impossible to get a nice formula analogous to the circular case. This brings us to Jacobi and the elliptic integrals and elliptic functions.
Because the elliptic integrals were so hard to understand anything about, Jacobi had a very clever idea. Instead of try to understand the function which eats a time parameter $t$ and tells you the arc length up to $t$, let's invert this function. Let's make a new function which eats arc length, and tells you the time parameter. You wouldn't think this would make a big difference, but it did. He defined
$$u = \int_0^\varphi \frac{d\theta}{\sqrt{1-m \sin \theta}}$$
as the arc length of an ellipse, which has been scaled to have unit length and a stretch factor of $m$. This writes $u$ as a function of $\varphi$. He then defined the elliptic sine to be $\sin \varphi$ and the elliptic cosine to be $\cos \varphi$. This relates them back to a standard circle. In particular, he called these functions $sn(u)$ and $cn(u)$. They satisfy a Pythagoras identity $sn^2 u + cs^2 u = 1$.
They also have addition theorems and half angle formulas similar to the usual ones for ordinary $\sin$ and $\cos$. They're a little ugly to look at, but basically how they work is because of Jacobi's clever idea, you can take derivatives of these functions, which interacts nicely with integrals, and then get a differential equation that they satisfy. You can then solve this equation to get addition formulas. I'll link the wikipedia page so you can see them. Jacobi's elliptic functions.
So if you wanted to make an analogue of sine and cosine for a general algebraic curve, you might ask yourself what sine and cosine are really supposed to represent, and which properties you want to capture. For the Jacobi elliptic functions, the goal is to make functions which satisfy a Pythagoras identity, and can be used when parametrizing the curve to understand its arc length. This suggests that a starting point for your own idea is to parametrize your curve, try to write down its arc length as a function of the parameter, and then, when you get an impossibly hard integral, instead of trying to solve it, invert the function and relate it back to the parameter, say, with a differential equation.