Are there parametric equations with trigonometric functions for the parabola $f(x)=x^2$?

Question

Parametric equations for the circle: $$ x(t)=\cos(t)\qquad y(t)=\sin(t) $$ Parametric equations for the ellipse: $$ x(t,a)=a\cos(t)\qquad y(t,b)=b\sin(t) $$ Parametric equations for the hyperbola: $$ x(t)=\cosh(t)\qquad y(t)=\sinh(t) $$ Are there equations for the parabola that follows this pattern?

Answer 1

Unfortunately, the parameterization that follows the pattern is simply $$y = \frac{t^2}4\quad x = \frac t2.$$ This is because trigonometric functions come from $e^{it}$ where $i^2 = -1$ in the elliptic case, $i^2 = 1$ in the hyperbolic case, and $i^2= 0$ in the parabolic case. They are known as complex, split complex and dual numbers, respectively.

However you can have trig parameterizations that doesn't follow the pattern. For example $$y = \frac{\cos t}{1-\cos t}\qquad x = \frac{\sin t}{1-\cos t}$$ up to some coefficient.

Answer 2

By using Equation (10) and Equation (11) from the paper https://www.researchgate.net/publication/327652282_Parabolic_Trigonometry, I tried the following parametric form of the e.g. parabola $x=-\frac{y^{2}}{4}+1$:

$$\left(x,y\right)=\left(3-2\cosh2t,+4\sinh t\right)$$

You may see its plot here https://www.desmos.com/calculator/cmsi6davzl.

You should see that these hyperbolic functions are in fact a parametrization of the parabola by, instead of making the canonical parametrization $y=t \Leftrightarrow x=-\frac{t^{2}}{4}+1$, making it $y=4\sinh t \Leftrightarrow x=-4\sinh^{2} t+1=3-2\cosh2t$.