Indeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$
When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor that depends on $a$. If you take a solution $f$ and draw the parametric plot $(x,y)=(f'(t), f(t))$, the result is an ellipse whose eccentricity depends on $a$. For $a=-1$ the ordinary sine and cosine are solutions, and you get a circle.
On the other hand, when $a$ is positive the solutions are either hyperbolic sines or hyperbolic cosines, again with a horizontal scaling factor that depends on $a$. A plot of $(x,y)=(f'(t), f(t))$ is one arm of a hyperbola with a central angle that depends on $a$. For $a=1$ the hyperbolic sine and cosine are solutions, and the hyperbola is right-angled.
Intuitively, then, since a parabola is the limiting case between an ellipse and a hyperbola, we should expect to get a "parabolic function" by setting $a=0$. Unfortunately the differential equation then becomes
$$f''(t)=0$$
whose solutions are first-degree polynomials, and it is hard to make those create a parabola. However, there's a way out (many thanks to Qiaochu Yuan for pointing this out!): Instead of $f''(t)=af(t)$ we can take the basic differential equation to be
$$f'''(t)=af'(t)$$
In the $a\ne 0$ case all this changes is to allow us to add a constant term to solutions, which just moves the conic about in the plane. But for $a=0$, the solutions are now all the polynomials of degree $\le 2$. And when we take any quadratic polynomial $f$ and plot $(x,y)=(f'(t), f(t))$, what we get is indeed a parabola centered around the $y$-axis!
If we take $f$ to be a first-degree polynomial, the parametric plot is just a straight (vertical) line, another limiting case of conic sections.
In all of the above cases, plotting $(f_1(t),f_2(t))$ for two unrelated solutions (for the same $a$) generally produces a conic of the same general kind, but perhaps moved and rotated. And the dependency on $a$ of the eccentricity/angle disappears; that was mediated through the derivative in the $x$ position.
So a "parabolic function" is simply another (redundant) term for a quadratic polynomial. It is not quite clear which should be counted as the parabolic sine and cosine, though. Cases could be made for either $\operatorname{sinp}(t) = t$ and $\operatorname{cosp}(t) = 1+\frac12 t^2$ or the other way around -- but worrying too much about that is just silly.
It depends.
Take an example.
Whatever major and minor axis are given to you, shift and rotate axes such that the major axis is new x-axis and minor axis is new y-axis. Now, if you are given points, say, $(3,0)$ and $(-3,0)$ (in new coordinate system). It may be ellipse with equation
$$\frac{x^2}{9}+\frac{y^2}{k^2}=1$$
(for some k) or it may be a hyperbola with equation
$$\frac{x^2}{9}-\frac{y^2}{k^2}=1$$
But, say, if your points are like $(3,0)$ and $(0,2)$, you know it is ellipse with equation,
$$\frac{x^2}{9}+\frac{y^2}{4}=1$$
Hope it helps:)
Best Answer
Not a stupid question at all.
I'd say that the hyperbola is usually parameterized with hyperbolic sines and cosines, so maybe not exactly "trig", but let's ignore that for a moment.
One thing you can do is say that there's a linear transformation from a circle to an ellipse (e.g., scale along one axis or the other or both, or even rotate a bit). In the simplest case, this converts the $(\cos t, \sin t)$ parameterization for the circle $x^2 + y^2 = 1$ into a $(\cos t, 2\sin t)$ parameterization for the ellipse $x^2 + (y/2)^2 = 1$.
If we could find a nice transformation from, say, a circle to a parabola, we could pull off the same trick. (It's actually more helpful to have a transformation from the parabola to the circle: you notice in my example how there were both a factor of $2$ and a factor of $1/2$? It'd be nicer if those were both $1/2$s, or both $2$s, and having the transform in the other direction makes that work.)
Let's look at the projective transformation $(x, y) \mapsto (\frac{x}{y-1}, \frac{y}{y-1})$. (You may well be observing that this isn't even defined on the line $y = 1$, and that's right --- projective transformations often have a domain that's 'missing a line' when written in standard $\mathbb R^2$ coordinates, and there's a whole complicated story there that you can learn about while looking at a projective geometry book; I love Hartshorn's "introduction to projective geometry" (or some name like that)).
Suppose we "substitute" that transformation into the equation of a circle, i.e., replace each $x$ with $x/(y-1)$, and similarly for $y$. We get $$ x^2 + y^2 = 1 \to \\ \frac{x^2}{(y-1)^2} + \frac{y^2}{(y-1)^2} = 1 \\ x^2 + y^2 = (y-1)^2 \\ x^2 + y^2 = y^2 - 2y + 1 \\ x^2 + 2y - 1 = 0 $$ which is the equation of a parabola.
Now if we knew an "inverse" to that projective transformation we'd be in good shape. Fortunately (for this contrived example), the inverse is exactly the same transformation! That means that since $(\cos t, \sin t)$ is a point of the circle, the point $$ (\frac{\cos t}{\sin t -1 }, \frac{\sin t}{\sin t -1 }) $$ is a point of our parabola. And there is a parameterization of the parabola by sines and cosines. You might object that when $\sin t$ is 1, the quotients are undefined, etc. That just means that the domain of the parameterization needs to be, say $-3\pi/2 < t < \pi/2$.