Usually the parmetric equations of a parabola, say $y^{2}=4ax$ are taken to be $y=2at$ and $x=at^2$. These do not involve any trigonometric functions, why? when all other conic sections (ellipse, circle, hyperbola) have parametric equations that involve trigonometric functions. Is it possible to have parameters for parabola too that inolve trig? If no then why not?
Also, is this a stupid question?
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Noting the double angle identity $\cos2\theta=2\cos^2\theta-1$, we can take
$$x = \frac{b^2}{8a}(1+\cos2t),$$ $$y = b\cos t,$$
for any constant $b$. However, this only covers a finite segment of the parabola.
This is a kind of degenerate Lissajous curve (shifted horizontally, and delayed if it's supposed to use $\sin$ instead of $\cos$).
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Here's a variation on an earlier answer, using a little algebra to reposition the parabola, followed by an inspired guess to parameterize the formula.
For a constant $k \geq 0$, when $k\sin t \neq 1$ let $$ (x, y) = \left(\frac{\cos t}{1 - k\sin t}, \frac{1 + \sin t}{1 - k\sin t}\right). $$
Then \begin{align} (1+k)x^2 &+ (1-k)y^2 - 2y \\ &= (1+k)\frac{\cos^2 t}{(1 - k\sin t)^2} + (1 - k)\frac{(1 + \sin t)^2}{(1 - k\sin t)^2} - \frac{2(1 + \sin t)}{1 - k\sin t} \\ &= \frac{(1+k)\cos^2 t + (1 - k)(1 + \sin t)^2 - 2(1 + \sin t)(1 - k\sin t)} {(1 - k\sin t)^2} \tag1 \end{align}
To simplify this, observe that \begin{align} (1 - k)(1 + \sin t) - 2(1 - k\sin t) &= 1 + \sin t - k - k\sin t - 2 + 2k\sin t \\ &= \sin t + k\sin t - k - 1 \\ &= (1 + k)(\sin t - 1), \\ \end{align}
so \begin{align} (1 - k)(1 + \sin t)^2 - 2(1 + \sin t)(1 - k\sin t) &= (1 + k)(\sin t - 1)(1 + \sin t) \\ &= (1 + k)(\sin^2 t - 1). \\ \end{align}
Then \begin{multline} (1+k)\cos^2 t + (1 - k)(1 + \sin t)^2 - 2(1 + \sin t)(1 - k\sin t) \\ = (1+k)\cos^2 t + (1 + k)(\sin^2 t - 1) = (1 + k)(\cos^2 t + \sin^2 t - 1) = 0, \end{multline}
that is, the right-hand side of Equation $(1)$ is zero. Therefore $$ (1+k)x^2 + (1-k)y^2 - 2y = 0, \tag2 $$ which is the equation of a conic. The parameter $k$ determines what kind of conic Equation $(2)$ describes.
When $k = 1$, Equation $(2)$ simplifies to $ 2x^2 - 2y = 0 $, or $y = x^2$, which is a simple parabola. So this is an answer to the main question.
For the remaining cases it is more convenient to use the equation $$ (1 - k^2)x^2 + (1-k)^2\left(y - \frac1{1-k}\right)^2 = 1, \tag3 $$ which is equivalent to Equation $(2)$ when $k \neq 1$.
When $k = 0$, Equation $(3)$ becomes $x^2 + (y - 1)^2 = 1$. This is the equation of a circle of radius $1$ with center $(0, 1)$.
When $0 < k < 1$, Equation $(3)$ is the equation of an ellipse with semi-major axis $\dfrac{1}{1-k}$ along the $y$-axis, semi-minor axis $\dfrac{1}{\sqrt{1 - k^2}}$ parallel to the $x$-axis, and center $\left(0, \dfrac{1}{1-k}\right)$.
When $k > 1$, because $k^2 - 1 > 0,$ it is convenient to rewrite Equation $(3)$ as $$ (k - 1)^2\left(y + \frac1{k-1}\right)^2 - (k^2 - 1)x^2 = 1. $$ This is the equation of a hyperbola with center $\left(0, -\dfrac{1}{k-1}\right)$.
Note that since the left hand side of Equation $(2)$ has no constant term, it becomes zero when $(x,y) = (0,0)$, so every one of these conics passes through the origin. In fact, each of these conics is tangent to the $x$-axis at $(0,0)$, and except for the circle (which has no vertex), every one of these conics has a vertex at $(0,0)$.
Follow this link for a demonstration. Note that the formula $$ (x, y) = \left(\frac{\cos t}{1 + k\sin t}, \frac{1 - \sin t}{1 + k\sin t}\right) $$ produces the same family of conics with the same mapping from $k$ to the type of conic, but in the case of the parabola ($k=1$) the unusable values of $t$ are $-\frac\pi2 + 2\pi n$ for integer $n$ rather than $\frac\pi2 + 2\pi n$, which some people might find preferable.
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 $-3\pi/2 < t < \pi/2$.