When the parabola is expressed in parametric form, it can be easily rotated to open toward any given angle if the parametric vector equation is multiplied by a trigonometric rotation matrix. Given how much algebra was previously used, this sounds too good to be true but it works. The trigonometric rotation matrix in $\mathbb{R}^{2}$ is given in equation $\eqref{1}$. We have seen this equation previously in the section on Axes Rotation. $$R=\left[\begin{array}{cc} cos\theta & -sin\theta\\ sin\theta & cos\theta \end{array}\right] \tag{1} \label{1}$$ Now suppose that we have a parabola with $p=1/4$ given by $$\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} 2pt+2\\ pt^{2}+4 \end{array}\right)\ -5\leq t \leq 5$$ If we multiply by the rotation matrix, we will get a new matrix, equation $\eqref{2}$ which will point in the direction of $\theta$, but as $\theta$ changes, it will rotate about the origin because no vertex (offset) was added to the equation. $$\left(\begin{array}{c} x\\ y \end{array}\right)=\left[\begin{array}{cc} cos\theta\ & -sin\theta\\ sin\theta\ & cos\theta \end{array}\right]\cdot\left(\begin{array}{c} 2pt+2\\ pt^{2}+4 \end{array}\right)\ -5\leq t\leq5 \tag{2} \label{2}$$ We can write a similar equation that rotates the parabola around its vertex by simply moving the Vertex $(2,4)$ outside of the rotation. Thus the equation becomes $$\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} 2\\ 4 \end{array}\right)+\left[\begin{array}{cc} cos\theta\ & -sin\theta\\ sin\theta\ & cos\theta \end{array}\right]\cdot\left(\begin{array}{c} 2pt\\ pt^{2} \end{array}\right)\ -5\leq t\leq5 .$$ When $\theta$ is zero or $2\pi$ these equations yield the same parabola.
Every vertical parabola can be parametrized as $$\left(\begin{array}{c} x\\ y \end{array}\right)=V+\left(\begin{array}{c} \pm2pt\\ pt^{2} \end{array}\right) \tag{3} \label{3}$$ where $p$ is the distance from the vertex to the focus and $t$ is the parameter. We have written $\pm2$ because it doesn't matter whether we traverse the distance as negative to positive or vice versa. Similarly, if $p$ is negative, then the parabola will open down instead of up. Going side to side is just a matter of reversing the vector elements to put $pt^{2}$ on top and $2pt$ on the bottom. For example, in Geogebra, to get an open right parabola, aligned with the axis, with $p=1.5$ and vertex $V=(1,2)$ we could input $$(p*t^2, 2*p*t)+V$$ and I have shown "*" in the input because the multiplication symbol must be explicitly typed to keep it from being considered a variable name "pt". Other graphics, CAS software works similarly.
Rotating $\eqref{3}$ is just a matter of multiplying the vector equation by the rotation matrix. Since a parametric equation is just a collection of $x,y$ points, we use $$R=\left(\begin{array}{cc} \cos(\theta)\,\, & -\sin(\theta)\\ \sin(\theta)\,\, & \cos(\theta) \end{array}\right)$$ to rotate it counter-clockwise. So the rotated parametric equation is $$\left(\begin{array}{c} x\\ y \end{array}\right)=V+\left(\begin{array}{cc} \cos(\theta)\,\, & -\sin(\theta)\\ \sin(\theta)\,\, & \cos(\theta) \end{array}\right)\left(\begin{array}{c} \pm2pt\\ pt^{2} \end{array}\right) \tag{4} \label{4}$$ Since this equation has variables for the vertex, the distance to the focal point, and the angle of rotation, it can be fit to any parabola in the $xy$ plane.
If we have a parabola written in general form, $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$, we know the directrix vector to be $\mathbf{e_{2}}=(2A,-B)$ and we can easily find the rotation angle $\theta$ as $\atantwo (\mathbf{e_{2y},\mathbf{e_{2x}}})$. We can then rotate by $-\theta$ to get a vertical parabola without an $xy$ or $y^{2}$ term. If it is “open down”, but we want “open up”, then use $\mathbf{e_{2}}=(-2A,B)$. Once we have a vertical parabola, we can put it into the form $4p(y-k)=(x-h)^{2}$ and identify the vertex, $(h,k)$ and the distance, $p$. Having done that, we have all of the parameters of equation $\eqref{4}$ and we can write the parametric form.