I don't know if this is useful, but I would proceed with the parametrization and the rotation matrix, anyway.
Let us rename $x-X\rightarrow x$. Then, notice that the equation of the parabola $y = a x^2$ can be parametrized by $x = t$, $y = a t^2$, as $t$ goes from $-\infty$ to $\infty$; or, as a vector,
$$ (x(t), y(t))=(t,a t^2) $$
To rotate the graph of the parabola about the origin, you must rotate each point individually. Rotation clockwise by an angle $\theta$ is a linear transformation with matrix
$$ \left( \begin{array}{ccc}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta \\ \end{array} \right) $$
Thus, if we apply this linear transformation to a point $(t, t^2)$ on the graph of the parabola, we get
$$\left( \begin{array}{ccc}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta \\ \end{array} \right)
\left( \begin{array}{ccc}
t \\
a t^2\\ \end{array} \right) = \left( \begin{array}{ccc}
t\cos\theta +a t^2\sin\theta\\
-t\sin\theta+a t^2\cos\theta\\ \end{array} \right)$$
So, as $t$ goes from $-\infty$ to $\infty$, this is a parametrization of the graph of the rotated parabola.Then you have to convert back to $x$ and $y$, put them in the equation $y=x^2$ and that's it.
To get a cartesian equation for the new parabola, you can just solve for $t$ in the first line $a t^2 + t cot\theta = x/\sin\theta$ and put the expression for $t$ in the second one. Doing this, you have an equation for $x$ and $t$ that corresponds to the "constraint" $x$ and $y$ must satisfy to be on the new parabola!
