Update:
Given
$x = r_1 \cos(t)$, $y = r_2 \sin(t)$
using the equation of ellipse in polar form we have : (could someone confirm this please?)
$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta)}+{r_2^2 \cos^2 (\theta)}}} $
to rotate by angle $\phi$ :
$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta-\phi)}+{r_2^2 \cos^2 (\theta-\phi)}}} $
The original incorrect answer is below:
$x = r_1 \cos(t + \phi)$, $y = r_2 \sin(t + \phi) \text {, where }\phi$ is your angle of tilt in radians
This was as pointed out was my mistake on thinking polar while working with parametric. I have tried converting this to polar form but adding angle of $\phi$ and reverting back to parametric form so far has not been possible.
Has anyone seen solution of this problem by converting to polar coordinates and performing an angle addition and converting back to parametric form?