I have the coefficients for the curves and trying to write python code do the same.
To define what I mean by 'parallel', consider the two lines for a lane on the road stretch with no intersection. Here both lines follow each other such that the separation between the two lines is nearly the same all the way. That what I mean by the term 'parallel' here.
Your problem is a bit underspecified, since separation being nearly the same all the way is a bit vague.
You could be looking for exactly the same distance everywhere. But as Jens' comment already hinted, the parallel curve of a quadric will in general be far more complicated, so with some very few special cases, the answer will always be “no”.
As another extreme, two lines are called parallel if they do not intersect. Applying the same to quadrics would mean testing for intersections. You could use my answer on intersecting conics, and check whether all points of intersection have complex coordinates. But this completely ignores the question of distance, apart from ensuring it never becomes zero.
So what else can you do? You could look at the difference between minimal and maximal distance. You could integrate distances in some way, and come up with some term for the distance variance. All of this is algebraically complicated, since for every point on one conic you'd have to intersect the perpendicular line in this point with the second conic, and from the two intersections pick the nearer one, which is tough to capture as a closed formula suitable for integration.
So I guess your best bet would be coming up with a somewhat clearer criterion of what you want, and then evaluating that numerically in a sufficient number of points on each conic. Lots of choices to make, including how to space these points, particularly since equidistant spacing is yet another tricky problem.
So how do you compute the distance between a given point $p$ on one conic and the nearest point $q$ on the other conic? One necessary condition is that the line from $p$ to $q$ must be orthogonal to the tangent in $q$. I like to express this in projective geometry using homogeneous coordinates. There the line joining them is $p\times q$. The tangent in $q$ is $Bq$, where $B$ is the matrix of the second conic, i.e. $q^TBq=0$ is the set of all points $q$ on that conic. Two lines $g$ and $h$ are perpendicular if $gFh=0$ where $F=\operatorname{diag}(1,1,0)$. So you want $(p\times q)FBq=0$. Writing $P=\hat p$ for the matrix notation of the cross product operation, you get $q^TP^TFBq=0$ which means $q$ must lie on a conic with matrix $P^TFB$. Intersect the second conic with this conic, and you get up to four points of intersection. At all of them, the line to $p$ is orthogonal to the second conic. One of them will be closest. COmpute distance for all then pick the minimum.
