Let $M$ be an elliptic element of $SL_2(\mathbb R)$. Then it is conjugate to a rotation $R(\theta)$. Note that we can calculate $\theta$ in terms of the trace of $M$; it means that we actually know $R(\theta)$ and we can write:
$$M=TR(\theta) T^{-1}$$
If $S^1$ is the unit circle in $\mathbb R^2$, it follows that $T(S^1)$ is the conic section $\mathcal C$ which is preserved by $M$.
Is there any explicit way to find the equation $\mathcal C$ in general?
My procedure is quite uneffective, because one has to find $T$ first (so non-linear system) and then write down $T(S^1)$, which is in general not obvious.
Here's an approach you might like: let $\lambda = \cos \theta + i \sin \theta$. Let $v \in \Bbb C^2$ denote the eigenvector of $M$ associated with $\lambda$. Then the matrix $T$ whose columns are the real and complex parts of $v$ are such that $$ M = TR(-\theta)T^{-1}. $$ So, this $T$ is such that $T(S^1)$ is the conic section preserved by $M$.
To see that this is the case, note that $$ Mv = \lambda v = (\cos \theta + i \sin \theta) (v_R + iv_I) = (\cos \theta v_R - \sin \theta v_I) + i(\sin \theta v_R + \cos \theta v_I). $$ It follows that $$ Mv_R = \operatorname{Re}[Mv] = \cos \theta \,v_R - \sin \theta v_I,\\ Mv_I = \operatorname{Im}[Mv] = \sin \theta v_R + \cos \theta v_I. $$ Once you have $T$, it is not too difficult to find $T(S^1)$.
We can make things even more direct. Note that $$ M^\top = T^{-\top}R(\theta) T^\top. $$ Find the eigenvector $v$ of $M^\top$ associated with $\lambda$. The real and imaginary parts of this eigenvector form the rows of $T^{-1}$. Now, the equation of $T(S^1)$ is given by $\|T^{-1}(x_1,x_2)\| = 1$. In other words, $$ (v_{R,1} x_1 + v_{R,2}x_2)^2 + (v_{I,1}x_1 + v_{I,2} x_2)^2 = 1. $$ Note that $v$ can be explicitly expressed in terms of the entries of $M$ as $$ v = (m_{21}, \lambda - m_{11}). $$