fixed point of a parabolic commutator

Question

For a hyperbolic isometry $g$ of the hyperbolic plane, denote by $ax(g)$ the oriented geodesic running from the repelling fixed point $p_g$ and the attracting one $q_g$. Let $G$ be the fundamental group of a hyperbolic once-punctured torus and suppose that $G$ is generated by two hyperbolic elements $a$ and $b$. In particular, the projections of their axes $ax(a)$ and $ax(b)$ are simple closed geodesics on the torus with intersection number equal to $1$. Assume also that the commutator $aba^{-1}b^{-1}$ is a parabolic element, hence it has a unique fixed point. The boundary of the disc model $\partial D$ is subdivided into two arcs by the points $q_a$ and $q_b$. Denote by $J$ the arc which does not contain the point $p_a$. If the pair $(a,b)$ is positively oriented (namely $ax(b)$ crosses $ax(a)$ from the right to the left), then the fixed point of $[a,b]$ lies in $J$. How can we prove this?

Answer 1

I think you can just prove this by computation for one explicit example, and then it will hold for all other such punctured tori by connectedness of Teichmuller space.