In two variables, type and characteristic directions are coordinate invariant

Write the quadratic symbol matrix as $M=\left(\begin{array}{cc}A& B\\ B& C\end{array}\right)$. By [L2], if $x=\mathrm{\Phi }(y)$ has Jacobian $J=D\mathrm{\Phi }(y)$, then the transformed principal symbol is ${\tilde{p}}_2(y,\eta )=p_2(\mathrm{\Phi }(y),J^{-T}\eta )={\eta }^T(J^{-1}M{J}^{-T})\eta$, so the new matrix is $M^{\mathrm{\prime }}=J^{-1}M{J}^{-T}$. Therefore $\det M^{\mathrm{\prime }}=(\det J{)}^{-2}\det M$, and because [L1] gives $\mathrm{\Delta }=-\det M$, the transformed discriminant is $(\det J{)}^{-2}\mathrm{\Delta }$; its sign is unchanged.

If a $C^1$ curve is written locally as $\varphi (x,y)=0$, then its tangent vector $v=(\dot{x},\dot{y})$ is annihilated by the normal covector $d\varphi =({\varphi }_x,{\varphi }_y)$, so in two dimensions $d\varphi$ is a quarter-turn of $v$ up to a nonzero scalar; the curve is characteristic exactly when $p_2(d\varphi )=0$ by [L3], equivalently when $A(dy{)}^2-2Bdxdy+C(dx{)}^2=0$, and because [L3] makes characteristic conormals coordinate invariant, the corresponding tangent directions are coordinate invariant as well.