I have a surface of the form $ X = (u,v,f(u,v)) $ around the non-planar parabolic point $p=(0,0)$, I take $f(u,v)$ as its Taylor expansion, then $X$ is of the form $$ X = (u,v,q_{00}+q_{10}u+q_{01}v+q_{11}uv+q_{20}u^2+q_{02}v^2+q_{21}u^2v+q_{12}uv^2+q_{22}u^2v^2+...+o(6))$$
I need $k^-1(0)$ (this refers to the set where the Gaussian curation is zero) to be a regular curve so I need to find the conditions that need to satisfy the coefficients of the Taylor expansion of $f(u,v)$ for this to be fulfilled.
I know that the Gaussian curvature is given by $k=\frac{LN-M^2}{EG-F^2}$ where $L=\langle q_{20},n \rangle$, $N=\langle q_{02},n \rangle$, $M=\langle q_{11},n \rangle$.
My attempt: I think I should make that at the point $(0,0)$ the Gaussian curvature must be $0$ so I make the following conditions $q_{20}=0, q_{02}=0, q_{11}=0$ with this the Gaussian curvature will be zero at the point $(0,0)$ and since the other coefficients are continuous and with continuous derivatives, I avoid the existence of singular points or cusps in $k^-1(0)$ resulting in $k^-1(0)$ is a regular curve.
That's what I thought, however I don't know if it's correct. Any advice or help would be appreciated.