Do we only get conics when we do the Dupin indicatrix procedure?

Assume a regular surface $S$ with the parametrization \begin{equation} \sigma: \left(u,v\right) \longmapsto \left(\sigma_1,\sigma_2,\sigma_3\right). \end{equation} Furthermore, let $N$ be the unit normal to $S$. If $T_p\,S$ is tangent to the surface at the point $p$ and $q = \sigma(u + du, v + dv)$ is another point on the surface, then the distance from $q$ to $T_p\,S$ is: \begin{equation} h := \langle\,q - p\,,\,N\,\rangle. \label{distance:to:tangent:plane} \end{equation} Note that in this way $q - p$ will be a vector in $\mathbb{R}^3$. Having in mind that each $\sigma_i:U\subset\mathbb{R}^2 \longrightarrow \mathbb{R}$ and using the multi-variable Taylor expansion we get \begin{equation*} q - p = \sigma\left(u + du, v + dv\right) - \sigma\left(u, v\right) =\\[1 cm] \underbrace{ \left(\begin{array}{>{\displaystyle}c} \frac{\partial \sigma_1}{\partial u}\,du + \frac{\partial \sigma_1}{\partial v}\,dv\\[0.5 cm] \frac{\partial \sigma_2}{\partial u}\,du + \frac{\partial \sigma_2}{\partial v}\,dv\\[0.5 cm] \frac{\partial \sigma_3}{\partial u}\,du + \frac{\partial \sigma_3}{\partial v}\,dv \end{array}\right)}_{\star} +\frac{1}{2}\,\underbrace{ \left(\begin{array}{>{\displaystyle}c} \frac{\partial^2\sigma_1}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_1}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_1}{\partial v^2}\,dv^2\\[0.5 cm] \frac{\partial^2\sigma_2}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_2}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_2}{\partial v^2}\,dv^2\\[0.5 cm] \frac{\partial^2\sigma_3}{\partial u^2}\,du^2 + 2\,\frac{\partial^2\sigma_2}{\partial u\,\partial v}\,du\,dv + \frac{\partial^2\sigma_3}{\partial v^2}\,dv^2 \end{array}\right)}_{\star\star} + \dots. \end{equation*} Now, in the light of the above equation, $h$ simplifies to ($\star$ disappears due to $N$ being normal to $T_p\,S$): \begin{equation} h = \frac{1}{2}\left(\,\langle\,N\,,\,\sigma_{uu}\,\rangle\,du^2 + \langle\,N\,,\,\sigma_{uv}\,\rangle\,du\,dv + \langle\,N\,,\,\sigma_{vv}\,\rangle\,dv^2\right) + \ldots = \frac{1}{2}\,\mathbf{II}\,(u,v) + \ldots \end{equation} Neglecting the higher order terms results in \begin{equation} 2\,h = \mathbf{II}\,(u,v). \label{eq:dis:2nd:fundamental:form} \end{equation} Now, if on $T_p\,S$ we choose the basis ${e_1}$ and ${e_2}$ such that they are the eigenvectors of $dN_p$ then this is equal to thinking of the surface (locally) to be parameterized by curvature lines ($F = f = 0$). This results in $k_1 = e/E$ and $k_2 = g/G$. Hence, $h$ simplifies to \begin{equation*} 2\,h = \kappa_1{E}\,du^2 + \kappa_2{G}\,dv^2, \end{equation*} where $\kappa_i$ are the principal curvatures. Now, thinking of $\sqrt{E}du$ and $\sqrt{G}dv$ as $x$ and $y$ then you'll see the result is a quadric unless if one of your principal curvatures vanish (parabolic points) or both of them vanish, which gives a planar point.

Now, having in mind that the Dupin indicatrix is the set of points $(x,y)$ fitting into the equation $$\kappa_1 x^2 + \kappa_2 y^2 = \pm 1$$ you can see the similarity between the two equations. Therefore, the Dupin indicatrix tells you how the surface "behaves" around the point under consideration. You can use this on all regular surfaces to know the local behavior of the surface around a point.