Well, I have the set
$X=\{(x,y,z) \in \mathbb{R}^3 | 3x^2+2x^3+y^2+z^2=1\}$
How can I calculate the tangent cone at the point $(-1,0,0)$ ? What are the standard ways to calculate the tangent cone to a subset of $\mathbb{R}^3$ at a given point? Thank you!
The surface $X$ has rotational symmetry around the $x$-axis, therefore it is sufficient to work in the $(x,y)$-plane. Let me illustrate one way that we can find tangent lines.
Point $(-1,0)$ is critical. Near it, the curve will resemble couple of tangent lines. Let us therefore use the ansatz $y = kt$, $x = -1 + t$. We obtain $$ \array{3(-1 + t)^2 + 2(-1 + t)^3 + k^2 t^2 = \cr 3 -6t + 3t^2 -2 +6t -6t^2 + 2t^3 + k^2 t^2 = \cr 1 + (k^2 - 3) t^2 + o(t^2)}$$ and this is supposed to be equal to $1$ as $t \to 0$. Therefore $k = \pm \sqrt{3}$. As a remark, note that in the above calculation the linear term vanished because $(-1,0)$ is critical.