The conditioning number of a projective curve

Let $C$ be a smooth algebraic curve in $\P^2$. That is, for some homogeneous $f\in \C[x_0,x_1,x_2]$ we let $C = \{x\in \P^2\ :\ f(x)=0\}$. Describe $C$ as a manifold via the usual open sets $U_i = \{x\in \P^2\ :\ x_i\neq 0\}$ and charts
$$\begin{array}{r c l} \varphi_0\ :\ U_0 & \to & \C^2, \\\ [x_0:x_1:x_2] & \mapsto & (\frac{x_1}{x_0},\frac{x_2}{x_0}), \end{array} \hspace{1cm} \begin{array}{r c l} \varphi_1\ :\ U_1 & \to & \C^2, \\\ [x_0:x_1:x_2] & \mapsto & (\frac{x_0}{x_1},\frac{x_2}{x_1}), \end{array} \hspace{1cm} \begin{array}{r c l} \varphi_2\ :\ U_2 & \to & \C^2, \\\ [x_0:x_1:x_2] & \mapsto & (\frac{x_0}{x_2},\frac{x_1}{x_2}). \end{array}$$
Let $w=[w_0:w_1:w_2]\in \P^2$ for which $f(w)=0$. The Jacobian of $C$ at $w$ is then
$$J_w = \left[ \left.\frac{\dy f}{\dy x_0}\right|_w \ :\  \left.\frac{\dy f}{\dy x_1}\right|_w \ :\  \left.\frac{\dy f}{\dy x_2}\right|_w \right] \in \P^2.$$
Assume that $\left.\frac{\dy f}{\dy x_0}\right|_w\neq 0$ and pass to $\varphi_0(U_0)$ to get the Jacobian to be
$$J_w^0 = \left( \frac{\dy f/\dy x_1|_w}{\dy f/\dy x_0|_w}\ ,\ \frac{\dy f/\dy x_2|_w}{\dy f/\dy x_0|_w}\right)  \in \C^2.$$
Assume that $w_0\neq 0$, so the tangent line to $\varphi_0(C)\subset \C^2$ at $\varphi_0(w)=(w_1/w_0,w_2/w_0)$ is
$$T_{\varphi_0(w)}= \{\varphi_0(w)+tJ_w^0\ :\ t\in \C\}\subset \C^2.$$
A vector orthogonal to the Jacobian $J_w^0$ is
$$\bar J_w^0 = \left(-\frac{\dy f/\dy x_2|_w}{\dy f/\dy x_0|_w}\ ,\ \frac{\dy f/\dy x_1|_w}{\dy f/\dy x_0|_w}\right) \in \C^2,$$
so the space space normal to $T_{\varphi_0(w)}$ is given by
$$T_{\varphi_0(w)}^\perp = \{\varphi_0(w)+t\bar J_w^0\ :\ t\in \C\}\subset \C^2.$$

Example: Let $C\subset \P^2$ be the zero locus of $f(x_0,x_1,x_2) = x_0^2+x_1x_2-x_1x_0$. The Jacobian is $J = [2x_0-x_1:x_2-x_0:x_1]$, and as $J=0$ implies $x_0=x_1=x_2=0$, but $0\not\in\P^2$, the curve $C$ is smooth. Consider two points $w=[1:1:0],z=[2:1:-2]\in C$, at which the Jacobian is
$$J_w = [1:-1:1] \hspace{1cm},\hspace{1cm} J_z = [3:-4:1].$$
Both $w_0$ and $z_0$ are non-zero, with $\varphi_0(w)=(1,0)$ and $\varphi_0(z)=(1/2,-1)$, giving the tangent and normal spaces to be
$$\begin{align*} T_{(1,0)} & = \{(1,0)+t(-1,1)\ :\ t\in \C\}, & T_{(1/2,-1)} & = \{(1/2,-1)+s(-4/3,1/3)\ :\ s\in \C\}, \\ T^\perp_{(1,0)} & = \{(1,0)+t(-1,-1)\ :\ t\in \C\}, & T_{(1/2,-1)}^\perp & = \{(1/2,-1)+s(-1/3,-4/3)\ :\ s\in \C\}. \end{align*}$$
The two normal spaces intersect at $(t,s)=(1/3,-1/2)$ at distances of $1/3\cdot ||(-1,-1)|| = \sqrt 2/3\approx 0.471$ and $1/2\cdot||(-1/3,-4/3)|| = \sqrt{17}/3\approx 1.374$ from the points $\varphi_0(w),\varphi_0(z)$, respectively. Hence the conditioning number of $C$ is at most $\sqrt 2/3$.

Given a smooth projective curve and a finite set of points, this Sage code will calculate the conditioning number from that collection of points.