[step:Choose mixed canonical coordinates on $\Lambda$]
Choose a coordinate chart on $X$ near $x_0$ with local coordinates $x=(x_1,\dots,x_n)$, and use the induced canonical cotangent coordinates $(x,\xi)$ on $T^*X$, where $\xi=(\xi_1,\dots,\xi_n)$. Let $L:=T_{\lambda_0}\Lambda\subset T_{\lambda_0}(T^*X)$ be the tangent space. Since $\Lambda$ is Lagrangian, $L$ is a Lagrangian subspace for the canonical symplectic form. We use the following mixed-coordinate linear algebra fact. In a symplectic [vector space](/page/Vector%20Space) with canonical coordinates $(x_1,\dots,x_n,\xi_1,\dots,\xi_n)$ and symplectic form $\sum_{i=1}^n d\xi_i\wedge dx_i$, every Lagrangian subspace $L$ admits a choice of exactly one coordinate from each pair $(x_i,\xi_i)$ such that the corresponding coordinate projection is injective on $L$. Indeed, if no such choice existed, the coordinate restrictions on $L$ would be linearly dependent for every transversal choice; applying the elementary exchange argument for the symplectic basis would produce a nonzero vector in $L\cap L^\omega{}^\perp$ annihilating all chosen coordinates, contradicting $L=L^\omega$ and $\dim L=n$. Since the target coordinate space has dimension $n$, injectivity is then an isomorphism onto that coordinate space.
Because $\Lambda$ is conic and $\xi_0\neq 0$, the radial vector $(0,\xi_0)$ belongs to $L$ and is nonzero. Hence the selected coordinate space cannot consist only of base coordinates, since all base coordinate differentials vanish on the radial vector. After relabelling the canonical coordinate pairs, the selected variables may therefore be written as $(x',\xi'')$, where $x'=(x_1,\dots,x_k)$ and $\xi''=(\xi_{k+1},\dots,\xi_n)$ with $k+N=n$ and $N\geq 1$. Equivalently, writing
\begin{align*}
x=(x',x'')\in \mathbb{R}^k\times\mathbb{R}^N,\qquad \xi=(\xi',\xi'')\in \mathbb{R}^k\times\mathbb{R}^N,
\end{align*}
the map
\begin{align*}
\rho:\Lambda \to \mathbb{R}^k\times \mathbb{R}^N,\qquad (x',x'',\xi',\xi'')\mapsto (x',\xi'')
\end{align*}
has rank $n$ at $\lambda_0$. By the [inverse function theorem](/theorems/51) applied to $\rho|_\Lambda$, after shrinking $\Lambda$ around $\lambda_0$, the map $\rho$ is a coordinate map onto an [open set](/page/Open%20Set). Since $\Lambda$ is invariant under positive fibre dilations and the variables $(x',\xi'')$ scale by $(x',\xi'')\mapsto (x',t\xi'')$, the shrink can be taken invariant under this scaling. Thus the image is an open conic set $V\subset \mathbb{R}^k\times\mathbb{R}^N_0$.
Thus there are smooth maps
\begin{align*}
a:V\to\mathbb{R}^N
\end{align*}
and
\begin{align*}
b:V\to\mathbb{R}^k
\end{align*}
such that the corresponding local parametrization of $\Lambda$ is
\begin{align*}
F:V\to T^*X\setminus 0,\qquad (x',\eta)\mapsto (x',a(x',\eta),b(x',\eta),\eta),
\end{align*}
where $\eta$ denotes the mixed fibre variable $\xi''$. Since $\Lambda$ is conic and $F(x',\eta)$ has fibre component $(b(x',\eta),\eta)$, after shrinking $V$ to a conic neighbourhood we have
\begin{align*}
a(x',t\eta)=a(x',\eta)
\end{align*}
and
\begin{align*}
b(x',t\eta)=t\,b(x',\eta)
\end{align*}
for all $(x',\eta)\in V$ and all $t>0$ for which $(x',t\eta)\in V$.
[/step]