= Solution
Since $A$ is invertible, the <quadratic form gradient> vanishes exactly at the origin. At every nonzero point of $C_\alpha$, the scalar derivative is surjective, so the <regular level set theorem> gives a smooth submanifold of dimension $n-1$.
If $\alpha\ne0$, the origin is absent. Therefore \b[the largest relatively open smooth subset is all of $C_\alpha$].
If $\alpha=0$, the nontriviality assumption forces the form to be an <indefinite quadratic form>: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use <Sylvester's law of inertia> to reduce to $|u|^2-|w|^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span $\mathbb R^n$. Each line in one of these directions lies in the cone.
If a relatively open manifold neighborhood of the origin existed, its <tangent space> would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in $\mathbb R^n$, impossible for the zero set of a nonzero quadratic form. Consequently
$$
\boxed{\text{smooth locus}=\begin{cases}C_\alpha,&\alpha\ne0,\\C_0\setminus\{0\},&\alpha=0.\end{cases}}
$$
This gives the <smooth locus of a nondegenerate real quadratic level set>, including maximality at the singular vertex.
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