Let $Q^4 \subset \mathbb{P}^5$ a smooth quadric over $\mathbb{C}$ which is via Pluecker map isomorphic to Grassmannian of lines $\mathbb{G}(1,\mathbb{P}^3)$ in $\mathbb{P}^3$.
Consider following Construction: Let $C'$ any smooth curve in $\mathbb{P}^3$ and $C \subset \mathbb{G}(1,\mathbb{P}^3)$ a curve obtained as image of $C'$ under the Gauss map $ \mathcal{G}_{C'}: C' \to \mathbb{G}(1,\mathbb{P}^3)$ maps $c \in C'$ to the tangent space $T_c C' $. We assume that $ \mathcal{G}: C' \to C$ is an isomorphism.
I would like to know how to prove pure algebraically that $Q^4 \subset \mathbb{P}^5$ (identified with $\mathbb{G}(1,\mathbb{P}^3)$ with Pluecker) contains the tangent variety $TC$ of $C$. Motivation: this is claimed here and I would like to understand why this is true.
Here an idea how one could argue using analytic methods:
Fact: a line $L \subset \mathbb{P}^5$ is contained in Grassmannian $\mathbb{G}(1,\mathbb{P}^3) \subset \mathbb{P}^5$ iff there exist a point $p \in \mathbb{P}^3$, and a hyperplane $H \subset \mathbb{P}^3$ such that $L= \{ \mathcal{l} \in \mathbb{G}(1,\mathbb{P}^3) \vert p \in \mathcal{l} \subset H \} $.
If we work in analytic setting we can locally parametrize $C' \subset \mathbb{P}^3$ via $t \mapsto v(t):=(v_1(t),v_2 (t),v_3(t)) \in \mathbb{A}^3 \subset \mathbb{P}^3 $. Then the Gauss map $G: C' \to \mathbb{G}(1,\mathbb{P}^3), c \mapsto T_c C'$ is explicitely given locally as $ v(t) \mapsto [v(t) \wedge v'(t)]$. The tangent at $[v(t) \wedge v'(t)]$ is given as the set $T_{[v(t) \wedge v'(t)]}C:=\{\mathcal{l} \in \mathbb{G}(1,\mathbb{P}^3) \vert v(t) \in \mathcal{l} \subset \langle v(t), v'(t), v''(t) \rangle \} $
And due to Fact above $T_{[v(t) \wedge v'(t)]}C$ is contained in $\mathbb{G}(1,\mathbb{P}^3)$, and therefore in $Q^4$. Since we showed this for arbitrary tangent line of $C$ we are done
Problem: This argument use analytic parametrization of $C'$ I would like to avoid and to give a different preferably pure algebraic proof of the statement above.
