The scroll and the conic bundle

Proof. The curve is nondegenerate. Indeed, if it were contained in a hyperplane section $T_0$ of $Q$, every quadric section containing $C$ would contain $T_0$: otherwise its intersection with $T_0$ would have degree four and could not contain $C$. This contradicts the stated ideal generation. By Riemann–Roch, $\mathcal L=\mathcal O_C(1)$ has five sections, so the given embedding is complete.

Let $f:C\to\mathbb P^1$ be the hyperelliptic double cover. The bundle $V=f_*\mathcal L$ has rank two and degree three. It is globally generated, since $$H^1(C,\mathcal L(-f^{-1}(t)))=0$$ for every $t\in\mathbb P^1$: the Serre-dual bundle has degree $-2$. Thus $$V\simeq\mathcal O(2)\oplus\mathcal O(1) \quad\text{or}\quad V\simeq\mathcal O(3)\oplus\mathcal O.$$ Evaluation embeds $C$ in $\mathbb P(V)$, with its map to $\mathbb P^4$ induced by the tautological divisor $h$. The image is the corresponding cubic scroll. If $r$ is a ruling fibre, then $h^2=3$, $h\cdot r=1$, and the lifted curve has class $2h+dr$. Its degree six gives $d=0$. In the cone case the negative section has zero intersection with this curve, and is disjoint from it.

The quadric $Q$ cannot contain the cubic scroll: every divisor on the smooth quadric threefold has even degree, since $\operatorname{Pic}(Q)=\mathbb Z[\mathcal O_Q(1)]$. Its pullback to $\mathbb P(V)$ is therefore a nonzero divisor of class $2h$ containing $C$, and consequently equals $C$. This proves the scheme-theoretic assertion and, in the cone case, excludes the vertex from $Q$. The displayed matrices are the usual parametrizations of the two scrolls: the first comes from $$[x_0:x_1:x_2:x_3:x_4] =[s^2a:sta:t^2a:sb:tb],$$ and the second is the twisted-cubic matrix with one unused coordinate. ◻

Proof. The rank-zero locus of $M$ is empty for the first matrix in (38.1); for the second it is the vertex, which misses $Q$. Where $M$ has rank two, the incidence variety is the graph of the map defined by its minors. At a point of $C$, invertible local row and column operations put the matrix in the form $$\begin{pmatrix}1&0&0\\0&a&b\end{pmatrix}.$$ The ideal of $C$ is $(a,b)$, and the incidence equations become $y_0=0$ and $ay_1+by_2=0$. These are the equations of the blow-up of this regular codimension-two ideal. Hence (38.2) is the blow-up, scheme-theoretically.

For every $y$, the two linear forms comprising $M(x)y$ are independent, as is immediate from either matrix in (38.1). They cut out a plane $\Pi_y$. A smooth quadric threefold contains no plane, so $q|_{\Pi_y}$ is nonzero and defines a plane conic. These fibres all have the same Hilbert polynomial, giving flatness. The three minors define $|2H-E|$, which identifies the second polarization with $B$. ◻

Proof. Twist the Koszul resolution of (38.2) by $\mathcal O(a,b)$. Every term has twists at least $-4$ on $\mathbb P^4$ and at least $-2$ on $\mathbb P^2$, so it has no higher cohomology. Taking sections gives the asserted quotient and its Hilbert series. Its diagonal gives the second formula. ◻

Proof. The class of the complete intersection in the product is $2H(H+B)^2$, giving the intersection numbers. The two contractions span the extremal rays, hence give the nef cone. For the last assertion intersect $D$ with a general conic fibre and with a general exceptional fibre $e$ of $E\to C$. These curves are not contained in $D$, and $$D\cdot B^2=2a\ge0,\qquad D\cdot e=b\ge0.$$ The coefficients are integral because $H,E$ form a basis of $\operatorname{Pic}(X)$ and $B=2H-E$. Since $H=(E+B)/2$, every such divisor lies in the cone generated by $E,B$, proving the effective-cone assertion. ◻