Return to Answer

The answer is yes in the case $g=2$.

In other words, any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

ADDENDUM. Just for completeness, let me add a deformation theory argument showing that the general smooth hyperelliptic curve of genus $g$ lies on a quadric $Q$. This will replace my previous imprecise naive count of parameters. Notice that Jack Huizenga showed in his answer that this is actually true for any such a curve.

Let us start with a curve $C \subset Q$ of bidegree $(2, g+1)$. This is clearly hyperelliptic and we have a short exact sequence $$0 \to T_C \to T_Q \otimes \mathcal{O}_C \to N_{C/Q} \to 0$$ and since $H^0(T_C)=0$ for $g \geq 2$ this gives in turn a sequence in cohomology $$0 \to H^0(T_Q \otimes \mathcal{O}_C) \to H^0(N_{C/Q}) \stackrel{\delta}{\to} H^1(T_C).$$

Now standard computations yield $$h^0(T_Q \otimes \mathcal{O}_C)=g+6, \quad h^0(N_{C/Q})=3g+5$$ hence the image of the map $\delta \colon H^0(N_{C/Q}) \to H^1(T_C)$ has dimension $3g+5-g-6=2g-1.$

This means that the embedded deformations of $C$ in $Q$ form a family of dimension $2g-1$. But this is exactly the dimension of the hyperelliptic locus in $\mathcal{M}_g$, since any hyperelliptic curve of genus $g$ is a double cover of $\mathbb{P}^1$ branched in $2g+2$ points.

The answer is yes in the case $g=2$.

In other words, any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

ADDENDUM. Just for completeness, let me add a deformation theory argument showing that the general smooth hyperelliptic curve of genus $g$ lies on a quadric $Q$. This will replace my previous imprecise naive count of parameters. Notice that Jack Huizenga showed in his answer that this is actually true for any such a curve.

Let us start with a curve $C \subset Q$ of bidegree $(2, g+1)$. This is clearly hyperelliptic and we have a short exact sequence $$0 \to T_C \to T_Q \otimes \mathcal{O}_C \to N_{C/Q} \to 0$$ and since $H^0(T_C)=0$ for $g \geq 2$ this gives in turn a sequence in cohomology $$0 \to H^0(T_Q \otimes \mathcal{O}_C) \to H^0(N_{C/Q}) \stackrel{\delta}{\to} H^1(T_C).$$

Now standard computations yield $$h^0(T_Q \otimes \mathcal{O}_C)=g+6, \quad h^0(N_{C/Q})=3g+5$$ hence the image of the map $\delta \colon H^0(N_{C/Q}) \to H^1(T_C)$ has dimension $3g+5-g-6=2g-1.$

This means that the embedded deformations of $C$ in $Q$, which are all hyperelliptic, form a family of dimension $2g-1$. But this is exactly the dimension of the hyperelliptic locus in $\mathcal{M}_g$, since any hyperelliptic curve of genus $g$ is a double cover of $\mathbb{P}^1$ branched in $2g+2$ points.

The answer is yes in the case $g=2$.

In other words, any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

The answer is yes in the case $g=2$.

In other words, any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

ADDENDUM. Just for completeness, let me add a deformation theory argument showing that the general smooth hyperelliptic curve of genus $g$ lies on a quadric $Q$. This will replace my previous imprecise naive count of parameters. Notice that Jack Huizenga showed in his answer that this is actually true for any such a curve.

Let us start with a curve $C \subset Q$ of bidegree $(2, g+1)$. This is clearly hyperelliptic and we have a short exact sequence $$0 \to T_C \to T_Q \otimes \mathcal{O}_C \to N_{C/Q} \to 0$$ and since $H^0(T_C)=0$ for $g \geq 2$ this gives in turn a sequence in cohomology $$0 \to H^0(T_Q \otimes \mathcal{O}_C) \to H^0(N_{C/Q}) \stackrel{\delta}{\to} H^1(T_C).$$

Now standard computations yield $$h^0(T_Q \otimes \mathcal{O}_C)=g+6, \quad h^0(N_{C/Q})=3g+5$$ hence the image of the map $\delta \colon H^0(N_{C/Q}) \to H^1(T_C)$ has dimension $3g+5-g-6=2g-1.$

This means that the embedded deformations of $C$ in $Q$ form a family of dimension $2g-1$. But this is exactly the dimension of the hyperelliptic locus in $\mathcal{M}_g$, since any hyperelliptic curve of genus $g$ is a double cover of $\mathbb{P}^1$ branched in $2g+2$ points.

The answer is yes in the case $g=2$.

Let's start with a naive count of parameters. The curves of genus $2$ depend on $3g-3=3$ moduli. On the other hand, a curve of genus $2$ on a quadric surface $Q \subset \mathbb{P}^3$ must necessary have bidegree $(2, 3)$. Then the curves of genus $2$ on $Q$ form a family of dimension $$h^0(\mathcal{O}_Q(2,3))-1- \dim \textrm{Aut}(\mathbb{P}^3)=12-1-8=3.$$

This suggest that the general curve of genus $2$ can be embedded in a quadric.

But actually more is true: any curve $C$ of genus $2$ has this property. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

The answer is yes in the case $g=2$.

In other words,   any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.