13 A line with a slope of 1 intersects the $x$-axis at point $C$, and intersects a parabola with the $x$-axis as its axis of symmetry, opening to the right, and passing through the point $(-1,1)$ at points $A$ and $B$. $|A B|=5 \sqrt{2}$, and the distance between point $C$ and the vertex of the parabola is $\frac{4}{3}$. Find the equation of the parabola.
13 Let the equation of the parabola be $y^{2}=p(x-q)(p>0)$, then the vertex is $(q, 0)$. Since the parabola passes through the point $(-1,1)$, we have $1=p(-1-q)$. Let the equation of the line be $y=x-c$, then its intersection point $C$ with the $x$-axis is $(c, 0)$, thus $|c-q|=\frac{4}{3}$, i.e., $c=q+\frac{4}{3}$ or $c=q-\frac{4}{3}$.
If $c=q+\frac{4}{3}$, substituting $x=y+q+\frac{4}{3}$ into the parabola equation yields $y^{2}=p\left(y+\frac{4}{3}\right)$, from which the $y$-coordinates of points $A$ and $B$ are
$$
y_{1,2}=\frac{p \pm \sqrt{p^{2}+\frac{16}{3} p}}{2} .
$$

Thus, the distance between points $A$ and $B$ is
$$
5 \sqrt{2}=|A B|=\sqrt{2\left(y_{1}-y_{2}\right)^{2}}=\sqrt{2\left(p^{2}+\frac{16}{3} p\right)},
$$

which simplifies to
$$
3 p^{2}+16 p-75=0,
$$

Solving this, we get $p=3$ or $-\frac{25}{3}$, but $p=-\frac{25}{3}$ is not valid, so we discard it. Thus, one of the equations of the parabola is $y^{2}=3\left(x+\frac{4}{3}\right)$ or $y^{2}=3 x+4$. Similarly, for $c=q-\frac{4}{3}$, we get $p=$ -3 or $\frac{25}{3}$, discarding $p=-3$, we obtain the other equation that meets the requirement:
$$
y^{2}=\frac{25}{3}\left(x+\frac{28}{25}\right)
$$

or
$$
3 y^{2}=25 x+28
$$