B. As shown in Figure 7, in the Cartesian coordinate system $x O y$,
$$
\begin{array}{l}
A O=8, A B=A C, \\
\sin \angle A B C=\frac{4}{5}
\end{array}
$$
$C D$ intersects the $y$-axis at $E$,
and $S_{\triangle C O E}=S_{\triangle A D E}$.
It is known that the image passing through points $B$, $C$, and $E$
is a parabola. Find the analytical expression of the quadratic function corresponding to this parabola.
B. From $\sin \angle A B C=\frac{A O}{A B}=\frac{4}{5}, A O=8$, we get $A B=10$.
By the Pythagorean theorem,
$B O=\sqrt{A B^{2}-A O^{2}}=6$.
Thus, $\triangle A B O \cong \triangle A C O$
$\Rightarrow C O=B O=6$.
Therefore, $A(0,-8), B(6,0), C(-6,0)$.
Let point $D(m, n)$.
From $S_{\triangle C O E}=S_{\triangle A D E}$
$\Rightarrow S_{\triangle C D B}=S_{\triangle A O B}$
$\Rightarrow \frac{1}{2} B C|n|=\frac{1}{2} A O \cdot B O$
$\Rightarrow \frac{1}{2} \times 12(-n)=\frac{1}{2} \times 8 \times 6$
$\Rightarrow n=-4$.
Thus, $D(3,-4)$ is the midpoint of $A B$.
Hence, $C D$ and $A O$ are the medians of $A B$ and $B C$, respectively, and $E\left(0,-\frac{8}{3}\right)$ is the centroid of $\triangle A B C$.

Let the quadratic function corresponding to the parabola passing through points $B$, $C$, and $E$ be
$$
y=a(x-6)(x+6) .
$$

Substituting the coordinates of point $E$ into the equation, we solve to get $a=\frac{2}{27}$.
Therefore, the quadratic function corresponding to the parabola passing through points $B$, $C$, and $E$ is
$$
y=\frac{2}{27} x^{2}-\frac{8}{3} .
$$