One. (20 points) As shown in Figure 3, in the Cartesian coordinate system, point $A$ is on the $y$-axis, points $B$ and $C$ are on the $x$-axis, $A O=8$, $A B=A C$, $\sin \angle A B C=\frac{4}{5}$, point $D$ is on $A B$, and $C D$ intersects the $y$-axis at point $E$, satisfying $S_{\triangle C O E}=S_{\triangle A D E}$. Find the equation of the parabola passing through points $B$, $C$, and $E$.
Given $\sin \angle A B C=\frac{A O}{A B} \Rightarrow A B=10$.
By the Pythagorean theorem,
$$
B O=\sqrt{A B^{2}-A O^{2}}=6 \text {. }
$$

It is easy to see that $\triangle A B O \cong \triangle A C O$.
Therefore, $C O=B O=6$.

Thus, $A(0,-8) 、 B(6,0) 、 C(-6,0)$.
Let point $D(m, n)$.
From $S_{\triangle C O E}=S_{\triangle A O E}$, we get $S_{\triangle C B D}=S_{\triangle M O B}$.
Then $\frac{1}{2} B C|n|=\frac{1}{2} A O \cdot B O$
$$
\Rightarrow \frac{1}{2} \times 12(-n)=\frac{1}{2} \times 8 \times 6
$$
$\Rightarrow n=-4$.
Thus, $D$ is the midpoint of $A B$, i.e., $D(3,-4)$.
Therefore, $C D 、 A O$ are two medians of $\triangle A B C$.
Hence, $E$ is the centroid of $\triangle A B C$, i.e., $E\left(0,-\frac{8}{3}\right)$.
Let the equation of the parabola passing through points $B 、 C 、 E$ be $y=a(x-6)(x+6)$.
Then $-\frac{8}{3}=-36 a \Rightarrow a=\frac{2}{27}$.
Therefore, the equation of the parabola passing through points $B 、 C 、 E$ is $y=\frac{2}{27} x^{2}-\frac{8}{3}$.