6. If the line $6 x-5 y-28=0$ intersects the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0$, and $a, b$ are integers $)$ at $A 、 C$, let $B(0, b)$ be the upper vertex of the ellipse, and the centroid of $\triangle A B C$ be the right focus $F_{2}$ of the ellipse, then the equation of the ellipse is $\qquad$
【Analysis】Let $A\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$, according to the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=3 c, \\
y_{1}+y_{2}+b=0,
\end{array}\right.
$$

Combining the ellipse equation and the line equation:
$$
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \\
6 x-5 y-28=0,
\end{array}\right.
$$

we get:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{336 a^{2}}{36 a^{2}+25 b^{2}}=3 c \text { (1), } \\
y_{1}+y_{2}=-\frac{280 b^{2}}{36 a^{2}+25 b^{2}}=-b(2),
\end{array}\right.
$$
From (1) and (2) we can derive:
$$
\frac{2 a^{2}}{5 b^{2}}=\frac{c}{b},
$$

which means: $2 a^{2}=5 b c$, squaring both sides, and using $c^{2}=a^{2}-b^{2}$, we get:
$$
4 a^{4}-25 a^{2} b^{2}+25 b^{4}=0,
$$

Solving this, we get:
$$
a^{2}=5 b^{2} \text { or } a^{2}=\frac{5}{4} b^{2} \text {, }
$$