7. Given the ellipse $\Gamma_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\Gamma_{2}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$. The left and right foci of the ellipse $\Gamma_{1}$ are $F_{1}$ and $F_{2}$, respectively, and the left and right foci of the hyperbola $\Gamma_{2}$ are $F_{3}$ and $F_{4}$, respectively. A line is drawn through $F_{4}$, intersecting the ellipse at two distinct points $A$ and $B$ (with $B$ between $A$ and $F_{4}$), and the line $F_{3} A$ intersects $F_{2} B$ at point $C$. If the lines $A F_{2}$, $B F_{3}$, and $C F_{1}$ are concurrent, then $\frac{a^{2}}{b^{2}}=$ $\qquad$.
7. $\frac{3 \sqrt{2}}{4}$.

By Ceva's Theorem, we have $\frac{F_{3} F_{1}}{F_{1} F_{2}} \cdot \frac{F_{2} B}{B C} \cdot \frac{C A}{A F_{3}}=1$. By Menelaus' Theorem, we have $\frac{F_{3} F_{4}}{F_{4} F_{2}} \cdot \frac{F_{2} B}{B C} \cdot \frac{C A}{A F_{3}}=1$.
Thus, $\frac{F_{3} F_{1}}{F_{1} F_{2}}=\frac{F_{3} F_{4}}{F_{4} F_{2}}$.
$$
\begin{array}{l}
\text { Also, } F_{3} F_{1}=\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}=F_{4} F_{2}, \\
F_{1} F_{2}=2 \sqrt{a^{2}-b^{2}}, F_{3} F_{4}=2 \sqrt{a^{2}+b^{2}}, \\
\text { then } \frac{\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}{2 \sqrt{a^{2}-b^{2}}} \\
=\frac{2 \sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}} .
\end{array}
$$

Simplifying, we get $\frac{a^{2}}{b^{2}}=\frac{3 \sqrt{2}}{4}$.