5. Through the right focus $F$ of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$

draw $B_{1} B_{2} \perp x$ axis, intersecting the hyperbola at points $B_{1}$ and $B_{2}$. The line connecting $B_{2}$ and the left focus $F_{1}$ intersects the hyperbola at point $B$. Connecting $B_{1} B$ intersects the $x$ axis at point $H$. Then the x-coordinate of $H$ is $\qquad$
5. $-\frac{a^{2}}{c}$.

From the given, we have point $B_{2}\left(c,-\frac{b^{2}}{a}\right)$.
By combining the equation of line $B_{2} F_{1}$ with the hyperbola equation, we get
$$
\left(1-\frac{b^{2}}{4 c^{2}}\right) x^{2}-\frac{b^{2}}{2 c} x-\frac{b^{2}}{4}-a^{2}=0 \text {. }
$$

By Vieta's formulas, we have
$$
c x_{B}=\frac{b^{2} c^{2}+4 a^{2} c^{2}}{b^{2}-4 c^{2}} \text {. }
$$

By Menelaus' theorem, we know
$$
\frac{F_{1} H}{H F}=\frac{B_{1} B_{2}}{F B_{1}} \cdot \frac{B F_{1}}{B B_{2}}=2 \cdot \frac{x_{B}+c}{c-x_{R}}=\frac{b^{2}}{2 c^{2}-b^{2}} \text {. }
$$

It is easy to see that, $x_{H}=-\frac{a^{2}}{c}$.