5. Take any three points $A, B, C$ on the hyperbola $x^{2}-y^{2}=2$, then the locus equation of the orthocenter $H$ of $\triangle A B C$ is $\qquad$.
5. $x^{2}-y^{2}=2$.

Let the coordinates of the three vertices of $\triangle ABC$ be $A\left(\frac{t_{1}^{2}+1}{\sqrt{2} t_{1}}, \frac{t_{1}^{2}-1}{\sqrt{2} t_{1}}\right)$, 
$$
B\left(\frac{t_{2}^{2}+1}{\sqrt{2} t_{2}}, \frac{t_{2}^{2}-1}{\sqrt{2} t_{2}}\right), C\left(\frac{t_{3}^{2}+1}{\sqrt{2} t_{3}}, \frac{t_{3}^{2}-1}{\sqrt{2} t_{3}}\right),
$$
where $t_{1}, t_{2}, t_{3}$ are all non-zero and distinct.
Since $\triangle ABC$ has at most one side parallel to the $y$-axis, we can assume that $BC$ and $AC$ are not parallel to the $y$-axis, so the slopes of lines $BC$ and $AC$ both exist, and are given by
$$
k_{BC}=\frac{t_{2} t_{3}+1}{t_{2} t_{3}-1}, \quad k_{AC}=\frac{t_{1} t_{3}+1}{t_{1} t_{3}-1}.
$$

Then the equation of $AH$ is
$$
y-\frac{t_{1}^{2}-1}{\sqrt{2} t_{1}}=-\frac{t_{2} t_{3}-1}{t_{2} t_{3}+1}\left(x-\frac{t_{1}^{2}+1}{\sqrt{2} t_{1}}\right).
$$

Rearranging, we get
$$
\begin{array}{l}
t_{1} t_{2} t_{3}(x+y)-t_{1}(x-y) \\
=\sqrt{2}\left(t_{1}^{2} t_{2} t_{3}-1\right).
\end{array}
$$

It is easy to verify that this equation includes the cases where the slope of $AH$ does not exist or is zero.
Similarly, the equation of $BH$ is
$$
\begin{array}{l}
t_{3} t_{3} t_{1}(x+y)-t_{2}(x-y) \\
=\sqrt{2}\left(t_{2}^{2} t_{3} t_{1}-1\right).
\end{array}
$$
$$
\left\{\begin{array}{l}
x-y=-\sqrt{2} t_{1} t_{2} t_{3}, \\
x+y=-\frac{\sqrt{2}}{t_{1} t_{2} t_{3}}.
\end{array}\right.
$$

Eliminating $t_{1} t_{2} t_{3}$, we obtain the equation of the locus of the orthocenter as $x^{2}-y^{2}=2$.