8. Let $O$ be the origin, $B(4,0), C(5,0)$, and draw a perpendicular line to the $x$-axis through point $C$. Let $M$ be a moving point on this perpendicular line. Construct a circle with $O$ as the center and $OB$ as the radius. $M T_{1}$ and $M T_{2}$ are tangents to the circle. Then the equation of the locus of the orthocenter of $\triangle M T_{1} T_{2}$ is $\qquad$
8. $\left(x-\frac{16}{5}\right)^{2}+y^{2}=\left(\frac{16}{5}\right)^{2}(x>0)$.

The equation of the circle with center $O$ and radius $O B$ is
$$
x^{2}+y^{2}=16 \text {. }
$$

As shown in Figure 2, connect $O T_{1}$ and $O T_{2}$. Let $H(x, y)$ be the orthocenter of $\triangle M T_{1} T_{2}$, $M(5, b)$, and $N$ be the intersection of $O M$ and $T_{1} T_{2}$.
It is easy to prove that quadrilateral $O T_{2} H T_{1}$ is a rhombus.
Then $N\left(\frac{x}{2}, \frac{y}{2}\right)$, and $\frac{b}{5}=\frac{y}{x}$.
Since $O M \perp T_{1} T_{2}$ and $O T_{1} \perp M T_{1}$, then
$$
\begin{array}{l}
O T_{1}^{2}=O N \cdot O M . \\
\text { Therefore }\left(x-\frac{16}{5}\right)^{2}+y^{2}=\left(\frac{16}{5}\right)^{2}(x>0) .
\end{array}
$$