10. (14 points) As shown in Figure 1, $A$ is the right vertex of the hyperbola $\frac{x^{2}}{4}-y^{2}=1$. Two perpendicular lines passing through $A$ intersect the right branch of the hyperbola at points $M$ and $N$, respectively. Is the line $MN$ guaranteed to pass through a fixed point on the $x$-axis? If such a fixed point does not exist, please explain the reason; if there is such a fixed point $P$, try to find the coordinates of this fixed point $P$.
10. Clearly, the right vertex $A(2,0)$.

Translate the $y$-axis to the right by 2 units, making $A$ the origin of the new rectangular coordinate system. In the new coordinate system, the equation of the hyperbola is
$$
\frac{\left(x^{\prime}+2\right)^{2}}{4}-y^{2}=1 \text {, }
$$

which simplifies to $4 y^{2}-x^{\prime 2}-4 x^{\prime}=0$.
If $M N \perp x$-axis, then the slope of $A M$ is 1, i.e., $l_{A M}: y=x^{\prime}$.
Substituting into equation (1) yields $M\left(\frac{4}{3}, \frac{4}{3}\right)$.
Consequently, $N\left(\frac{4}{3},-\frac{4}{3}\right)$.
Thus, the intersection point of $M N$ with the $x$-axis is $P\left(\frac{4}{3}, 0\right)$.
If $M N$ is not perpendicular to the $x$-axis, let
$l_{M N}: y=k x^{\prime}+t(t \neq 0)$.
Then $\frac{y-k x^{\prime}}{t}=1$.
Thus, equation (1) can be rewritten as
$$
4 y^{2}-x^{\prime 2}-4 x^{\prime} \cdot \frac{y-k x^{\prime}}{t}=0 \text {, }
$$

which simplifies to $4 t\left(\frac{y}{x^{\prime}}\right)^{2}-4\left(\frac{y}{x^{\prime}}\right)+4 k-t=0$.
The two roots $k_{1} 、 k_{2}$ of the quadratic equation in $\frac{y}{x^{\prime}}$ are the slopes of $A M$ and $A N$.
Since $A M \perp A N$, we have
$$
k_{1} k_{2}=\frac{4 k-t}{4 t}=-1 \text {. }
$$

Solving for $t$ gives $t=-\frac{4}{3} k$.
Thus, $l_{M N}: y=k x-\frac{4}{3} k$, which also passes through the point $P\left(\frac{4}{3}, 0\right)$.
In summary, in the new coordinate system, the line $M N$ passes through the fixed point $P\left(\frac{4}{3}, 0\right)$ on the $x$-axis, i.e., in the original coordinate system, the line $M N$ passes through the fixed point $P\left(\frac{10}{3}, 0\right)$ on the $x$-axis.