* An ellipse on the $x y$ plane, with foci at $(9,20)$ and $(49,55)$, and tangent to the $x$-axis. Find the length of the major axis.
Analysis one: The foci of the ellipse are $F_{1}(9,20), F_{2}(49,55)$. The point $F_{2}$ reflected over the $x$-axis is $F_{2}^{\prime}(49,-55)$. We prove: the length of the major axis $k$ of the ellipse is equal to the length of the line segment $F_{1} F_{2}^{\prime}$. Let $P$ be the point where the ellipse touches the $x$-axis, and the line segment $F_{1} F_{2}^{\prime}$ intersects the $x$-axis at $P^{\prime}$, and assume $P^{\prime}$ and $P$ are different points. By the triangle inequality, we have
$$
\left|P F_{1}\right|+\left|P F_{2}^{\prime}\right|>\left|F_{1} F_{2}^{\prime}\right|=\left|P^{\prime} F_{1}\right|+\left|P^{\prime} F_{2}^{\prime}\right|=\left|P^{\prime} F_{1}\right|+\left|P^{\prime} F_{2}\right| \text {. }
$$

By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is equal to $k$, and the sum of the distances from any point outside the ellipse to the two foci is greater than $k$. By the definition of a tangent, $P$ is on the ellipse, and $P^{\prime}$ is outside the ellipse, so we have
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=k \text {, (1) }\left|P^{\prime} F_{1}\right|+\left|P^{\prime} F_{2}\right|>k \text {. }
$$
(2) From (1) and (2), we get
$$
\begin{aligned}
\left|P F_{1}\right|+\left|P F_{2}^{\prime}\right| & =\left|P F_{1}\right|+\left|P F_{2}\right| \\
& =k\left|P^{\prime} F_{1}\right|+\left|P^{\prime} F_{2}\right| \text{, a contradiction. } \text{So } P \text{ and } P^{\prime} \text{ must be the same point, thus } k=\left|F_{1} F_{2}^{\prime}\right|. \text{ By the distance formula, } k=85. \text{ Therefore, the length of the major axis of the ellipse is 85. }
\end{aligned}
$$

Analysis two: Let the ellipse touch the $x$-axis at $P(x, 0)$. Then $k_{P F_{2}}=-k_{P F_{1}}$. So $\frac{55-0}{49-x}=$ $-\frac{20-0}{9-x}$, which gives $x=\frac{59}{3}$. Let the length of the major axis be $k$, then
$$
k=\left|P F_{1}\right|+\left|P F_{2}\right|=\sqrt{\left(9-\frac{59}{3}\right)^{2}+20^{2}}+\sqrt{\left(40-\frac{59}{3}\right)^{2}+55^{2}}=85 .
$$