jmc

Denote the ellipse by $\mathcal{E}.$ Let $F_1=(9,20)$ and $F_2=(49,55)$ be its foci, and let $X$ be the point where it touches the $x$-axis. By definition, $\mathcal{E}$ is the set of all points $P$ for which the quantity $P{F}_1+P{F}_2$ is equal to a particular (fixed) constant, say $k.$ Furthermore, letting $A$ and $B$ be the endpoints of the major axis, we observe that $$AB=A{F}_1+F_{1}B=F_{2}B+F_{1}B=k$$since $A{F}_1=F_{2}B$ by symmetry. That is, $k$ is the length of the major axis. Therefore, it suffices to compute the constant $k,$ given that $\mathcal{E}$ is tangent to the $x$-axis.

Note that for points $P$ strictly inside $\mathcal{E},$ we have $P{F}_1+P{F}_2<k,$ and for points $P$ strictly outside $\mathcal{E},$ we have $P{F}_1+P{F}_2>k.$ Since the $x$-axis intersects $\mathcal{E}$ at exactly one point $X$ and $X{F}_1+X{F}_2=k,$ it follows that $k$ is the smallest possible value of $P{F}_1+P{F}_2$ over all points $P$ on the $x$-axis.

Now reflect $F_1$ over the $x$-axis to point $F_1^{\prime },$ as shown: For a point $P$ on the $x$-axis, we have $P{F}_1+P{F}_2=P{F}_1^{\prime }+P{F}_2.$ Then, by the triangle inequality, $P{F}_1^{\prime }+P{F}_2\ge F_1^{\prime }{F}_2,$ and equality holds when $P$ lies on segment $\overline{F_1^{\prime }{F}_2}.$ Therefore, the smallest possible value of $P{F}_1+P{F}_2$ over all points $P$ on the $x$-axis is $F_1^{\prime }{F}_2,$ and so it follows that $k=F_1^{\prime }{F}_2.$ Then we compute $$\begin{array}{rl}{F}_1^{\prime }{F}_2& =\sqrt{(49-9{)}^2+(55-(-20){)}^2}\\ & =\sqrt{40^2+75^2}\\ & =5\sqrt{8^2+15^2}\\ & =5\cdot 17\\ & =\boxed{85}.\end{array}$$