An [ellipse](https://artofproblemsolving.com/wiki/index.php/Ellipse) has [ foci](https://artofproblemsolving.com/wiki/index.php/Focus) at $(9,20)$ and $(49,55)$ in the $xy$-plane and is [ tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_line) to the $x$-axis. What is the length of its [major axis](https://artofproblemsolving.com/wiki/index.php/Major_axis)?
An ellipse is defined to be the [locus](https://artofproblemsolving.com/wiki/index.php/Locus) of points $P$ such that the sum of the distances between $P$ and the two foci is constant.  Let $F_1 = (9, 20)$, $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$-axis.  Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal.  (The last claim begs justification:  Let $F'_2$ be the reflection of $F_2$ across the $x$-axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse.  We will show that $X=Y$. Note that  $X F_2 = X F’_2$ since $X$ is on the $x$-axis.  Also, since the entire ellipse is on or above the $x$-axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$-axis, we must have $F_2 Y \leq F’_2 Y$ with equality if and only if $Y$ is on the $x$-axis.  Now, we have
\[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \leq F_1 Y + F’_2 Y\]
But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$., so this is only possible if we have equality and thus $X = Y$). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$, we can reflect (as above) the second leg of this path (from $X$ to $F_2$) across the $x$-axis.  Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$-axis, and this path will have shortest length exactly when our original path has shortest length.  This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$axis.  

[asy] size(200); pointpen=black;pathpen=black+linewidth(0.6);pen f = fontsize(10); pair F1=(9,20),F2=(49,55); D(shift((F1+F2)/2)*rotate(41.186)*scale(85/2,10*11^.5)*unitcircle); D((-20,0)--(80,0)--(0,0)--(0,80)--(0,-60)); path p = F1--(49,-55); pair X = IP(p,(0,0)--(80,0)); D(p,dashed);D(F1--X--F2);D(F1);D(F2);D((49,-55)); MP("X",X,SW,f); MP("F_1",F1,NW,f); MP("F_2",F2,NW,f); MP("F_2'",(49,-55),NE,f); [/asy]
The sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$, which by the [distance formula](https://artofproblemsolving.com/wiki/index.php/Distance_formula) is just $d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85$.  
Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse.  Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $\boxed{085}$.