1985 AIME Problem 11

At tangency, the constant sum of distances is the minimum such sum for a point on the $x$-axis

Reflect one focus across the $x$-axis to turn the broken path into a straight segment

Reflect $(49,55)$ across the $x$-axis to $(49,-55).$ For a point $P$ on the $x$-axis, the sum of its distances to the original foci equals the length of a broken path from $(9,20)$ through $P$ to $(49,-55).$ Its minimum is the straight-line distance $$\begin{aligned} &\sqrt{(49-9)^2+(-55-20)^2}\\ &\qquad{}=\sqrt{40^2+75^2}=85. \end{aligned}$$ Tangency means that the ellipse’s constant distance sum equals this minimum. That sum is the major-axis length, so the answer is $85.$