Geometrically interpreting the answer to a vector calculus question involving tangent line segments to ellipses. Let E be an ellipse centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question:

Can you give a geometric reason for why the length is a+b?

This was originally asked to me by Frank Jones a few years ago.
 A: By working simultaneously in the 4 quadrants, this becomes a question of minimizing the perimeter of enclosing rhombi with diagonals on the coordinate axes.  Proving the inequality was Problem of the Week No. 13 in Spring 2005 at Purdue.  Here is Steven Landy's solution to the problem.  The proof is geometric in the sense of Descartes rather than Euclid, and shows that the minimum is at least a+b.  There's no calculus, so maybe this is close to what you're looking for.  (Edit: I entertained the delusion that this might be close to answering your question only before akopyan's answer was posted.)
A: There is a geometric way to show that $n$-gon circumscribed around an ellipse has minimal perimeter if it is inscribed in a confocal ellipse. From Poncelet porism (and generalization of optical property) it follows that we have continuous family of "minimal" polygons.
If we know it, then it is easy to understand that the circumscribed rhomb (from your question) and the circumscribed rectangular (with perimeter $4(a+b)$) are minimal polygons. So, side of the rhomb equals $a+b$.
