Hello, killasnake,

as you may know an ellipse is constructed by mapping the circle

x² + y² = a²

perpendicularly to the x-axis with the ratio .

So first I was looking for the triangle which is build by the coordinate axes and a tangent to thecirclewith the smallest area. It is the right isoscele triangle which area is a².

When you map the circle to the ellipse you are mapping this triangle too. The base of the triangle is constant, only the height of the triangle changes with the same ratio as all points of the coordinate plane.

Therefore the original smallest triangle has the area , thus

the area of the smallest triangle at the ellipse is

you asked for the equation of the tangent:

EB

PS.: The preview doesn't work properly so I send you this reply without checking my text. There are of course some ugly mistakes but I hope you can use this post as a starter.