If we write the covector (finally we understand this notation) (careful, these L's may look like 1's on some screens)
k dx + l dy
then the tensor maps this covector to k^2 + 2 l^2. Thus we have
k^2 + 2 l^2 = 1
This looks just like a dispersion relation, albeit homogeneous of degree two (that is what makes it a tensor).
A single line representing one of these covectors will have the equation
k x + l ly = 1
To find the envelope, we want this line to intersect the line with slightly different k and l values, consistent with the above relation. This gives us another condition
- 2 l x + k y = 0
To derive that, think of k and l depending upon some parameter, and differentiate both the dispersion relation and the equation of the line with respect to this parameter and simplify.
Now if we square the above two equations and add twice the first to the second we find that we can eliminate both k and l, to find
2 x^2 + y^2 = 1.
Thus the envelope of the above dispersion relation, and the above tensor, is an ellipse.
2 k l = 1
k x + l y = 1
-k x + l y = 0
And now square and subtract the two equations to find
2 x y = 1
as the equation of the envelope. This is a rectangular hyperbola.