Ok Madyl,
I've given it lots of thought.
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) and you know that a tangent to this is y+x=k
Substitute y=-x+k into the ellipse formula and simplify it.
It will simplify down to \((a^2+b^2)x^2+(-2a^2k)x+(a^2k^2-a^2b^2)=0\\ \)
Now this is a quadratic but since there is only one point of intersection (since one is a tangent to the other) there can be only 1 solution.
So that means the discriminant must equal =0
so find the discriminant and set it equal to 0.
Solve it
You will end up with \(a^2+b^2=k^2\)
Since a, b and k are all positive
The restriction on a and b is \(\bf{k=\sqrt{a^2+b^2}}\)
Here is the Desmos graph: https://www.desmos.com/calculator/p2ywsrcrfu