+288 Let \(k\) be a positive real number. The square with vertices \((k,0),(0,k),(-k,0),\) and \((0,-k)\) is plotted in the coordinate plane.
Find conditions on \(a>0\) and \(b>0\) such that the ellipse
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
is contained inside the square (and tangent to all of its sides).
A hint came:
Suppose that the line \(x+y=k\) is tangent to the ellipse
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Algebraically, what can we say about the solutions? In particular, the number of solutions.
Sorry for the big picture...
+118687 Hi Madyl,
I like the big picture, and the title.
I'm not good on ellipses though.......
+118687 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 
