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...

madyl Mar 6, 2020

#1**+1 **

Hi Madyl,

I like the big picture, and the title.

I'm not good on ellipses though.......

Melody Mar 7, 2020

#3**+1 **

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

Melody Mar 8, 2020