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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...

 Mar 6, 2020
 #1
avatar+108688 
+1

Hi Madyl,

I like the big picture, and the title.

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

 Mar 7, 2020
 #2
avatar+150 
0

What should I do?

madyl  Mar 7, 2020
 #3
avatar+108688 
0

Ok  Madyl, 

 

I've given it lots of thought.     indecision

 

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)           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               laugh

 Mar 8, 2020
 #4
avatar+150 
+1

Melody, I actually finished the problem and just forgot to tell you, and I got the same answer. Thanks for your imput though!! laugh

madyl  Mar 8, 2020
 #5
avatar+108688 
-1

You are welcome. 

Melody  Mar 9, 2020

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