SOLUTION: For this question, (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. It is possible to draw an ellips

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: For this question, (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. It is possible to draw an ellips      Log On


   

Question 1173061: For this question, (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. It is possible to draw an ellipse so that it is tangent to all sides of the square; several examples are shown below.
(https://latex.artofproblemsolving.com/6/e/6/6e67d2c4f7235dab1e7d931aa1228b4ef76678e2.png)
Find necessary and sufficient conditions on a > 0 and b > 0 such that the ellipse
{x^2}/{a^2} + {y^2}/{b^2} = 1 is contained inside the square (and tangent to all of its sides).
Make sure to prove that your conditions are both necessary (the ellipse being tangent to all the square's sides implies your conditions) and sufficient (your conditions imply the ellipse being tangent to all the square's sides))
I have gotten to the part where the I have substituted y+x=k into {(x^2)/a^2)}+ {(y^2)/(a^2)} and simpified it down to (b^2+a^2)x^2-2a^2kx+a^2k^2-a^2b^2=0. Can someone show me what to do from here? I really, really appreciate the help!
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!


Solving the system:

system%28x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2=1%2Cx%2By=k%29 

gives

%28b%5E2%2Ba%5E2%29x%5E2-2a%5E2kx%2Ba%5E2k%5E2-a%5E2b%5E2=0

which you have found.  That is the quadratic equation

Ax%5E2%2BBx%2BC=0, where

A=b%5E2%2Ba%5E2, B=-2a%5E2k, C=a%5E2k%5E2-a%5E2b%5E2

Its discriminant is B%5E2-4AC. Substituting,

%28-2a%5E2k%29%5E2-4%28b%5E2%2Ba%5E2%29%28a%5E2k%5E2-a%5E2b%5E2%29

which simplifies to 

4a%5E2b%5E2%28a%5E2+%2B+b%5E2+-+k%5E2%29

The discriminant must be 0 in order for there to be only one solution
where the ellipse just touches each side of the square at only one
point.

4a%5E2b%5E2%28a%5E2+%2B+b%5E2+-+k%5E2%29=0

can only be true if and only if

a%5E2%2Bb%5E2=k%5E2

Any of the four lines x+y=k, x-y=k, -x+y=k,-x-y=k 
gives the same discriminant, when solved simultaneously with
the ellipse.

So a%5E2%2Bb%5E2=k%5E2 is the necessary and sufficient condition.

Edwin