TITLE: The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
QUESTION [5 upvotes]: The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $       

It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it looks like this:

The problem is that I'm not sure that an ellipse here can be a finite area
I will be glad to any hint or solution

REPLY [4 votes]: Consider the case of the curves $y= \pm\frac{1}{x^2+1}$ (the general case can reduced to this case). The ellipse of largest area is not $2 x^2 + y^2=1$ but $ \frac{x^2}{2} + 2 y^2 =1$, tangent to the graphs at points $(\pm 1, \pm \frac{1}{2})$.

There is a family of ellipses tangent to the graph 
$$\frac{x^2}{u}+\frac{y^2}{v}=1$$ where $v= \frac{27 u}{4(u+1)^3}$. The maximum of $uv = \frac{27 u^2}{4(u+1)^3}$ occurs at $u=2$, with $v=\frac{1}{2}$.
So, take your ellipse, double the horizontal diameter, and divide the vertical diameter by $\sqrt{2}$ to get the maximal ellipse. Its area is $\pi$,  $1/2$ the area between the curves.  Note that the common tangent to the ellipse and the curve passes through the vertex of the curve.