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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: enter image description here

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

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  • $\begingroup$ Is it important that $c>0$? $\endgroup$
    – John Glenn
    May 9 '18 at 6:34
  • $\begingroup$ I realized that my decision is exactly wrong, because at some with an ellipse will not be contained in curves ... $\endgroup$ May 9 '18 at 6:35
  • $\begingroup$ Yes, this is important, otherwise it turns out to be a completely different task $\endgroup$ May 9 '18 at 6:36
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Given the function for the ellipse as: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}$$ It is clear that $b$ must be equal to $\frac1c$ so that the ellipse must be below the curve $\frac{\pm1}{x^2+c}$ Solving for $y$ in $(1)$ gives us: $$\left\{y\to\frac{b \sqrt{a^2-x^2}}{a}\right\},\left\{y\to -\frac{b \sqrt{a^2-x^2}}{a}\right\}$$ We know that $b=\frac1c$, therefore, half of the ellipse can be written as: $$y=\frac{\sqrt{a^2-x^2}}{a c}$$ Now, in order for the ellipse to be under the curve, we must ensure that: $$\frac{\sqrt{a^2-x^2}}{a c}<\frac{1}{c+x^2}$$ Which can be rewritten as: $$\frac{1}{c+x^2}-\frac{\sqrt{a^2-x^2}}{a c}>0$$ Now the equality is achieved when for the following solutions in terms of $x$: $$\left\{\{x\to 0\},\left\{x\to -\frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to -\frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\}\right\}$$ Meaning, these are values of $x$ in the $xy$-plane where the ellipse and the curves "intersect". Now, we want this solutions equal to $0$, so that the graphs of the functions only ever intersect at $x=0$ (if that is meaningful). Now, we get the solutions as: $$a\to\pm\frac{\sqrt{c}}{\sqrt{2}}$$ Since $a$ must be greater than 0, we reject the negative answer. Now we have $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, now half of the ellipse is defined by: $$y=\frac{\sqrt{2} \sqrt{\frac{c}{2}-x^2}}{c^{3/2}}$$ Square both sides, and you get the function for the ellipse as: $$y^2=\frac{2 \left(\frac{c}{2}-x^2\right)}{c^3}$$


The area the ellipse is given by: $$A=2\int_{-a}^{a}\frac{b \sqrt{a^2-x^2}}{a}dx$$ Given $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, the maximum area of the ellipse should be: $$A=2\int_{-\sqrt{\frac{c}{2}}}^{\sqrt{\frac{c}{2}}}\frac{\sqrt2\sqrt{c/2-x^2}}{c^{3/2}}dx=2\frac\pi{2\sqrt{2c}}=\frac\pi{\sqrt{2c}}$$

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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 $(\pm 1, \pm \frac{1}{2})$.

enter image description here

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.

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Consider a point on the curve $\frac1{x^2+c}$, $(p, \frac1{p^2+c}$). A ellipse is drawn which is contained in the curves and it is tangent at this point. This means that $$a\cos\theta=p\\ b\sin\theta=\frac1{p^2+c} \\ \implies A=\pi ab=\frac{2p\pi}{(p^2+c)\sin2\theta}$$ For maximum and minimum area we should take the derivative wrt to $p$ and $\theta$, and set them to $0$.

$$\frac{\partial A}{\partial\theta}=\frac{4p\pi \cos2\theta}{(p^2+c)\sin^22\theta}=0 \\ \frac{\partial A}{\partial p}=\frac{2(c-p^2)\pi}{(p^2+c)^2\sin2\theta}=0$$

This gives $\theta=\pi/4$ from the first equation and the $p= \sqrt c$. So the area is $$A=\pi/\sqrt c$$

To show that this ellipse is completely contained in the curves we must show that tangent of this ellipse at $\theta=\pi/4$ is same as tangent at for the curve at $p=\sqrt c$ which is true in this case. So this is the completely contained ellipse.

Incomplete:

Although what this is missing is, how to tell if this is maxima or minima. Plus it feels like the cases of $x=0, \infty$ aren't well accommodated here.

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Expanding in Taylor series around $x=0$

$$ y = b\sqrt{1-\left(\frac{x}{a}\right)^2} = b-\frac{b x^2}{2 a^2}-\frac{b x^4}{8 a^4}+O\left(x^5\right) $$

and

$$ y = \frac{1}{(x^2+c)} = \frac{1}{c}-\frac{x^2}{c^2}+\frac{x^4}{c^3}+O\left(x^5\right) $$

Now solving $$ \left\{ \begin{array}{rcl} b & = & \displaystyle\frac{1}{c}\\ \displaystyle\frac{b}{2 a^2} & = & \displaystyle\frac{1}{c^2} \end{array} \right. $$

we get

$$ a = \sqrt{\frac{c}{2}}, b = \frac{1}{c} $$

and the area is

$$ S = \pi a b = \frac{\pi}{\sqrt{2 c}{}} $$

NOTE

In the series expansion the third term is negative for the ellipse.This guarantees the inclusion.

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  • $\begingroup$ If you considered the Taylor expansions of $1/y$, one has just two terms, and the other only positive terms, so the inequality is obvious. $\endgroup$
    – orangeskid
    May 9 '18 at 11:52
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I don't think that your ellipse is maximal. In order to decrease the number of parameters I take $y=\pm{1\over 1+x^2}$ as bounding curves and then look for the largest ellipse. Experimenting has shown that the largest ellipse looks as in the following figure. Its area is about $40\%$ larger than the area of your ellipse.

enter image description here

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