<s>[INST] Maximum ellipse inscribed in Witch of Agnesi curve An ellipse with variable $(2a,2b)$ axes parallel to the $(x,y)$ coordinate axes is inscribed inside fixed curve of equation.
$$ y=\pm\dfrac{1}{1+x^2}$$

Show that maximum ellipse area occurs when it touches the curve at its inflection point.
I am looking to generalizing a variable ellipse contact point with a curve having an inflection, like in the recent Bell Curve post. My intuition needs to be validated or disproved later using simple methods of differential calculus.
 [/INST] Take $ f $ an even function and consider $ \mathcal E $ the ellipse that touches $ f $ in $ (c, f(c)) $. Suppose it has equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ (and thus an area of $ ab $). You get, from $ (c, f(c)) \in \mathcal E $ and from $ f $ tangent to $ \mathcal E $,
$$ \frac{c^2}{a^2} + \frac{f(c)^2}{b^2} = 1 $$
$$ \frac c{a^2} + \frac{f(c)f'(c)}{b^2} = 0 $$
Solving this, you find $ a^2 = c^2 - \frac{c f(c)}{f'(c)} $ and $ b^2 = f(c)^2 - cf(c)f'(c) $. Thus you want to find the maximum of
$$ a^2b^2 = \left(c^2 - \frac{c f(c)}{f'(c)}\right)\left(f(c)^2 - cf(c)f'(c)\right) = 2c^2f(c)^2 - c^3f(c)f'(c) - \frac{cf(c)^3}{f'(c)} $$
Deriving with respect to $ c $, you get
$$ \begin{eqnarray}
\frac{\mathrm da^2b^2}{\mathrm dc}
& = & 4cf(c)^2 + 4c^2f(c)f'(c) - 3c^2f(c)f'(c) - c^3f'(c)^2 - c^3f(c)f''(c) - \frac{(f(c)^3 + 3cf(c)^2f'(c))f'(c) - cf(c)^3f''(c)}{f'(c)^2} \\
& = & \frac{cf(c)^2f'(c)^2 + c^2f(c)f'(c)^3 - c^3f'(c)^4 - c^3f(c)f'(c)^2f''(c) - f(c)^3f'(c) + cf(c)^3f''(c)}{f'(c)^2}
\end{eqnarray} $$
There's no connection between this derivative vanishing and $ f''(c) = 0 $, so your conjecture is wrong.
In the case of the Witch of Agnesi, the inflexion points are $ c = \pm \frac 1{\sqrt 3} $ and this doesn't correspond to the ellipse of maximal area. Indeed,
$$ a^2b^2
 = 2c^2f(c)^2 - c^3f(c)f'(c) - \frac{cf(c)^3}{f'(c)}
 = \frac{2c^2}{(1 + c^2)^2} + \frac{2c^4}{(1 + c^2)^3} + \frac 1{2(1 + c^2)}
 = \frac{4c^2(1 + c^2) + 4c^4 + (1 + c^2)^2}{2(1 + c^2)^3}
 = \frac{(1 + 3c^2)^2}{2(1 + c^2)^3} \le 1 $$
with equality iff $ c = \pm 1 $. (the last inequality is equivalent to $ \frac{c^6 + c^6 + 1}3 \ge c^4 $ which is true iff $ c^6 = 1 $ by the arithmetico-geometric inequality)
Surprisingly, the zeroes of $ f''' $ are $ 0, \pm 1 $ and correspond to the extrema of $ ab $. This is not the case in general.
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