An ellipse with variable $(2a,2b)$ axes parallel to the $(x,y)$ coordinate axes is inscribed inside fixed curve of equation.

$$ y=\pm\dfrac{1}{1+x^2}$$ WofA

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.

  • $\begingroup$ Thanks. it was a vague feeling of expectation or a hunch before I posted the proposition without calculations.. and later found to be erroneous, not quite as per earlier expectation. Even if reputation compromise I aired what I felt in order to verify it / get verified qualitatively ( that is also a mathematical direction)... and am overwhelmed with your encouragement. $\endgroup$ – Narasimham Jun 5 at 3:08

A Witch of Agnesi of height $2a$, and an ellipse of radii $p$ and $q$, are parameterized by $$(x,y) = (2a\tan\theta,2a\cos^2\theta) \qquad (x,y) = (p \cos\phi, q \sin\phi) \tag{1}$$ Respective tangent vectors are given by $$(x',y') = (2 a\sec^2\theta,-4a\cos\theta\sin\theta) \qquad (x',y')=(-p\sin\phi,q\cos\phi) \tag{2}$$ Inscribing the ellipse in the witch requires that the points in $(1)$ match and the vectors in $(2)$ are proportional, so we have this system $$\begin{align} 2 a \tan\theta &= \phantom{-}p \cos\phi \\ 2 a \cos^2\theta &= \phantom{-}q\sin\phi \\ 2 a k \sec^2\theta &= -p \sin\phi \\ 4 a k \cos\theta\sin\theta &= -q \cos\phi \end{align}\tag{3}$$ We can solve the first three equations as a linear system in $p$, $q$, $k$: $$ k =-\frac{2 a \cos\theta\sin\theta \sin\phi}{\cos\phi} \qquad p =\frac{2 a \sin\theta}{\cos\theta\cos\phi} \qquad q =\frac{2 a \cos^2\theta}{\sin\phi} \tag{4}$$ Substituting into the fourth equation of $(4)$ we find (after discarding an extraneous factor of $\cos\theta$) $$\sin^2\phi = \frac{1}{1+2\sin^2\theta}\quad\to\quad \cos^2\phi = \frac{2\sin^2\theta}{1+2\sin^2\theta} \tag{5}$$ Therefore, the area of the ellipse is given by $$\pi p q = \frac{4\pi a^2 \sin\theta\cos\theta}{\sin\phi\cos\phi} = 2\pi a^2 \sqrt2 \cos\theta (1 + 2 \sin^2\theta) \tag{6}$$ To find critical points of $(6)$ we equate its derivative to zero: $$\cos2\theta\sin\theta = 0 \quad\to\quad \theta=\frac\pi4 \quad\to\quad \pi p q = 4 \pi a^2 \tag{7}$$ That's all well and good (and surprisingly simple), but note that the witch's inflection point corresponds to $\theta=\pi/6$, so the ellipse of maximal area does not touch that point. $\square$

Here's a walk-through of the general case. Let a curve be parameterized as $$(x,y) = (u(t),v(t)) \qquad (x',y') = (u'(t),v'(t)) \tag{1',2'}$$ (where I'll suppress the parameter going forward). Solving the corresponding $p$-$q$-$k$ system gives $$p = u \sec\phi \quad q = v \csc\phi \quad k = -\frac{u'}{u}\cot\phi \tag{4'}$$ and from the fourth equation we get $$\cos^2\phi =\frac{uv'}{uv'-u'v} \qquad \sin^2\phi = -\frac{u'v}{uv'-u'v} \tag{5'}$$ $$(\pi pq)^2 = -\pi^2 \frac{uv}{u'v'}\left(uv'-u'v\right)^2 \tag{6'}$$ Differentiating, and assuming $uv'-vu'\neq 0$, yields these conditions for the critical values of $(6')$: $$u v' + u'v = 0 \qquad\text{or}\qquad u v(u' v''-v'u'') = u'v'( u v'-u' v) \tag{7'}$$ that is, $$(uv)' = 0 \qquad\text{or}\qquad \left(\frac{uv'}{u'v}\right)' = 0 \tag{7''}$$ so that, respectively, $$\pi p q = 2\pi u v \qquad\text{or}\qquad (\pi pq)^2 = -\pi^2 \frac{(uv'-v'u)^3}{u'v''-u''v'} \tag{8'}$$ In the case of the witch, the second condition of $(7')$ gives extraneous or minimizing values, so that we rely on the first condition to get the first value of $(8')$ as the maximum area. It's not clear to me if we can always discount the second condition of $(7')$.

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  • $\begingroup$ So neat this !! It establishes that the proposition does not hold good. How the Bell satisfies the condition and whether there are other parallel examples, one could still investigate. The witch has third derivative vanishing and not the second. Hope calculation was enjoyable. $\endgroup$ – Narasimham Jun 4 at 19:19
  • $\begingroup$ Marvellous.. Now did you check how the Bell rings in well, while the witch does n't? $\endgroup$ – Narasimham Jun 5 at 5:52
  • $\begingroup$ It's straightforward parameterize the Bell via $(u,v) = (t,\exp(-t^2))$ and find from the first condition in $(7')$ (here again, the second condition gives discardable values) that the critical points occur when $t=\pm 1/\sqrt{2}$, which happen to correspond exactly with the Bell's inflection points. ... To perhaps characterize curves that have this (or almost-this) property, we'd want the curvature to vanish. (continued) $\endgroup$ – Blue Jun 5 at 6:19
  • $\begingroup$ (continuing) Vanishing curvature requires $u′v′′=u′′v′$. This makes the left-hand side of the second condition of $(7′)$ zero, which (since $uv'\neq u'v$) means $u′$ or $v′$ is zero: that is, we have a horizontal or vertical tangent, which (probably) means a "flat" ellipse of zero area; not a maximum. Thus, we (probably) need only consider the first condition in $(7′)$, so that the curves in question are (probably) characterized by $(uv)'=0=u'v''−u''v'$ for some $t$. I'm not sure there's anything particularly profound in that characterization, but I haven't given it much thought. $\endgroup$ – Blue Jun 5 at 6:39
  • $\begingroup$ Next steps are simple, differentials recognizable.. Let me see why the Bell is or is not unique.. $\endgroup$ – Narasimham Jun 5 at 6:44

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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  • $\begingroup$ Thanks for calculation, finding that the third derivative vanishes at maximum area is new. It was a vague feeling in the direction of intuition, but not a conjecture..I made no calculation before the post while seeking your help. Thanks once again. $\endgroup$ – Narasimham Jun 4 at 19:26
  • $\begingroup$ Please copy your answer here to this link: math.stackexchange.com/questions/3798002/…. In a week, I may copy your answer into the link so that the answer is preserved. The answer will be a Community Wiki post, and you can remove the part that belongs to you, and put it in your own separate answer. (I'm not sure if you were notified from my comment in the deleted post.) $\endgroup$ – Batominovski Aug 21 at 14:29

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