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I know that it is possible to construct an approximation of ellipses using 4 arcs, but are there any ellipses that can be constructed from arcs with 100 percent accuracy?

Usually, the construction would be slightly off: approximation vs real

Is it possible the there exists an ellipse where this is not the case? And it can be perfectly represented using arcs?

I imagine this is impossible. But in my research, I never managed to find a definitive answer.

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    $\begingroup$ Yes, they are called circles. :-) $\endgroup$ Commented Aug 15, 2021 at 23:14
  • $\begingroup$ ...indeed ONLY circles. $\endgroup$ Commented Aug 15, 2021 at 23:20
  • $\begingroup$ The orthogonal lines to the tangents of a circle all intersect at a single point. This is not true for an arc of an ellipse. $\endgroup$ Commented Aug 15, 2021 at 23:27

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We claim that an arc on an ellipse is the arc of a circle if and only if that ellipse is a circle. Indeed, look at the curvature: An arc of a circle has constant curvature. By direct calculation, though, you can check (see how to calculate the curvature of an ellipse) that in a non-circular ellipse, the curvature varies in such a way that it is never locally constant. (To see this, look at the linked answer's formula. Specifically, focus on $\sqrt{a^2\sin^2 t + b^2\cos^2 t}$. Calculus tells us this radical expression has derivative $$(a^2-b^2)\frac{\sin(2t)}{2\sqrt{a^2\sin^2 t + b^2\cos^2 t}}.$$ Note that as long as $a \neq b,$ this will only be zero at isolated points, so there can't be any small arc of an ellipse on which the curvature is constant.)

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    $\begingroup$ See here for a more geometric construction of the radius of curvature of an ellipse: math.stackexchange.com/questions/3455731/… $\endgroup$ Commented Aug 16, 2021 at 8:26
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    $\begingroup$ Ah, thinking about it in terms of curvature is what confirmed it for me. Thanks $\endgroup$ Commented Aug 16, 2021 at 12:10

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