Is the Oval ( based on Ptolemy inequality) known?

Question

It has a property of enclosing quadrilaterals so the ratio of their diagonals product and sum of their opposite sides pair products is constant $(e<1)$. The curve is from a family defined by the Ptolemy Inequality

In order to rope in the Ptolemy Inequality Oval took three points on a unit generating circle radius $ a=1 $ and the fourth one outside the circle

$$(-1,0),(0,-1),(1,0),(x,y)$$

as particular vertices of a non-cyclic quadrilateral. The ratio $e$ defines its equation.

$$ \dfrac{\sqrt 2 \sqrt{x^2+(1+y)^2}}{\sqrt{y^2+(x+1)^2} + \sqrt{y^2+(x-1)^2}} =e<1 \tag 1 $$

Special case $e=1$ is the circle enclosing cyclic quadrilaterals that have the property given by Ptolemy theorem. A set of non-cyclic quadrilaterals can be inscribed in this oval shape. In this drawing $ e=0.95; $

Some shapes for Other $e$ values

Further simplification yields a fourth degree algebraic curve:

$$\left(-a^4-2 a^3 y+a^2 \left(2 \left(e^2-1\right) x^2-2 y^2\right)-2 a y \left(x^2+y^2\right)-\left(x^2+y^2\right)^2\right)+\frac{\left((a+y)^2+x^2\right)^2}{2 e^2}=0$$

Answer 1

Just a comment because I want to insert a picture

You can find it here https://mathcurve.com/courbes3d/crepe/crepe.shtml . Sorry if the link is in french .

Hope it helps you !

Answer 2

Not another answer. For comments/ achille hui with sketch:

h = 1.4; a = .5; b = 0.3; A = 1;
mirror = ParametricPlot[{A Cos[t], A Sin[t]}, {t, 0, 2 Pi}];
ell = ParametricPlot[{h + a Cos[t], b Sin[t]}, {t, 0, 2 Pi}, 
   PlotStyle -> {Thick, Purple}];
Oval = ParametricPlot[
   A^2 {h + a Cos[t], b Sin[t]}/((h + a Cos[t])^2 + (b Sin[t])^2), {t,
     0, 2 Pi}, PlotStyle -> {Thick, Red}];
Show[{ell, Oval, mirror}, PlotRange -> All, GridLines -> Automatic]