MHB How can the equation be modified to create a perfect ellipse on the graph?

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The equation cos(a) + cos(b) = cos(a + b) does not produce a perfect ellipse on the graph due to its inherent mathematical properties. Instead, it resembles an ellipse under certain conditions, particularly when the parameters are adjusted. To create a perfect ellipse, modifications to the equation are necessary, focusing on ensuring that the second derivatives meet specific criteria at an extremum point. The discussion highlights that for smooth functions, the appearance of an ellipse can be achieved if certain derivative conditions are satisfied. Ultimately, achieving a perfect ellipse requires careful alteration of the original equation's parameters.
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The graph of the equation:
cos(a) + cos(b) = cos(a + b)
show when the equation is valid.

The graph show an ellipses-like and not like a "perfect" ellipse. Why?
If I want to change the equation, what can I do to get a perfect ellipse?!
 
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This equation is discussed on Math.StackExchange. The accepted answer says that it is indeed not an ellipse, but the solutions to any equation $f(x,y)=\text{const}$ around $(x_0,y_0)$ looks like an ellipse for a smooth function $f$ if $\frac{\partial f}{\partial x}(x_0,y_0)=\frac{\partial f}{\partial y}(x_0,y_0)=0$ and $$\begin{vmatrix}\frac{\partial^2f}{\partial x^2}f(x_0,y_0)&\frac{\partial^2f}{\partial x\partial y}f(x_0,y_0)\\\frac{\partial^2f}{\partial x\partial y}f(x_0,y_0)&\frac{\partial^2f}{\partial y^2}f(x_0,y_0)\end{vmatrix}>0$$ (i.e., if $(x_0,y_0)$ is an extremum point of $f$).
 
If you move the slider for this Desmos graph, you will see that the graph looks very like an ellipse until the parameter $a$ is quite close to $1$.

[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-1.3747197653766552,"ymin":-0.9204796366940666,"xmax":6.9623451644014835,"ymax":7.416585293084072}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"\\cos x\\ +\\ \\cos y\\ -\\ \\cos\\left(x+y\\right)\\ =\\ a","style":"SOLID"},{"type":"expression","id":"2","color":"#388c46","latex":"a=0","hidden":true,"sliderHardMin":true,"sliderHardMax":true,"sliderMin":"-3","sliderMax":"1","sliderInterval":"0.25","style":"SOLID"}]}}[/DESMOS]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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