I found a post claiming that if $\alpha,\beta,\gamma,\theta$ are the eccentric angles of points $X,Y,Z,T$ we can parameterize those points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as $(a\sec(t),b\tan(t))$ which makes sense. Further in the post is proven that if the four points are concyclic we have $\tan(\frac{\alpha+\beta+\gamma+\theta}{2})=0$. My question is whether this works backwards. I checked on desmos and almost never points with eccentric angles $\alpha,\beta,\gamma, -\alpha-\beta-\gamma$ are concyclic. Can someone provide a reason for this.
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1The post that you referenced is talking about an ellipse, not a hyperbola ??? – Hosam Hajeer Jan 09 '24 at 08:03
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The same holds for hyperbolahttps://math.stackexchange.com/questions/3577217/concyclic-points-on-hyperbola – Helixglich Jan 09 '24 at 09:01
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In classical mathematical terms : "works backwards" = "reciprocal proposition is true". – Jean Marie Jan 09 '24 at 13:14
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@JeanMarie I am asking whether there is a not so ugly if and only if condition for four points to be concyclic. – Helixglich Jan 09 '24 at 15:40
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I've written a small code, intersecting the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1 $ with the circle $ (x - 1)^2 + (y - 2)^2 = 25 $, and calculated the eccentric angles of the points of intersection, and as it turns out, the condition $\tan \left( \dfrac{\alpha + \beta+ \gamma + \delta }{2} \right) = 0 $ does not hold. – Hosam Hajeer Jan 16 '24 at 01:36
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However, the condition $\tan \left( \dfrac{\alpha + \beta + \gamma + \delta}{2} \right) = 0 $ does indeed hold for intersection between an ellipse and a circle. – Hosam Hajeer Jan 16 '24 at 08:11
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Interesting. Thank you very much – Helixglich Jan 16 '24 at 10:22