If $\;\alpha, \; \beta,\; \gamma,\; \delta\;$ are eccentric anlges of four conclyclic points on the standard ellipse $\; \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . Then $\alpha + \beta + \gamma + \delta =\; ?$
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See https://goo.gl/mmxCrL – lab bhattacharjee Aug 16 '15 at 06:19
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You have a better and easier method – kichapps Sep 24 '15 at 16:02
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Well solved here. – Ng Chung Tak Feb 17 '22 at 20:03
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Hint:
The points with eccentric angles $\alpha$, $\beta$, $\gamma$ & $\delta$ are $(a\cos\alpha, b\sin \alpha)$, $(a\cos\beta, b\sin \beta)$, $(a\cos\gamma, b\sin \gamma)$ & $(a\cos\delta, b\sin \delta)$
I hope you can solve further by applying the condition of con-cyclic points which are the points of intersection of ellipse & its auxiliary circle.
Harish Chandra Rajpoot
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Alright, auxiliary circle is the circumcircle with center coincident with that of the ellipse & having a radius equal to the semi major axis of the ellipse. – Harish Chandra Rajpoot Aug 24 '15 at 19:12
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I know that the auxiliary circle is the locus of foot of perpendiculars from the focus to any tangent of the ellipse. I asked whether the condition will prove the question or whether it'll give the answer... – kichapps Sep 24 '15 at 16:01