The ellipse

$$ \frac{p}{r}= 1 - e \cos \theta, \quad p = r- e\cdot z $$ with Cartesian coordinates in $zy $ plane $ (z= r \cos \theta,\, y= r \sin \theta) $ is tilted by angle $ \gamma$ about its latus rectum as a hinge/axis and then fully rotated about $z$ axis sweeping out a surface of revolution.

What is its meridian and what degree do planar sections which are not passing through the origin have?

  • $\begingroup$ Some personal misunderstandings 1) Do you mean that $a$ and $b$ are the same for all ellipses 2) Do we agree that in your first paragraph you only consider planar sections passing through the origin (it should be said) 3) Shouldn't the existence of such a surface be proved before proceeding to the question you finally ask ? $\endgroup$ – Jean Marie Sep 23 '16 at 18:52
  • $\begingroup$ Answer to 1), 2) is yes. As for 3) it is intuitively obvious I thought that an arc symmetric to $zx$ plane sweeps out a surface of revolution.when rotated about $z$ axis. But I shall re-write it ,if it helps. $\endgroup$ – Narasimham Sep 23 '16 at 23:00
  • $\begingroup$ The question before edit1 //A surface of revolution has all its planar sections as an ellipse (major, minor axes $2a,2b$) its latus rectum in $x-y$ plane and its focus at the origin. (It is formed by z-axis piercing an elliptic lamina at its focus and rotated around $z$ axis).// $\endgroup$ – Narasimham Sep 23 '16 at 23:01
  • $\begingroup$ Thanks for your answers. $\endgroup$ – Jean Marie Sep 23 '16 at 23:23
  • $\begingroup$ Now, with your new formulation, everything is clearer. $\endgroup$ – Jean Marie Sep 24 '16 at 6:32

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