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I have an ellipse in $x$-$y$ plane with eccentricity $e$, whose semi-major (along $x$-axis) and semi-minor (along $y$-axis) axes lengths are known to be $A$ and $B$. I rotate this ellipse about $y$-axis such that it makes an angle $\epsilon$ with the $x$-$y$ plane. Let's call this new ellipse $E_1$ (shown in red). Further, I rotate $E_1$ by $120^\circ$ anticlockwise about the $z$-axis, and I call this ellipse $E_2$ (shown in blue).

The problem is to find the point on $E_2$ that is at a distance of $L$ from the end-point of the semi-major axis of $E_1$.

The known quantities are: $e$ (eccentricity), the angle $\epsilon$, the coordinates $(x_0,y_0,z_0)$ of the end of semi-major axis of $E_1$, and the distance $L$. The ellipses are shown in the diagram.

enter image description here

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  • $\begingroup$ $y_0=0$ am I right? $\endgroup$ – Djura Marinkov Sep 23 '17 at 18:07
  • $\begingroup$ @DjuraMarinkov yes, y0 = 0 $\endgroup$ – Komal Gupta Sep 23 '17 at 18:07
  • $\begingroup$ but if you take $(x_0,y_0,z_0)$ as given, what is the purpose of knowing that it is on the semimajor axis of $E_1$? $\endgroup$ – G Cab Sep 23 '17 at 18:10
  • $\begingroup$ @GCab That is part of the problem statement. $\endgroup$ – Komal Gupta Sep 23 '17 at 18:15
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Let rotate coordinates around Z first so the blue ellipse get along with the new axes.

$p=-\frac 1 2 x+\frac {\sqrt 3}2 y$

$q=\frac {\sqrt 3} 2 x+\frac 1 2 y$

Now we can easily define ellipsoid around the blue ellipse by squizing sphere $p^2+q^2+z^2=A^2$ along q axis. Its equation is $p^2+(qA/B)^2+z^2=A^2$

This ellipsoid intersects with a plane to make the blue ellipse. Equation of the plane is $z=\frac {z_0}{x_0}p$

Now it's left to find intersection with the sphere $(p+\frac{x_0}{2})^2+(q-\frac{x_0\sqrt3}{2})^2+(z-z_0)^2=L^2$

Good luck with solving this system...

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  • $\begingroup$ Can you please explain the solution? Am I right in assuming that you are rotating the coordinates of blue ellipse and not the axes? How would the steps change if the angle of rotation was 240 degrees instead of 120 degrees? $\endgroup$ – Komal Gupta Sep 23 '17 at 21:20
  • $\begingroup$ @KomalGupta I was rotating axes, instead x and y axes there are p and q. Actually it is not rotation as I see, just switching to new axes. You can see p and q don't have same orientation as x and y. $\endgroup$ – Djura Marinkov Sep 23 '17 at 22:01
  • $\begingroup$ Okay. And how will the solution change if the angle of rotation was 240 degrees instead of 120 degrees? $\endgroup$ – Komal Gupta Sep 24 '17 at 20:56
  • $\begingroup$ @KomalGupta draw your new axes whatever you like. Relation to old axes is when you multiply old axis with cosine of angle that it makes with the new axis. So if x and p make angle of 120 degrees, I would multiply x with -1/2. y and p make angle of 30 so it's $\sqrt 3/2$. If I rotate it by 240 than p and x again make angle of 120 but p and y make angle of 150 now. $\endgroup$ – Djura Marinkov Sep 24 '17 at 21:16

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