How do I find the equation of an ellipse formed by a projection of a tilted rotating circle? Suppose I have a circle with radius $R$ on $xy$ plane with it's center at $(0,0,0)$. Then I rotate this circle about the y-axis by $0 \le \varphi \le {\pi \over 2}$, and then start rotating it about z-axis by $0 \le \alpha \le 2\pi$. What is the expression for the projection of this circle onto a $yz$ plane?
I know the projection is an ellipse with a semi-major axis always staying equal to $R$, and semi-minor axis ranging from $R\sin \varphi$ at $\alpha=\pi n$ to $0$ at $\alpha= {\pi\over2}n $. When semi-minor axis equals $0$ the projection becomes a segment of a line $y=\pm x\tan\varphi$ with the length $2R$. It's similar to the most basic Lissajoux curve, but not quite the same.
I need a non-parametric expression for this projection that is a function of $R, \alpha, \varphi$. Would greatly appreciate the expression itself as well as it's derivation. Thank you in advance.
 A: The circle is initially (before any rotation) spanned by two vectors
$ u_1 = (R, 0, 0) $ and $u_2 = (0, R, 0) $
So that points on the circle are parametrically given by
$ P(t) = \cos t u_1 + \sin t u_2 $
Apply the rotation about the $y$ axis gives
$ Q(t) = \cos t R_y(\varphi) u_1 + \sin t R_y(\varphi) u_2 $
where
$ R_y(\varphi) = \begin{bmatrix} \cos \varphi && 0 && \sin \varphi \\
0 && 1 && 0 \\ -\sin \varphi && 0 && \cos \varphi \end{bmatrix} $
Thus
$Q(t) = \cos t v_1 + \sin t v_2 $
where
$ v_1 = \begin{bmatrix} R \cos \varphi \\ 0 \\ - R \sin \varphi \end{bmatrix} , \hspace{50pt} v_2 = \begin{bmatrix} 0 \\ R \\ 0 \end{bmatrix} $
Next, apply the rotation about the $z$-axis by an angle $\alpha$, this we give a new parametric curve,
$ S(t) = \cos t R_z(\alpha) v_1 + \sin t R_z(\alpha) v_2 $
where
$ R_z(\alpha) = \begin{bmatrix} \cos \alpha && - \sin \alpha && 0 \\ \sin \alpha && \cos \alpha && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Hence
$ S(t) = \cos t \ w_1 + \sin t \ w_2 $
where
$ w_1 = R \begin{bmatrix} \cos \alpha \cos \varphi \\ \sin \alpha \cos \varphi \\ - \sin \varphi  \end{bmatrix} $
$ w_2 = R \begin{bmatrix} - \sin \alpha \\ \cos \alpha \\ 0 \end{bmatrix} $
Now projecting $S(t)$ onto the $xy$ plane gives
$ r(t) = \cos t \ V_1 + \sin t \ V_2 $
where $V_1 = R \begin{bmatrix} \cos \varphi  \cos \alpha \\ \cos \varphi \sin \alpha \end{bmatrix} , V_2 = R \begin{bmatrix} -\sin \alpha \\ \cos \alpha \end{bmatrix} $
Clearly, $V_1 \perp V_2 $, therefore these two vectors are the semi-minor and semi-major axes of the ellipse of projection, respectively.
Writing r(t) in matrix-vector form , we get
$ r(t) = [x, y]^T = [V_1, V_2] u(t) = V u(t)$
where $ u(t) = [\cos t , \sin t ]^T $
From the above equation, we get
$u(t) = V^{-1} r(t) $
Now,
$V^{-1} = \dfrac{1}{R \cos \varphi} \begin{bmatrix}  \cos \alpha &&  \sin \alpha \\ - \cos \varphi \sin \alpha && \cos \varphi \cos \alpha \end{bmatrix} = \dfrac{1}{R} \begin{bmatrix} \sec \varphi \cos \alpha && \sec \varphi \sin \alpha \\ - \sin \alpha &&  \cos \alpha \end{bmatrix}$
Now since $u^T(t) u(t) = 1 $ then
$ r^T V^{-T} V^{-1} r = 1 $
what remains is to find $V^{-T} V^{-1}$ and this is given by
$  \dfrac{1}{R^2} \begin{bmatrix} 1 + \tan^2 \varphi \cos^2 \alpha && \sin \alpha \cos \alpha \tan^2 \phi \\ \sin \alpha \cos \alpha \tan^2 \varphi && 1 + \tan^2 \varphi \sin^2 \alpha \end{bmatrix} $
Thus the final equation of the projected ellipse is
$ \bigg(1 + \tan^2 (\varphi)  \cos^2 (\alpha) \bigg) x^2 + \bigg(\sin(2 \alpha) \tan^2( \varphi) \bigg) x y + \bigg(1 + \tan^2(\varphi) \sin^2(\alpha) \bigg) y^2 = R^2 $

If you're looking the projection onto the $yz$ plane then
$ r(t) = [y, z]^T = \cos t V_1 + \sin t V_2  $
where $V_1 = R \begin{bmatrix} \sin \alpha \cos \varphi \\ - \sin \varphi \end{bmatrix} , \hspace{50pt} V_2 = R \begin{bmatrix} \cos \alpha \\ 0 \end{bmatrix} $
Writing r(t) in matrix-vector form , we get
$ r(t) = [y, z]^T = [V_1, V_2] u(t) = V u(t)$
where $ u(t) = [\cos t , \sin t ]^T $
From the above equation, we get
$u(t) = V^{-1} r(t) $
Now,
$V^{-1} = \dfrac{1}{R \cos \alpha \sin \varphi} \begin{bmatrix}  0 &&  - \cos \alpha \\  \sin \varphi && \sin \alpha \cos \varphi \end{bmatrix} = \dfrac{1}{R} \begin{bmatrix} 0 && -\csc \varphi  \\  \sec \alpha &&  \tan \alpha \cot \varphi \end{bmatrix}$
Now since $u^T(t) u(t) = 1 $ then
$ r^T V^{-T} V^{-1} r = 1 $
what remains is to find $V^{-T} V^{-1}$ and this is given by
$  \dfrac{1}{R^2} \begin{bmatrix} \sec^2 \alpha && \sec \alpha \tan \alpha \cot \varphi \\ \sec \alpha \tan \alpha \cot \varphi && 
1 + \cot^2 \varphi \sec^2 \alpha \end{bmatrix} $
Thus the final equation of the projected ellipse on the $yz$ plane is
$ \bigg(\sec^2 \alpha \bigg) y^2 + \bigg(2\sec \alpha \tan \alpha \cot \varphi \bigg)  yz + \bigg(1 + \cot^2 \varphi \sec^2 \alpha \bigg) z^2 = R^2 $
