Proving that a parameterized curve in 3D is an ellipse and calculating its shape parameters

Question

Consider the following closed curve in 3D, parameterized by $\theta$: $$ \begin{align} x &= A\cos(\theta) \\ y &= -2A\sin(\theta) \tag{1}\label{xyz} \\ z &= -C\sin(\theta-\phi) \end{align} $$ Two questions:

1. Is this curve an ellipse? When I graph it, it certainly looks like one. How can I prove that?
2. Assuming the answer to (1) is affirmative, what are the shape parameters (e.g. semi-major axis, semi-minor axis) of this ellipse? How do I calculate them?

Here's what I've done so far.

1. It's easy to verify that the curve lies in the plane normal to $(2C\sin\phi, C\cos\phi, -2A)$.
2. The curve lies on the surfaces $$ 4x^2+y^2=4A^2 \\ (\sin^2\phi)x^2 + (\sin\phi\cos\phi)xy + (\cos^2\phi/4)y^2 - (A^2/C^2)z^2 = 0 $$ so presumably it is their intersection. But I don't know why the intersection of these surfaces must be an ellipse.
3. The distance from the origin is $$ \begin{align*} r^2 &= x^2+y^2+z^2 \\ &= A^2\cos^2\theta + 4A^2\sin^2\theta + C^2(-\sin\theta\cos\phi+\cos\theta\sin\phi)^2 \\ &= (A^2+C^2\sin^2\phi)\cos^2\theta + (4A^2+C^2\cos^2\phi)\sin^2\theta - 2C^2\sin\phi\cos\phi\sin\theta\cos\theta \\ &= \frac12 (A^2+C^2\sin^2\phi)(1+\cos 2\theta) + \frac12 (4A^2+C^2\cos^2\phi)(1-\cos 2\theta) - \frac12 C^2 \sin 2\phi \sin 2\theta \\ &= \frac12 (5A^2+C^2) + \frac12 (-3A^2 + C^2(\sin^2\phi-\cos^2\phi))\cos 2\theta - \frac12 C^2 \sin 2\phi \sin 2\theta \\ &= \frac12 (5A^2+C^2) - \frac12 (3A^2+C^2\cos 2\phi)\cos 2\theta - \frac12 C^2 \sin 2\phi \sin 2\theta \\ \end{align*} $$ so $$ \frac{d(r^2)}{d\theta} = (3A^2+C^2\cos 2\phi)\sin 2\theta - C^2\sin 2\phi\cos 2\theta. $$ Setting $d(r^2)/d\theta = 0$ gives $$ (3A^2+C^2\cos 2\phi)\sin 2\theta - C^2\sin 2\phi\cos 2\theta = 0 $$ or $$ \tan 2\theta = \frac{\sin 2\phi}{3A^2/C^2+\cos 2\phi}. $$ This gives the local extrema of the distance function and should be a step towards finding the semi-major and semi-minor axes. In theory, I should use this equation to substitute back into Eq. \eqref{xyz} and be able to get a semi-major axis and semi-minor axis vector. But it looks computationally hard and I decided not to try to proceed further without asking this question first.

Am I heading in the right direction? Is there a way to proceed that is not as computational? (At least for the proof of (1); I expect that (2) is going to get computational at some point no matter what.)

Answer 1

Here is a way which is indeed non-computational.

You surely mean $4x^2+y^2=4$ which is in 3D the equation of an elliptical-based cylinder, belonging to the category of quadric surfaces. It is well known that the intersection of a quadric with a plane is a conic curve. As this curve is bounded $(|x|\leq A, |y|\leq 2A, |z|\leq C)$, it can be neither an hyperbola nor a parabola ; it remains only the case of an ellipse.