Finding the center and axes of an ellipse in 3D I am studying ellipses and need to prove the intersection of a plane and an elliptical cylinder is still an ellipse in 3D. My approach is to find its center and axes and have been struggling.
I have a parametric equation, $\left(x,\, y,\, z\right) = \left(\frac{1}{4}-\frac{3}{2}\cos t-\frac{1}{4}\sin t,\, 2\cos t,\, \sin t \right)$ . This is an intersection between a plane $4x+3y+z=1$ and an elliptical cylinder $y^2+4z^2=4$.
How do I find its center and two axes? Equivalently, how can I express it in the vector form:
$$\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$$
where $\mathbf u$ and $\mathbf v$ are orthogonal vectors from the center $\mathbf c$ whose norms represent the lengths of the axes and whose directions represent the directions of the axes.

The center of the ellipse is $C=(1/4,0,0)$. Given a generic point $P(t)$ on the ellipse, you can write the squared distance $(P-C)^2$ as a function of $t$ and find for which values of $t$ it is a minimum or maximum (one can do that with or without calculus). That will give you the vertices of the ellipse.
I can write the explicit result, if needed, but it isn't particularly appealing.
EDIT.
To prove that a curve is an ellipse, it isn't necessary to use the "canonical" parametrization you are looking for, in the sense that $\mathbf u$ and $\mathbf v$ needn't be orthogonal. One can show that the curve is still an ellipse for any pair of (non parallel) vectors $\mathbf u$ and $\mathbf v$, which then define two conjugate semidiameters.
In fact you can check that the line tangent to the ellipse at $\mathbf c+\mathbf u$ (an endpoint of the first diameter, corresponding to $t=0$) is parallel to $\mathbf v$, while the line tangent to the ellipse at $\mathbf c+\mathbf v$ (an endpoint of the second diameter, corresponding to $t=\pi/2$) is parallel to $\mathbf u$, which is the definition of conjugate diameters.
If you want then to find the axes of the ellipse, you can use the formulas explained here.
