Finding the center and axes of an ellipse in 3D

Question

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.

Answer 1

$$\left(x,\, y,\, z\right) = \left(\frac{1}{4}-\frac{3}{2}\cos t-\frac{1}{4}\sin t,\, 2\cos t,\, \sin t \right)=(\frac{1}{4},0,0)+(\frac{-3}{2},2,0)\cos t+(\frac{-1}{4},0,1)\sin t$$ So $c=(\frac{1}{4},0,0)$, $u=(\frac{-3}{2},2,0)$ and $v=(\frac{-1}{4},0,1)$

Answer 2

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.

Answer 3

Let $ r = [x, y, z]^T $

Then

$ r = r_0 + \cos(t) v_1 + \sin(t) v_2 = r_0 + [v_1, v_2] U(t) $

where $r_0 = \begin{bmatrix} \frac{1}{4} \\ 0 \\ 0 \end{bmatrix},\hspace{20pt} v_1 = \begin{bmatrix} -1.5 \\ 2 \\ 0 \end{bmatrix} , \hspace{20pt} v_2 = \begin{bmatrix} -0.25 \\ 0 \\ 1 \end{bmatrix} ,\hspace{20pt} U(t) = \begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} $

Using Gram-Schmidt orthogonalization, we get

$ [v_1, v_2] = [w_1, w_2] G $

where

$ w_1 = \begin{bmatrix} -0.6 \\ 0.8 \\ 0 \end{bmatrix} , w_2 = \begin{bmatrix} -0.156892908 \\ -0.117669681 \\ 0.980580676 \end{bmatrix} $

and

$ G = \begin{bmatrix} 2.5 && 0.15 \\ 0 && 1.019803903 \end{bmatrix} $

Note that the set $\{ w_1, w_2 \} $ is an orthonormal set, therefore, $r$ can be described by a $2D$ coordinate vector $ u $, i.e.

$ r = r_0 + W u $

where $W = [w_1, w_2]$ and $ u = G U(t) \Longleftrightarrow U(t) = G^{-1} u $

To find the equation governing $u$, we use the fact that

$ U^T(t) U(t) = 1 $

From which it follows that

$ u^T G^{-T} G^{-1} u = 1 \hspace{20pt} (1a)$

And we have

$ G^{-T} G^{-1} =\begin{bmatrix} 0.16 && -0.023533936 \\ -0.023533936 && 0.965 \end{bmatrix} $

Diagonalizing $G^{-T}G^{-1} $ into $ R D R^T $, we find that

$ D = \begin{bmatrix} 0.159312579 && 0 \\ 0 && 0.965687421 \end{bmatrix}$

$ R = \begin{bmatrix} 0.999573668 && -0.029197307 \\ -0.029197307 && 0.999573668 \end{bmatrix} $

Now equation $(1a)$ can be re-written as

$ u^T R D R^T u = 1 \hspace{20pt} (1b) $

So that

$ D^{1/2} R^T u = U(s) $

where $U(s)$ is the unit vector

$U(s) = \begin{bmatrix} \cos(s) \\ \sin(s) \end{bmatrix} $

Thus, finally, the coordinate vector $u$ can be expressed as,

$ u = R D^{-1/2} U(s) $

And with that,

$ r = r_0 + W u = r_0 + W R D^{-1/2} U(s) $

So the orthogonal vectors that represent the semi-major and semi-minor axes of the ellipse $r$ are the columns of the $3 \times 2 $ matrix $ W R D^{-1/2} $ which is given by

$ W R D^{-1/2} = \begin{bmatrix} -1.514068635 && -0.14176096 \\ 1.994848166 && -0.14346009 \\ 0.071730043 && 0.997424083 \end{bmatrix} $

The first column of the above matrix has a length of $2.505387841 $ and the second column has a length of $1.017610813 $

Therefore, the semi-major axis is along the the first column, and the semi-minor axis is along the second column.

The center, of course, is $r_0$.

---

Note that

$ r - r_0 = [v_1, v_2] U(t) = W G U(t) = W R D^{-1/2} U(s) $

Hence,

$ U(t) = G^{-1} R D^{-1/2} U(s) $

But $G^{-T} G^{-1} = R D R^T $, so it follows that $G^{-1} = R' D^{1/2} R^T $ for some rotation matrix $R'$, which is given by $R' = G^{-1} R D^{-1/2} $

Therefore,

$ U(t) = R' U(s) $

i.e. $\begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} = \begin{bmatrix} \cos(\theta) && - \sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(s) \\ \sin(s) \end{bmatrix} $

From this it follows that $ t = s + \theta + 2 k \pi , \ k $ integer.