The projection of an ellipse is still an ellipse

Question

While studying the problem of determining the orbit of a binary star system, my astronomy notes say: "the projection of an ellipse onto a plane is still an ellipse (except the special case in which it's a segment)". How could I prove this?

EDIT: As suggested i'm adding what I know about ellipses. I know the basics of ellipses in the plane, the equation of an ellipse in a plane in cartesia coordinates, the equation of an ellipse in polar coordinates, the fact that ellipses can be obtained by the intersection of planes with a cone. In my geometry course we defined ellipses as the the points of the plane such that the ratio between the distance from a point and a line is constant and less than 1. I also know that the sum of the distances of the points of the ellipse from the foci is constant.

Answer 1

Let $AB$, $CD$ be the principal axes of the ellipse, intersecting at center $O$, and let $P$ be any point on the ellipse, $H$ and $K$ its projections on the axes (see figure below). We have: $$ {OH^2\over OB^2}+{OK^2\over OD^2}=1. $$ Let now $A'$, $B'$, ... be the perpendicular projections of points $A$, $B$, ... on a given plane. As perpendicular projections preserve the ratios of segments on a line, we have: $$ {O'H'^2\over O'B'^2}+{O'K'^2\over O'D'^2}=1. $$ But this equation means that point $P'$ belongs to the ellipse having $A'B'$ and $C'D'$ as conjugate diameters. You can find the principal axes of that ellipse, for instance, following the method explained here.

Answer 2

Here is a simple explanation if you are familiar with vectors.

An ellipse in 3D space can be written in a parametric form in this way :

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\underbrace{\begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}}_{\text{center} \ C}+a \cos(t)\underbrace{\begin{pmatrix}p\\q\\r\end{pmatrix}}_{\vec{U}}+b\sin(t)\underbrace{\begin{pmatrix}p'\\q'\\r'\end{pmatrix}}_{\vec{V}}$$

where $\vec{U}$ and $\vec{V}$ are unit vectors giving the directions of the principal axes of the ellipse.

Now projecting for example on $xOy$ plane means forgetting coordinate $z$ ; the first two equations are indeed parametric equations of an ellipse in the plane.

Answer 3

We might consider this a generic parameterization of an ellipse in $\mathbb R^3$

$(x,y,z) = \mathbf u \cos t + \mathbf v \sin t +\mathbf P$

Where $\bf{u,v}$ are independent vectors in $\mathbb R^3$ and $\mathbf P$ is a point in representing the center of the ellipse.

If we project $\bf{u,v}$ and $\bf{P}$ into a two-dimensional subspace, we will still have $(x,y,z) = \mathbf u_\perp \cos t + \mathbf v_\perp \sin t +\mathbf P_\perp$

If the projection is in just the right direction, $\mathbf u_\perp, \mathbf v_\perp$ will cease to be independent, giving us the line-segment solution.

i.e. $\mathbf u = (u_x,u_y, u_z), \mathbf v = (v_x,v_y, v_z), \mathbf P = (P_x,P_y, P_z)$

$x = u_x \cos t + v_x\sin t + P_x\\ y = u_y \cos t + v_y\sin t + P_y\\ z = u_z \cos t + v_z\sin t + P_z\\ $

Projected onto the xy plane gives.

$x = u_x \cos t + v_x\sin t + P_x\\ y = u_y \cos t + v_y\sin t + P_y\\ z = 0\\ $

Answer 4

Problem

If you take a photograph of an eclipse, is it still a conic curve on the image?

Solution

Yes.

To project a point in 3D space into a 2D canvas, the canonical way of doing that in computer graphics is:

Step 1. make a "view-projection transformation", which maps the points $(x,y,z)$ from the "world space" into the points in the "clip space" $(X,Y,Z)$. In particular, it is a projective transformation that maps a viewing frustum (the region your camera is looking at) to a cube.

It is done by using homogeneous coordinates:

\begin{equation} \begin{bmatrix} \tilde{X} \\ \tilde{Y} \\ \tilde{Z} \\ \tilde{W} \end{bmatrix} = P \cdot \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} \end{equation}

\begin{equation} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} = \begin{bmatrix} \tilde{X}/\tilde{W} \\ \tilde{Y}/\tilde{W} \\ \tilde{Z}/\tilde{W} \end{bmatrix} \end{equation}

Where $P$ is an invertible matrix based on the position, orientation, and FOV of the camera.

Step 2. in the "clip space", make an orthogonal projection onto the $Z=0$ plane:

\begin{equation} (X,Y,Z) \mapsto (X,Y) \end{equation}

it is easy to show that step 2 maps a 3D conic curve into a 2D one. So let's take a look at whether step 1 maps a 3D conic curve to another 3D conic curve.

A conic curve in 3D space can be written as the intersection of a quadric surface with a plane:

\begin{equation} \left\{ \begin{array}{c} v^T \cdot M \cdot v=0 \\ N \cdot v=0 \end{array} \right., v= \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} \end{equation}

Where $M$ as a 4x4 matrix, N is a 1x4 vector

by replacing

\begin{equation} V= \tilde{W}(x,y,z)^{-1} P \cdot v, V= \begin{bmatrix} X \\ Y \\ Z \\ 1 \end{bmatrix} \end{equation}

The equation defining the set of points on the curve becomes:

\begin{equation} \left\{ \begin{array}{c} \tilde{W}(X,Y,Z)^{2} V^T \cdot P^{-1,T} \cdot M \cdot P^{-1} \cdot V=0 \\ \tilde{W}(X,Y,Z) N \cdot P^{-1} \cdot v=0 \end{array} \right. \end{equation}

The key step is that, although $\tilde{W}$ has a complex dependence on $(X,Y,Z)$, it can be removed from the equation as a scalar multiplier, since RHS is zero. The remaining part is a constant matrix. Then we reached a similar form:

\begin{equation} \left\{ \begin{array}{c} V^T \cdot M' \cdot V=0 \\ N' \cdot v=0 \end{array} \right., M'= P^{-1,T} \cdot M \cdot P^{-1}, N'= N \cdot P^{-1} \end{equation}

So the conic curve is still a conic curve after the "view-project" transformation, which remains a conic curve after the final orthogonal projection.