# The projection of an ellipse is still an ellipse

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.

• I would start by recalling the precise definition of "ellipse" and then proving, under that definition, that an ellipse is the affine image of a circle. Feb 4 '20 at 16:53
• I don't know what an affine image is Feb 4 '20 at 16:59
• Think of it in this way - the original ellipse is created by a cone intersecting a plane. Now using this cone you can create the other ellipse. Thus two ellipses define a unique cone(?)
– Moti
Feb 4 '20 at 17:07
• @Rhino: You should edit your question to include what you know about ellipses and projections. This will help people avoid wasting time (theirs and yours) explaining things you already know or glossing-over things with which you aren't familiar.
– Blue
Feb 4 '20 at 17:36

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.

• Nice explanation. Feb 4 '20 at 19:41
• I still don't understand why ${O'H'^2\over O'B'^2}+{O'K'^2\over O'D'^2}=1$ represents an ellipse even if the axis are not perpendicular Feb 4 '20 at 20:26
• @Rhino That is the "general" equation of an ellipse, as written by Apollonius of Perga many years ago. Apollonius proved that equation directly from the definition of ellipse as section of a cone. Probably the simplest way to see why it holds is that of regarding an ellipse as a dilation of a circle and conjugate diameters in the ellipse as the transformed of a pair of perpendicular diameters in the circle. Feb 4 '20 at 20:55

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.

• This is correct. Notice, however, that the projections of $\vec U$ and $\vec V$ are not, in general, perpendicular. Hence they don't give the directions of the principal axes of the projected ellipse, but rather the directions of two conjugate diameters. Feb 4 '20 at 19:23
• @Aretino I fully agree, I should have mentionned it, Feb 4 '20 at 19:38

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.

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