When is the projection of an ellipsoid a circle?

Question

Consider an ellipsoid in the three dimensional Euclidean space, say $$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$ where $a$, $b$, $c$ are positive reals. I'm counting the number of planes through the origin so that the image is a perfect circle. There may be divergent cases if we consider the case that some of $a$, $b$, $c$ are coincide. But at first, let us focus on the case that $a$, $b$, $c$ are all different, say $a>b>c$.

I guess the answer would be $4$. I have made many efforts but failed. What I have observed is the that at least two such planes exists and the radius of the circle is $b$. Just consider rotating plane possesses $y$ axis and apply intermediate value theorem.

Causion! We are concerning projection, not intersection.

PS. Now I guess there are infinitely many...

PS2. According to one suggested answer, there are just two such planes for the non-degenerate case. I'm checking if it is correct.

PS3. Another opinion appeared that the selected answer may have fault. And it seems making sense. I think somewhat stronger analysis is required.

PS4. The above PS3 is about another answer which now have disappeared.

Answer 1

Considering an enveloping cylinder, namely

$$ \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right) \left( \frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2} \right)= \left( \frac{l x}{a^2}+\frac{m y}{b^2}+\frac{nz}{c^2} \right)^2$$

The eigenvalues $\lambda$ are given by

$$0=\det \begin{pmatrix} \frac{1}{a^2} \left( \frac{m^2}{b^2}+\frac{n^2}{c^2} \right)-\lambda & -\frac{l m}{a^2 b^2} & -\frac{l n}{a^2 c^2} \\ -\frac{m l}{b^2 a^2} & \frac{1}{b^2} \left( \frac{n^2}{c^2}+\frac{l^2}{a^2} \right)-\lambda & -\frac{m n}{b^2 c^2} \\ -\frac{n l}{c^2 a^2} & -\frac{n m}{c^2 b^2} & \frac{1}{c^2} \left( \frac{l^2}{a^2}+\frac{m^2}{b^2} \right)-\lambda \\ \end{pmatrix}$$

where $\lambda=0$ is one of the roots.

The discriminant (with last factor in cyclic permutations of $a$, $b$ and $c$) is given by

$$\Delta =\frac{(l^2+m^2+n^2)^2}{a^8 b^8 c^8} \left( \frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2} \right) [(\alpha l^2-\gamma n^2)^2- 2\beta m^2 (\alpha l^2+\gamma n^2)+ \beta^2 m^4 ]$$

where $\alpha=b^2-c^2$, $\beta=c^2-a^2$, $\gamma=a^2-b^2$.

Note that for $a>b>c>0$,

$$\fbox{$\alpha,-\beta,\gamma>0$}$$

For circular cylinder, there are two equal eigenvalues implying $\Delta=0$, hence

$$\fbox{$\frac{l^2}{n^2}=\frac{\gamma}{\alpha} \land m=0$}$$

The enveloping circular cylinder is given by

$$ \fbox{$\frac{b^2}{a^2 c^2} \left( \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1 \right)= \left( \sqrt{\frac{a^2-b^2}{a^2-c^2}} \, \frac{x}{a^2} \pm \sqrt{\frac{b^2-c^2}{a^2-c^2}} \, \frac{z}{c^2} \right)^2 $}$$

Further points to be noticed

• The touching conic is not circular and lying on the plane

$$\color{green}{z=\pm \frac{c^2}{a^2} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

• The uniform cross section of the cylinder lies on the plane

$$\color{blue}{z=\pm \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

• The circular section for the ellipsoid lies on the plane

$$\color{red}{z=\pm \frac{c}{a} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

Answer 2

An outline of an answer …

Think about the silhouette curve of the ellipsoid. This is the locus of points where the ellipsoid’s surface normal is perpendicular to the projection direction. It’s the outline of the ellipsoid that you see when viewing along the projection direction.

It turns out that this silhouette curve is planar, and hence is an ellipse. The plane that it lies on is the so-called polar plane of the direction vector. If the projection direction is $(u,v,w)$, then its polar plane has equation $$ \frac{ux}{a^2} + \frac{vx}{b^2} + \frac{wx}{c^2} =0 $$ You can find the silhouette ellipse just by intersecting the ellipsoid with this polar plane.

So, now you have an ellipse, and I’m sure you can figure out the two directions from which this ellipse appears circular. Then just equate these directions to the projection direction, and solve.

I learned about polar lines and planes in high school, sixty years ago, but nowadays they are mostly forgotten. If you Google “polar plane of quadric” you’ll find some references, but I couldn’t find any very good ones.

Answer 3

For convenience, we use $$ax^2+by^2+cz^2=1.$$

Consider the parametric lines of constant direction $(u,v,w)$ through some point $(x,y,z)$ on the ellipsoid.

$$a(x+tu)^2+b(y+tv)^2+c(z+tw)^2=1$$

or, after simplification,

$$(au^2+bv^2+cw^2)t^2+2(aux+bvy+cwz)t=0.$$

We will obtain the silhouette line on the ellipsoid by expressing that the rays are tangent, i.e. that the $t$ equation has a double root at $t=0$, which simply gives us the plane

$$aux+bvy+cwz=0.$$

That plane intersects the ellipsoid along an ellipse and one can always find an oblique projection that makes it a circle !

If you mean an orthogonal projection, then we must find the plane(s) that intersect along a circle. We can do this by rotating the ellipsoid so that the plane goes to $Z=0$.

We use the rotation relations

$$\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}0&\frac{B^2+C^2}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{A}{\sqrt{A^2+B^2+C^2}}\\\frac{C}{\sqrt{B^2+C^2}}&\frac{-AB}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{B}{\sqrt{A^2+B^2+C^2}}\\\frac{-B}{\sqrt{B^2+C^2}}&\frac{-AC}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{C}{\sqrt{A^2+B^2+C^2}}\end{pmatrix} \begin{pmatrix}X\\Y\\Z\end{pmatrix}.$$

When plugged in the ellipsoid equation, this gives the equation of a conic in the $XY$ plane. This conic is a circle when the quadratic coefficients are equal and the is no mixed term. This is more conveniently done by rewriting the matrix in spherical coordinates.