Solving for Ellipsoidal shape with only one ellipsoidal slice

Question

Suppose we have a slice of an ellipsoid parallel to the major axis (but not on the major axis) so that we get a concave ellipsoidal mirror. I have knowledge of just the slice and nothing about the ellipsoid itself with measurements of the slice's 'minor/major' axis (say $\mathit{a'}$ and $\mathit{b'}$ ) and the depth of the slice (from the center of the ellipsoidal slice to the plane where it is sliced) say $\mathit{z'}$ .

Is it possible to reconstruct the original ellipsoid from just this information?

My intuition is telling me that it should be possible to reconstruct the ellipsoid so that I can find the foci (in one plane) of the ellipsoid since there will be only one solution to fit the 3 measurements made on the slice.

A depiction of my application

Answer 1

Let's take $x$-axis perpendicular to the plane of the slice and $z$-axis along major axis. The equation of the ellipsoid can be written as: $$ {x^2\over b^2}+{y^2\over b^2}+{z^2\over a^2}=1. $$ If $d$ is the distance from the plane of the slice to the center of the ellipsoid, then the equation of the slicing plane is $x=d$ and substituting that into the previous equation one gets: $$ {y^2\over b^2}+{z^2\over a^2}=1-{d^2\over b^2}, $$ which is the equation of an ellipse with semi-axes $a'$ and $b'$ obeying: $$ a^2\left(1-{d^2\over b^2}\right)=a'^2,\quad b^2\left(1-{d^2\over b^2}\right)=b'^2. $$ Substitute here the given values of $a'$, $b'$ and $d$ to find $a$ and $b$.

Answer 2

If it is a prolate spheroid, than a cross section of it normal to the revolution axis would be a circle, of which the portion you sliced out represents an arc.
Three points, or actually two, taking into account the symmetry, are sufficient to determine the equatorial arc, thus the polar axis of the spheroid.
Thereafter, one (off-center) point is enough to determine the longitudinal ellipse.

Answer 3

Yes, certainly possible to reconstruct the entire ellipsoid.

When intersecting ellipsoid

$$ \dfrac{x^2}{a^2}+ \dfrac{y^2+z^2}{b^2} = 1 $$

with plane $z=z'$

we have

$$ \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} = 1- \dfrac{z'^2}{b^2}$$

that is

$$ \dfrac{x^2}{\big(a\sqrt{1- \dfrac{z'^2}{b^2}}\big)^2}+ \dfrac{x^2}{\big(b\sqrt{1- \dfrac{z'^2}{b^2}}\big )^2} = 1 $$

Reduced major minor axes $(a',b')$ are seen in the denominators having reduction factor with radical sign. It will be geometrically similar with same central section eccentricity.

When $z'=b$ it is a tangential section resulting in zero size cut ellipse.