Obtaining an ellipse when a sphere is cut with a plane

Question

Let us assume that there is an ellipse whose major axis length is 2a and minor axis length is 2b with y=mx+c as it's principal axis, which is lying on xy plane now let us assume that there is a sphere with radius r and its center is at origin can we obtain the given ellipse at exact position with same major and minor axis lengths when we cut this sphere with a plane (when the cut portion is seen from the top, is it always possible with any ellipse or is it restricted to some special cases).

Also if there are infinite such spheres and plane that can yield the required ellipse at required position can we find atleast one such possibility.

Is there any mathematical approach to solve this kind of problems.

What if we have only the equation of ellipse on xy plane then can we find the sphere and plane equations?

Answer 1

Take the $xy$ plane and pivot around the ellipse's major axis. The vertical projection of the ellipse on the plane will be another ellipse with the same major axis but longer minor axis. At some angle (positive or negative) the ellipse on the tilted plane will be a circle. That circle is the intersection of the plane with a sphere of appropriate radius if and only if the centre of the sphere is on the line $L$ through the centre of the circle orthogonal to the plane. The only other freedom you have is to move everything vertically. This changes the position of the intersection of $L$ with the plane, but it stays on the vertical projection of $L$ on the plane. If that projection doesn't intersect the origin, you can't have the centre of your sphere at the origin.

Answer 2

What follows is very similar to Robert Israel's answer, from a calculus point of view.
Suppose that the 3-dimensional euclidian space is equipped with an origin $O$ and an orthonormal basis $(\vec\imath,\vec\jmath,\vec k)$. Consider a plane $\mathscr{P}$ not parallel to the $(Oz)$ axis.
By applying a (same) rotation around this axis to $\vec\imath$ and $\vec\jmath$, we may suppose without loss of generality that $\vec\jmath$ is parallel to $\mathscr{P}$. Then this plane has a cartesian equation of the form

$z=ax+b$, where $a,b\in\mathbb{R}$.

Consider also a sphere $\mathscr{S}$ centered at $O$, having radius $r$; a cartesian equation for $\mathscr{S}$ is

$x^2+y^2+z^2=r^2$.

We assume that $\mathscr{P}\cap\mathscr{S}$ is a circle. Since the distance from $O$ to $\mathscr{P}$ is $d=\frac{|b|}{\sqrt{a^2+1}}$ this assumption is equivalent to $r>d$, or $r^2-\frac{b^2}{1+a^2}>0$.

Replacing $z$ by $ax+b$ in $\mathscr{S}$' equation we get $x^2+y^2+a^2x^2+2abx+b^2=r^2$.

After some straightforward calculations this is rewritten as $$\bigg( \frac{\sqrt{a^2+1}}{k}(x-x_0)\bigg)^2+\Big(\frac{y}{k}\Big)^2=1,$$ where we have set $x_0=-\frac{ab}{1+a^2}$ and $k=\sqrt{r^2-\frac{b^2}{1+a^2}}\cdot$

We have obtained the cartesian equation in the plane $(xOy)$ of the orthogonal projection to this plane of the circle $\mathscr{P}\cap\mathscr{S}$.
We recognize then that the minor axis of this ellipse is $(Ox)$, and the major axis is the line parallel to $(Oy)$ passing at $C\mathord:(x_0,0)$. This implies that the center of $\mathscr{S}$ belongs to the minor axis of the ellipse; thus it cannot be any ellipse.