I want to find the parametric equation of the ellipse in 3d space which is formed by the intersection of a known ellipsoid and a known plane. The ellipsoid has the Cartesian equation: $(x/a)^2+(y/b)^2+(z/a)^2=1$.
While the plane has the equation: $mx+ny+kz=0$. I have substituted one equation in the other but what I get is an elliptical cylinder since I had eliminated the $z$-components. How can I get the exact equation of the ellipse of intersection in parametric form?
[Math] Intersection of an ellipsoid and plane in parametric form
algebraic-geometrydifferential-geometrygeometry
Related Solutions
$\newcommand{\Vec}[1]{\mathbf{#1}}$Generalities: Let $S$ be the sphere in $\mathbf{R}^{3}$ with center $\Vec{c}_{0} = (x_{0}, y_{0}, z_{0})$ and radius $R > 0$, and let $P$ be the plane with equation $Ax + By + Cz = D$, so that $\Vec{n} = (A, B, C)$ is a normal vector of $P$.
If $\Vec{p}_{0}$ is an arbitrary point on $P$, the signed distance from the center of the sphere $\Vec{c}_{0}$ to the plane $P$ is $$ \rho = \frac{(\Vec{c}_{0} - \Vec{p}_{0}) \cdot \Vec{n}}{\|\Vec{n}\|} = \frac{Ax_{0} + By_{0} + Cz_{0} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
The intersection $S \cap P$ is a circle if and only if $-R < \rho < R$, and in that case, the circle has radius $r = \sqrt{R^{2} - \rho^{2}}$ and center $$ \Vec{c} = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|} = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
Now consider the specific example $$ S = \{(x, y, z) : x^{2} + y^{2} + z^{2} = 4\},\qquad P = \{(x, y, z) : x - z\sqrt{3} = 0\}. $$ The center of $S$ is the origin, which lies on $P$, so the intersection is a circle of radius $2$, the same radius as $S$.
When you substitute $x = z\sqrt{3}$ or $z = x/\sqrt{3}$ into the equation of $S$, you obtain the equation of a cylinder with elliptical cross section (as noted in the OP). However, you must also retain the equation of $P$ in your system. That is, each of the following pairs of equations defines the same circle in space: \begin{align*} x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, \\ x^{2} + y^{2} + z^{2} &= 4; & \tfrac{4}{3} x^{2} + y^{2} &= 4; & y^{2} + 4z^{2} &= 4. \end{align*} These may not "look like" circles at first glance, but that's because the circle is not parallel to a coordinate plane; instead, it casts elliptical "shadows" in the $(x, y)$- and $(y, z)$-planes.
Note that a circle in space doesn't have a single equation in the sense you're asking.
Just an idea - I don't know if it will fit your problem or not, cause actually you didn't describe it with sufficient level of details.
Let's think in terms of a cylindrical coordinate system - the equations below describe your cylindrical surface, ignoring that fact it has limited height (if I understood your question correctly).
$$x = R \cdot \sin(\phi), y = R \cdot \cos(\phi), z = z$$
Then you can substitute these $x$ and $y$ into the ellipsoid equation:
$$\frac {R^2} {a^2} \cdot \sin^2(\phi) + \frac {R^2}{b^2}\cdot \cos^2(\phi) + \frac{z^2}{c^2} = 1$$
You can express the $z$ as a function of $\phi$ - this will give you the intersection curve equation in cylindrical coordinates.
I hope it'll help.
Best Answer
Hint. Find vectors of the semiaxes of the ellipse, meaning find (x,y,z) which satisfies both equations and its length is longest(shortest). $$(x/a)^2+(y/b)^2+(z/a)^2=1$$ $$mx+ny+kz=0$$ $$x^2+y^2+z^2=\min \lor \max$$ And then parametric equation will be $(x,y,z)=\vec v_1\sin \theta+\vec v_2\cos \theta$
Edit - More than a hint As yota suggested this hint isn't too helpful since problem is still hard to solve I will try to push it till the end:
$$x^2+y^2+z^2=x^2+y^2+z^2-a^2((x/a)^2+(y/b)^2+(z/a)^2-1)=y^2(1-a^2/b^2)+a^2$$ As we can see one of the semiaxes vectors has $y$ coordinate equal to $0$, since the expression above reaches its max/min when $y=0$
We can put it in the equations to find $x$ and $z$: $$z=-mx/k$$ $$(x/a)^2+(-mx/k/a)^2=1$$ $$x^2k^2+x^2m^2=k^2a^2$$ $$x^2=k^2a^2/(k^2+m^2)$$ So we have one semiaxis $v_1=(ka/\sqrt{k^2+m^2},0,-ma/\sqrt{k^2+m^2})$
Another should be colinear to $v_1\times(m,n,k)$ or even better $(k,0,-m)\times(m,n,k)$ which is: $$v_2=t(mn,-m^2-k^2,kn)$$ where $$t=1/\sqrt{(mn/a)^2+((m^2+k^2)/b)^2+(kn/a)^2}$$