# How to find the new semi-axis of an ellipse as result of intersection between a spheroid an a plane

A plane (P) passing always by origin with an equation $αx + βy + γz = 0$.
A oblate spheroid (S) centered at origin $(x/a)^2 + (y/a)^2 + (z/b)^2 = 1$ where $a$: equatorial semi-axis & $b$: polar semi-axis.

Intersection of (P) & (S), is always an ellipse (except when it is onto equatorial plane then a circle) yields:

$[(b\gamma)^2+(aα)^2]*x^2 + 2aαβ*xy + [(bγ)^2+(aβ)^2]*y^2 - (abγ)^2 = 0$

General conic equation of an ellipse $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$
Here $D$, $E$ are both zero bcz the ellipse is centered at origin

My question, how to find these semi-axis (of this intersection) major and minor, $m$ & $n$ respectively in terms of $a, b, α, β$ and $γ$.

the semi-axis major $m$ will be always equal to $a$, so $m=a$, but what it happens with semi-axis minor $n$, $n=??$

I found here similar question, but it gives the semi-axis of the PROJECTION of this intersection onto xy plane, unfortunately it's not my case :(

• do you know about eigenvectors of a matrix ? – G Cab Feb 6 '17 at 0:01
• @G Cab, yes I know a little but not enough – user412889 Feb 6 '17 at 6:10
• ok, so tell what you know about conics (e.g. can you transform $Ax^2+Bxy+Cy^2$ in polar form ?) – G Cab Feb 6 '17 at 18:16
• $m=a$ that is ok at any case, but $n$ fails, I tested some planes equations for example $y=0$ , $n$ should be equal to b but no, it seems wrong – user412889 Feb 6 '17 at 18:39
• To tell what I did, I rotated this $Ax^2+Bxy+Cy^2+F=0$ to eliminate the $xy$ term, I got $Ax'^2+Cy'^2=-F$ then dividing both sides by $-F$ to get $(A/-F)x'^2+(C/-F)y'^2 = 1$ then I got semi-axis $m=a$ and $n= abγ/sqrt[a^2(α^2+β^2)+(bγ)^2]$ – user412889 Feb 6 '17 at 18:41

You have two important directions in your setup. $(0,0,1)$ is the direction of the axis connecting the poles. Any vector perpendicular to that will lie in the equatorial plane of $S$. $(\alpha,\beta,\gamma)$ is the normal of the plane $P$. Any vector perpendicular to this will lie within the plane. Now you can use the cross product of these two vectors to obtain a vector which is perpendicular to both. It will therefore be a vector in the plane $P$ and in the equatorial plane of $S$. This points in the direction of the semi-axis $m$, so if you scale that vector to length $a$ you'll get a point on the intersection.

To obtain the other direction, you need the perpendicular to the direction of $m$ within the plane. You can compute another cross product, this time with that direction and the normal of $P$ (again). Doing the math:

$$v=\left(\begin{pmatrix}0\\0\\1\end{pmatrix}\times \begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}\right)\times \begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}= \begin{pmatrix}-\beta\\\alpha\\0\end{pmatrix}\times \begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}= \begin{pmatrix}\alpha\gamma\\\beta\gamma\\-\alpha^2-\beta^2\end{pmatrix}$$

Now plug $s$ times this into the equation for $S$:

\begin{align*} \left(\frac{s\alpha\gamma}a\right)^2+ \left(\frac{s\beta\gamma}a\right)^2+ \left(\frac{-s(\alpha^2+\beta^2)}b\right)^2&=1 \\ s^2(\alpha^2+\beta^2) \left(\frac{\gamma^2}{a^2}+\frac{\alpha^2+\beta^2}{b^2}\right)&=1 \end{align*}

Solve for $s$ picking the positive solution and you get $s\,\lVert v\rVert$ as the length of the semiaxis that does not lie in the equatorial plane.

$$n^2=s^2\,\lVert v\rVert^2 =\frac{\alpha^2\gamma^2+\beta^2\gamma^2+(\alpha^2+\beta^2)^2}{(\alpha^2+\beta^2)\left(\frac{\gamma^2}{a^2}+\frac{\alpha^2+\beta^2}{b^2}\right)} =\frac{a^2b^2(\alpha^2+\beta^2+\gamma^2)}{a^2(\alpha^2+\beta^2)+b^2\gamma^2}$$

In a comment you indicate that you did a sanity check with the plane $y=0$, i.e. $\alpha=0,\beta=1,\gamma=0$. In that case you get $n^2=\frac{a^2b^2}{a^2}=b^2$ as expected. For $\beta=2$ you get the same, which is important as $2y=0$ is the same plane.

• thank you so much for your help, regards! – user412889 Feb 7 '17 at 20:06