find equations of an ellipsoid axes

Question

I have an ellipsoid with the center point at the Origin and the following equation: $$\alpha_1 x^2+\alpha_2 y^2+\alpha_3 z^2+2\beta_1 zy+2\beta_2 xz+2\beta_3 xy=1$$ How can I find the equations of lines correspond to its three main axes?

Answer 1

I'd like to fill in Jack's remark that this is simply an eigenvector problem, though I won't do so from a variational perspective. Let's write the ellipse in bilinear form using matrices:

\begin{align} \mathbf{x}^T M \mathbf{x} &=\begin{pmatrix} x&y&z \end{pmatrix}\begin{pmatrix}\alpha_1&\beta_3&\beta_3\\ \beta_{2}&\alpha_2 & \beta_1\\ \beta_2 &\beta_1 & \alpha_3\end{pmatrix}\begin{pmatrix}{x\\y\\z}\end{pmatrix}\\\\ &=\alpha_1 x^2+\alpha_2 y^2+\alpha_3 z^2+2\beta_1 zy+2\beta_2 xz+2\beta_3 xy=1 \end{align}

Suppose the matrix $M$ has full rank i.e. three orthonormal eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ with eigenvalues $\lambda_1,\lambda_2,\lambda_3$ respectively. Then $M$ may be diagonalized using the matrix $S=(\mathbf{v}_1\,\mathbf{v}_2\,\mathbf{v}_3)$ as

\begin{align} S^TMS &=\begin{pmatrix}\mathbf{v}_1^T\\\mathbf{v}_2^T\\\mathbf{v}_3^T\end{pmatrix} \begin{pmatrix}M \mathbf{v}_1&M\mathbf{v}_2&M \mathbf{v}_3\end{pmatrix}\\ &=\begin{pmatrix}\mathbf{v}_1^T\\\mathbf{v}_2^T\\\mathbf{v}_3^T\end{pmatrix} \begin{pmatrix}\lambda_1 \mathbf{v}_1&\lambda_2 \mathbf{v}_2 &\lambda_3 \mathbf{v}_3 \end{pmatrix}=\begin{pmatrix} \lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{pmatrix} \end{align} where the last equality follows from orthonormality. Hence $S^T MS=D$ where $D$ is diagonal. But $S^T S=I$ (since $(S^T S)_{ij}=\mathbf{v}_i^T \mathbf{v}_j=\delta_{ij})$ by orthonormality), so we may write $M=S D S^T$ and therefore the ellipse becomes $$\mathbf{x}^T M \mathbf{x}=\mathbf{x}^T (S D S^T)\mathbf{x}=\mathbf{X}^T D \mathbf{X}=\lambda_1 X^2+\lambda_2 Y^2+\lambda_3 Z^2=1$$ where $\mathbf{X}=S^T \mathbf{x}=(\mathbf{v}^T_1 \mathbf{x},\mathbf{v}^T_2 \mathbf{x},\mathbf{v}^T_2 \mathbf{x})^T=(X,Y,Z)^T$.

In these coordinates, we just have an ellipsoid with principal $XYZ$-axes; in other words, its axes are lines whose tangent vectors are just the basis vectors $\hat{e}_X,\hat{e}_Y,\hat{e}_Z$. We can map this back to the original coordinates via $\mathbf{x}=S\mathbf{X}$. But $S\cdot (\hat{e}_X,\hat{e}_Y,\hat{e}_Z)=SI=S$, so the column vectors of $S$ are the images of the basis $XYZ$-vectors.

So we arrive at a cute conclusion: The tangent vectors of the principal axes of the ellipse are the eigenvectors of the coordinate matrix of the ellipse. Glancing at the $XYZ$-form of the ellipse, we conclude that we can also read off the length of these axes from the cooresponding eigenvalues. So the analytic-geometry problem of describing the ellipse is indeed equivalent to an eigenvalue-eigenvector problem in linear algebra.

Answer 2

The vertex of an ellipse are the points for which the distance from the centre is locally stationary. So, find the centre (it is the origin), then the stationary points for the quadratic function giving the squared distance from the centre for a point on the ellipse. You have just an eigenvectors problem.