I'm trying to write a SVG drawing program, and it is necessary to render skewed ellipses as rotated ellipses with correctly recalculated axis lengths. This shows the problem I'm trying to solve:
The dark blue points are easily calculated from the skew geometry. I know the center of the new ellipse, but I don't know how to calculate the angle of rotation of the new ellipse and its new axis lengths. I did this by eye in a drawing program, but I need to do this mathematically in my code.
Doable, but the result is a bit messy.
First, observe that the pictured transformation is equivalent to a shear that leaves the center of the ellipse fixed followed by a translation. Translations don’t affect the shape or orientation of an ellipse, so we can simplify things a bit by considering an ellipse in standard position and a shear that fixes the origin.
A general equation for an ellipse centered on the origin is $Ax^2+Bxy+Cy^2=1$, which can be written in vector form as $$\mathbf x^T\begin{bmatrix}A&\frac B2\\\frac B2&C\end{bmatrix}\mathbf x=1.$$ For an ellipse, the eigenvalues of this matrix are both positive. If $\lambda_1$ and $\lambda_2$ are these eigenvalues, with $\lambda_1\le\lambda_2$, then the semimajor and semiminor axis lengths of the ellipse are given by $a^2=1/\lambda_1$ and $b^2=1/\lambda_2$, respectively. The related eigenvalues point in the directions of these axes, so the rotation of the ellipse is given by the angle of the line defined by the eigenspace of $\lambda_1$. Solving for the eigenvalues, we get $$\lambda_{1,2}=\frac12\left(A+C\mp\sqrt{(A+C)^2+B^2-4AC}\right)\tag{1}$$ with associated eigenvectors $$\left[A-C\mp\sqrt{(A+C)^2+B^2-4AC},B\right]^T.\tag{2}$$ (You can check that these two vectors are orthogonal.)
In this case, we start with the ellipse $x^2/a^2+y^2/b^2=1$, or in vector form $\mathbf x^T\mathbf Q\mathbf x=1$, with $\mathbf Q=\operatorname{diag}(a^{-2},b^{-2})$. The equation of the transformed ellipse is obtained by substituting $\mathbf S^{-1}\mathbf x$ for $\mathbf x$, where $\mathbf S$ is the matrix of the shear transformation:$$\mathbf S=\begin{bmatrix}1&\tan\beta\\0&1\end{bmatrix},$$ i.e., $(\mathbf S^{-1}\mathbf x)^T\mathbf Q(\mathbf S^{-1}\mathbf x)=\mathbf x^T((\mathbf S^{-1})^T\mathbf Q\mathbf S^{-1})\mathbf x=\mathbf x^T\mathbf Q'\mathbf x$. (Note that inverting $\mathbf S$ is a simple matter of replacing $\tan\beta$ with $-\tan\beta$.) Multiplying this out, we get $$\mathbf Q'=\begin{bmatrix}\frac1{a^2}&-{\tan\beta\over a^2}\\-{\tan\beta\over a^2}&\frac1{b^2}+{\tan^2\beta\over a^2}\end{bmatrix}.\tag{3}$$ Substitute these values for $A$, $B$ and $C$ in equations (1) and (2) to find the semi-axis lengths and rotation of the transformed ellipse. (Instead of doing all of this, you could have looked up formulas for the semi-axis lengths and rotation of a general-form ellipse in sources such as Wikipedia, but it’s not very difficult to work it out for this relatively simple case.)
For example, let’s say that for your ellipse, $a=4$, $b=2$ and, as in the illustration, $\beta=\frac\pi6$. Substituting these values into $\mathbf Q'$ we get $A=\frac1{16}$, $B=-\frac1{8\sqrt3}$ and $C=\frac{13}{48}$. Plugging these coefficients into equation (1) produces ${4\mp\sqrt7\over24}$ for the eigenvalues, which gives (after a bit of work) $\frac2{\sqrt3}(\sqrt7\pm1)$ (i.e., approximately $4.21$ and $1.90$) for the semi-axis lengths and $\arctan\left({2\sqrt{21}-5\sqrt3\over3}\right)\approx9.55°$ for the ellipse’s rotation.
As a sanity check, observe that the shear transformation $\mathbf S$ has determinant $1$, so $\det{\mathbf Q'}=\det{\mathbf Q}=(ab)^{-2}$. But the determinant of a matrix is equal to the product of its eigenvalues, so the product of the semi-axis lengths of these ellipses is constant. If we multiply the semi-axis lengths of the transformed ellipse computed in the previous paragraph together, we indeed get $8=4\cdot2$.
It’s interesting to plot these values as functions of the shear angle $\beta$. The semi-axis lengths are monotonic, as one might expect, but the rotation angle reaches a maximum—at around $60°$ of shear for the above example—and then decreases as the shear angle is further increased.
Addendum: It turns out that the peak rotation angle is achieved when the shear angle $\beta$ satisfies $\tan\beta={\sqrt{a^2-b^2}\over b}=\frac fb$, where $f$ is the distance of the foci from the center of the original ellipse. This quantity is also known as the second eccentricity of the ellipse and the corresponding angle its angular eccentricity.