Finding the tangents common to two rotated ellipses? Is there a way to find the four tangents that two rotated ellipses share?
I believe that if two ellipses do not intersect and do not contain one another, they will have four tangents in common and I wish to find these tangents.
I know how to complete the same process with circles and wish to do the same with ellipses.
 A: Assuming, for the sake of argument, the ellipse is centered and rotated about the origin (if it weren't, the same principle would apply, but the math would be more complex), the original, un-rotated ellipse can be modeled as a parametric equation:
$$x(t)=a\cos(t),y(t)=b\sin(t)\space 0\le t<2\pi$$
Now rotate it by multiplying it by the rotation matrix:
$$
        \begin{bmatrix}
        \cos(\theta) & -\sin(\theta) \\
        \sin(\theta) &  \cos(\theta) \\
        \end{bmatrix}\begin{bmatrix}
        a\cos(t) \\
        b\sin(t) \\
        \end{bmatrix}
$$
The resulting parametric equation for the rotated ellipse is
$$x(t)=a\cos(\theta)\cos(t)-b\sin(\theta)\sin(t)$$
$$y(t)=a\sin(\theta)\cos(t)+b\cos(\theta)\sin(t)$$
The tangent to any parametric curve is equal to $\frac{y'(t)}{x'(t)}$, so take the derivatives of each set of parametric equations, equate their ratios, and solve for t (remember that $\theta$ is a constant). That should give you the point where both their tangents are equal.
A: Expanding on WimC’s comment, this problem is dual to the problem of finding the intersection of a pair of conics. That is, if the two nondegenerate conics are represented by $3\times3$ homogeneous matrices $A$ and $B$, and lines are represented by homogeneous vectors $\mathbf l = [\lambda, \mu, \tau]^T$, so that the implicit equation of the line is $\mathbf l^T\mathbf x = 0$, the common tangents to the two conics are the solutions of the system $\mathbf l^TA^*\mathbf l = \mathbf l^TB^*\mathbf l = 0$. For nondegenerate conics, the dual conic matrix is any nonzero multiple of the inverse of the conic’s matrix (such as the adjugate). Of course, since any multiple of $\mathbf l$ represents the same line (these are homogeneous vectors, after all), if there are any solutions at all, then there will be an infinite number of them, so it’s helpful to add another simple constraint such as $\lambda+\mu+\tau = 1$. If you know that none of the tangents can pass through the origin, another simple way to constrain the solutions is to set $\tau=-1$.  
Using the example from Mark McClure’s answer, we have $$A = \begin{bmatrix} 1 & -\frac12 & 0 \\ -\frac12 & 1 & 0 \\ 0&0&-1\end{bmatrix} \\ B = \begin{bmatrix} 2 & \frac12 & 0 \\ \frac12 & 3 & 0 \\ 0&0&-4 \end{bmatrix}.$$ (Note that I’m using the constant terms from the Mathematica code, which are swapped from those in the text of the answer.) Inverting these and multiplying by some arbitrary-looking scalars to make things look tidier, we get $$A^* = 3 A^{-1} = \begin{bmatrix} 4 & 2 & 0 \\ 2 & 4 & 0 \\ 0&0&-3 \end{bmatrix} \\ B^* = \frac{23}2 B^{-1} = \begin{bmatrix} 6&-1&0 \\ -1&4&0 \\ 0&0&-\frac{23}8 \end{bmatrix}.$$ The resulting system of equations is $$4 \lambda ^2+4 \lambda  \mu +4 \mu ^2-3 \tau ^2 = 0 \\ 6 \lambda ^2-2 \lambda  \mu +4 \mu ^2-\frac{23}8\tau ^2 = 0.$$ In this case both ellipses surround the origin, so we set $\tau = -1$ and solve. Turning to Mathematica, we obtain the solutions
{{lambda -> 0.0246484, mu -> 0.853438},
  {lambda -> -0.0246484, mu -> -0.853438},
  {lambda -> -0.703267, mu -> -0.264046},
  {lambda -> 0.703267, mu -> 0.264046}}

which you can verify are the same lines as those computed in Mark McClure’s answer.  
If you don’t have handy software for computing solutions to systems of second-degree equations, Jürgen Richter-Gebert presents an algorithm for finding the intersections of two conics in his Perspectives on Projective Geometry, which is outlined in this answer.
