Finding the intersection points of common tangents on a pair of non-intersecting ellipses I'm having some trouble with this, I don't know why but for some reason it is giving me a lot of trouble. Ultimately I intend to implement it into a program for modelling something, but I cannot even get past this first step!
Let's say two ellipses (non-rotated):
$\frac{(x+x_{1})^{2}}{a^2} + \frac{(y+y_1)^2}{b^2} =\,1   $, and
$\frac{(x+x_{2})^{2}}{c^2} + \frac{(y+y_2)^2}{d^2} =\,1   $
I understand that at the tangents the ellipses will share identical gradients and so differentiating the above equations:
$\frac{x+x_1}{a^2} + \frac{y+y_1}{b^2} \frac{\text{d}y}{\text{dx}} = 0$, and
$\frac{x+x_2}{c^2} + \frac{y+y_2}{d^2} \frac{\text{d}y}{\text{dx}} = 0$, and now rearranging for $\frac{\text{d}y}{\text{dx}}$, and setting equal to one another:
$\frac{b^2}{a^2}\frac{x+x_1}{y+y_1}=\frac{d^2}{c^2}\frac{x+x_2}{y+y_2}$
Now by a quick sketch I can see that there will be 4 tangents, two running parallel to each other and two intersecting with one another, so I ultimately expect 4 solutions to the sets of equations, I only really care about a single pair of points but I suppose that is neither here nor there.  
What is there to do?  I must admit my geometry game is a bit off.  I don't know why this confuses me so, but it does!  Can anyone please offer me some insight?  Thnx.
 A: You need more than identical gradients for a tangent line, you need identical lines, which also means identical $y$-intercepts (or $x$-intercepts if the line is vertical). That will give you two non-linear equations in the two variables $x$ and $y$, which you can solve to find the anchor point of each tangent line.
I think you would be better off using a parametrization of each ellipse:
$$\text{Ellipse $1$:}\quad x=x_1+a\cos t,\quad y=y_1+b\sin t,\quad 0\le t\le 2\pi$$
$$\text{Ellipse $2$:}\quad x=x_2 +c\cos u,\quad y=y_2+d\sin u,\quad 0\le u\le 2\pi$$
For the first ellipse you get
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac ba\cot t$$
and for the second
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac dc\cot u$$
The equation of the tangent line for ellipse $1$ is
$$y=-\frac ba\cot t\left( x-x_1-a\cos t \right)+y_1+b\sin t$$
which simplifies to
$$y=\left(-\frac ba\cot t\right)x+ y_1 + b\sin t + \frac ba x_1 \cot t + b \frac{\cot t}{\sin t}$$
The equation of the tangent line for ellipse $2$ is similar:
$$y=\left(-\frac dc\cot u\right)x+ y_2 + d\sin u + \frac dc x_2 \cot u + d \frac{\cot u}{\sin u}$$
Equate the slopes and the $y$-intercepts and solve. That looks pretty horrendous, probably best done with graphical or numerical techniques.
