Tangent Points for Common Tangent to Two Ellipses This is somewhat similar to my other question here.
Consider the two ellipses given by the equations
\begin{equation}
  \frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1
\end{equation}
and
\begin{equation}
  \frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1.
\end{equation}
How do I find the coordinates for the two tangent points of their common tangent at the left side of the ellipses? (I hope the question makes sense.) 

Let's introduce the following new variables:
$$2u=x\ \text{ and }\ y-1=v.$$
With these new variables, we have
$$u^2+v^2=1\ \text{ and }\ u^2+(v-3)^2=\frac14$$ 
that is, we have two circles as shown in the figure below.

We have similar triangles and we can see that $OD=6$. Also, by the Pythagorean theorem $DC=\frac{\sqrt{35}}2$ and by the similarity of $OBD$ and $O'CD$: $OB=\frac6{\sqrt{35}}$. So, the slope of the red straight lien is $-\sqrt{35}$. Don't forget a about the other tangent, not shown, whose slope is $\sqrt{35}$.
The two tangent lines in the $u,v$ system are
$$v=-\sqrt{35}u+6\ \text{ and } \ v=\sqrt{35}u+6.$$
Returning to the $x,y$ coordinate system, we get
$$y=-\frac{\sqrt{35}}2x+7\ \text{ and } \ y=\frac{\sqrt{35}}2x+7.$$
EDIT
Unforgivable! I forgave the other pair of tangents:

After similar calculations we get the equations of the other pair of tangent lines.
$$y=-\frac{\sqrt3}{2}x+3\ \text{ and }y=\frac{\sqrt3}{2}x+3 \ $$
