<s>[INST] 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.) 
 [/INST] 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.
$$y=-\frac{\sqrt3}{2}x+3\ \text{ and }y=\frac{\sqrt3}{2}x+3 \ $$
</s>