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 such a tangent line be $y=mx+c$
Solving simultaneously with the ellipses gives two quadratics in $x$ which are $$(4m^2+1)x^2+8mx(c-1)+4c^2-8c=0$$ and $$(4m^2+1)x^2+8mx(c-4)+4c^2-32c+63=0$$
Each of these must have double roots at the point of tangency and therefore the discriminant is zero, so this leads to two equations in $m$ and $c$ which are $$64m^2(c-1)^2=4(4m^2+1)(4c^2-8c)$$ and $$64m^2(c-4)^2=4(4m^2+1)(4c^2-32c+63)$$
Dividing these eliminates the $m$ terms and we get an equation for $c$ which is $$3c^2-30c+63=0\implies c=3,7$$
A diagram will confirm that the value $c=3$ gives us the tangents which pass between the ellipses, so we choose $c=7$ from which we get the gradient for the common tangent on the left as $$m=+\frac{\sqrt{35}}{2}$$ 
Using "$x=-\frac{B}{2A}$" the $x$-coordinates of the points of tangency which are $$x_1=-\frac{\sqrt{35}}{3}$$ and $$x_2=-\frac{\sqrt{35}}{6}$$
