Concentric and Tangent Ellipse from 2 Hyperbolas Find the equation of the ellipse that is concentric and tangent to the following hyperbolas:
$$\begin{align}
-2x^2 + 9y^2 - 20x - 108y + 256 &= 0 \\
x^2 - 4y^2 + 10x + 48y - 219 &= 0 
\end{align}$$
I did the math for both equations and the center is the same: $(-5,6)$. 
I have the equations of each:
For the first:
$$-2x^2 + 9y^2 - 20x - 108y + 256 = 0$$  
$$-\frac{(x+5)^2}{9} + \frac{(y-6)^2}{2} = 1$$
For the 2nd:
$$x^2 - 4y^2 + 10x + 48y - 219 = 0$$ 
$$\frac{(x+5)^2}{100} - \frac{(y-6)^2}{25} = 1$$
I know concentric means the new Ellipse has to have the same center, but I don't know how to make it tangent. Please help. Thanks

Not a complete solution, but this approach will work:
An ellipse with centre at the point $(-5,6)$ would be 
$$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$
Now change to a new set of axes ($u, v$) parallel to the $x,y$ axes, but with origin at the point $(-5,6)$. In other words, put $u = x+5$ and $v = y-6$.
Referred to the new axes, the equations of the three conics become:
$$-\frac{u^2}{9} + \frac{v^2}{2} = 1,$$
$$\frac{u^2}{100} - \frac{v^2}{25} = 1,$$
$$\frac{u^2}{a^2}+\frac{v^2}{b^2} = 1$$
(Note that the values $a$ and $b$ are unchanged by the change in coordinates.)
Now, looking at a hyperbola $u^2/a^2-v^2/b^2 = \pm 1$ and an ellipse $u^2/c^2+v^2/d^2 = 1$, you can see (from a sketch) that the ellipse will be a tangent to the hyperbola only when they meet at a point on the $u$ or $v$ axis.
Can you finish it from there?
