Ensuring tangency between two ellipses we know that two tangent circles of centers $(x_{1},y_{1})$ and $(x_{2},y_{2})$, radii $r_{1}$ and $r_{2}$ will have to respect this formula :
$$
(x_{1}-x_{2})^2+(y_{1}-y_{2})^2=(r_{1} \pm r_{2})^2
$$
Is there a similar formula for two tangent ellipses ?
Assuming that we have two ellipses with the following equations :
$$ 
\frac{(x-x_{1})^2 }{a^2} + \frac{(y-y_{1})^2}{b^2} = 1
$$
and
$$ 
\frac{(x-x_{2})^2 }{c^2} + \frac{(y-y_{2})^2}{d^2} = 1
$$
Let's say that we do not know  $x_{2}$ in the last equation. How could it be defined to ensure  tangency  ?
EDIT : Dear everyone, thank you for your answers. I wanted to add some more details :
With a CAD software, I drew these two ellipses :
The first one is centered :
$$
\frac{x^2}{50^2} + \frac{y^2}{25^2} = 1 
$$
The other one has this equation :
$$
\frac{(x+35)^2}{20^2} + \frac{(y-y_{2})^2}{50^2} = 1 
$$
with the constraint $y_{2} > 0$  , the software's solver find the coordinate of the center with $y_{2}$ for tangency of $(-35;30.732)$ and the coordinates of the tangency point $(-48.99,-5.01)$
As illustrated in the following pic :

Angles are badly displayed but are bound to the respective lines between the tangency point and the foci
Using Amogh's elegant answer I tried the following :
Assuming that there exist a point $(x,y)$ such that the two previous ellipses are tangent, this point must satisfy the following system ( setting $ \omega = \theta $  due to the assumed tangency ) :
$$
(1)\; \;   x_{1} + a \, cos(\theta) - x_{2} - c \, cos(\theta) = 0
$$
$$
(2)\; \;   y _{1} + b \, sin(\theta) - y_{2} - d \, sin(\theta) = 0
$$
Or in our case :
$$
(1)\; \;   50 cos(\theta) + 35 - 20 cos(\theta) = 0
$$
$$
(2)\; \;    25 sin(\theta) - y_{2} - 50 sin(\theta) = 0
$$
the first line gives :
$$
\theta =  arccos(\frac{-35}{50-20}) 
$$
which is not possible.
Can someone point out what I'm doing wrong ?
PS : I still have to put some time in studying the other (really good) answers and suggestion, which are more complex but would like to thank the posters
 A: Two curves are defined to be tangent at a point if and only if their respective tangent lines at that point have the same slope.
Differentiating the equation of the first ellipse gives us
$$\frac{dy}{dx} = -\frac{b}{a}\frac{(x-x_1)}{(y-y_1)}$$
Similarly, for the second ellipse we have
$$\frac{dy}{dx} = -\frac{d}{c}\frac{(x-x_2)}{(y-y_2)}$$
For tangency, we must have
$$-\frac{b}{a}\frac{(x-x_1)}{(y-y_1)} = -\frac{d}{c}\frac{(x-x_2)}{(y-y_2)}$$
where $(x, y)$ is the point of tangency
An arbitrary point on the first ellipse has coordinates of the form $(x_1 + a\cos{\theta}, y_1 + b\sin{\theta})$, where $\theta$ is a parameter.
Similarly, an arbitrary point of the second ellipse can be written $(x_2 + c\cos{\omega}, y_2 + d\sin{\omega})$, where $\omega$ is a parameter.
Substituting this into the condition for tangency, we get
$$-\frac{b}{a}\frac{a\cos{\theta}}{b\sin{\theta}} = -\frac{d}{c}\frac{c\cos{\omega}}{d\sin{\omega}}$$
$$\cot\theta = \cot\omega$$
For two ellipses to be tangent, the values of $\theta$ and $\omega$ must be equal, or differ by an integral multiple of $\pi$.
These parameters $\theta$ and $\omega$ are called the eccentric angles of the points.
Before we check for tangency, we must first check whether the two ellipses intersect. Then, if they intersect at a point where
$$\theta - \omega = n\pi$$
where n is an integer, then the two ellipses are tangent to each other.
