Recently, I've been trying to work out a closed formula for the Mandart inellipse of a triangle, and I made a little plaything on Desmos to streamline the process. So far I've successfully located the splitters of the triangle and the Mittenpunkt, so I think I'm good to go, but I wanted to make sure just in case.
So, with that said:
If one knows an ellipse is tangent to a triangle at three points and has a defined center, then is it sufficient to say that only one such ellipse exists?
Here, when I refer to the "center" of the ellipse, I am referring to the midpoint between the two foci, to be clear. Any help is appreciated! Thank you!
Using skew (axes) transformation,
$$(x',y')=(x+y\cos \omega,y\sin \omega)$$
You could always generate a two-parameter family of ellipses touching the three lines defining $\Delta ABC$,
$$ \left( \frac{x}{\lambda a}+\frac{y}{\mu b}-1 \right)^2=\frac{4xy}{ab} \left( \frac{1}{\lambda}-1 \right) \left( \frac{1}{\mu}-1 \right)$$
which touches the axes at $A(\lambda a,0)$ and $B(0,\mu b)$ and the line $\dfrac{x}{a}+\dfrac{y}{b}=1$.
For inscribed ellipse, $$\lambda,\mu \in (0,1)$$
Now the centre is
$$(X,Y)=\frac{(\lambda a,\mu b)}{2(\lambda+\mu-\lambda \mu)} \implies (\lambda,\mu)= \left( 1+\frac{bX}{aY}-\frac{b}{2Y}, 1+\frac{aY}{bX}-\frac{a}{2X} \right)$$
If $a,b>0$, then
$$\frac{X}{2a}+\frac{Y}{2b}>1 \land 0<X<\frac{a}{2} \land 0<Y<\frac{b}{2}$$
which is the interior of the medial triangle of $\Delta ABC$.
Please also refer to my older post here.