Parameters of the Mandart inellipse What are the parameters of the Mandart ellipse
in terms of its associated triangle?
It looks that information on the Mandart ellipse
is less known than its Steiner counterpart. 
Easily found references
wiki,
MathWorld
look incomplete, 
so the provided answer collects the missing parts together.
 A: 
The Mandart inellipse of a triangle
is an ellipse inscribed within the triangle,
tangent to its sides
at the contact points
$X_a,X_b,X_c$
of its excircles.
Its parameters
are expressed in terms of the coordinates of vertices $A,B,C$,
side lengths $a,b,c$,
semiperimeter $\rho=\tfrac12\,(a+b+c)$,
inradius $r$ and
circumradius $R$
of the associated $\triangle ABC$.
The Mandart inellipse is a generalized Steiner inellipse
with parameters
\begin{align}
u&=
\frac{(b-c)^2-a^2}{(b-c)^2-a\,(-a+2b+2c)}
,\\
v&=
\frac{(c-a)^2-b^2}{(c-a)^2-b\,(-b+2c+2a)}
,\\
w&=
\frac{(a-b)^2-c^2}{(a-b)^2-c\,(-c+2a+2b)}
\end{align}
or
\begin{align}
u&=1-\frac{a\,(\rho-a)}{r\,(r+4\,R)}
,\\
v&=1-\frac{b\,(\rho-b)}{r\,(r+4\,R)}
,\\
w&=1-\frac{c\,(\rho-c)}{r\,(r+4\,R)}
.
\end{align}
The foci are the roots of quadratic equation
\begin{align}
z^2-((1-u)A+(1-v)B+(1-w)C)\,z+u\,B\,C+v\,C\,A+w\,A\,B
&=0
,
\end{align}
\begin{align}
F_{1,2}&=
\tfrac12\,((1-u)\,A+(1-v)\,B+(1-w)\,C)
\\
&\pm
\sqrt{\tfrac14\,((1-u)\,A+(1-v)\,B+(1-w)\,C)^2-u\,B\,C-v\,C\,A-w\,A\,B}
.
\end{align}
The center of the Mandart inellipse is the mittenpunkt of the triangle,
\begin{align}
M&=\tfrac12\,((1-u)\,A+(1-v)\,B+(1-w)\,C)
.
\end{align}
The distance between the foci
\begin{align}
|F_1F_2|&=\frac{\rho}{1+\displaystyle\frac r{4R}}\,\sqrt[4]{1-\frac{2\,r}R}
.
\end{align}
Its area is found as
\begin{align}
S_{M}&=\frac{\pi((a+b-c)(b+c-a)(c+a-b))}{(2(ab+bc+ca)-(a^2+b^2+c^2))^{3/2}}\,S_{ABC}
\\
&=\frac{\pi\,\rho^2\,r^3}{(r^2+4\,r\,R)^{3/2}}
=\frac{\pi\,\rho\,r^2}{(r^2+4\,r\,R)^{3/2}}
\,S_{ABC}
=\frac{\pi\,r}{(r^2+4\,r\,R)^{3/2}}
\,S_{ABC}^2
, 
\end{align}
Major and minor semi-axes are
\begin{align}
s_a&=\frac{\rho}{4+\displaystyle\frac rR}\,\sqrt{3-\Big(1+\frac rR\Big)^2
+2\,\sqrt{1-\frac{2\,r}R}}
,\\
s_b&=\frac{\rho}{4+\displaystyle\frac rR}\,\sqrt{3-\Big(1+\frac rR\Big)^2
-2\,\sqrt{1-\frac{2\,r}R}}
.
\end{align}

Edit:
Eccentricity of the Mandart inellipse is completely defined by
the ratio $V=\frac rR$:
\begin{align}
e^2=1-\frac{s_b^2}{s_a^2}
&=
\frac{4\Big(4V-2+\sqrt{1-2V}(3-(V+1)^2)\Big)}
{V^3(4+V)}
.
\end{align}
And vise versa, given the eccentricity $e$ of
the Mandart inellipse, the ratio $V=\frac rR$
is found as
a solution of the quartic
\begin{align}
V^4+4V^3+2qV-q&=0
,\quad
q=\tfrac{16}{e^4}\,(1-e^2)
,
\end{align}
\begin{align} 
V=\frac rR&=
-1-\tfrac12\,\sqrt{3+t}+\tfrac12\,\sqrt{9-t+\frac{16+4q}{\sqrt{3+t}}}
,\quad \text{where}\quad t=(\sqrt[3]{2q}-1)^2
.
\end{align}
