I arrive at the general formula for these ellipses (with your parameter $A$):
$\tag{0}\dfrac{(x + y)^2}{A} + \dfrac{(x - y)^2}{2-A}= 2$
with height=$\sqrt{A}$ and width=$\sqrt{2-A}$.
in a still different manner compared with @Christian Blatter and Parcly Taxel.
I first consider on the family $F$ of ellipses with axes $Ox$ and $Oy$ that are inscribed in the rotated square $C$ with vertices $(1,0), (0,1), (-1,0), (0,-1)$. (as Parcly Taxel does). It means that unit circle is their common "orthoptic circle": a theorem (established by Monge) says that the locus of points from which the 2 tangents that one can draw to the ellipse are orthogonal is a circle. See this. Remark : the name "orthoptic" comes from the fact that from any point of this circle, one "sees" the ellipse under a right angle.
Thus, using this property for a given ellipse $E \in F$, the rectangle $R$ with sides parallel to the axes which is circumscribed to E is inscribed in the unit circle. It should be observed that ratio length/width for $R$ is the same as the ratio of semi-axes of E. The coordinates of the vertices of $R$ are $(\pm \cos(t),\pm \sin(t))$, for a certain $t$. Thus, setting $a:=\cos(t)$, we have $\sqrt{1-a^2}=\sin(t)$ and the general ellipse $E \in F$ has equation:
$$\tag{1}\dfrac{X^2}{a^2}+\dfrac{Y^2}{1-a^2}=1$$
Now, by using a $\pi/4$ rotation of coordinates associated with a $\sqrt{2}$ enlargement (homothety), i.e. transformation $\cases{X=x+y\\Y=x-y}$ (in order that square C is transformed into the desired square), we obtain:
$$\tag{2}\dfrac{(x+y)^2}{a^2}+\dfrac{(x-y)^2}{1-a^2}=4$$
It remains to express (2) with parameter $A$ instead of parameter $A$. This is easily done by imposing, that (2) is verified for point $(x,y)=(A-1,1)$. This gives $a^2=\dfrac{A}{2}$. Plugging this expression into (2), one obtains equation (0).
The height and width are deduced from equation (0) by computing the coordinates of points $(x,x)$ and $(-x,x)$ of the ellipse.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/dFHC9.jpg)
(A-1,0)
,(1-A,0)
,(0,A-1)
and(0,1-A)
. $\endgroup$(A-1,1), (1-A,-1), (1,A-1), (-1,1-A)
. $\endgroup$