Suppose you are given three arbitrary points in the $xy$ plane. You want to generate all possible ellipses that pass through the three points, and at the same time have a given area of $S$. If $S$ is greater than the minimum possible area, then is it true that there will be exactly six possible ellipses?
My progress:
I first transformed the three given points (which are $A, B, C$) to the vertices of an equilateral triangle, call them $A',B',C'$, with $A' = (0,0), B' = (1, 0), C' = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$. From symmetry, we can pass three congruent ellipses through $A',B', C'$, the first with one axis parallel to $A'B'$, the second with one axis parallel to $B'C'$ and the third with axis parallel to $A'C'$.
The area of each of these three ellipses is $S \ |\det(T)| $ where $T$ is the transformation matrix that sends $A,B,C$ to $A', B', C'$.
Since the ellipses are congruent, and symmetrical about the centroid of $\triangle A'B'C'$ it suffices to find the first one, and then rotate it by $120^\circ$ and $240^\circ$ about the centroid to generate the second and third.
The equation of the ellipses with an axis parallel to the $x$ axis (and the other one to the $y$ axis), that passes through $A',B',C'$ takes the form
$ \dfrac{(x - \dfrac{1}{2})^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 $
Substituting $A'$ or $B'$ gives
$\dfrac{1}{4 a^2} + \dfrac{k^2}{b^2} = 1 \dots (1)$
Substituing $C'$ gives
$ \dfrac{\sqrt{3}}{2} - k = b \dots (2)$
And finally, $S' = S \ | \det(T) | $ will be the area of the ellipse. So
$ S' = S \ | \det(T) | = \pi a b \dots (3) $
Substituting for $k$ from $(2)$ in $(1)$:
$\dfrac{1}{4 a^2} + \dfrac{3}{4 b^2} - \dfrac{\sqrt{3}}{b} = 0 \dots (4)$
Next, I checked numerically the solution of $(3)$ and $(4)$.
And it turns out there are two possible solutions.
The figure below shows the two ellipses for an area that is $1.5 \times S_{min} $
Hence the total number of possible ellipses is $3 \times 2 = 6$ ellipses.
Putting this all together, I picked three points: $A = (3, 1), B = (16, 5), C = (7, 13) $ and drew the possible ellipses through them, each having an area of $1.5 S_{min}$ where $S_{min} $ is the minimum possible area of an ellipse passing through $A,B,C$. This is shown below:
I would appreciate it if someone could confirm my findings.
Let $ABC$ be an equilateral triangle, $D$, $E$ and $F$ the midpoints of its sides and $G$ its centre. Consider then any ray $r$ with origin $G$ and a point $O$ on $r$ inside triangle $DEF$.
The ellipse through $ABC$ centred at $O$ has its minimum area when $O=G$ (the ellipse is the circumcircle of $ABC$), but as $O$ approaches the boundary of triangle $DEF$ the area of the ellipse tends to $+\infty$, because the ellipse degenerates into a pair of parallel lines when $D$ is on a side of $DEF$.
By continuity, we can then choose $O$ on the ray such that the area of the ellipse has any fixed value $S$ greater than the minimum area. Hence there are infinitely many ellipses with area $S$, passing through $ABC$.
EDIT.
Here's a plot (made with Mathematica) of the loci of centres (blue), when the related ellipse has area equal to (from left to right) $1.1$, $1.2$, $1.5$, $2$ times minimum area, for an equilateral triangle centred at the origin with radius $1$. Loci are shown only for the inner triangle on the right and should be rotated by $\pm120°$ about the origin to get their full extent.