By the inelegant phrase " (added ) distances from foci definition" I simply mean this definition : an ellipse is the set of all points $P$ such that the sum of the distances from $P$ to two fixed points $F_1$ and $F_2$ is equal to a constant value $L$ ".
My attempt stops at the end of a reasoning aiming at recovering the length $m$ of the minor axis, and apparently fails.
Desmos construction : https://www.desmos.com/calculator/qcmoz3iiqt
Let point $F_1 = (a,b) $ and $ F_2 = ( c,d)$.
With these points, one can define an ellipse , as the set of all points $(x,y)$ such that
$$ \sqrt{( x-a)²+(y-b)²} + \sqrt {(x-c)² + ( y-d)²} = L $$ .
The center $C= (M_1 , M_2)$ of this ellipse is the midpoint of segment $[F_1, F_2]$, and the distance of each focus from this center is $D/2$ ( half the distance from $F_1$ to $F_2$.
Knowing the angle $A (= \arctan { (d-b)/(c-a)}$ ) that the line $(F_1, F_2)$ makes with the X-axis , one can define the coordinate system in which this ellipse has a standard equation:
- lime $X': y-b = tan(A) ( x-a)$
- line $Y' : y-b = (-1/ \tan(A) ) ( x-a) $
My first goal is to recover the length of the minor axis , that is the length of segment $[CP]$, point $P$ being the intersection of the ellipse with the new $Y$ axis , namely line $ Y'$.
$P$ must be at equal distance from both foci, hence, the length of segment $[PF_1]$ is $L/2$ . Also the length of $[CF_1]$ is $D/2$.
Let $\alpha$ be the angle $ \angle {PF_1C}$.
We have $ \alpha = \arccos { CF_1 / PF_1} = \arccos {(D/2)/(L/2) = \arccos {D/L}} $.
Hence finally the length $m$ of the minor axis should be " hypothenuse times sine($\alpha$)" , namely :
$ m = (L/2) \sin{(\alpha)}$.
But I can see the value I get with Desmos does not match the actual length of the minor axis on the graph.
I've always wondered this, and meant to work it out, so this question is a great opportunity. But I'm going to take a slightly different approach. Let me first rephrase things a little.
The geometric description of an ellipse is that it's the set of all points $P$ with the property that the sum of (i) the distance from $P$ to $F_1$ and (ii) the distance from $P$ to $F_2$ is some positive constant, which I'm going to call $2d$ to make the algebra nicer later. I'll use the letter G to refer to this description, as in "a G-ellipse".
The coordinate or analytic description is that it's the set of points $P = (x, y)$ satisfying an equation of the form $\frac{ (u-p)^2}{a^2} + \frac{(v - q)^2}{b^2} = 1$, where $u = x \cos t - y \sin t, v = x \sin t + y \cos t$ are coordinates in some rotated coordinate system, and (for nondegenerate ellipses), $a$ and $b$ are both nonzero; by negation, we can choose both to be positive.. In that rotated system, the ellipse-center is at $(u, v) = (p, q)$ and its major and minor axes have lengths $a$ and $b$ (possibly swapped depending on which of $a$ and $b$ is greater. I'll use C to refer to this description.
So the question becomes "Why do these two things define the same entity?", or perhaps, "Why is every G-ellipse also a C-ellipse, and vice-versa?"
You've taken the approach of (more or less) finding the equation of the C-ellipse from the G-ellipse description. I'm going to insert a couple of steps to make things easier.
This means that if we can prove that every G-ellipse with horizontally-aligned foci is a C-ellipse, then we will also know this is true for any G-ellipse. I'll actually show this claim, and show that the C-ellipse has the simple form where $t = 0$. I'll call this an axis-aligned C-ellipse, which has the coordinate form $\frac{(x-p)^2}{a^2} + \frac{(y-q)^2}{b^2} = 1$.
With these two "transformation rules" in hand, we can consider your question for the simple case where $F_1$ and $F_2$ lie on a horizontal line (using observation 1), and where their midpoint is the origin (observation 2).
So at this point, we have \begin{align} F_1 = (f, 0) \\ F_2 = (-f, 0) \end{align} Points with summed distances $2d$ from these are points $(x, y)$ satisfying $$ \sqrt{(x-f)^2 + y^2} + \sqrt{(x + f)^2 + y^2} = 2d $$ Note here that we must have $d > f$, because for any point other than the origin, the distances to $(f, 0)$ and $(-f, 0)$ sum to more than $2f$.
In particular, when $x= 0$, this gives \begin{align} \sqrt{f^2 + y^2} &+ \sqrt{f^2 + y^2} = 2d\\ 2\sqrt{f^2 + y^2} & = 2d\\ f^2 + y^2 & = d^2\\ y^2 & = (d^2 - f^2)\\ y & = \pm \sqrt{d^2 - f^2}\\ \end{align}
In the generic axis-aligned C-ellipse centered at the origin, whose equation is $$ \frac{x^2}{a^2} + \frac{y^2}{ b^2} = 1 $$ when $x = 0$, we get $y = b$. So (as a kind of foreshadowing), if our G-ellipse is to be a C-ellipse, we'll need to have $$ b = \sqrt{d^2 - f^2}. $$ (I've chosen the positive root in order to ensure that $b$ is positive.)
Doing a similar thing for the case where $y = 0$, we get \begin{align} \sqrt{(x-f)^2} &+ \sqrt{(x+f)^2} = d \\ (x-f) &+ (x+f) = 2d \\ 2x&= 2d \\ x = d \end{align}
which similarly leads us to the requirement that \begin{align} \frac{x^2}{a^2} = 1 \\ \frac{d^2}{a^2} = 1\\ a = d \end{align}
This means that if our G-ellipse, defined by $$ \sqrt{(x-f)^2 + y^2} + \sqrt{(x + f)^2 + y^2} = d $$ is actually a C-ellipse, then that C-ellipse must have the equation \begin{align} \frac{x^2 }{d^2}+ \frac{ y^2}{d^2 - f^2} &= 1 \\ \end{align} defined for $-d \le x \le d$.
We can do a little algebra here to express $y$ in terms of $x$: \begin{align} \frac{ y^2}{d^2 - f^2} &= 1 - \left( \frac{x^2 }{d^2} \right) \\ y^2 &= (d^2 - f^2)\left(1 - \left( \frac{x^2 }{d^2}\right) \right) \tag{*}\\ y &= \pm \sqrt{(d^2 - f^2)\left(1 - \left( \frac{x^2 }{d^2}\right)\right)} \\ \end{align}
Now let's see whether points of the form described by this formula (or by formula (*), which will be the one we use in practice) in fact satisfy the equation of the G-ellipse. We'll substitute ( * ) into the left side of $$ \sqrt{(x-f)^2 + y^2} + \sqrt{(x + f)^2 + y^2} = 2d $$ and see whether it reduces to $2d$. Buckle up!
Let's start with the first term, $\sqrt{(x-f)^2 + y^2}$: \begin{align} &\sqrt{(x-f)^2 + y^2} \\ &= \sqrt{(x-f)^2 + (d^2 - f^2)\left(1 - \left( \frac{x^2 }{d^2}\right) \right)} & \text{, after substituting in (*)} \\ &= \sqrt{x^2 - 2fx + f^2 + (d^2 - f^2) - \left( (d^2 - f^2)\frac{x^2 }{d^2}\right) } \\ &= \sqrt{x^2 - 2fx + d^2 - d^2 \frac{x^2 }{d^2} + f^2\frac{x^2 }{d^2} } \\ &= \sqrt{x^2 - 2fx + d^2 - x^2 + \frac{(fx)^2 }{d^2} } \\ &= \sqrt{- 2fx + d^2 + \left(\frac{fx}{d}\right)^2} \\ &= \sqrt{d^2 - 2d\frac{fx}{d} + \left(\frac{fx}{d}\right)^2} \\ &= \sqrt{(d- \frac{fx}{d})^2} \\ &= d- \frac{fx}{d} \end{align} where that last simplification needs some explanation -- how did I know which square root to take? You'll see above that $x$ lies in the range $-d$ to $d$, so the fraction term in the square root is between $-f$ and $f$. Because $d > f$, we know that $(d- \frac{fx}{d})$ is positive. So the choice of square root is justified.
Similarly, the second term becomes \begin{align} &\sqrt{(x+f)^2 + y^2} \\ &= \sqrt{(x+f)^2 + (d^2 - f^2)\left(1 - \left( \frac{x^2 }{d^2}\right) \right)} & \text{, after substituting in (*)} \\ &= \sqrt{x^2 + 2fx + f^2 + (d^2 - f^2) - \left( (d^2 - f^2)\frac{x^2 }{d^2}\right) } \\ &= \sqrt{x^2 + 2fx + d^2 - d^2 \frac{x^2 }{d^2} + f^2\frac{x^2 }{d^2} } \\ &= \sqrt{x^2 + 2fx + d^2 - x^2 + \frac{(fx)^2 }{d^2} } \\ &= \sqrt{ 2fx + d^2 + \left(\frac{fx}{d}\right)^2} \\ &= \sqrt{d^2 + 2d\frac{fx}{d} + \left(\frac{fx}{d}\right)^2} \\ &= \sqrt{(d + \frac{fx}{d})^2} \\ &= d + \frac{fx}{d} \end{align} Summing these two, we get $2d$, as required.
So our G-ellipse is in fact exactly the same as our axis-aligned C-ellipse. And the minor radius of our C-ellipse (the $y$-value when $x = 0$) is $\sqrt{d^2 - f^2}$, as expected.