Prove that the directrix-focus and focus-focus definitions are equivalent (NOTE: This is my attempt at answering this question and  this question, but I rewrote it in order to make it easier for me to solve. Also, I've made a YouTube video explaining this whole proof. Note that @Blue has a proof for this question with Dandelin spheres.)
We have two definitions of an ellipse and a hyperbola.


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*An ellipse is the locus of points that has a constant ratio of distance between a focus (point) and a directrix (line), where that constant ratio is between 0 and 1.



Here, the eccentricity is $\frac C A$, which, by this definition, must be a constant less than 1 for every point on the ellipse.


*

*A hyperbola is the locus of points that has a constant ratio of distance between a focus (point) and a directrix (line), where that constant ratio is greater than 1.



Here, the eccentricity is $\frac{C_1}{A_1}=\frac{C_2}{A_2}$, which, by this definition, must be a constant greater than 1 for each point on the hyperbola.


*An ellipse is the locus of points which has a constant sum of distances between two foci.



Here, the eccentricity is $\frac{F}{D_1+D_2}$. According to this definition, $D_1+D_2$ is a constant for every point on the ellipse, so we’ll just call it $D$. Obviously, $F$ is constant since it is just the distance between the foci, which doesn’t change.


*The hyperbola is the locus of points which has a constant difference of distances between two foci.



Here, the eccentricity is $\frac{F}{D_2-D_1}$. According to this definition, $D_2-D_1$ is a constant for every point on the hyperbola, so again, we’ll just call it $D$. Obviously, $F$ is constant since it is just the distance between the foci, which doesn’t change.
My question is:
$$\text{How do I prove these definitions are equivalent?}$$
 A: Proof of Focus Focus: 

Sphere $G_1$ is tangent to the plane the ellipse is contained on at point $F_1$. It is also tangent to the cone at circle $k_1$ 
Sphere $G_2$ is tangent to the same plane at point $F_2$. It is also tangent to the cone at $k_2$.
Connect the apex of the cone, $S$, to any point on the ellipse, $P$, with a line. Mark the intersection points of this line with circles $k_1$ and $k_2$ $P_1$ and $P_2$ respectively.
Since segments $PP_1$ and $PF_1$ are both tangent to sphere $G_1$, they are the same length. 
Similarly $PP_2$ and $PF_2$ are both the same length.
So $PF_1+PF_2=PP_1+PP_2=P_1P_2$. Since $P_1P_2$ remains constant as $P$ moves around the ellipse, $PF_1+PF_2$ is also constant.
A: If it can help here is a pure geometric demonstration (rather complex but it uses plane geometry and not the Dandelin spheres) that I devised some years ago for the ellipse (reference: https://www.physicsforums.com/threads/ellipse-geometric-equivalence-of-two-definitions.247417/).
I'm not much satisfied with it (especially in the second part that is just sketched) and I'm positive it could be significantly improved.

First part
Given the locus of points for which is constant the sum of the distances from two fixed points $F_1$ and $F_2$ then for any point $P$ belonging to the locus it is $PF_1=e\cdot PH_1$, where $PH_1$ is the distance form an appropriate fixed line (directrix) perpendicular to the line joining $F_1$ and $F_2$.
Demonstration
Let it be $F_1F_2=2c; \; \; PF_1+PF_2=2a; \; \; c/a=e; \; \; PF_1=f_1; \; \; PF_2=f_2; \; \; PH_1=d_1; \; \; PH_2=d_2$;
First, consider the triangle $F_1F_2P$, and construct the circumscribed circle around it.
The perpendicular bisector of the segment $F_1F_2$ meets the circle in $V$ (and the quadrilateral $VF_1F_2P$ is a cyclic quadrilateral).
Let’s call $N$ the intersection of this line with the line joining $F_1$ and $F_2$, and we’ll call $NF_1=m_1$; $NF_2=m_2$;
Since $F_1V=F_2V$, it follows, for the chord properties, that the segment $PV$ bisects the angle $F_1PF_2$, so that $F_1PN=NPF_2=F_1F_2V=\alpha$
Applying the angle bisector theorem to the triangle $F_1PF_2$ we have $$f_1:m_1=f_2:m_2$$ It's also $$(f_1+f_2):f_1=(m_1+m_2):m_1 \; \; \rightarrow \; \; 2a:f_1=2c:m_1 \; \; \rightarrow \; \; m_1/f_1=c/a$$ and $$(f_1+f_2):f_2=(m_1+m_2):m_2 \; \; \rightarrow \; \; 2a:f_2=2c:m_2 \; \; \rightarrow \; \; m_2/f_2=c/a$$
The triangles $H_1PF_1$ and $F_1PN$ are similar (they have the same angle $\alpha$ and two alternate interior angles). So, also the triangles $H_2PF_2$ and $F_2PN$ are similar.
Thence it is $$m_1:f_1=f_1:d_1=c/a$$ and $$m_2:f_2=f_2:d_2=c/a$$ whence
$$f_1:d_1=f_2:d_2=c/a$$ that proves the statement.
Second part
Given the locus of points for which the ratio between the distances from a fixed point $F_1$ (focus) and a fixed line (directrix) has a constant value that is larger than $0$ and less than $1$, then 
1) There exists a second focus $F_2$ and a second directrix for which the same relation applies 
2) For any point $P$ belonging to the locus the sum of the distances of $P$ to the foci $F_1$ and $F_2$ has a constant value. 
Demonstration of 1)
Basically, starting from the focus $F_1$, the directrix, a point $P$ belonging to the locus, and calling $H_1$ the perpendicular projection of $P$ on the directrix it’s possible to build the triangle $PF_1H_1$, and the similar triangle $PF_1N$. The lines $PN$ and $H_1F_1$ meets at a point $V$. The second focus $F_2$ is such that the line $PV$ bisects the angle $F_2PF_1$.
Having built the same figure as in the demonstration of the first part, it follows that the focus $F_2$ is the second focus and the line $VF_2$ intersects the line $H_1P$ in the point $H_2$ that defines the position of the second directrix. 
Demonstration of 2)
Since $$f_1:d_1=f_2:d_2=e$$ it follows that $$f_1=e\cdot d_1$$ and $$f_2=e\cdot d_2$$Summing the terms it is $$f_1+f_2=e \cdot (d_1+d_2)=e \cdot H_1H_2$$ so that the sum of the distances of $P$ to the foci $F_1$ and $F_2$ has a constant value.
A: This is an old question, but I found a very short proof for an ellipse (shown in Heinemann Modular Mathematics P5) which I haven't seen elsewhere on the internet. It requires both foci and directrices.
For a general ellipse (I don't have enough reputation to display it.)
The equations of the directrices upon which F and E lie are $x=-\frac ae$ and $x=\frac ae$ respectively.
From the focus-directrix property:
$$ PS = ePE, \; PT = ePF $$
Summing them gives:
$$ PS + PT = e(PE+PF) $$
But,
$$ PE+PF = EF = \frac{2a}{e} $$
So:
$$ PS + PT = e\left(\frac{2a}{e}\right) $$
Which results in the focal distances property:
$$ PS + PT = 2a $$
