I came across the fact that the following function defines a family of ellipses with focal distance $f$, parameterized by the value of the function:
$$\frac{x-f+\sqrt{(x-f)^2+y^2}}{x+f+\sqrt{(x+f)^2+y^2}}=\text{const.}$$
I was able to convert this expression algebraically into the canonical form of the ellipse (very lengthy calculation), and I found a short neat proof that uses the geometric definition of the ellipse, as such:
Define the projected distances of the point to the foci: $$x-f=z_1,\quad x+f=z_2$$ And real distances of the point to the foci: $$\sqrt{(x-f)^2+y^2}=r_1,\quad \sqrt{(x+f)^2+y^2}=r_2$$ By properties of ellipse, we have $$r_2+r_1=2a,\quad z_2-z_1=2f$$
Add and subtract these two equations: $$(r_2+z_2)+(r_1-z_1)=2(a+f)$$ $$(r_2-z_2)+(r_1+z_1)=2(a-f)$$ Multiply top by $r_1+z_1$ and bottom by $r_2+z_2$: $$(r_2+z_2)(r_1+z_1)+\color{red}{(r_1-z_1)(r_1+z_1)}=2(a+f)(r_1+z_1)$$ $$\color{red}{(r_2-z_2)(r_2+z_2)}+(r_1+z_1)(r_2+z_2)=2(a-f)(r_2+z_2)$$ Left hand sides are identical because the red parts equal through Pythagoras' theorem to the same perpendicular distance of the point $(x,y)$ to the $x$ axis, so we retrieve $$\frac{r_1+z_1}{r_2+z_2}=\frac{a-f}{a+f}$$ QED.
Can you help me find a constructional geometric proof (no algebraic manipulations, such as difference of squares and addition/subtraction of equations)?
I'm having difficulty finding a construction that puts $r+z$ on the same line, and to prove a ratio of two such distances are constant (similar triangles?). I tried with circular directrix, but I can't relate $r_1+z_1$ to $r_2+z_2$.