Is there geometric proof for the equation of hyperbola using only constant distance from two foci definition? Is there geometric or visual (but rigorous) proof for the equation of hyperbola whose foci are on the $x$-axis using only the traditional definition of hyperbola: a hyperbola is a set of points such that the difference of distances from each point to focus is constant $(|r_1 - r_2| = 2a)$?
Also, can you maybe geometrically explain asymptotes of hyperbola (why do they exist)?
 A: Let $S$, $S'$ be the foci of the hyperbola, $P$ a point on it and $H$ the projection of $P$ on line $SS'$ (see figure below). I'll suppose WLOG that $S$ is the focus nearest to $P$ and that $\angle PSS'>90°$; the reasoning must be slightly rephrased if $\angle PSS'\le90°$ but the same conclusions still hold.
By the Pythagorean theorem we have:
$$
PH^2=PS^2-HS^2=PS'^2-HS'^2,
$$
that is:
$$
(PS'-PS)(PS'+PS)=(HS'-HS)(HS'+HS).
$$
Observe that $HS'-HS=SS'=2c$ (here we define constant $c$ as usual) and $HS'+HS=2OH$, while by hypothesis: $PS'-PS=AB=2a$ (here we define constant $a$ as usual). Plugging these equalities into the above formula we get:
$$
\tag{1}
PS'+PS={2c\over a}OH.
$$
We can then compute the area of triangle $PSS'$ in two ways: either considering $SS'$ as base and $PH$ as the related altitude, or by Heron's formula. Equating the resulting expressions and squaring, we then obtain:
$$
(c\cdot PH)^2={1\over16}(PS+PS'+2c)(PS+PS'-2c)(-2a+2c)(2a+2c),
$$
that is:
$$
c^2PH^2={1\over4}\big((PS+PS')^2-4c^2\big)(c^2-a^2).
$$
If we now introduce, as usual, the constant $b^2=c^2-a^2$, this gives:
$$
\tag{2}
(PS+PS')^2=4c^2\left({PH^2\over b^2}+1\right)
$$
We can then combine equations $(1)$ and $(2)$ to eliminate $PS+PS'$ and finally get:
$$
{OH^2\over a^2}={PH^2\over b^2}+1,
$$
which is the cartesian equation of the hyperbola, once you set $x^2=OH^2$ and $y^2=PH^2$.

To understand geometrically the origin of asymptotes, consider the case when point $P$ is very far from the center: $OP\gg c$. Choose point $C$ on $PS'$ such that $PC=PS$ and consequently $CS'=2a$ (see figure below). Angles at $S$ and $C$ in isosceles triangle $PSC$ get nearer and nearer to $90°$ the farther $P$ is, while line $OP$ intersects the hyperbola at $P$ and $P'$, the symmetric of $P$ with respect to $O$.
In the limit $OP\to\infty$ lines $SP$, $OP$ and $S'P$ become parallel and triangle $SCS'$ becomes right-angled. The angle $\alpha$ line $OP$ forms with $SS'$ tends to a constant value such that $\cos\alpha=a/c$, and line $OP$ is tangent to the hyperbola "at infinity", i.e. it is an asymptote.

