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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)?

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  • What have you tried? Have you searched wikipedia? – David G. Stork Apr 11 '20 at 21:56
  • I searched probably over a hundred papers, links, videos, posts. Nothing helps. Wikipedia was one of my first attempts. – 1b3b Apr 11 '20 at 22:02
  • https://en.wikipedia.org/wiki/Dandelin_spheres – brainjam Apr 11 '20 at 22:16
  • I tried :) The proof which use Dandelin spheres, however beautiful it is, unfortunately "just" shows the properties of the cross section of a cone with a plane, not their equations. – 1b3b Apr 11 '20 at 22:24
  • What sort of geometric proof of an equation would satisfy you? You should be able to do this for yourself: write out your definition of a hyperbola in terms of the coordinates of the foci and of an arbitrary point and then simply the resulting equation. – amd Apr 11 '20 at 22:35
  • I am curious that there is geometric proof using the basic results: the similarity of triangles, Pythagorean theorem, etc. (so without the distance formula and analytical methods) showing that $x^2b^2 - y^2a^2 = a^2b^2$ is valid for every point of hyperbola. – 1b3b Apr 11 '20 at 22:56
  • How can you geometrcialy proof something what is analiticaly?!?! – nonuser Apr 16 '20 at 15:34
  • @Aqua, because there are obvious connections and equivalences that bind that two geometries together. It's hard to believe that you don't understand the given proof. – 1b3b Apr 16 '20 at 21:35

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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 $y^2=PH^2$.

enter image description here

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.

enter image description here

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