Editing Hyperbola (section)

===Hyperbola with equation {{math|1=''y'' = ''A''/''x''}}===
[[File:Hyperbel-gs-hl.svg|thumb|Rotating the coordinate system in order to describe a rectangular hyperbola as graph of a function]]
[[File:Hyperbeln-gs-3.svg|thumb|Three rectangular hyperbolas <math>y = A / x</math> with the coordinate axes as asymptotes<br/>
red: ''A'' = 1; magenta: ''A'' = 4; blue: ''A'' = 9]]
If the ''xy''-coordinate system is [[rotation matrix|rotated]] about the origin by the angle <math>+45^\circ</math> and new coordinates <math>\xi,\eta</math> are assigned, then <math>x = \tfrac{\xi+\eta}{\sqrt{2}},\; y = \tfrac{-\xi+\eta}{\sqrt{2}} </math>.<br/>
The rectangular hyperbola <math>\tfrac{x^2-y^2}{a^2} = 1</math> (whose [[semi-major and semi-minor axes|semi-axes]] are equal) has the new equation <math>\tfrac{2\xi\eta}{a^2} = 1</math>.
Solving for <math>\eta</math> yields <math>\eta = \tfrac{a^2/2}{\xi} \ . </math>

Thus, in an ''xy''-coordinate system the graph of a function <math>f: x \mapsto \tfrac{A}{x},\; A>0\; , </math> with equation
<math display="block">y = \frac{A}{x}\;, A>0\; ,</math> is a ''rectangular hyperbola'' entirely in the first and third [[quadrant (plane geometry)|quadrants]] with
*the coordinate axes as ''asymptotes'',
*the line <math>y = x</math> as ''major axis'' ,
*the ''center'' <math>(0,0)</math> and the ''semi-axis'' <math> a = b = \sqrt{2A} \; ,</math>
*the ''vertices'' <math>\left(\sqrt{A},\sqrt{A}\right), \left(-\sqrt{A},-\sqrt{A}\right) \; ,</math>
*the ''semi-latus rectum'' and ''radius of curvature '' at the vertices <math> p=a=\sqrt{2A} \; ,</math>
*the ''linear eccentricity'' <math>c=2\sqrt{A}</math> and the eccentricity <math>e=\sqrt{2} \; ,</math>
*the ''tangent'' <math>y=-\tfrac{A}{x_0^2}x+2\tfrac{A}{x_0}</math> at point <math>(x_0,A/x_0)\; .</math>

A rotation of the original hyperbola by <math>-45^\circ</math> results in a rectangular hyperbola entirely in the second and fourth quadrants, with the same asymptotes, center, semi-latus rectum, radius of curvature at the vertices, linear eccentricity, and eccentricity as for the case of <math>+45^\circ</math> rotation, with equation
<math display="block">y = -\frac{A}{x} \; , ~~ A>0\; ,</math>
*the ''semi-axes'' <math> a = b = \sqrt{2A} \; ,</math>
*the line <math> y = -x</math> as major axis,
*the ''vertices'' <math>\left(-\sqrt{A},\sqrt{A}\right), \left(\sqrt{A},-\sqrt{A}\right) \; .</math>

Shifting the hyperbola with equation <math>y=\frac{A}{x}, \ A\ne 0\ ,</math> so that the new center is {{nowrap|<math>(c_0,d_0)</math>,}} yields the new equation
<math display="block">y=\frac{A}{x-c_0}+d_0\; ,</math>
and the new asymptotes are <math>x=c_0 </math> and <math>y=d_0</math>. The shape parameters <math>a,b,p,c,e</math> remain unchanged.