This question is a follow-up question to half of a hyperbola.
When constructing a hyperbola in TikZ, how can I specify the eccentricity to be 1.44022?
2 Answers
Two quantities out of a, b, and e (the eccentricity) determine the hyperbola. You can, for example, define e(>1) and a(>0), from which you can then derive b. See below.
EDIT: see this link, which explains my parameterisation.
\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=1]
\pgfmathsetmacro{\e}{1.44022} % eccentricity
\pgfmathsetmacro{\a}{1}
\pgfmathsetmacro{\b}{(\a*sqrt((\e)^2-1)}
\draw plot[domain=-2:2] ({\a*cosh(\x)},{\b*sinh(\x)});
\draw plot[domain=-2:2] ({-\a*cosh(\x)},{\b*sinh(\x)});
\end{tikzpicture}
\end{document}
-
I would recommend using
pgfplotsfor these types of problems- you don't have to change your code very much to do it :)– cmhughesApr 16, 2013 at 3:10 -
-
1@dustin trigonometric functions are used for circles and ellipses:
(x/a)^2+(y/b)^2=1because of the Pythagorean identitysin^2(\theta)+cos^2(\theta)=1. Hyperbolic functions are used for Hyperbolae because they satisfy the identitycosh^2(x)-sinh^2(x)=1– cmhughesApr 16, 2013 at 3:29 -
@Jubobs
postaction = decoratedoesn't work on this type of tikz picture? I just tried decorating the hyperbola with arrows at 20 and 60 percent but nothing happened.– dustinApr 16, 2013 at 4:12 -
@dustin Your original post does not mention
position=decorate. You should either edit this question to explain yourposition=decorateproblem or post another question dedicated to it.– jub0bsApr 16, 2013 at 8:46
With PSTricks.
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot,pst-math}
\usepackage[nomessages]{fp}
\FPset\E{1.440}% 3 digits should be enough
\FPset\A{1}
\FPeval\B{round(A*root(2,E^2-1):3)}
\def\X(#1){\A*COSH(#1)}
\def\Y(#1){\B*SINH(#1)}
\psset{algebraic}
\begin{document}
\begin{pspicture}(-6,-4)(6,4)
\psaxes{->}(0,0)(-6,-4)(5.5,3.5)[$x$,0][$y$,90]
\psset{linecolor=blue}
\psparametricplot{-2}{2}{\X(t)|\Y(t)}
\psparametricplot{-2}{2}{-\X(t)|\Y(t)}
\end{pspicture}
\end{document}
answered Apr 16, 2013 at 9:15
$x^2/a^2 - y^2/b^2 = 1$, then the eccentricity is$\sqrt{a^2 + b^2}/a$. Choose$a$and$b$appropriately.