I want to draw hyperbola with pgfplots:
(x+1)^2/4-(y-2)^2/9=1
- 2Have you tried anything? This site usually works best when you present your best attempt. It's also polite to give us a minimal working example that we can start from.Teepeemm– Teepeemm2016-08-13 02:29:10 +00:00Commented Aug 13, 2016 at 2:29
- @Teepeemm I'm starting to use TikZ , so I do not show any initial work.casio– casio2016-08-13 02:31:45 +00:00Commented Aug 13, 2016 at 2:31
- The easiest solution is to use (y-2)^2/9=(x+1)^2/4-1 and take the square root (plus or minus). Note that x<-3 or x>1, so you will need four \addplots total.John Kormylo– John Kormylo2016-08-13 03:35:59 +00:00Commented Aug 13, 2016 at 3:35
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3 Answers
As stated in comments, you can reformulate the equation as "y = f(x)" with the problem of roots.
An alternative is to interprete it as "f(x,y)= (x+1)^2/4-(y-2)^2/9" and draw the contour "f(x,y) = 1" :
\documentclass{standalone} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \begin{document} \begin{tikzpicture} \begin{axis}[view={0}{90}] \addplot3[domain=-10:10, contour gnuplot={labels=false,levels={1}} ] {(x+1)^2/4-(y-2)^2/9}; \end{axis} \end{tikzpicture} \end{document} 
The value domain=-10:10 determines the computed range. Since domain y is missing, pgfplots assumes that it should also use -10:10.
Note that you need to compile this by means of pdflatex -shell-escape file.tex and you need gnuplot installed.
Note that the approach as such works for any kind of plot program, not just pgfplots.
answered Aug 13, 2016 at 12:43
Christian Feuersänger
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I would strongly recommend googling around for a tikz/pgf tutorial to learn at least the basics on your own. You'll learn much more that way.
The most straightforward approach here is to solve for y=2±√(9*(x+1)^2/4-9) with x < -3 or 1 < x. Plotting this requires four commands, and results in
\begin{tikzpicture} \begin{axis}[xmin=-10,xmax=10,ymin=-10,ymax=10] \addplot[domain=-10:-3] {2+sqrt(9*(x+1)^2/4-9)}; \addplot[domain=-10:-3] {2-sqrt(9*(x+1)^2/4-9)}; \addplot[domain=1:10] {2+sqrt(9*(x+1)^2/4-9)}; \addplot[domain=1:10] {2-sqrt(9*(x+1)^2/4-9)}; \end{axis} \end{tikzpicture} 
This is not a great plot (there's the obvious gap on the left, and if you look closer, you'll see that the right has a few straight segments and angles near the vertex). To fix this, we can change to a parametric plot: x=2*sec(t)-1, y=3*tan(t)+2. Plotting this still requires two commands (to avoid the vertical asymptotes of sec and tan) and results in:
\documentclass{standalone} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \begin{document} \begin{tikzpicture} \begin{axis}[xmin=-10,xmax=10,ymin=-10,ymax=10] \addplot[domain=-89:89] ({2*sec(x)-1},{3*tan(x)+2}); \addplot[domain=91:269] ({2*sec(x)-1},{3*tan(x)+2}); \end{axis} \end{tikzpicture} \end{document} 
\documentclass[border=2pt]{standalone} \usepackage{pgfplots} \begin{document} \begin{tikzpicture} \begin{axis}[xmin=-10,xmax=10, ymin=-15, ymax=15, restrict x to domain=-20:20]% remove crossing lines at t=90 and t=270 \addplot[variable=t,domain=0:360,samples=200] ({2*sec(t)-1}, {3*tan(t)+2}); \end{axis} \end{tikzpicture} \end{document} 
