The source of the difficulty is that ellipses are constructed in a particular way in TikZ. They are paths that start from the x-axis and proceed counter-clockwise around their centre. The vast majority of the time, the exact parametrisation doesn't matter. You appear to have found the one situation where it does!
In the actual question, you only want to be able to mirror the ellipse, and so draw it starting from the negative x-axis (the title of the question suggests a more flexible approach). That's actually not too hard since we can exploit the symmetry of the ellipse. The key is to provide it with a negative x-radius, since then it will start from the negative x-axis (and proceed clockwise, but we could correct for that by negating the y-radius as well). To do this, we interrupt the call from the node shape to the drawing command and flip the sign of the x-radius. The simplest way to do this is to redefine the \pgfpathellipse
macro to do the negation and then call the original macro. The following code does this.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,shapes,decorations.markings}
\makeatletter
\let\origpgfpathellipse=\pgfpathellipse
\def\revpgfpathellipse#1#2#3{%
#2%
\pgf@xa=-\pgf@x
\origpgfpathellipse{#1}{\pgfqpoint{\pgf@xa}{0pt}}{#3}}
\makeatother
\tikzset{
reversed ellipse/.style={
ellipse,
reverse the ellipse%
},
reverse the ellipse/.code={
\let\pgfpathellipse=\revpgfpathellipse
}
}
\begin{document}
\begin{tikzpicture}
\node[ellipse,
draw,
postaction={
decorate,
decoration={
markings,
mark=at position 1 with {
\arrow[line width=5pt,blue]{>}
}
}
}
] at (0,0) {hello world};
\node[reversed ellipse,
draw,
postaction={
decorate,
decoration={
markings,
mark=at position 1 with {
\arrow[line width=5pt,blue]{>}
}
}
}
] at (0,-2) {hello world};
\end{tikzpicture}
\end{document}
Here's the result:

(the arrow got clipped, but you can see where it lies)
I think that the best way to establish node-like anchors is to use a node. I'm guessing that the annoyance of using an actual node to draw the rectangle is that specifying a node rectangle by its coordinates is a little more complicated than just saying (0,0) rectangle (3,2)
. So here's some code that puts an invisible (rectangular - but that's only because the default is a rectangle) node around the current path. If the path is more complicated then the node is guaranteed to contain the rectangular bounding box.
Here's the code:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{fit}
\makeatletter
\tikzset{
fitting node/.style={
inner sep=0pt,
fill=none,
draw=none,
reset transform,
fit={(\pgf@pathminx,\pgf@pathminy) (\pgf@pathmaxx,\pgf@pathmaxy)}
},
reset transform/.code={\pgftransformreset}
}
\makeatother
\begin{document}
\begin{tikzpicture}[line width=.5cm]
\draw[thick,green] (-2,-2) -- (4,4);
\draw (0,0) rectangle (3,2) node[fitting node] (rect) {};
\draw[->,red] (5,5) -- (rect.north east);
\draw[->,red] (0,5) -- (rect.north);
\end{tikzpicture}
\end{document}
The green line is to show that the bounding box used is that of the path and not the current picture. The thick lines are to show that the node anchors are on the proper border of the path, not the "theoretical" path[1].
Here's the result:

[1]: With your method of specifying anchors via coordinates, the anchors would be on the "theoretical" path, namely the north west anchor would be at (3,2)
not (3 + half line width, 2 + half line width)
. If you prefer this, it's easy to modify this method to do that.
Edit Now copes with scale=2
as Altermundus asked about. With more complicated transformations then it gets increasingly difficult to keep track since nodes work differently, and it is working on the actual bounding box rather than the path itself. So in those cases, caveat texer.
Best Answer
You can specify the
minimum height
andminimum width
for a node. In conjunction withdraw
, you'll end up with a rectangle with the specified dimensions: