1 files changed, 460 insertions, 13 deletions
diff --git a/graphics/pstricks/contrib/pst-eucl/doc/pst-eucl-doc.tex b/graphics/pstricks/contrib/pst-eucl/doc/pst-eucl-doc.tex
index dc97768b52..b9000b9750 100644
--- a/graphics/pstricks/contrib/pst-eucl/doc/pst-eucl-doc.tex
+++ b/graphics/pstricks/contrib/pst-eucl/doc/pst-eucl-doc.tex
@@ -6,7 +6,7 @@
 \usepackage{multicol}
 \usepackage{ntheorem}
 \newtheorem{theorem}{Theorem}
-\usepackage{pst-plot,paralist}
+\usepackage{pst-func,pst-plot,paralist}
 \usepackage[mathscr]{eucal}
 \lstset{pos=l,wide=false,language=PSTricks,
     morekeywords={multidipole,parallel},basicstyle=\footnotesize\ttfamily}
@@ -403,6 +403,53 @@ for each points: \Lkeyword{PointSymbolA}, \Lkeyword{PointSymbolB} and
 \Lkeyword{PointSymbol}. The management of the default value followed the
 same rule.
 
+The macros \Lcs{pstTriangleSSS}, \Lcs{pstTriangleSAS}, \Lcs{pstTriangleAAS} and
+\Lcs{pstTriangleASA} are used to draw the triangle according the specified sides
+or angles.
+
+\begin{BDef}
+\Lcs{pstTriangleSSS}\OptArgs\Largr{pos}\Largb{A}\Largr{a,b,c}\Largb{B}\Largb{C}\\
+\Lcs{pstTriangleSAS}\OptArgs\Largr{pos}\Largb{A}\Largr{b,$\angle{A}$,c}\Largb{B}\Largb{C}\\
+\Lcs{pstTriangleAAS}\OptArgs\Largr{pos}\Largb{A}\Largr{$\angle{C}$,$\angle{A}$,c}\Largb{B}\Largb{C}\\
+\Lcs{pstTriangleASA}\OptArgs\Largr{pos}\Largb{A}\Largr{$\angle{A}$,c,$\angle{B}$}\Largb{B}\Largb{C}
+\end{BDef}
+
+- Macro \Lcs{pstTriangleSSS} create a triangle $ABC$ with given $A(x_1,y_1)$, and the three sides $a,b,c$,
+it output $B(x_2,y_2)$ and $C(x_3,y_3)$.\\
+- Macro \Lcs{pstTriangleSAS} create a triangle $ABC$ with given $A(x_1,y_1)$, the angle of $\angle{A}$, and the other two sides $b,c$,
+it output $B(x_2,y_2)$ and $C(x_3,y_3)$.\\
+- Macro \Lcs{pstTriangleAAS} create a triangle $ABC$ with given $A(x_1,y_1)$, the angle of $\angle{C}$, the angle of $\angle{A}$, and the side of $AB=c$,
+it output $B(x_2,y_2)$ and $C(x_3,y_3)$.\\
+- Macro \Lcs{pstTriangleASA} create a triangle $ABC$ with given $A(x_1,y_1)$, the angle of $\angle{A}$, the angle of $\angle{B}$, and the side of $AB=c$,
+it output $B(x_2,y_2)$ and $C(x_3,y_3)$.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,0)(4,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=-90](0,1){A}
+\pstTriangleSSS[linecolor=red!60,PosAngle={-90,90}]{A}(3,4,5){B}{C}
+\pstTriangleSSS[linecolor=blue!60,PosAngle={-90,90}]{A}(2,4.5,3.8){D}{E}
+\pstTriangleSAS[linecolor=green!60,PosAngle={-90,90}]{A}(3,40,2.8){F}{G}
+\pstTriangleAAS[linecolor=black!60,PosAngle={-90,90}]{A}(40,50,1.8){H}{I}
+\pstTriangleASA[linecolor=purple!60,PosAngle={-90,90}]{A}(70,1.0,60){J}{K}
+\end{pspicture}
+\end{LTXexample}
+
+The optional parameter \texttt{pos} setup the position of the first node $A$, it should be 'L' for left, 'R' for right, 'U' for up and 'D' for down.
+If you don't input this parameter, the default value is 'L'. The following example explains how to draw an isoceles triangle with the given isoceles sides
+and the vertex angle.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,0)(4,4)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=205](2,2){A}
+\pstTriangleSAS[linecolor=red!60,PosAngle={-90,90}]{A}(2,40,2){B}{C}
+\pstTriangleSAS[linecolor=blue!60,PosAngle={-90,-90}](U){A}(2,40,2){B}{C}
+\pstTriangleSAS[linecolor=purple!60,PosAngle={90,90}](D){A}(2,40,2){B}{C}
+\pstTriangleSAS[linecolor=green!60,PosAngle={90,-90}](R){A}(2,40,2){B}{C}
+\end{pspicture}
+\end{LTXexample}
+
 The macros \Lcs{pstTriangleIC} and \Lcs{pstTriangleOC} are used to draw the inner circle
 and outer circle of triangle $ABC$.
 
@@ -612,6 +659,26 @@ Here are some examples:
 
 \vspace{10pt}
 
+The macros \Lcs{pstLineCoef} is used to draw a line $ax+by+c=0$ with the given coefficents $a,b,c$,
+and create two new node $A,B$ on the line.
+
+\begin{BDef}
+\Lcs{pstLineCoef}\OptArgs\Largb{a,b,c}\Largb{A}\Largb{B}
+\end{BDef}
+
+Here are some examples:
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-2,-2)(2,2)
+\pstLineCoef[linecolor=red!60, PosAngle={210,0}]{3,-2,1}{A}{B}
+\pstLineCoef[linecolor=blue!60, PosAngle={180,0}]{4,3,2}{C}{D}
+\pstLineCoef[linecolor=green!60, PosAngle={90,90}]{0,3,-3}{E}{F}
+\pstLineCoef[linecolor=purple!60, PosAngle={180,180}]{4,0,4}{G}{H}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
 The macro \Lcs{pstLineAbsNode} creates a new node $C$ whose abscissa
 is the given value $x_1$ on the line $AB$. The macro \Lcs{pstLineOrdNode} creates a new node $C$ whose ordinate is the given value $y_1$ on the line $AB$.
 You can input $x_1$ or $y_1$ as any number(e.g, 2.0),
@@ -678,6 +745,30 @@ we will introduce \Lcs{pstDistDiv} later.
 \end{pspicture}
 \end{LTXexample}
 
+One application of \Lcs{pstProportionNode} is used to find the bisector and out bisector of a given angle.
+So we define the macro \Lcs{pstBisectorAOB} to do this work, it is more friendly than the macros
+\Lcs{pstBissectBAC} and \Lcs{pstOutBissectBAC}, as it put the new node $T_1$ and $T_2$ on line $AB$, not arc $AB$.
+
+\begin{BDef}
+\Lcs{pstBisectorAOB}\OptArgs\Largb{A}\Largb{O}\Largb{B}\Largb{$T_1$}\Largb{$T_2$}
+\end{BDef}
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,-1)(5,3)
+\psset{unit=0.6cm}\footnotesize\psset{PointSymbol=*}
+\pstGeonode[PosAngle=-90](0,0){A}
+\pstGeonode[PosAngle=90](3,3){C}
+\pstGeonode[PosAngle=-90](2,0){B}
+\pstBisectorAOB[PosAngle={-90,-90}]{A}{C}{B}{T_1}{T_2}
+\pstLineAB[linecolor=black!50]{A}{C}
+\pstLineAB[linecolor=black!50]{B}{C}
+\pstLineAB[linecolor=black!50]{A}{B}
+\pstLineAB[linestyle=dashed,linecolor=gray]{C}{T_1}
+\pstLineAB[linestyle=dashed,linecolor=gray]{C}{T_2}
+\pstLineAB[linestyle=dashed,linecolor=gray]{B}{T_2}
+\end{pspicture}
+\end{LTXexample}
+
 \vspace{10pt}
 
 The four collinear points $A,B,C,D$ are called \texttt{Harmonic Conjugation Points} if their cross ratio is $-1$,
@@ -838,11 +929,34 @@ of a point on the inversion circle is itself.
 
 \vspace{10pt}
 
+If you want to find the node $C$ from $A$ to $B$, such that $C$ is the golden section of the
+given segments $AB$, that is,
+$$|AC|^2=|AB|\times|BC|\quad\text{or}\quad{}AC:AB=BC:AC\quad\text{or}\quad{}AC=\dfrac{\sqrt{5}-1}{2}AB$$
+you can use the macro \Lcs{pstGoldenMean} to do this work.
+
+\begin{BDef}
+\Lcs{pstGoldenMean}\OptArgs\Largb{A}\Largb{B}\Largb{C}
+\end{BDef}
+
+In fact, we use the macro \Lcs{pstLocateAB} to implement this macro
+by passing the value $\dfrac{\sqrt{5}-1}{2}|AB|$ to parameter length.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,0)(4,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=90](0,1){A}(4,2){B}
+\pstGoldenMean[PosAngle=90,PointSymbol=o]{A}{B}{C}
+\pstLineAB{A}{B}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
 If you want to find the node $C$ from $A$ to $B$, such that $AC$ is the geometric mean of two
 given segments $DE$ of $FG$, that is,
 $$|AC|^2=|DE|\times|FG|$$
 you can use the macro \Lcs{pstGeometricMean} to do this work.
-It also can be used to draw a circle when given two points on the circle, 
+It also can be used to draw a circle when given two points on the circle,
 and a line tangents to the circle.
 
 \begin{BDef}
@@ -1107,15 +1221,15 @@ and the parameter to fill the circle.
 \psset{unit=0.65cm}\psset{dotscale=0.5}\psset{PointSymbol=*}
 \pstGeonode[PosAngle=90,CurveType=polyline](0,0){A}(1,0){B}
 \pstGeonode[PosAngle=90,CurveType=polyline](0,1){A'}(2,1){B'}
-\pstCircleOA[linecolor=gray,Radius=\pstDistAdd{A}{B}{A'}{B'}]{A}{}
-\pstCircleOA[linecolor=red,Radius=\pstDistAddVal{A}{B}{1.0}{\pstDistConst{0.5}}]{A}{}
-\pstCircleOA[linecolor=blue,Radius=\pstDistAddCoef{A}{B}{0.5}{A'}{B'}{1.5}]{A}{}
-\pstCircleOA[linecolor=green,Radius=\pstDistSub{A}{B}{A'}{B'}]{B'}{}
-\pstCircleOA[linecolor=brown,Radius=\pstDistSubCoef{A}{B}{1.8}{A'}{B'}{0.5}]{A}{}
+\pstCircleOA[linecolor=gray,Radius=\pstDistAdd{A}{B}{A'}{B'}]{A}{} % R=|AB|+|A'B'|
+\pstCircleOA[linecolor=red,Radius=\pstDistAddVal{A}{B}{1.0}{\pstDistConst{0.5}}]{A}{} % R=|AB|+0.5
+\pstCircleOA[linecolor=blue,Radius=\pstDistAddCoef{A}{B}{0.5}{A'}{B'}{1.5}]{A}{} % R=0.5|AB|+1.5|A'B'|
+\pstCircleOA[linecolor=green,Radius=\pstDistSub{A}{B}{A'}{B'}]{B'}{} % R=|AB|-|A'B'|
+\pstCircleOA[linecolor=brown,Radius=\pstDistSubCoef{A}{B}{1.8}{A'}{B'}{0.5}]{A}{} R=1.8|AB|-0.5|A'B'|
 \pnode(-1.5,-2){D}
-\pstCircleOA[linecolor=pink,fillstyle=solid,fillcolor=pink!40,Radius=\pstDistMul{A}{B}{0.8}]{D}{}
+\pstCircleOA[linecolor=pink,fillstyle=solid,fillcolor=pink!40,Radius=\pstDistMul{A}{B}{0.8}]{D}{} % R=0.8|AB|
 \psdot(D)\uput{0.2}[-45](D){$D$}
-\pstCircleOA[linecolor=purple,Radius=\pstDistConst{\pstAbscissa{D}} abs]{D}{}
+\pstCircleOA[linecolor=purple,Radius=\pstDistConst{\pstAbscissa{D}} abs]{D}{} % R=|D.x|
 \end{pspicture}
 \end{LTXexample}
 
@@ -1200,8 +1314,8 @@ for example,
 \pstGeonode[PosAngle=60](1.5,1.5){O}
 \pstGeonode[PosAngle=-30](2.5,0){A}
 \pstCircleOA[linecolor=red]{O}{A}
-\pstCircleAbsNode[PosAngleA=-60,PosAngleB=60,PointSymbol=*]{O}{A}{1.0}{C}{D}
-\pstCircleOrdNode[PosAngleA=150,PosAngleB=30,PointSymbol=*]{O}{A}{1.0}{E}{F}
+\pstCircleAbsNode[PosAngle={-60,60},PointSymbol=*]{O}{A}{1.0}{C}{D}
+\pstCircleOrdNode[PosAngle={150,30},PointSymbol=*]{O}{A}{1.0}{E}{F}
 \pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{C}{D}
 \pstLineAB[linestyle=dashed,linecolor=gray!40,nodesep=-0.5]{E}{F}
 \end{pspicture}
@@ -1468,6 +1582,25 @@ the macro will draw the whole ellipse.
 \end{pspicture}
 \end{LTXexample}
 
+Like as the coordinates, the parameters $a,b$ can be got by the raw PostScript commands too,
+where you can use the macros \Lcs{pstDist}*, for example,
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-3,-3)(3,3)
+\psset{dotscale=0.5}\psset{PointSymbol=*}\footnotesize
+\pstGeonode[PosAngle=-90,PointSymbol=*](0,0){O}(0,-2.5){R}
+\pstCircleOA[linecolor=blue!60]{O}{R}
+\pstGeonode[PosAngle=0,PointName=O_1,PointSymbol=*](0,0.8){O1}
+\pstGeonode[PosAngle=0,PointName=O_2,PointSymbol=*](0,1.6){O2}
+\pstCircleOrdNode[PointName={P,Q},PosAngle={180,0}]{O}{R}{\pstOrdinate{O1}}{P}{Q}
+\pstCircleOrdNode[PointName={P',Q'},PosAngle={180,0}]{O}{R}{\pstOrdinate{O2}}{P'}{Q'}
+\pstEllipse[linecolor=green!60,linestyle=dashed](O1)(! \pstUserDist{\pstDist{O1}{P}} 0.7)[0][180]
+\pstEllipse[linecolor=green!60](O1)(! \pstUserDist{\pstDist{O1}{P}} 0.7)[180][360]
+\pstEllipse[linecolor=red!60,linestyle=dashed](O2)(! \pstUserDist{\pstDist{O2}{P'}} 0.5)[0][180]
+\pstEllipse[linecolor=red!60](O2)(! \pstUserDist{\pstDist{O2}{P'}} 0.5)[180][360]
+\end{pspicture}
+\end{LTXexample}
+
 Now you can draw some points on this Ellipse using macro \Lcs{pstEllipseNode} or \Lcs{pstEllipseRotNode}.
 The macro \Lcs{pstEllipseNode} requires an explicit parameter $t$ as $\alpha$ in equation (\ref{ParametricFunctionOfEllipse})
 to calculate the point; but the macro \Lcs{pstEllipseRotNode} requires an implicit parameter \Lkeyword{RotAngle}
@@ -1746,7 +1879,120 @@ That is, \Lcs{pstGeneralEllipse} is more complex than \Lcs{pstEllipse}!
 \end{pspicture}
 \end{LTXexample}
 
-Similarly, we can location the points on the General Ellipse using the macros
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralEllipseFle} is used to define a General Ellipse with Focus $F$, directrix line $l$,
+and the eccentricity $e$, where $0\le{}e<1$. It just calculate the center $O$, major radius $a$, minor radius $b$ and the rotation
+angle $\theta$ of the major axis, then you can pass them into macro \Lcs{pstGeneralEllipse} to draw this ellipse.
+
+\begin{BDef}
+\Lcs{pstGeneralEllipseFle}\OptArgs\Largb{F}\Largb{A}\Largb{B}\Largb{$e$}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O} is a node name to store the center point, its label and symbol can
+be controlled by the options for \PST\ node, such as \Lkeyword{PosAngle}.
+The output parameter \texttt{Rab} is a PostScript key to store the pair of major radius and minor radius,
+it just use \PST\ node coordinate to store a pair of value, but not a geometrical point.
+The output parameter \texttt{$\theta$} is also a PostScript key to store the rotation angle of major axis,
+when you pass it to \Lcs{pstGeneralEllipse}, PostScript will lookup the value of this key in current dictionary.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-2,-2)(2,2)
+\psset{unit=1.0cm}\psset{dotscale=0.5}\footnotesize
+\psset{CodeFig=true,CodeFigColor=gray!50}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=30](1,-1){F_1}
+\pstGeonode[PosAngle=30](-1,1){F_2}
+\pstGeonode[PosAngle=-60](-2,-1){A}
+\pstGeonode[PosAngle=-60](2,1){B}
+\pstGeneralEllipseFle[PosAngle=30]{F_1}{A}{B}{0.5}{O_1}{R_1}{MajorRotAngle1}
+\pstGeneralEllipse[linecolor=red!60](O_1)(R_1)[MajorRotAngle1]
+\pstGeneralEllipseFle[PosAngle=30]{F_2}{A}{B}{0.5}{O_2}{R_2}{MajorRotAngle2}
+\pstGeneralEllipse[linecolor=blue!60](O_2)(R_2)[MajorRotAngle2]
+\pstGeneralEllipseFle[PosAngle=20]{F_1}{A}{B}{0.6}{O_3}{R_3}{MajorRotAngle3}
+\pstGeneralEllipse[linecolor=green!60](O_3)(R_3)[MajorRotAngle3]
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralEllipseCoef} is used to define a General Ellipse by the quadratic curve equation $ax^2+bxy+cy^2+dx+ey+f=0$,
+it just calculate the center $O$, major radius $a$, minor radius $b$ and the rotation angle $\theta$ of the major axis,
+then you can pass them into macro \Lcs{pstGeneralEllipse} to draw this ellipse.
+The package \texttt{pst-func} provides macro \Lcs{psplotImp} to draw an implicit defined functions too,
+but it can't tell you the geometrical elements like as center or radii, and it will take more time to
+calculate the function value point by point.
+
+\begin{BDef}
+\Lcs{pstGeneralEllipseCoef}\OptArgs\Largb{a,b,c,d,e,f}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, the output parameter \texttt{Rab} and the output parameter \texttt{$\theta$}
+are same with \Lcs{pstGeneralEllipseFle}. They are set to zero if the coeffients are invalid to construct an ellipse.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-4,-3)(2,3)
+\psset{unit=1.0cm}\footnotesize\psset{dotscale=0.5}
+\psset{CodeFig=true}\psset{PointSymbol=*}
+%2x^2-2xy+3y^2+6x+5y+8=0
+\pstGeneralEllipseCoef[PosAngle=-100,CodeFigColor=red!50]{2,-2,3,6,5,8}{O_1}{R_1}{MajorRotAngle1}
+\pstGeneralEllipse[linecolor=red!60](O_1)(R_1)[MajorRotAngle1]
+%3x^2-2xy+2y^2-3x+6y+3=0
+\pstGeneralEllipseCoef[PosAngle=-80,CodeFigColor=purple!50]{3,-2,2,-3,5,2}{O_2}{R_2}{MajorRotAngle2}
+\pstGeneralEllipse[linecolor=purple!60](O_2)(R_2)[MajorRotAngle2]
+%x^2-xy+y^2+x-3y+1=0
+\pstGeneralEllipseCoef[PosAngle=-90,CodeFigColor=green!50]{1,-1,1,1,-3,1}{O_3}{R_3}{MajorRotAngle3}
+\pstGeneralEllipse[linecolor=green!60](O_3)(R_3)[MajorRotAngle3]
+%2x^2+4xy+3y^2+8x+6y+8=0
+\pstGeneralEllipseCoef[PosAngle=-80,CodeFigColor=blue!50]{2,4,3,8,6,8}{O_4}{R_4}{MajorRotAngle4}
+\pstGeneralEllipse[linecolor=blue!60](O_4)(R_4)[MajorRotAngle4]
+\end{pspicture}
+\end{LTXexample}
+
+You can verify the output figures with \Lcs{psplotImp} as following:
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-4,-3)(2,3)
+\psset{unit=1.0cm}\footnotesize\psset{dotscale=0.5}
+%2x^2-2xy+3y^2+6x+5y+8=0
+\psplotImp[linecolor=red!60](-4,-3)(2,3){ 2 x dup mul mul -2 x mul y mul add 3 y dup mul mul add 6 x mul add 5 y mul add 8 add }
+%3x^2-2xy+2y^2-3x+6y+3=0
+\psplotImp[linecolor=purple!60](-4,-3)(2,3){ 3 x dup mul mul -2 x mul y mul add 2 y dup mul mul add -3 x mul add 6 y mul add 3 add }
+%x^2-xy+y^2+x-3y+1=0
+\psplotImp[linecolor=green!60](-4,-3)(2,3){ 1 x dup mul mul -1 x mul y mul add 1 y dup mul mul add 1 x mul add -3 y mul add 1 add }
+%2x^2+4xy+3y^2+8x+6y+8=0
+\psplotImp[linecolor=blue!60](-5,-3)(2,3){ 2 x dup mul mul 4 x mul y mul add 3 y dup mul mul add 8 x mul add 6 y mul add 8 add }
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralEllipseABCDE} is used to define a General Ellipse by the given five points $A,B,C,D,E$,
+it just calculate the center $O$, major radius $a$, minor radius $b$ and the rotation angle $\theta$ of the major axis,
+then you can pass them into macro \Lcs{pstGeneralEllipse} to draw this ellipse.
+
+\begin{BDef}
+\Lcs{pstGeneralEllipseABCDE}\OptArgs\Largb{A}\Largb{B}\Largb{C}\Largb{D}\Largb{E}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, the output parameter \texttt{Rab} and the output parameter \texttt{$\theta$}
+are same with \Lcs{pstGeneralEllipseFle}. They are set to zero if the points are invalid to construct an ellipse.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-2)(5,4)
+\psset{unit=1.0cm}\footnotesize\psset{PointSymbol=*}
+\psset{CodeFig=true,CodeFigColor=gray!50}
+\pstGeonode[PosAngle=180](0,0){A}
+\pstGeonode[PosAngle=-90](2,-1){B}
+\pstGeonode[PosAngle=90](3,3){C}
+\pstGeonode[PosAngle=-90](4,0){D}
+\pstGeonode[PosAngle=0](5,2){E}
+\pstGeneralEllipseABCDE[PosAngle=0]{A}{B}{C}{D}{E}{O}{R}{MajorRotAngle}
+\pstGeneralEllipse[linecolor=red!60](O)(R)[MajorRotAngle]
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+We can location the points on the General Ellipse using the macros
 \Lcs{pstGeneralEllipseNode}, \Lcs{pstGeneralEllipseRotNode}, \Lcs{pstGeneralEllipseAbsNode}
 and \Lcs{pstGeneralEllipseOrdNode} as following.
 
@@ -2212,6 +2458,7 @@ We also use the theorem \ref{ParabolaTangentPointTheorem} to find the tangent no
 The General Parabola $P$ with coordinate translation and rotation is defined by vertex $O(x_0,y_0)$,
 the half of the focus chord axis $abs(p)$, the sign of $p$ indicates the direction of the parabola,
 and the rotation angle $\theta$ of the symmetrical axis.
+The symmetrical axis is perpendicular to x-axis when $\theta=0^\circ$, and perpendicular to y-axis when $\theta=90^\circ$.
 
 The equation can be got from the parametric function of the parabola equation (\ref{ParametricFunctionOfStandardParabola}),
 using the rotation transform formula (\ref{RotationTransformFormula}), then we have
@@ -2271,6 +2518,111 @@ at most time.
 \end{pspicture}
 \end{LTXexample}
 
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralParabolaFl} is used to define a General Parabola with Focus $F$, and the directrix line $l$. It just calculate the vertex $O$, half focal chord $p$, and the rotation
+angle $\theta$ of the symmetrical axis, then you can pass them into macro \Lcs{pstGeneralParabola} to draw this parabola.
+
+\begin{BDef}
+\Lcs{pstGeneralParabolaFl}\OptArgs\Largb{F}\Largb{A}\Largb{B}\Largb{O}\Largb{p}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O} is a node name to store the vertex point, its label and symbol can
+be controlled by the options for \PST\ node, such as \Lkeyword{PosAngle}.
+The output parameter \texttt{p} is a PostScript key to store the value of half focal chord.
+The output parameter \texttt{$\theta$} is also a PostScript key to store the rotation angle of symmetrical axis,
+when you pass it to \Lcs{pstGeneralParabola}, PostScript will lookup the value of this key in current dictionary.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-2,-2)(2,2)
+\psset{unit=1.0cm}\psset{dotscale=0.5}\footnotesize
+\psset{CodeFig=true,CodeFigColor=gray!50}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=-60](1,1){F_1}
+\pstGeonode[PosAngle=-60](-1,-1){F_2}
+\pstGeonode[PosAngle=60](1,-1){B}
+\pstGeonode[PosAngle=60](-1,1){A}
+\pstGeneralParabolaFl[PosAngle=60]{F_1}{A}{B}{O_1}{semifocalchordp1}{SymAxisRotAngle1}
+\pstGeneralParabola[linecolor=red!60](O_1){semifocalchordp1}[SymAxisRotAngle1]{-2}{2}
+\pstGeneralParabolaFl[PosAngle=60]{F_2}{A}{B}{O_2}{semifocalchordp2}{SymAxisRotAngle2}
+\pstGeneralParabola[linecolor=blue!60](O_2){semifocalchordp2}[SymAxisRotAngle2]{-2}{2}
+\pstGeneralParabolaFl[PosAngle=60]{B}{F_1}{F2}{O_3}{semifocalchordp3}{SymAxisRotAngle3}
+\pstGeneralParabola[linecolor=green!60](O_3){semifocalchordp3}[SymAxisRotAngle3]{-2}{2}
+\pstGeneralParabolaFl[PosAngle=60]{A}{F_1}{F2}{O_4}{semifocalchordp4}{SymAxisRotAngle4}
+\pstGeneralParabola[linecolor=green!60](O_4){semifocalchordp4}[SymAxisRotAngle4]{-2}{2}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralParabolaCoef} is used to define a General Parabola by the quadratic curve equation $ax^2+bxy+cy^2+dx+ey+f=0$,
+it just calculate the vertex $O$, half focal chord $p$ and the rotation angle $\theta$ of the symmetrical axis,
+then you can pass them into macro \Lcs{pstGeneralParabola} to draw this parabola.
+The package \texttt{pst-func} provides macro \Lcs{psplotImp} to draw an implicit defined functions too,
+but it can't tell you the geometrical elements like as center or radii, and it will take more time to
+calculate the function value point by point.
+
+\begin{BDef}
+\Lcs{pstGeneralParabolaCoef}\OptArgs\Largb{a,b,c,d,e,f}\Largb{O}\Largb{p}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, \texttt{p} and \texttt{$\theta$}are same with \Lcs{pstGeneralParabolaFl}.
+They are set to zero if the coeffients are invalid to construct a parabola.
+If you pass the zero $p$ into macro \Lcs{pstGeneralParabola}, it will abort with the exception of dividing by zero.
+
+In the following example, we use \Lcs{psplotImp} to draw the same parabolas, just to check the results
+given by macros \Lcs{pstGeneralParabolaCoef} are correct.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-1)(4,4)
+\psset{unit=0.40cm}\footnotesize\psset{dotscale=0.5}
+\psset{CodeFig=true}\psset{PointSymbol=*}
+%x^2-2xy+y^2-8x+16=0
+\psplotImp[linecolor=green!30](-10,-10)(10,10){ 1 x dup mul mul -2 x mul y mul add 1 y dup mul mul add -8 x mul add 0 y mul add 16 add }
+\pstGeneralParabolaCoef[PosAngle=0,CodeFigColor=red!50]{1,-2,1,-8,0,16}{O_1}{P1}{SymAxisRotAngle1}
+\pstGeneralParabola[linecolor=red!60](O_1){P1}[SymAxisRotAngle1]{-3}{3}
+%x^2+2xy+y^2+2x-2y-5=0
+\psplotImp[linecolor=green!30](-10,-10)(10,10){ 1 x dup mul mul 2 x mul y mul add 1 y dup mul mul add 2 x mul add -2 y mul add -5 add }
+\pstGeneralParabolaCoef[PosAngle=-90,CodeFigColor=black!60]{1,2,1,2,-2,-5}{O_2}{P2}{SymAxisRotAngle2}
+\pstGeneralParabola[linecolor=black!60](O_2){P2}[SymAxisRotAngle2]{-3}{3}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralParabolaABCDE} is used to define a General Parabola by the given five points $A,B,C,D,E$,
+it just calculate the vertex $O$, half focal chord $p$ and the rotation angle $\theta$ of the symmetrical axis,
+then you can pass them into macro \Lcs{pstGeneralParabola} to draw this parabola.
+
+\begin{BDef}
+\Lcs{pstGeneralParabolaABCDE}\OptArgs\Largb{A}\Largb{B}\Largb{C}\Largb{D}\Largb{E}\Largb{O}\Largb{p}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, \texttt{p} and \texttt{$\theta$} are same with \Lcs{pstGeneralParabolaFl}.
+They are set to zero if the points are invalid to construct a parabola.
+If you pass the zero $p$ into macro \Lcs{pstGeneralParabola}, it will abort with the exception of dividing by zero.
+
+Note the algorithm may fit a hyperbola quadratic curve from the given five points,
+in order to get the right parabola curve, you must input the point coordinates very precisely.
+In the following example, if you input point $A$ as $(3,1.732)$, it will fail as no such parabola
+can fit these five points.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](0,-2)(4,2)
+\psset{unit=0.5cm}\footnotesize\psset{PointSymbol=*}
+\psset{CodeFig=true,CodeFigColor=gray!50}
+% five points from y^2-2x+3=0
+\pstGeonode[PosAngle=90](3,1.73205){A}
+\pstGeonode[PosAngle=90](2,1){B}
+\pstGeonode[PosAngle=-90](2,-1){C}
+\pstGeonode[PosAngle=90](6,3){D}
+\pstGeonode[PosAngle=-90](6,-3){E}
+\pstGeneralParabolaABCDE[PosAngle=235]{A}{B}{C}{D}{E}{O}{P}{SymAxisRotAngle}
+\pstGeneralParabola[linecolor=red!60](O){P}[SymAxisRotAngle]{-3}{3}
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
 The macro \Lcs{pstGeneralParabolaFocusNode} is used to find the focus of the parabola,
 and the macro \Lcs{pstGeneralParabolaDirectrixLine} is used to find the directrix line of the parabola.
 
@@ -2969,6 +3321,101 @@ it should be setup from 0 to 90.
 \end{pspicture}
 \end{LTXexample}
 
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralHyperbolaFle} is used to define a General Hyperbola with Focus $F$, directrix line $l$,
+and the eccentricity $e$, where $e>1$. It just calculate the center $O$, real radius $a$, imaginary radius $b$ and the rotation
+angle $\theta$ of the real axis, then you can pass them into macro \Lcs{pstGeneralHyperbola} to draw this hyperbola.
+
+\begin{BDef}
+\Lcs{pstGeneralHyperbolaFle}\OptArgs\Largb{F}\Largb{A}\Largb{B}\Largb{$e$}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O} is a node name to store the center point, its label and symbol can
+be controlled by the options for \PST\ node, such as \Lkeyword{PosAngle}.
+The output parameter \texttt{Rab} is a PostScript key to store the pair of real radius and imaginary radius,
+it just use \PST\ node coordinate to store a pair of value, but not a geometrical point.
+The output parameter \texttt{$\theta$} is also a PostScript key to store the rotation angle of real axis,
+when you pass it to \Lcs{pstGeneralHyperbola}, PostScript will lookup the value of this key in current dictionary.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-2,-2)(2,2)
+\psset{unit=1.0cm}\psset{dotscale=0.5}\footnotesize
+\psset{CodeFig=true,CodeFigColor=gray!50}\psset{PointSymbol=*}
+\pstGeonode[PosAngle=-60](1,1){F_1}
+\pstGeonode[PosAngle=-60](-1,-1){F_2}
+\pstGeonode[PosAngle=-60](1,-1){B}
+\pstGeonode[PosAngle=-60](-1,1){A}
+\pstGeneralHyperbolaFle[PosAngle=-60]{F_1}{A}{B}{2.4}{O_1}{R_1}{RealAxisRotAngle1}
+\pstGeneralHyperbola[linecolor=red!60](O_1)(R_1)[RealAxisRotAngle1][65]
+\pstGeneralHyperbolaFle[PosAngle=-60]{F_2}{A}{B}{2.4}{O_2}{R_2}{RealAxisRotAngle2}
+\pstGeneralHyperbola[linecolor=blue!60](O_2)(R_2)[RealAxisRotAngle2][65]
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralHyperbolaCoef} is used to define a General Hyperbola by the quadratic curve equation $ax^2+bxy+cy^2+dx+ey+f=0$,
+it just calculate the center $O$, real radius $a$ and imaginary radius $b$ and the rotation angle $\theta$ of the real axis,
+then you can pass them into macro \Lcs{pstGeneralHyperbola} to draw this hyperbola.
+The package \texttt{pst-func} provides macro \Lcs{psplotImp} to draw an implicit defined functions too,
+but it can't tell you the geometrical elements like as center or radii, and it will take more time to
+calculate the function value point by point.
+
+\begin{BDef}
+\Lcs{pstGeneralHyperbolaCoef}\OptArgs\Largb{a,b,c,d,e,f}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, the output parameter \texttt{Rab} and the output parameter \texttt{$\theta$}~
+are same with \Lcs{pstGeneralHyperbolaFle}. They are set to zero if the coeffients are invalid to construct a hyperbola.
+
+In the following example, we use \Lcs{psplotImp} to draw the same hyperbolas, just to check the results
+given by macros \Lcs{pstGeneralHyperbolaCoef} are correct.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-4,-4)(4,4)
+\psset{unit=0.40cm}\footnotesize\psset{dotscale=0.5}
+\psset{CodeFig=true}\psset{PointSymbol=*}
+%x^2+3xy-2y^2+10x-5y+6=0
+\psplotImp[linecolor=green!30](-10,-10)(10,10){ 1 x dup mul mul 3 x mul y mul add -2 y dup mul mul add 10 x mul add -5 y mul add 6 add }
+\pstGeneralHyperbolaCoef[PosAngle=-30,CodeFigColor=red]{1,3,-2,10,-5,6}{O_1}{R_1}{RealAxisRotAngle1}
+\pstGeneralHyperbola[linecolor=red](O_1)(R_1)[RealAxisRotAngle1][60]
+%x^2-3xy+y^2+10x-10y+21=0
+\psplotImp[linecolor=blue!30](-10,-10)(10,10){ 1 x dup mul mul -3 x mul y mul add 1 y dup mul mul add 10 x mul add -10 y mul add 21 add }
+\pstGeneralHyperbolaCoef[PosAngle=-10,CodeFigColor=black]{1,-3,1,10,-10,21}{O_2}{R_2}{RealAxisRotAngle2}
+\pstGeneralHyperbola[linecolor=black](O_2)(R_2)[RealAxisRotAngle2][60]
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
+The Macro \Lcs{pstGeneralHyperbolaABCDE} is used to define a General Hyperbola by the given five points $A,B,C,D,E$,
+it just calculate the center $O$, real radius $a$ and imaginary radius $b$ and the rotation angle $\theta$ of the real axis,
+then you can pass them into macro \Lcs{pstGeneralHyperbola} to draw this hyperbola.
+
+\begin{BDef}
+\Lcs{pstGeneralHyperbolaABCDE}\OptArgs\Largb{A}\Largb{B}\Largb{C}\Largb{D}\Largb{E}\Largb{O}\Largb{Rab}\Largb{$\theta$}
+\end{BDef}
+
+The output parameter \texttt{O}, the output parameter \texttt{Rab} and the output parameter \texttt{$\theta$}~
+are same with \Lcs{pstGeneralHyperbolaFle}. They are set to zero if the points are invalid to construct a hyperbola.
+
+\begin{LTXexample}[width=6cm,pos=l]
+\begin{pspicture}[showgrid=true](-1,-3)(4,2)
+\psset{unit=0.5cm}\footnotesize\psset{PointSymbol=*}
+\psset{CodeFig=true,CodeFigColor=gray!50}
+\pstGeonode[PosAngle=180](0,0){A}
+\pstGeonode[PosAngle=-90](2,-1){B}
+\pstGeonode[PosAngle=-90](3,-3){C}
+\pstGeonode[PosAngle=-90](4,0){D}
+\pstGeonode[PosAngle=0](5,2){E}
+\pstGeneralHyperbolaABCDE[PosAngle=0]{A}{B}{C}{D}{E}{O}{R}{RealAxisRotAngle}
+\pstGeneralHyperbola[linecolor=red!60](O)(R)[RealAxisRotAngle][80]
+\end{pspicture}
+\end{LTXexample}
+
+\vspace{10pt}
+
 The macro \Lcs{pstGeneralHyperbolaNode} is used to draw a node whose parameter is the given value $t$ on General Hyperbola,
 please refer to equation (\ref{ParametricFunctionOfGeneralHyperbola}).
 The macro \Lcs{pstGeneralHyperbolaAbsNode} is used to draw the nodes whose abscissa are the given value $x_1$ on General Hyperbola.
@@ -4613,7 +5060,7 @@ set of all the mediator lines of segments defined by $A$ and the
 circle points, create two conics depending of the position of $A$:
 
 \begin{compactitem}
-\item inside the circle: an hyperbola;
+\item inside the circle: a hyperbola;
 \item outside the circle: an ellipse.
 \end{compactitem}