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This is my code.

\documentclass[pstricks]{standalone}
\usepackage{pst-intersect}
\usepackage{pst-eucl,pstricks-add,pst-calculate}
\usepackage{pstricks-add,pst-calculate}
\usepackage{auto-pst-pdf}
\begin{document}
    \multido{\ry=0.3+0.1}{13}{%
    \begin{pspicture}[showgrid,algebraic](-5,-5)(5,5)
    \psset{PointName=none,PointSymbol=none}
    \def\a{1}
    \def\xo{\ry}
    \def\func{\a*x^2}
    \def\derivative{\a*2*x}
    \def\derivativexo{\pscalculate{\xo*\a*2}}
    \def\yo{\pscalculate{\a*\xo^2}}
    \pnodes(\xo,\yo){A}(5,0){B}(0,0){O}
\pssavepath[linestyle=none]{pathA}{\psplot{-2}{2}{\derivativexo*(x-\xo)+\yo}} %% y=f'(x0)(x-x0)+y0
\pssavepath[linestyle=none]{pathB}{\psline(0,0)(5,0)}
    \psintersect[name=C]{pathA}{pathB}
    \pstBissectBAC[linestyle=none]{B}{C1}{A}{A'}
    \pstRotation[RotAngle=90]{A}{C1}[N]
    \pstInterLL{A}{N}{C1}{A'}{E}
    \pstProjection{O}{B}{E}[K]
    \pscustom[fillstyle=solid,fillcolor=gray!50!orange,linestyle=none]{%
        \psplot{\xo}{0}{\func}
        \psline(0,0)(K)
        \pstArcnOAB[linestyle=none]{E}{K}{A}
        \closepath}
\psplot{-2}{2}{\func}
\pstCircleOA[Radius=\pstDistAB{E}{A}]{E}{}
\psline{->}(-2,0)(4,0)
\psline{->}(0,-2)(0,4.25)
\rput(0.75,2.5){$y=ax^2$}
\end{pspicture}
}
\end{document}

enter image description here

Note: You can reproduce my code if you want to say several new things!

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  • 2
    Welcome to TeX.SX! Could you please clarify what is your question? A general "please help me" is a bit vague. Oct 4, 2019 at 17:19

1 Answer 1

3

Given is the point P0(x0,y0) on the parabola. The slope is given by the derivation. The slope of the line P0-Mcircle is the negative inverse: -1/(2ax0).

enter image description here

The origin of the circle is O(x0+Delta,radius). Delta is radius*sin(alpha):

\documentclass[pstricks]{standalone}
\usepackage{pstricks-add,xfp}% xfp is the same as pst-calculate
\usepackage{auto-pst-pdf}
\begin{document}
\def\a{1}
\def\func{\a*x^2}
\multido{\rx=0.3+0.1}{15}{%
  \begin{pspicture}[showgrid,algebraic](-1,-1)(7.5,7.5)
    \psplot{-1}{3}{\func}%
    \def\Alpha{\fpeval{atan(2*\a*\rx)}}%  in radiant
    \def\radius{\fpeval{\a*\rx*\rx/(1+sin(\Alpha)/(2*\a*\rx))}}%
    \pnode(!\rx\space\radius\space\Alpha\space RadtoDeg sin mul add \radius){O}%\psdot(O)%
    \pscircle(O){\radius}%
    \pscustom[fillstyle=solid,fillcolor=gray!50!orange,linestyle=none,opacity=0.6]{%
      \psarc(O){\radius}{!180 \Alpha\space sub}{270}
      \psline(0,0)
      \psplot{0}{\rx}{\func}
    }%
    \psline{->}(-1,0)(7,0)
    \psline{->}(0,-1)(0,7.25)
    \rput(0.75,2.5){$y=ax^2$}
  \end{pspicture}%
}
\end{document}

enter image description here

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  • Many thanks for your explanation and illustration ...
    – user197952
    Oct 5, 2019 at 16:11

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