Parallel arrows along a curve with tikz

Question

I want to draw parallel arrows (for example 10 arrows) from a line to a curve. It is not good idea that dividing the line and curve to 10 parts and then join them to each other because of mathematical reason the resulting arrows are not parallel.

What I have done is the following.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows,positioning}
\begin{document}
\begin{tikzpicture}[allow upside down]

\draw  (0,0) circle (3.5);
\draw (0,3.5) arc (90:-90:1.75 and 3.5) \foreach \p in {0,5,...,100} {
    node[inner sep=0cm,above,pos=\p*0.01,
    anchor=center,
    minimum height=\p*0.03cm,minimum width=(10+\p)*0.3cm]
    (N \p){}
};

\draw (0,3.5) --(0,-3.5) \foreach \p in {0,5,...,100} {
    node[inner sep=0cm,above,pos=\p*0.01,
    anchor=center,
    minimum height=\p*0.03cm, minimum width=(10+\p)*0.3cm]
    (M \p){a}
};  

\foreach \p in {5,10,15,...,95} {
    \draw[-latex,blue] (N \p.center) -| (0,0);
}
\end{tikzpicture}
\end{document}

Answer 1

I do not really know if I understand what you want but a slight modification of your code produces "parallel arrows from a line to a curve".

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\begin{tikzpicture}[allow upside down]

\draw  (0,0) circle (3.5);
\draw (0,3.5) arc (90:-90:1.75 and 3.5) \foreach \p in {0,5,...,100} {
    node[inner sep=0cm,above,pos=\p*0.01,
    anchor=center,
    minimum height=\p*0.03cm,minimum width=(10+\p)*0.3cm]
    (N \p){}
};
\draw (0,3.5) -- (0,-3.5);

\foreach \p in {5,10,15,...,95} {
    \draw[latex-,blue] (N \p.center) -- (0,0 |- N \p.center);
}
\end{tikzpicture}
\end{document}

ADDENDTUM: To reverse the arrows, you only need to replace \draw[latex-,blue]... but \draw[-latex,blue]. However, making the distance equal, requires slightly more effort

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{intersections}
\begin{document}
\begin{tikzpicture}[allow upside down]

\draw  (0,0) circle (3.5);
\draw[name path=arc] (0,3.5) arc (90:-90:1.75 and 3.5);
\draw (0,3.5) -- (0,-3.5);

\foreach \p in {-3.25,-3,...,3.25} {
\path[name path=line] (0,\p) -- (3,\p);
\draw[latex-,blue,name intersections={of=line and arc}] 
(0,\p) -- (intersection-1);
}
\end{tikzpicture}
\end{document}

You do not need intersections in this case, you could just use

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\begin{tikzpicture}[allow upside down]

\draw  (0,0) circle (3.5);
\draw (0,3.5) arc (90:-90:1.75 and 3.5);
\draw (0,3.5) -- (0,-3.5);

\clip  (0,3.5) arc (90:-90:1.75 and 3.5) -- cycle;
\foreach \p in {-3.25,-3,...,3.25} {
\draw[latex-,blue] 
(0,\p) -- (3,\p);
}
\end{tikzpicture}
\end{document}

Answer 2

A PSTricks solution only for fun purposes.

\documentclass[pstricks]{standalone}
\usepackage{pst-plot}

\begin{document}
\begin{pspicture}[algebraic](-4,-4)(4,4)
    \pscircle{3}\psline(0,-3)(0,3)
    \psellipticarc(0,0)(1,3){-90}{90}
    \curvepnodes[plotpoints=20]{-3}{3}{sqrt(1-(t/3)^2)|t}{A}
    \foreach \i in {1,...,\numexpr\Anodecount-1}{\pcline[nodesepB=.4pt]{<-}(0,0|A\i)(A\i)}
\end{pspicture}
\end{document}