TikZ: normal and tangent vectors added to ellipse 2

Question

In Tikz: unit tangent vectors to a curve, Jake helped with adding a tangent vector to the ellipse. However, in that question, the ellipse was oriented with the semi-major and minor axis in line with the x and y axis.

In this question TikZ: Drawing an ellipse through two points, Percusse helped with the constructing of a random elliptical arc which is what I am using here. So we don't know the alignment of the semi-major and minor axis. Maybe we can find the rotation of it, but I am not 100% on that yet.

In the absence of this information, my question is how can I add a normal and tangent vector to the path ending at P2?

Can Jake's answer be adapted, or do we need a different method due to the ambiguity of the elliptical orientation?

  \documentclass[tikz]{standalone}
  \usetikzlibrary{arrows,calc}

  \begin{document}

  \begin{tikzpicture}[
    dot/.style = {outer sep = +0pt, inner sep = +0pt, shape = circle, draw = black, label = {#1}},
    small dot/.style = {minimum size = 1pt, dot = {#1}},
    big dot/.style = {minimum size = 2pt, dot = {#1}},
    line join = round, line cap = round, >=triangle 45
    ]
    \node[ fill = black, big dot = {below: \(F\)}] (F) at (0, 0) {};
    \node[ fill = black, big dot = {below: \(P_1\)}] (P1)  at (2, 0) {};
    \node[ fill = black, big dot = {above right=.25cm:\(P_2\)}] (P2) at (-2, 2) {};    

    \begin{pgfinterruptboundingbox}                  
    \begin{scope}[decoration = {markings,
        mark = at position 0.5 with {\arrow{>}}
      }]
      \clip (2, 0) -- (-2, 0) -- (-2, 4) -- (2, 4) -- cycle;
      \draw[name path global = ellp, postaction = decorate] let
        \p0 = ($(P2) - (F)$),
        \p1 = ($(P1) - (P2)$)
      in (P2|-P1) ++ (\x1, 0) arc (0:100: \x1 and \y0);                                         
  \end{scope}
\end{pgfinterruptboundingbox}

\path[name path = aux1] (P2) circle [radius = 1bp];
\draw[name intersections = {of = ellp and aux1}, -latex] (P2) --
($(intersection-2)!.75cm!(intersection-1)$);


  \end{tikzpicture}
  \end{document} 

Answer 1

You can draw a normal vector using a similar approach to what was done in Tikz: unit tangent vectors to a curve for getting a tangential vector:

If you've defined a very small circle path called aux1, say, around your tangential/normal point, you can find the intersections between this circle and the curve. The connecting line between the two intersections will be approximately tangential to the curve. To draw a normal line, you can use the calc syntax for a rotated line:

(P2) -- ($(P2)!0.75cm!-90:($(intersection-2)!.75cm!(intersection-1)$)$)

---

  \documentclass[tikz, border=5mm]{standalone}
  \usetikzlibrary{arrows,calc, decorations.markings, intersections}

  \begin{document}

  \begin{tikzpicture}[
    dot/.style = {outer sep = +0pt, inner sep = +0pt, shape = circle, draw = black, label = {#1}},
    small dot/.style = {minimum size = 1pt, dot = {#1}},
    big dot/.style = {minimum size = 2pt, dot = {#1}},
    line join = round, line cap = round, >=triangle 45
    ]
    \node[ fill = black, big dot = {below: \(F\)}] (F) at (0, 0) {};
    \node[ fill = black, big dot = {below: \(P_1\)}] (P1)  at (2, 0) {};
    \node[ fill = black, big dot = {above right=.25cm:\(P_2\)}] (P2) at (-2, 2) {};    

    \begin{pgfinterruptboundingbox}                  
    \begin{scope}[decoration = {markings,
        mark = at position 0.5 with {\arrow{>}}
      }]
      \clip (2, 0) -- (-2, 0) -- (-2, 4) -- (2, 4) -- cycle;
      \draw[name path global = ellp, postaction = decorate] let
        \p0 = ($(P2) - (F)$),
        \p1 = ($(P1) - (P2)$)
      in (P2|-P1) ++ (\x1, 0) arc (0:100: \x1 and \y0);                                         
  \end{scope}
\end{pgfinterruptboundingbox}

\path[name path = aux1] (P2) circle [radius = 1bp];
\draw[name intersections = {of = ellp and aux1}, -latex] (P2) --
($(intersection-2)!.75cm!(intersection-1)$);
\draw [-latex] (P2) --
($(P2)!0.75cm!-90:($(intersection-2)!.75cm!(intersection-1)$)$);


  \end{tikzpicture}
  \end{document}