You can use path decorations:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}
\begin{document}
\begin{tikzpicture}[decoration={
markings,
mark=at position 0.2 with {\arrow{<}}}
]
\draw[postaction={decorate}] (0,0) ellipse (1 and 2);
\end{tikzpicture}
\end{document}
If we were dealing with circles, then the radius would be perpendicular to the tangent, so we would need to find a right triangle connecting the center C, point P, and the circle. If you recall your geometry, any triangle inscribed in a circle where one side of the triangle passes through the center will form a right triangle. So if we construct a circle whose center is the midpoint between P and C, the intersection of the two circles is the point(s) of tangency.
Since ellipses can be thought of as circles seen from an angle, instead of a circle we need an ellipse with the same aspect ratio.
For the purposes of illustration, I made the construct visible (pink).
\documentclass{standalone}
\usepackage{tikz,pgfplots}
\usetikzlibrary{calc,intersections,arrows}
\usepackage{times}
\begin{document}
\begin{tikzpicture}[scale=1,>=latex',scale=1.5,outer sep=0,inner sep=0,line width=0.7pt]
\def\a{2} \def\b{1} % radii of Ellipse
\def\cx{0} \def\cy{0} % determines center of Ellipse
\def\xp{0} \def\yp{-5} % coordinates of point P
\coordinate (C) at (\cx,\cy);% center of ellipse
\coordinate (P) at (\xp,\yp);% focal point
\draw[name path=ellipse,fill=gray!30,opacity=0.5](C)circle[x radius=\a,y radius=\b];
\pgfmathparse{abs(\cx-\xp)/\a}% compute normalized distance from C to P
\edef\dx{\pgfmathresult}
\pgfmathparse{abs(\cy-\yp)/\b}
\edef\dy{\pgfmathresult}
\pgfmathparse{0.5*ifthenelse(\dx,ifthenelse(\dy,sqrt(\dx*\dx+\dy*\dy),\dx),\dy)}
\edef\d{\pgfmathresult}% half the distance from C to P
\pgfmathparse{\a*\d}% compute ellipse radii
\edef\rx{\pgfmathresult}
\pgfmathparse{\b*\d}
\edef\ry{\pgfmathresult}
\draw[color=pink,name path=construct] ($(C)!0.5!(P)$) circle[x radius=\rx, y radius=\ry];
\path [name intersections={of = ellipse and construct}];
\coordinate (X) at (intersection-1);
\coordinate (Y) at (intersection-2);
\draw (X)--(P)--(Y);
\end{tikzpicture}
\end{document}
![tangents](https://i.stack.imgur.com/Vl0bQ.png)
Best Answer
I very much like Altermundus' solution, but if you want exactly what you drew, here's how I would do it:
Result:
I think that if given a little latitude in the design, then I would add a gentle curve to the lines of latitude.