Defining/constructing an ellipse Years ago I was confronted with a (self imposed) problem, which unexpectedly resurfaced just recently... I don't know whether it makes sense to explain the background or not, so I'll be brief.
If I was given two points in a plane, and two lines passing through them would I be able to construct (or get a equation of) an ellipse to which those lines are tangents in given points? At a first glance, four equations (each point belonging to the ellipse, and each of the lines being a tangent in respective point) would be insufficient to solve for five unknowns (major and minor axis, coordinates of the ellipse centre and the angle for coordinate system transform)- so there would be infinite number of ellipses that could be drawn with those constraints. But it was intuitive to me that only finite number would indeed be possible... So, what is the correct answer to this?
Hopefully, I wont be considered rude- but there is a twist to this... Firstly, I believe to have solved this problem, but recently I was working in a top of the line CAD program, where those constraints seemed to yield an infinite number of solutions. This bit puzzled me greatly- because it doesn't seem correct (so I'm assuming the ellipses they use are not proper ellipses). Secondly, I hope it won't be taken against me because I'm trying to pick your brains and having a theory of my own... which I will disclose a bit later.

OK, if it's of any interest, here it goes...
Assuming the equation of the ellipse $ b^2x^2+a^2y^2=a^2b^2 $, we get:
$$ y=\pm\frac{b}{a}\sqrt{a^2-x^2}$$
$$y´=-\frac{b^2x}{a^2y}=\mp\frac{b}{a}\frac{x}{\sqrt{a^2-x^2}} $$
For two points $ i=1,2 $ equation of tangent is:
$$ t_i ... y=y´(x-x_i)+y_i=\pm\frac{b}{a}\frac{a^2-x \centerdot x_i}{\sqrt{a^2-x_i^2}} $$
Intersection of those tangents will be ( $ x_I $ is calculated first by equating right sides of tangent equations):
$$ \pm\frac{b}{a}\frac{a^2-x_Ix_1}{\sqrt{a^2-x_1^2}}=\pm\frac{b}{a}\frac{a^2-x_Ix_2}{\sqrt{a^2-x_2^2}} $$
$$ \Downarrow $$
$$ \left( a^2-x_Ix_1\right)\sqrt{a^2-x_2^2}=\left( a^2-x_Ix_2\right)\sqrt{a^2-x_1^2} $$
$$ \Downarrow $$
$$ x_I=a^2\frac{\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}} $$
$$ y_I=\pm\frac{b}{a}\frac{a^2-x_Ix_1}{\sqrt{a^2-x_1^2}}=\text{ ... }=\pm ab\frac{x_1-x_2}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}} $$
Coordinates of midpoint of a secant:
$$ x_S=\frac{x_1+x_2}{2}\text{ ,  }y_S=\frac{y_1+y_2}{2}=\pm \frac{b}{a}\frac{\sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}}{2} $$
If we assume the equation of the line connecting intersection of tangents and midpoint of secant to be: $ y=\frac{y_I-y_S}{x_I-x_S}x+l $.
In order to prove my claim that that line must pass through the centre of the ellipse, it will be: $ l=0 $- so let's calculate l:
$$ l=y_S-\frac{y_I-y_S}{x_I-x_S}x_S=\frac{x_Iy_S-x_Sy_I}{x_I-x_S} $$
Concentrating on the numerator:
\begin{eqnarray*}
x_Iy_S-x_Sy_I&=&a^2\frac{\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\left(\pm \frac{b}{a}\frac{\sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}}{2}\right)-\frac{x_1+x_2}{2}\left(\pm ab\frac{x_1-x_2}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\right)\\
&=&\pm\frac{1}{2} ab \frac{\left(\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}\right) \left( \sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}\right)-\left(x_1+x_2 \right)\left(x_1-x_2 \right)}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\\
&=&\pm\frac{1}{2} ab \frac{\left[ \left(a^2-x_2^2 \right)-\left(a^2-x_1^2 \right) \right]-\left(x_1^2-x_2^2 \right)}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}=0
\end{eqnarray*}
