This question already has an answer here:
There are infinitely many solutions.
I will only consider the generic case where the lines are not parallel and neither line contains both points. Let the points be $P$ and $Q$ and the intersection of the lines be $O$. This looks like the following (ignore the ellipses for now).
There exists an invertible affine transformation which maps $O$ to the origin and the vectors $OP$ and $OQ$ to the standard basis $(1,0)$ and $(0,1)$, turning the above into
and vice versa. In this space, the points $P$ and $Q$ lie at $(1,0)$ and $(0,1)$ and the lines are $y=0$ and $x=0$. Here the ellipse
$$(x-1)^2+(y-1)^2+2cxy=1$$
passes through $P$ and $Q$ with the desired tangents for any $c \in (-1, 1)$. Shown above are the ellipses corresponding to $c=\frac12$, $c=0$, and $c=-\frac12$ in blue, green, and yellow. Transform these back into the original space via the inverse of the affine transformation, and you get the ellipses you want. This works because an ellipse remains an ellipse under an affine transformation, and of course incidences and tangencies do not change.
There might be special cases where there are only finitely many solutions, but in general there will be infinitely many ellipses satisfying the constraints. This is trivial to see in the case where the lines are parallel to each other and perpendicular to the line joining the two points, for example.
What’s your reasoning behind there being only a finite number of solutions?
It was a ‘gut feeling’ at first, and in case the points and tangents were not ‘connected’ I’d agree about infinite number of solutions… Something about the points and tangents at those points screamed ‘special case’ to me.
So, I racked my brain for quite a while, until I came up with fifth equation. I don’t know how it occurred to me, but I detected a property of the ellipse that would help solve the problem. After that it was a trivial matter of proving it, which I did. The frustrating bit was not only finding out that I have not discovered anything new- but that I had this bit of information in an old book ‘Matematische Formelsammlung’ I had lying on my table, but must have missed it when looking in it (actually, I just looked it up and must have missed it, again).
The missing equation is that the centre of the ellipse must lie on the line passing through the intersection of tangents and a midpoint of a secant connecting the points.
If it’s of any interest, I can do the proof again and post it here. So, you can understand my astonishment when that CAD program behaved he way it di…
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*}