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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.

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4 Answers 4

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).

enter image description here

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

enter image description here

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.

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Thanks @Rahul, but if I may ask you something... You lost me with non-orthogonal CS, with which I've had zero expirience- could you pls elaborate on the ellipse equation you used (what are the numbers in it, &c). The interesting part in your graph is that I notice ellipse centres seem to be on the quadrant diagonal (y=x), as my 'fifth equation' predicts, so I look forward to investigating it a bit more. –  Tomislav Petričević Dec 8 '12 at 12:36
    
Please see my edit. –  Rahul Dec 8 '12 at 19:10

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?

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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...

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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*}

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