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I have three points, $A$, $B$, and $C$, on an ellipse; two tangent lines at $A$ and $B$ are also known. Are these enough to determine the whole ellipse?

Blue
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fosuwxb
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  • It is well-known that you need five points to uniquely determine a conic section. Replacing two of the points with two slopes sounds like it ought to be fine. So from a pure degrees of freedom perspective, the answer is "probably". – Arthur Oct 31 '20 at 17:00
  • Points of tangency count twice, so you effectively have five points, determining a unique conic. (If the tangent lines at $A$ and $B$ have direction vectors $u$ and $v$, then you can define fourth and fifth points $A':=A+\alpha u$ and $B':=B+\beta v$, find the unique conic containing these, then let $\alpha$ and $\beta$ tend to $0$.) – Blue Oct 31 '20 at 17:06
  • Related (duplicate?): "What is the equation for an ellipse given 3 points and the tangent line at those points?". That question asks about specific points and tangents, but the general principle holds. My answer walks-through the process of determining the equation of the conic from the three given points and just two of the tangents. – Blue Oct 31 '20 at 17:13
  • Thanks! Actually, the tangent lines at A,B and C are all known and they form a three-sided open figure that is tangent at A,B and C with the ellipse. I want a proof for the conclusion that the condition can determine uniquely an ellipse. – fosuwxb Oct 31 '20 at 17:57
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    @fosuwxb: Five points determine a unique conic, and as I mentioned (and effectively demonstrate in my linked answer) points of tangency count twice, so your given information definitely determines a unique conic. Whether the conic is specifically an ellipse depends upon how the given points and tangents are arranged. Unless/until you provide details about the arrangement, the specific nature of the conic cannot be predicted. – Blue Oct 31 '20 at 18:14
  • To get an ellipse, point $C$ must lie inside the parabola touching the given lines at $A$ and $B$. – Intelligenti pauca Oct 31 '20 at 23:13
  • Thanks. I get a paper as follows:

    Paris Pamfilos . A Gallery of Conics by Five Elements,Forum Geometricorum, Volume 14 (2014) 295–348. My problem was completely solved.

     – fosuwxb Nov 01 '20 at 10:41
  • Go ahead and post your solution as an Answer. It is totally ok for an Asker to post an answer to their own question. – brainjam Nov 01 '20 at 17:28

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Answered in comments by OP:

[See] Paris Pamfilos, A Gallery of Conics by Five Elements, Forum Geometricorum, Volume 14 (2014) 295–348. My problem was completely solved

brainjam
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A conic by five distinct points can be constructed as follows: consider the lines $d_{12}=0,d_{34}=0,d_{13}=0$ and $d_{24}=0$ and form the pencil of conics

$$(1-\lambda)d_{12}d_{34}+\lambda d_{13}d_{24}=0.$$ these are all the conics through $1,2,3,4$. Then plug the coordinates of $5$ in the equation and determine $\lambda$ to get the conic by $1,2,3,4,5$.

Now if $1$ and $2$ get closer and closer, $d_{12}$ tends to a tangent, let $t_1$, and the pencil becomes

$$(1-\lambda)t_{1}d_{34}+\lambda d_{13}d_{14}=0.$$

If $4$ also migrates to $3$, defining the tangent $t_3$,

$$(1-\lambda)t_{1}t_3+\lambda d_{13}^2=0.$$