Parametric equation of a conic passing through 5 coplanar points in space

Question

I would need to calculate the formula to generate the conic passing through $5$ coplanar points in space.

I know this question could be long and boring, in fact I was hoping to find some "smart" method to solve it rather than wasting hours of calculation. Since the solution is a parametric equation the question can be simplified as "what is the parametric equation of a conic passing through 5 points in the plane?" (the condition that the $5$ points are coplanar is automatically respected by rotating and translating the solution).

I know there is this formula: $${\displaystyle \det {\begin{bmatrix}x^{2}&xy&y^{2}&x&y&1\\x_{1}^{2}&x_{1}y_{1}&y_{1}^{2}&x_{1}&y_{1}&1\\x_{2}^{2}&x_{2}y_{2}&y_{2}^{2}&x_{2}&y_{2}&1\\x_{3}^{2}&x_{3}y_{3}&y_{3}^{2}&x_{3}&y_{3}&1\\x_{4}^{2}&x_{4}y_{4}&y_{4}^{2}&x_{4}&y_{4}&1\\x_{5}^{2}&x_{5}y_{5}&y_{5}^{2}&x_{5}&y_{5}&1\end{bmatrix}}=0}$$

is there a way to transform it into parametric form?

Something like: $$\Gamma(t): (x_0,y_0)+(a_1\cos(t)+b_1\sin(t),a_2\cos(t)+b_2\sin(t))$$

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I know that given a conic of the form: $$\Gamma(x,y): ax^2+2bxy+cy^2+2dx+2ey+f=0$$ the center and angle of rotation are: $$\mathbf{C}=\left(\frac{cd-be}{b^2-ac^2},\frac{ae-bd}{b^2-ac}\right)\qquad \theta=\frac{1}{2}\arctan(2b,a-c)$$ and the conic form is $$\lambda_1 x^2+\lambda_2 y^2=-\frac{\det(\mathbf{A})}{\det(\mathbf{B})}$$ Where $\mathbf{A}$ and $\mathbf{B}$ are the matrices referred to the conic in matrix form and $\lambda_{1,2}$ are the eigenvalues ​​of matrix $\mathbf{B}$.

Honestly I would like to avoid developing the determinant of a $6\times 6$ matrix and calculating this type of formula. Literally the "brute" method would be:

1. Calculate the determinant of the $6\times 6$ matrix;
2. Take the coefficients $a$, $b$, $c$, $d$ and $e$;
3. Calculate the eigenvalues of $\mathbf{B}$ ​​to transform it into canonical form;
4. Transform the canonical form into parametric form: $(x,y)=(\cos(t),\sin(t))$ or $(x,y)=(\cosh(t), \sinh(t))$;
5. Apply the inverse rototranslation (of vector $\mathbf{C}$ and angle $\theta$) to obtain the starting conic.

Answer 1

Let $ABCDE$ be the five given points, lying on the same plane. We can use Pascal's theorem to find the line tangent to the ellipse at $A$: if $F$ is the intersection of $AB$ with $CD$, and $G$ is the intersection of $AC$ with $BE$, then the intersection $T$ of $FG$ with $DE$ is a point on the tangent at $A$ to the conic $ABCDE$ (see here for a detailed proof). Repeat the same construction to find the tangents at points $B$ and $C$.

Let then $N$ be the intersection of the tangents at $A$ and $B$, $P$ the intersection of the tangents at $B$ and $C$. If $L$ and $M$ are the midpoints of segments $AB$ and $BC$, then the center $O$ of the ellipse is the intersection of lines $PM$ and $LN$ (this is due to a another well-wknon theorem: center, midpoint of a chord and intersection of the tangents at the endpoints of the chord are collinear).

Construct $Q$ on ray $ON$ such that $OQ=\sqrt{OL\cdot ON}$ (see figure below). Draw then ray $OU$ parallel to $AB$, meeting $AN$ at $U$; construct on it point $S$ such that $AS\parallel ON$ and point $R$ such that $OR=\sqrt{OS\cdot OU}$. Segments $OQ$ and $OR$ are a pair of conjugate diameters. If the conic is an ellipse, a parametric equation can be written as: $$ O+(Q-O)\cos t + (R-O)\sin t. $$

If the conic is a hyperbola an analogous equation can be written, ask if interested.

Answer 2

Intelligenti's answer mentions using Pascal's theorem but seems to talk about tangent lines in what seems to be an application of the dual result (Brianchon's theorem), though I am not entirely sure. At any rate, since what the OP is looking for is the conic rather than tangent lines to a conic, one can proceed as follows. Consider points $A,B,C,D,E$. One would like to construct a point $X$ such that the six points lie on a common conic (the point $X$ would correspond to the point $F$ in a standard illustration of Pascal's theorem). Let $o$ be an arbitrary line through $A$. Then one forms the points $L=AE\cap BD$, $M=CD\cap o$, and $N=CE\cap LM$. Then $X= NB\cap o$. As the line $o$ ranges through all the lines through $A$ lying in the plane, the corresponding point $X$ describes all the points on the conic (of course, taking care of the cases of parallelism in a suitable fashion).

Answer 3

If you have a conic $C$ and a point $P$ on $C$, then you can parametrise it as follows: let $L$ be a line through $P$ with slope $t$, so the equation of $L$ is

$$y=t(x-x_0)+y_0$$

where $P=(x_0,y_0)$. The intersection of $C$ with $L$ contains $P$ and one other point, call it $Q(t)$. Then you have defined for each $t$ a point $Q(t)$ on $C$. This is a parametrisation!

Ex.

$C:\,xy=1$, $P=(1,1)$ gives $L:\,y=t(x-1)+1$. $C$ and $L$ intersect at $P$ and $(-1/t,-t)$. $t\mapsto (-1/t,-t)$ is the parametrisation.