Equation of conic: Given five points on it

Question

How do we find the equation of a conic given five points on it?

Is there a quick method, such as some determinant; which may help solve for the equation easily?

Of course, one may just substitute the given 5 points in the general equation of the conic, and solve for the unknowns - but that method is very time consuming and exhausting.

Could someone please help me generate a matrix, whose determinant gives the required equation of the conic? Data available: Coordinates of 5 points on the conic.

Thanks in advance!

P.S. I've successfully generated a determinant for the equation of a circle, given three points on it.

Here's the link to where I first found it, and later proved it - Get the equation of a circle when given 3 points

Answer 1

Given distinct points $P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4), P_5=(x_5,y_5)$, the equation of the conic through them is $$\det{\begin{pmatrix}x^2&xy&y^2&x&y&1\\x_1^2&x_1y_1&y_1^2&x_1&y_1&1\\x_2^2&x_2y_2&y_2^2&x_2&y_2&1\\x_3^2&x_3y_3&y_3^2&x_3&y_3&1\\x_4^2&x_4y_4&y_4^2&x_4&y_4&1\\x_5^2&x_5y_5&y_5^2&x_5&y_5&1\end{pmatrix}}=0.$$ The equation is not identically zero since the five points are distinct, and it is satisfied for the five points, since then the first row repeats the row of the point in question.

Answer 2

You can try to use this matrix, although you must first replace the points in the general formula, to obtain the constants.

In this case, you can use the determinant of this matrix square A to that you know who is the conic that represent this equation, and use this table for that you can identificate what sort of conic is.

where, for example, the value of A00, A11 are found so.

Answer 3

Jan-Magnus Økland's answer gives a beautiful and memorable big determinant formula. Now, big determinants are great as long as you do not need to expand them; otherwise a formula with smaller determinants might be preferable. Here is one. Let $\color{blue}{P}=(x,y)$, $P_i=(x_i,y_i)$ and $$\langle P_i,P_j,P_k\rangle = \begin{vmatrix} x_i & x_j & x_k \\ y_i & y_j & y_k \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x_j & x_k \\ y_j & y_k \end{vmatrix} + \begin{vmatrix} x_k & x_i \\ y_k & y_i \end{vmatrix} + \begin{vmatrix} x_i & x_j \\ y_i & y_j \end{vmatrix}$$ Then the conic through $P_1,\ldots,P_5$ can be described as $$\begin{align} \langle P_1,P_3,P_5\rangle \langle P_2,P_4,P_5\rangle \langle P_1,P_2,\color{blue}{P}\rangle \langle P_3,P_4,\color{blue}{P}\rangle \\{} - \langle P_1,P_3,\color{blue}{P}\rangle \langle P_2,P_4,\color{blue}{P}\rangle \langle P_1,P_2,P_5\rangle \langle P_3,P_4,P_5\rangle &= 0 \tag{1} \end{align}$$ The proof strategy is the same as indicated in Jan-Magnus Økland's comment: The equation is of total degree at most $2$ in $x,y$ and fulfilled for $\color{blue}{P}\in\{P_1,\ldots,P_5\}$.

As to where this equation comes from: There is a nice way to parametrize the family of conics through four given points: The ratio $$ \langle P_1,P_2,\color{blue}{P}\rangle \langle P_3,P_4,\color{blue}{P}\rangle : \langle P_1,P_3,\color{blue}{P}\rangle \langle P_2,P_4,\color{blue}{P}\rangle$$ is constant for each conic. Giving a fifth point determines that constant except in some cases of collinearity. Equation $(1)$ is the result.