In Parabola Problem, Michael introduces the problem of finding a conic section given 5 of its points.

The conic section will have the form

\[ ax^2 + bxy + cy^2 + dx + ey + f = 0, \]

where the constants are determined up to multiplicity. WLOG, we may set \( f = 1 \). This is why we have a well determined system of 5 equations and 6 unknowns. This then becomes an ugly (to me) system of equations to solve.

I made the observation that we can just use the equation:

\[ \left|\begin{pmatrix} x^2 & xy & y^2 & x & y & 1 \\ p_1^2 & p_1q_1 & q_1^2 & p_1 & q_1 & 1 \\ p_2^2 & p_2q_2 & q_2^2 & p_2 & q_2 & 1 \\ p_3^2 & p_3q_3 & q_3^2 & p_3 & q_3 & 1 \\ p_4^2 & p_4q_4 & q_4^2 & p_4 & q_4 & 1 \\ p_5^2 & p_5q_5 & q_5^2 & p_5 & q_5 & 1 \end{pmatrix}\right| = 0 \]

Why does this work? What is the more general principle that is applied in this scenario?

**Hint:** What is the matrix three point form of a plane? Given 3 points in 3 dimensions in general position (aka not all 3 points lie on the same line), the unique plane that passes through \( (x_i, y_i, z_i ) \) is

\[ \left | \begin{pmatrix} x & y & z & 1 \\ x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ \end{pmatrix} \right| = 0 \]

This generalizes to \(n\) dimensions.

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TopNewestI have made one such problem based on the '5 points determine a conic' for people to try as well. It is rather interesting how conics on a 2D plane can be determined by 5 points only. I do believe there are more points required for higher dimensional graphing.

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