5 points determine a conic

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

The conic section will have the form

ax2+bxy+cy2+dx+ey+f=0, ax^2 + bxy + cy^2 + dx + ey + f = 0,

where the constants are determined up to multiplicity. WLOG, we may set f=1 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:

(x2xyy2xy1p12p1q1q12p1q11p22p2q2q22p2q21p32p3q3q32p3q31p42p4q4q42p4q41p52p5q5q52p5q51)=0 \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 (xi,yi,zi) (x_i, y_i, z_i ) is

(xyz1x1y1z11x2y2z21x3y3z31)=0 \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 nn dimensions.

Note by Calvin Lin
5 years, 3 months ago

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I 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.

Sharky Kesa - 5 years, 3 months ago

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