Matrix Representation of Conic Sections - Center

Center

In the center of the conic, the gradient of the quadratic form Q vanishes, so: \nabla Q = = .

We can calculate the center by taking the first two rows of the associated matrix, multiplying each by (x, y, 1)T, setting both inner products equal to 0, and solving the system.

S \ \stackrel{\mathrm{def}}{=}\ \left\{ \begin{matrix} a_{11} + a_{12}x + a_{13}y & = & 0 \\ a_{21} + a_{22}x + a_{23}y & = & 0 \end{matrix} \right. \ \stackrel{\mathrm{def}}{=}\ \left\{\begin{matrix} D/2 + Ax + (B/2)y & = & 0 \\ E/2 + (B/2)x + Cy & = & 0 \end{matrix} \right.

This becomes

\begin{pmatrix} x_c \\ y_c \end{pmatrix} = \begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}^{-1} \begin{pmatrix} -D/2 \\ -E/2 \end{pmatrix} = \begin{pmatrix} (BE-2CD)/(4AC-B^2) \\ (DB-2AE)/(4AC-B^2) \end{pmatrix}

Note that in the case of a parabola, defined by (4AC-B2) = 0, there is no center since the above denominators become zero.

Read more about this topic:  Matrix Representation Of Conic Sections

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