Thread: A problem about singular points of conic
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Old January 13th, 2009, 03:30 PM
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Originally Posted by chipai View Post

Let ax^2+bxy+cy^2+dx+ey+f=0 be a conic. Prove that if M= \begin{bmatrix} 2a & b & d \\ b & 2c & e \\ d & e & 2f \end{bmatrix}, and \det M \neq 0, then this conic has no singular point.
you can look at the conic as a projective curve if you put x=\frac{X}{Z}, \ y=\frac{Y}{Z}. then you'll get F=aX^2 + bXY+cY^2 + dXZ + eYZ + fZ^2=0. now singular points are non-zero solutions of

\frac{\partial F}{\partial X}=\frac{\partial F}{\partial Y}=\frac{\partial F}{\partial Z}=0, which gives us: 2aX + bY + dZ = bX+2cY + eZ=dX + eY + 2fZ=0, which can be written as: M \tilde{X}=\bold{0}, where \tilde{X}=[X \ Y \ Z]^T. but \det M \neq 0, and hence

\tilde{X}=\bold{0}. thus the curve has no singular point.
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