A problem about singular points of conic

chipai · #1 $Old$ January 13th, 2009, 04:43 AM

Let $a(x)^2+bxy+c(x)^2+dx+ey+f=0$ be a conic.
Prove that if
( 2a b d )
( b 2c e ) = M , det(M) is nonzero
( d e 2f )

,then this conic has no singular point.

thanks!!

NonCommAlg · #2 $Old$ January 13th, 2009, 03:30 PM

Quote:

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.