1. ## A problem about singular points of conic

Let $\displaystyle a(x)^2+bxy+c(x)^2+dx+ey+f=0$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!!

2. Originally Posted by chipai

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

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

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