There's a matrix C about a conic and a point vector x. Why the line l=Cx tangent to the conic on x?

Question

Here's a Conic: $ax^2+bxy+cy^2+dx+ey+f=0$, and it is denoted by a matrix C: $\left| \begin{matrix} a \ \frac{b}{2} \ \frac{d}{2} \\ \frac{b}{2} \ c \ \frac{e}{2} \\ \frac{d}{2} \ \frac{e}{2} \ f \end{matrix} \right|.$

x is a point on the plane.

and the line $l=Cx$

Someone said that the l is tangent to the Conic and contact with the conic on x.

Can you tell me why?

Answer 1

If you define $r = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$

Then the equation of the conic is $r^T C r = 0 $

The normal vector to the conic at the point $r_0 = [x_0, y_0, 1] $ is $ n = 2 C r_0 $

Therefore, the equation of the tangent plane at $r_0$ is $r_0^T C (r - r_0 ) = 0 $

And this reduces to $ r_0^T C r = 0 $ because $r_0^T C r_0 = 0 $

Finally, we have to intersect this tangent plane with the plane $z = 1 $. This can be done by letting the third element of $r$ be $1$, then the tangent line is

$r_0^T C r = 0 $

Answer 2

Such a thing that when i post my question then i know the answer. The slope of the tangent line $l$ is $\frac{\frac{\partial{C}}{\partial{x}}}{\frac{\partial{C}}{\partial{y}}}$

and the a in l:(a,b,c) equals the first row dot and x. we know the value about artial derivative of square will doubled. and in the matrix ,b is halfed.

so it only a Partial derivative