# Proof that the line $\textbf{l}$ tangent to a conic $C$ is $\textbf{l}=C\textbf{x}$

Prove that the line $$\textbf{l}$$ tangent to a conic $$C$$ at a point $$\textbf{x}$$ on C is given by $$\textbf{l}=C\textbf{x}$$. The equation of a conic in matrix form is defined as $$\textbf{x}^TC\textbf{x}=0$$.

The proof in the textbook says:

The line $$\textbf{l}=C\textbf{x}$$ passes through $$\textbf{x}$$, since $$\textbf{l}^T\textbf{x}=\textbf{x}^TC\textbf{x}=0$$. If $$\textbf{l}$$ has a one-point contact with the conic, then it is a tangent, and we are done. Otherwise suppose that $$\textbf{l}$$ meets the conic in another point $$\textbf{y}$$. Then $$\textbf{y}^TC\textbf{y}=0$$ and $$\textbf{x}^TC\textbf{y}=\textbf{l}^T\textbf{y}=0$$. From this it follows that $$(\textbf{x}+\alpha \textbf{y})^TC(\textbf{x}+\alpha \textbf{y})=0$$ for all $$\alpha$$, which means that the whole line $$\textbf{l}=C\textbf{x}$$ joining $$\textbf{x}$$ and $$\textbf{y}$$ lies on the conic C, which is therefore degenerate.

What I am confused about is where this:

From this it follows that $$(\textbf{x}+\alpha \textbf{y})^TC(\textbf{x}+\alpha \textbf{y})=0$$ for all $$\alpha$$

comes from, why is this statement true? I don't get how the preceding sentence implies this.

We simply have $$(x+\alpha y)^TC(x+\alpha y)=x^TCx+\alpha x^TCy +\alpha y^TCx+\alpha^2y^TCy$$ and each term on the right side is $$0$$, using $$y^TCx=x^TCy$$.
• Thanks! That makes sense. One more question I have is why does this statemen timply that the whole line joining $\textbf{x}$ and $\textbf{y}$ lies on the conic C? Isn't the line joining $\textbf{x}$ and $\textbf{y}$ defined by $(1-\alpha)\textbf{x} + \alpha\textbf{y}$ rather than $\textbf{x} + \alpha\textbf{y}$?
• A vector $v$ selects the same point in the projective space as $\lambda v$ for any nonzero $\lambda$ (homogeneous coordinates), so $\alpha x+(1-\alpha)y$ is the same point as $x+(\frac1\alpha-1)y$ provided $\alpha\ne 0$. Commented Feb 14, 2021 at 22:45