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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.

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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 $y^TCx=x^TCy$.

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  • $\begingroup$ 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 $\textbf{x} + \alpha\textbf{y}$? $\endgroup$
    – jlcv
    Commented Feb 14, 2021 at 22:24
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    $\begingroup$ 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 $\alpha\ne 0$. $\endgroup$
    – Berci
    Commented Feb 14, 2021 at 22:45

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