# Line tangent to a conic at a point proof (verification of proof step)

The book provides the following result and proof on pg. 31 (verbatim):

Result: The line $\textbf{l}$ tangent to $C$ at a point $\textbf{x}$ on $C$ is given by $\textbf{l} = C\textbf{x}$.

Proof: The line $\textbf{l} = C\textbf{x}$ passes through $\textbf{x}$, since $\textbf{l}^\intercal\textbf{x} = \textbf{x}^\intercal C \textbf{x} = 0$. If $\textbf{l}$ has 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 $\textbf{y}^\intercal C \textbf{y} = 0$ and $\textbf{x}^\intercal C \textbf{y} = \textbf{l}^\intercal \textbf{y} = 0$. From this it follows that $(\textbf{x} + \alpha\textbf{y})^\intercal C (\textbf{x} + \alpha\textbf{y}) = 0$ 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 (degenerate conics being covered in the next section of the book).

This particular section is focusing on the projective space $\mathbb{P}^2$ of $\mathbb{R}^2$, so the vectors $\mathbb{x}, \mathbb{y}, \mathbb{l}$ are in $\mathbb{P}^2$ and $C$ is the conic coefficient matrix $$C = \begin{bmatrix} a & b/2 & d/2 \\ b/2 & c & e/2 \\ d/2 & e/2 & f \end{bmatrix}$$ of the conic $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$.

Just looking for verification on what I believe to be how $(\textbf{x} + \alpha\textbf{y})^\intercal C (\textbf{x} + \alpha\textbf{y}) = 0$ for all $\alpha$ follows from $\textbf{y}^\intercal C \textbf{y} = 0$ and $\textbf{x}^\intercal C \textbf{y} = 0$. I haven't looked at any math for over 2 years so my recollection of even basic facts is pretty rusty.

Let $\alpha \in \mathbb{R}^\ast$. $\textbf{x}^\intercal C \textbf{y} = 0$ and $\textbf{y}^\intercal C \textbf{y} = 0$ provides that $$\textbf{x}^\intercal C \textbf{y} = -\alpha\textbf{y}^\intercal C \textbf{y}$$ and so \begin{align} 0 & = \textbf{x}^\intercal C \textbf{y} +\alpha\textbf{y}^\intercal C \textbf{y} \\ & = (\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{y}. \end{align} As $\textbf{y}$ is a point where $\textbf{l}$ and $C$ intersect, $\textbf{l} = C\textbf{y}$. Again using the fact that $0 = 0$, \begin{align} 0 & = (\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{y} + \alpha(\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{y} \\ & = (\textbf{x} +\alpha\textbf{y})^\intercal \textbf{l} + \alpha(\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{y} \\ & = (\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{x} + \alpha(\textbf{x} +\alpha\textbf{y})^\intercal C \textbf{y} \\ & = (\textbf{x} +\alpha\textbf{y})^\intercal C (\textbf{x} + \alpha\textbf{y}) \end{align}

I'm not sure why you'd want to do this in two steps. And I have serious doubts how you'd argue $\mathbf l=C\mathbf y$ without assuming the very thing you want to prove. Actually even if $\mathbf x$ and $\mathbf y$ were the same point, they need not be the same vector (as homogeneous coordinates identify scalar multiples), so that equality is really tricky.
$$(\mathbf{x} + \alpha\mathbf{y})^\intercal C (\mathbf{x} + \alpha\mathbf{y}) = \mathbf x^\intercal C\mathbf x + \alpha\mathbf x^\intercal C\mathbf y + \alpha\mathbf y^\intercal C\mathbf x + \alpha^2\mathbf y^\intercal C\mathbf y = 0$$
The first of these summands is zero as $\mathbf x$ lies on $C$. The second is zero as $\mathbf y$ is supposed to also be a point on $\mathbf l$. Likewise the third, as $C$ is symmetric so $(\alpha\mathbf x^\intercal C\mathbf y)=(\alpha\mathbf y^\intercal C\mathbf x)$. And the last is because you assumed $\mathbf y$ to lie on $C$. Hence you get $0+0+0+0=0$. As $\alpha$ is a scalar, you can reorder it freely in each summand, which is what allowed me to move them to the very front.