# $(\mathbf{x} + \alpha \mathbf{y})^T C (\mathbf{x} + \alpha \mathbf{y}) = 0$ for all $\alpha$

I'm doing some work with conics. I am told that the line $$\mathbf{l}$$ tangent to a conic coefficient matrix $$C$$ at a point $$\mathbf{x}$$ on $$C$$ is given by $$\mathbf{l} = C\mathbf{x}$$. It is then said that the line $$\mathbf{l} = C \mathbf{x}$$ passes through $$\mathbf{x}$$, since $$\mathbf{l}^T \mathbf{x} = \mathbf{x}^T C \mathbf{x} = 0$$. And so, if $$\mathbf{l}$$ has one-point contact with the conic, then it is a tangent, and we are done. Otherwise suppose that $$\mathbf{l}$$ meets the conic in another point $$\mathbf{y}$$. Then $$\mathbf{y}^T C \mathbf{y} = 0$$ and $$\mathbf{x}^T C \mathbf{y} = \mathbf{l}^T \mathbf{y} = 0$$. It is then said that, from this, it follows that $$(\mathbf{x} + \alpha \mathbf{y})^T C (\mathbf{x} + \alpha \mathbf{y}) = 0$$ for all $$\alpha$$, which means that the whole line $$\mathbf{l} = C\mathbf{x}$$ joining $$\mathbf{x}$$ and $$\mathbf{y}$$ lies on the conic $$C$$, which is therefore degenerate.

I'm confused how it follows that $$(\mathbf{x} + \alpha \mathbf{y})^T C (\mathbf{x} + \alpha \mathbf{y}) = 0$$ for all $$\alpha$$, which means that the whole line $$\mathbf{l} = C\mathbf{x}$$ joining $$\mathbf{x}$$ and $$\mathbf{y}$$ lies on the conic $$C$$, which is therefore degenerate. I was wondering if someone could please explain this. Thank you.

As far as the algebra goes, it comes down, basically, to carefully expanding your quadratic form and repeatedly using that $$C\mathbf x = \mathbf 1$$, the symmetry of $$C$$ and the fact that $$\mathbf 1^T\mathbf x = \mathbf 1^T\mathbf y = 0$$
$$(\mathbf{x} + \alpha \mathbf{y})^T C (\mathbf{x} + \alpha \mathbf{y})$$
$$=\mathbf{x}^T C (\mathbf{x} + \alpha \mathbf{y}) + \alpha \mathbf{y}^T C (\mathbf{x} + \alpha \mathbf{y})$$
$$=\mathbf{x}^T C \mathbf{x} + \alpha \big(\mathbf{x}^T C\big)\mathbf{y} + \alpha\mathbf{y}^T \big(C \mathbf{x}\big) + \alpha^2 \mathbf{y}^T C\mathbf{y}$$
$$= 0 + \alpha \big(\mathbf 1^T\big)\mathbf y +\alpha\mathbf{y}^T \big(\mathbf 1\big) + \alpha^2 0$$
$$= 0 + \alpha 0 +\alpha 0 + 0$$
$$=0$$
• You used $C\mathbf x = \mathbf 1$, but don't you mean $C\mathbf x = \mathbf l$? Commented Feb 21, 2020 at 8:14
• haha, no. In the textbook, $\mathbf{1}$ is said to be the vector of all 1's, whereas $C\mathbf{x} = \mathbf{l}$ is a line. Never mind, though. Commented Feb 21, 2020 at 20:15