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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.

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1 Answer 1

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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 $\mathbf 1^T\mathbf x = \mathbf 1^T\mathbf y = 0$

$(\mathbf{x} + \alpha \mathbf{y})^T C (\mathbf{x} + \alpha \mathbf{y})$
$=\mathbf{x}^T C (\mathbf{x} + \alpha \mathbf{y}) + \alpha \mathbf{y}^T C (\mathbf{x} + \alpha \mathbf{y})$
$=\mathbf{x}^T C \mathbf{x} + \alpha \big(\mathbf{x}^T C\big)\mathbf{y} + \alpha\mathbf{y}^T \big(C \mathbf{x}\big) + \alpha^2 \mathbf{y}^T C\mathbf{y}$
$= 0 + \alpha \big(\mathbf 1^T\big)\mathbf y +\alpha\mathbf{y}^T \big(\mathbf 1\big) + \alpha^2 0$
$= 0 + \alpha 0 +\alpha 0 + 0$
$=0$

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  • $\begingroup$ You used $C\mathbf x = \mathbf 1$, but don't you mean $C\mathbf x = \mathbf l$? $\endgroup$ Commented Feb 21, 2020 at 8:14
  • $\begingroup$ this feels like a trick question...? $\endgroup$ Commented Feb 21, 2020 at 19:46
  • $\begingroup$ 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. $\endgroup$ Commented Feb 21, 2020 at 20:15

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