I'm reading the book "Multiple View Geometry in Computer Vision", and there is a proof in there that I need help understanding. The following is all in $\mathbb{P}^2$ space and in homogeneous representation.

The theorem in question is:

The line $l$ tangent to a conic $C$ at a point $x$ on $C$ is given by $l = Cx$.

I can understand the proof for the non-degenerate case:

Given the line $l = Cx$, we know it passes through $x$ since $l^Tx = x^TCx = 0$. This comes from the fact that $x$ is a point that lies on $C$. If $x$ is the only intersection, the conic is non-degenerate and we're done.

However, for the degenerate case, the proof goes on to say that, suppose $l$ also intersects the conic at another point $y$, then $y^TCy = 0$, and $x^TCy = 0$ since $x^TC = l^T$ and $l^Ty = 0$. From this it follows that $(x + \alpha y)^TC(x + \alpha y) = 0$ for all $\alpha$, which means that the whole line $l = Cx$ joining $x$ and $y$ lies on the conic $C$.

The part in bold is verbatim and I can't figure out how that conclusion was made. Could someone explain how $x^TCy \iff (x + \alpha y)^TC(x + \alpha y) = 0$, and how to visualize such a degenerate case on a 2D projection?


1 Answer 1


From $x^TCy=0$ it also follows $y^TCx=0$ and $$ (x+\alpha y)^TC(x+\alpha y)= x^TCx+\alpha x^TCy+\alpha y^TCx+\alpha^2 y^TCy=0. $$

  • $\begingroup$ Not sure how I missed that, thanks! $\endgroup$
    – eclmist
    Apr 1, 2021 at 6:17

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