From the book https://www.amazon.com/Multiple-View-Geometry-Computer-Vision/dp/0521540518

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}


1 Answer 1


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.

Actually this is far simpler if you just do it in one go, using the distributive law twice to turn the single quadratic form into a sum of four simpler ones like this:

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

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.


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