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So I'm trying to find a function that sends $x^2-y^2=z^2$ to $yz=x^2$. I know it can be done because all conics are projectively equivalent. I think I have to use a matrix of some kind but I don't know how.

Thanks for any help.

Art
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1 Answers1

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In general you have three degrees of freedom when mapping one conic to another. One approach is the following: choose three distinct points $A,B,C$ resp. $A',B',C'$ on each conic. Then construct a point $D$ as follows: take the tangents to the first conic in $A$ and $B$, intersecting these in a point $P$ and join that point of intersection with $C$. That last line will intersect the conic in a second point, which is that point $D$. Do the same to construct $D'$ on the second conic. Now you can compute the projective transformation which maps these four preimages to their corresponding images.

Applied to your situation, where you can use any matching transformation, you could choose simple coordinates on each conic. E.g.

\begin{align*} && M &= \begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix} & && M' &= \begin{pmatrix}-2&0&0\\0&0&1\\0&1&0\end{pmatrix} \end{align*} \begin{align*} A &= \begin{pmatrix}1\\0\\1\end{pmatrix} & MA &= \begin{pmatrix}1\\0\\-1\end{pmatrix} & A' &= \begin{pmatrix}0\\1\\0\end{pmatrix} & M'A' &= \begin{pmatrix}0\\0\\1\end{pmatrix} \\ B &= \begin{pmatrix}1\\0\\-1\end{pmatrix} & MB &= \begin{pmatrix}1\\0\\1\end{pmatrix} & B' &= \begin{pmatrix}0\\0\\1\end{pmatrix} & M'B' &= \begin{pmatrix}0\\1\\0\end{pmatrix} \\ C &= \begin{pmatrix}1\\1\\0\end{pmatrix} & P &= \begin{pmatrix}0\\1\\0\end{pmatrix} & C' &= \begin{pmatrix}1\\1\\1\end{pmatrix} & P' &= \begin{pmatrix}1\\0\\0\end{pmatrix} \end{align*} \begin{align*} D &= (P^TMP)C-(2C^TMP)P & D' &= (P'^TM'P')C'-(2C'^TM'P')P' \\ D &= -C+2P = \begin{pmatrix}-1\\1\\0\end{pmatrix} & D' &= -2C'+4P' \sim \begin{pmatrix}-1\\1\\1\end{pmatrix} \end{align*}

If you now follow the steps from my other post you will find that the projective transformation which maps $A,B,C,D$ to $A',B',C',D'$ can be described by the matrix

$$\begin{pmatrix}0&1&0\\1&0&1\\1&0&-1\end{pmatrix}$$

Of course, this is ony one suitable transformation from the three-parameter family of possible transformations.

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