Projection of from a point to a line of a rational conic

Question

I am trying to follow the projection of the conic $C = x^2 - y^2 = z^2$ from the point $(1:0:1)$ to the line $x=0$

I follow the generation of the regular function $\phi$ for the projection and can see why this is represented by $(y:z-x)$

I also see how this has a regular representation on the conic itself as $(z+x: -y)$ which is regular when restricted to $C$.

This gives us a regular map of the projection at all points of $C$.

The next step is to show that it's surjective, and for this I'm given that we have a point $$ Q = \left(1 - \frac{2}{1-a^2}: \frac{2a}{1-a^2}:1\right) $$ such that $Q$ lies in the affinisation of $C$ in $\mathbb{A}^2_{z \neq 0}$ and $\phi(Q) = (1:a)$ for $a \neq \pm 1$. The remaining points are dealt with separately and I can see how that works as well, and I'm comfortable with the proof of injectivity.

The derivation of $Q$, however, ruins me. I have sat here with pen and paper for hours. I can, of course, see that it works, but I cannot reverse engineer it in any natural and obvious way that would help me complete a similar exercise. What am I missing?

Answer 1

Consider the line through $(1,0)$ having slope $-a$. The equation of the line is $y=-a(x-1)$. We'd like to find the second point of intersection of the line with the hyperbola $x^2-y^2=1$, the first being $(1,0)$.

\begin{align*} x^2 - \left(-a(x-1)\right)^2 &= 1 \\ x^2 - a^2 x^2 + 2a^2x-a^2 & =1 \\ (1-a^2)x^2 +2a^2x -(1+a^2) & = 0 \\ x^2 +\frac{2a^2}{1-a^2}x - \frac{1+a^2}{1-a^2} &= 0 \\ \end{align*} Now $x=1$ is a root of the quadratic, so the other root must be $-\dfrac{1+a^2}{1-a^2} = 1-\dfrac{2}{1-a^2}$ Plugging this back into $y=-a(x-1)$ \begin{align*} y &= -a(x-1) \\ y &= -a\left(1-\frac{2}{1-a^2} -1\right) \\ y &= \frac{2a}{1-a^2} \\ \end{align*}

Answer 2

Let $\omega=(1:0:1)$ be the projection centre.
I) We have a rational map from the projective plane $\mathbb P^2$ to to the $Y$ axis $L:X=0$ which is regular outside of $\omega$ and is described by the morphism $$\pi:\mathbb P^2\setminus \{\omega\}\to L:(x:y:z)\mapsto (0:y:z-x)$$ This morphism cannot be extended through its pole $\omega$, but the restriction $\pi\vert C\setminus \{\omega\}$ can be extended to a morphism, actually an isomorphism, $\; p:C\stackrel {\sim}{\to} L$ satisfying $$ p(x:y:z)= (0:y:z-x)=(0:v:w) \; \operatorname {for} \; (x:y:z)\neq (1:0:1)=\omega \\ p(x:y:z)=(0:x+z:-y)=(0:v:w) \; \operatorname {for}\; (x:y:z)\neq (-1:0:1)=\omega' $$ In particular $p(\omega)=(0:2:0)=(0:1:0)$.

II) In order to calculate $p^{-1}:L \stackrel {\sim}{\to} C$ we compute that the line through $(0:v:w)$ and $\omega$ has equation $$vX+wY-vZ=0$$ Its intersection with $C$ is obtained by solving the system $$vX+wY-vZ=0\\ X^2-Y^2-Z^2=0$$ a) If $w=1$ we get the system $$Y=v(Z-X)\\ X^2-v^2(Z-X)^2-Z^2=(X-Z)[X+Z-v^2(X-Z)] =0$$ Its solutions are
(i) $(X:Y:Z)=(1:0:1)=\omega$ (of course!) for $X-Z=0$
(ii) For $X-Z\neq0$ we get the desired formula $$p^{-1}(0:v:1)=(v^2+1:-2v:v^2-1)$$ b) And finally for $w=0$, we have $p^{-1}(0:1:0)=\omega=(1:0:1)$