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Last edited by hbghlyj 2023-1-10 13:22 jstor.org/stable/pdf/2968245.pdf
A Quadratic Cremona Transformation Defined by a Conic
Author(s): Leonard E. Dickson
在任意圆锥曲线上选择四个固定点 A、B、C、D。将平面上的点 P 对应于由以下构造确定的点 R：
设 PA 和 PD 分别再次交圆锥曲线于 F 和 G ，则 R 是 BF 与 CG 的交点。
设圆锥曲线的一般方程为\begin{equation}A x^{2}+2 H x y+B y^{2}+2 G x+2 F y+C=0\label1\end{equation}
点$A(-a,0),B(0,b),C(c,0),D(0,-d)$在圆锥曲线上,我们得到\begin{array}{l}A a^{2}-2 G a+C=0 \\ B b^{2}+2 F b+C=0 \\ A c^{2}+2 G c+C=0 \\ B d^{2}-2 F d+C=0\end{array}从第一式和第三式,$2G=A(a-c)$,从第二式和第四式,$2F=-B(b-d)$.所以$C=-Aac=-Bbd$.
代入\eqref{1},除以$A$,记$h=\frac{2 H}{A}$,\begin{equation}b d x^{2}+h b d x y+a c y^{2}+b d(a-c) x+a c(d-b) y-abcd=0\label2
\end{equation}令\eqref{2}的判别式等于0得到$h=\frac{a b+c d}{b d}$或$h=-\frac{a b+c d}{b d}$.
在前一种情形,曲线方程\eqref{2}变为$(b x+c y-b c)(d x+a y+a d)=0$,代表两条直线$AD$与$BC$.我们不考虑这种平凡的情况,因平面上的每个点都对应于直线$BC$. 量$ab+cd-hbd$在下面重复出现.
在$h=-\frac{a b+c d}{b d}$的情形,$(b x-a y+a b)(d x-c y-c d)=0$,代表两条直线$AB$与$DC$.这种情况不是平凡的.
求出对应于点$P(x',y')$的点$R$的坐标$x_1,y_1$ AP的方程为$y^{\prime}(x+a)=y\left(x^{\prime}+a\right)$,DP的方程为$x^{\prime}(y+d)=x\left(y^{\prime}+d\right)$
AP和圆锥曲线的第二个交点F的坐标为$\begin{array}{l}\frac{c\left\{b d\left(x^{\prime}+a\right)^{2}+a y^{\prime}(b-d)\left(x^{\prime}+a\right)-a^{2} y^{\prime 2}\right\}}{b d\left(x^{\prime}+a\right)^{2}+h b d y^{\prime}\left(x^{\prime}+a\right)+a c y^{\prime 2}} \\ \frac{y^{\prime}\left\{a y^{\prime}(b c-c d+h b d)+b d(a+c)\left(x^{\prime}+a\right)\right\}}{b d\left(x^{\prime}+a\right)^{2}+h b d y^{\prime}\left(x^{\prime}+a\right)+a c y^{\prime 2}}\end{array}$
DP和圆锥曲线的第二个交点G的坐标为$\array{\frac{x^{\prime}\left\{x^{\prime} b d(c-a+d h)+a c(b+d)\left(y^{\prime}+d\right)\right\}}{b d x^{\prime 2}+bdh x^{\prime}\left(y^{\prime}+d\right)+a c\left(y^{\prime}+d\right)^{2}}\\
\frac{-b\left\{d^{2} x^{\prime 2}+d x^{\prime}(a-c)\left(y^{\prime}+d\right)+a c\left(y^{\prime}+d\right)^{2}\right\}}{b d x^{\prime 2}+b d h x^{\prime}\left(y^{\prime}+d\right)+a c\left(y^{\prime}+d\right)^{2}}}$
BF的方程为$\frac{x}{y-b}=\frac{-c\left(a y^{\prime}+d x^{\prime}+a d\right)\left(a y^{\prime}-b x^{\prime}-a b\right)}{d\left\{b x^{\prime}+(h b-c)y^{\prime}+a b\right\}\left\{a y^{\prime}-b x^{\prime}-a b\right\}}=\frac{-c\left(a y^{\prime}+d x^{\prime}+a d\right)}{d\left\{b x^{\prime}+(h b-c) y^{\prime}+a b\right\}}$,记为$A\over B$
CG的方程为$\frac{x-c}{y}=\frac{\left\{d x^{\prime}-c y^{\prime}-c d\right\}\left\{b(a-d h) x^{\prime}-a c y^{\prime}-a c d\right\}}{b\left(d x^{\prime}-c y^{\prime}-c d\right)\left(d x^{\prime}+a y^{\prime}+a d\right)}=\frac{b(a-d h) x^{\prime}-a c y^{\prime}-a c d}{b\left(d x^{\prime}+a y^{\prime}+a d\right)}$,记为$D\over E$
消元得,R的坐标为$x_{1}=\frac{A(b D+c E)}{A E-B D}, \quad y_{1}=\frac{E(b A+c B)}{A E-B D}$
Now $A E-B D$ expanded gives
$$
\begin{gathered}
-(a b+c d-b d h)\left(b d x^{\prime 2}+a c y^{\prime 2}+b d h x^{\prime} y^{\prime}+a b d x^{\prime}+a c d y^{\prime}\right) ; \\
\quad b D+c E=b x^{\prime}(a b+c d-b d h) ; \quad b A+c B=-c y^{\prime}(a b+c d-b d h) . \end{gathered}
$$
\begin{multline}
\therefore x_{1}=\frac{b c x^{\prime}\left(d x^{\prime}+a y^{\prime}+a d\right)}{b d x^{\prime 2}+a c y^{\prime2}+b d h x^{\prime} y^{\prime}+a b d x^{\prime}+a c d y^{\prime}} \\
y_{1}=\frac{b c y^{\prime}\left(d x^{\prime}+a y^{\prime}+a d\right)}{b d x^{\prime 2}+a c y^{\prime2}+b d h x^{\prime} y^{\prime}+a b d x^{\prime}+a c d y^{\prime}} \label3\end{multline}
Hence\begin{equation}\frac{x_{1}}{y_{1}}=\frac{x^{\prime}}{y^{\prime}}
\end{equation}
or $P R$ always passes through $O$, Pascal's Theorem for a hexagon inscribed in a conic.
Solving \eqref{3} for $x^{\prime}$ and $y^{\prime}$,
\begin{multline}\label{5}
x^{\prime}=\frac{a d x_{1}\left(-b x_{1}-c y_{1}+b c\right)}{b d x_{1}^{2}+a c y_{1}^{2}+b d h x_{1} y_{1}-a b c y_{1}-b c d x_{1}}, \\
y^{\prime}=\frac{a d y_{1}\left(-b x_{1}-c y_{1}+b c\right)}{b d x_{1}^{2}+a c y_{1}^{2}+b d h x_{1} y_{1}-a b c y_{1}-b c d x_{1}}
\end{multline}
The reciprocity between $P$ and $R$ shown by \eqref{3} and \eqref{5} is evident geometrically.
If $P$ describes a straight line $y=m x+l$, the locus of $R$ is a conic.
Substituting the value \eqref{5} in $y^{\prime}=m x^{\prime}+l$, and dropping the subscripts to the co-ordinates of $R$, we find its locus: \begin{equation}\label{6}b d x^{2}(l-m a)+d x y(a b-m a c+b l h)+a c y^{2}(l+d)-a b c y(l+d)-b c d x(l-m a)=0 \end{equation}Its discriminant is $-\frac{1}{4} a b^{2} c^{2} d l(l+d)(l-m a)(a b+c d-b d h)$. Rejecting as before the trivial case $h=\frac{a b+c d}{b d}$, this can be zero only if $l=0,-d$, or $ma$.
Hence \eqref{6} will degenerate to a pair of straight lines in just three cases:
If $l=0$, \eqref{6} becomes $a d(y-m x)(b x+c y-b c)=0$. Hence if $P$ describes a straight line through $O$, the locus of $R$ is this same line through $O$ and the line $B O$, given when $P$ is at $O$ as the indeterminate intersection of $B C$ with itself.
If $l=-d$, \eqref{6} becomes $-x\{b d x (m a+d)+d y(m a c-a b+b d h)-b c d(m a+d)\}=0$. The second factor gives the equation to the line through $C$ and the second intersection of $y=m x-d$ with the base conic \eqref{2}. Hence the transform of any line $D G$ through $D$ is $C G$ and the $y$-axis $BD$, the latter being given when $P$ is at $D$ as the intersection of $B D$ with an indeterminate line through $C$.
If $l=m a$, \eqref{6} becomes $y\{a d x(b-m c+m b h)+a c y(d+m a)-a b c(d+m a)\}=0$. The second factor gives the equation to the line through $B$ and the second intersection of $y=m(x+a)$ with the base conic \eqref{2}. Hence the transform of any line $A F$ through $A$ is $B F$ and the $x$-axis $AC$, the latter being given when $P$ is at $A$ as the intersection of $A C$ with an indeterminate line through $B$.
The conic \eqref{6} passes through the points $O, B, C$, and, since every point on the base conic is self corresponding, the points in which $y=m x+l$ intersects the base conic.
The line $B C$ whose equation is $b x+c y-b c=0$ transforms into
\begin{equation}\label7
b^{2} d x^{2}(a+c)+a c^{2} y^{2}(b+d)+b c d x y(2 a+b h)-b^{2} c d x\left(a+c\right)-a b c^{2} y\left(b+d\right)=0 \end{equation}
The tangent to it at the origin, $b d x ( a+c)+a c y(b+d)=0$, passes through the intersection $\left(\frac{a c(b+d)}{a b-dc}, \frac{-b d(a+c)}{a b-d c}\right)$ of $A D$ and $B C$. Further, \eqref{7} is tangent to the base conic \eqref{2} at the points $B$ and $C$. Thus, the tangent at $B$ to either \eqref{2} or \eqref{7} is $b d x(a-c+b h)+a c y(b+d)-a b c(b+d)=0$.
Applying \eqref{3}, the equation to the curve which transforms into the line $A D$, or $d x+a y+a d=0$, is $b d^{2} x^{2}(a+c)+a^{2} c y^{2}(b+d)+a b d x y(2 c+d h)+a b d^{2} x(a+c)+a^{2} c d y(b+d)=0$, which has the same tangent at the origin as \eqref{7}.
The line at infinity transforms by \eqref{5} into the conic. \begin{equation}\label{8}
b d x^{2}+a c y^{2}+b d h x y-b c d x-a b c y=0\end{equation}
Since \eqref{2} and \eqref{8} differ only by the linear expresssion $a d(b x+c y-b c)$, the points of intersection of $B C$ with \eqref{2} lie on \eqref{8}; also \eqref{2} and \eqref{8} are simultineously ellipses, parabolas, or hyperbolas. The discriminant of \eqref{8}, $-\frac{1}{4} a b^{2} c^{2} d(a b+c d-b d h)$, shows that it breaks up into two right lines only in trivial case above excluded.
The conic which transforms into the line at infinity, given by the vanishing of the denominator of \eqref{3}, is\begin{equation}\label{9}b d x^{2}+a c y^{2}+b d h x y+a b d x+a c d y=0\end{equation}passing through $O, A, D$. Subtracting \eqref{8} from \eqref{9}, we find their intersections lie on the line $b d x(x+c)+a c y(b+d)=0$, which passes through $K$, the intersection of $A D$ and $B C$.
Note that the conics \eqref{2}, \eqref{8} and \eqref{9} are similar.
Generally, the intersections of the conic which transforms into any straight line with the conic into which that straight line transforms lie two on the straight line $OK$ and two on the line itself, the latter two being real and distinct, coincident, or imaginary, according as the line intersects the base conic in two real, coincident, or imaginary points.
The conic which transforms into $y=m x+l$, given by substituting from \eqref{3} into $y_{1}=m x_{1}+l$, is\begin{equation}\label{10}b d x^{2}(l+m c)+a c y^{2}(l-b)+b x y(m a c-c d+l d h)
+a b d x(l+m n c)+a c d y(l-b)=0
\end{equation}
This intersects the conic \eqref{6} into which $y=m x+l$ transforms in four points, which, if we subtract \eqref{6} from \eqref{10} and factor, are seen to lie on
$$
\{a c y(b+d)+b d x(a+c) \}\{m x-y+l\}=0 \text {. }
$$
$O$ is one of the two intersections lying on $OK$. Call the other $H$. Then the point in which $y=m x+l$ meets $O K$ and the point $H$ mutually correspond. We thus have an involution marked out on $OK$.
We saw above that the points $A, D, O$ transform into the lines $A C$, $D B, B C$ respectively. Now we can prove either geometrically or analytically that the lines $A D, A O, D O$ transform into the points $O, C, B$ respectively. Thus the sides and vertices of $\triangle A D O$ transform into the vertices and sides of $\triangle O B C$. With this exception the correspondence between the points in the two systems is one to one. The projective treatment of this transformation and its dual will be given elsewhere.
[For Projective Treatment, see my paper in the Rendiconti del Circolo di Palermo.] | 
