求射影变换将二次曲线映为圆且保持顶点在P的角度不变

hbghlyj

math.stackexchange.com/questions/3608272/prove-that-the-line-xy- ... x-y-are-on-fixed-con
It's well known that there are homographies that take C to a circle.
Let's assume that we can find a specific homography T that maps angles at P to equal angles at P′. I.e. C′ is a circle, and for all points A,B

∠APB=∠A′P′B′.

T怎么构造的？

hbghlyj

过$S$作点$O$的极线的垂线，在这条直线上任取一点$T$（固定点）.
对于二次曲线上每个点$P$，直线$TP$与点$O$的极线交于$z$，过$T$作$SP$的平行线与直线$Sz$交于$p$.
则当点$P$在二次曲线上运动时，点$p$的轨迹为圆。
$P\mapsto p$就是所求的映射：因为$Tp$总是平行于$SP$.

右边是 Frégier's Theorem的证明：$\angle\mathrm{PSP'}=90$°
Then $\mathrm{P P}^{\prime}$ and $\mathrm{QQ}^{\prime}$ project into diameters $p p^{\prime}$ and $q q^{\prime}$, and, since the angles $p \mathrm{T} p^{\prime}$ and $q \mathrm{T} q^{\prime}$ are right angles, the three diameters $o p, o q, o\mathrm T$ are all equal; hence the conic projects into a circle; and since every diameter of the circle subtends a right angle at $\mathrm T$, therefore every chord of the conic through $\mathrm O$ subtends a right angle at $\mathrm S$. Again, since $o\mathrm T$ is perpendicular to the tangent at $\mathrm T$, therefore $\rm OS$ is perpendicular to the tangent at $\mathrm{S}$; that is, $\mathrm{OS}$ is normal.

Corollary.—Hence any conic can be projected into a circle by taking a point on its circumference as focus of projection, and the tangent and normal at that point project into tangent and normal.

hbghlyj

如果去掉“将二次曲线映为圆”，只看“保持顶点在S的角度不变”，那它应该是形如$\pmatrix{\cosα&-\sinα&0\\\sinα&\cosα&0\\A&B&C}$
旋转一个角$\alpha$，变成“保持过S的所有直线不变”，问题变成：
T保持S且保持过S的所有直线不变，则T形如$\pmatrix{1&0&0\\0&1&0\\A&B&C}$.

可以验证给出的变换确实保持过S[0,0,1]的所有直线不变：

1. {u, v, 0} . Adjugate[{{1, 0, 0}, {0, 1, 0}, {A, B, C}}]

Copy the Code

{C u, C v, 0}

可以直接计算来证明。把问题和证明又发到了math.stackexchange.com/q/4739105

hbghlyj

hbghlyj 发表于 2024-3-12 13:56
把问题和证明又发到了math.stackexchange.com/q/4739105
Given a fixed line $L$ and two fixed points $S,S'$.
For any point $P$ on the plane, let the line $PS'$ intersect $L$ at $Z$.
Draw the line through $S'$ parallel to $SP$, intersecting the line $SZ$ at $P'$.
The map $P\mapsto P'$ maps lines through $S$ to parallel lines through $S'$: the line $S'P'$ is parallel to $SP$.

当$S=S'$时，这样构造出来的$P$是直线SP上满足$ZP':ZS=ZS:ZP$的点
这只是“保持过S的所有直线不变”的射影变换的一部分（这些是对合，所以这些是C=-1的情况）。
如何构造出一般的“保持过S的所有直线不变”的射影变换$\pmatrix{1&0&0\\0&1&0\\A&B&C}$呢？
可以把对合右乘一个“关于S缩放$-C^{-1}$倍”：
$\pmatrix{1&0&0\\0&1&0\\A&B&C}=\pmatrix{1&0&0\\0&1&0\\A&B&-1}\pmatrix{1&0&0\\0&1&0\\0&0&-C}$
这样就能构造出所有的“保持过S的所有直线不变”的射影变换。