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/prov … 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.