从一点作椭圆的四条法线的截距的关系

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C. Smith, An Elementary Treatise on Conic Sections, Macmillan, London, 1885. pp. 226 (EXAMPLES ON CHAPTER X.)

61. If from any point four normals be drawn to an ellipse meeting an axis in $G_1,G_2,G_3,G_4$, then will$$\frac{1}{C G_{1}}+\frac{1}{C G_{2}}+\frac{1}{C G_{3}}+\frac{1}{C G_{4}}=\frac{4}{C G_{1}+C G_{2}+C G_{3}+CG_{4}}$$

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设椭圆的方程为$$mx^2+ny^2=1$$在$Q(x_0,y_0)$处的切线$mxx_0+nyy_0=1$与$y$轴的交点的横坐标为$\frac1{ny_0}$.
设$PQ$为法线, 点$P$的坐标为$(x_P,y_P)$, 则\begin{cases}\frac{m x_0}{n y_0}=\frac{x_0-x_P}{y_0-y_P}\\mx_0^2+ny_0^2=1\end{cases}消去$x_0$得\begin{multline}y_0^4 \left(m^2 n-2 m n^2+n^3\right)+y_0^3 \left(2 m n^2 y_P-2 m^2 n y_P\right)\\+y_0^2 \left(m^2 n y_P^2-m^2+m n^2 x_P^2+2 m n-n^2\right)+y_0 \left(2 m^2 y_P-2 m n y_P\right)-m^2 y_P^2=0\end{multline}在(1)中$y_0$的四个根是$CG_1,CG_2,CG_3,CG_4$, 所以
$$\frac{1}{C G_{1}}+\frac{1}{C G_{2}}+\frac{1}{C G_{3}}+\frac{1}{C G_{4}}=\frac{2 m^2 y_P-2 m n y_P}{m^2 y_P^2}=\frac{2 (m-n)}{m y_P}$$
$$\frac{4}{C G_{1}+C G_{2}+C G_{3}+CG_{4}}=-\frac{4 \left(m^2 n-2 m n^2+n^3\right)}{2 m n^2 y_P-2 m^2 n y_P}=\frac{2 (m-n)}{m y_P}$$Q.E.D