谢惠民下册P238第21章的一个参考题:

证明与曲面$ax^2+by^2+cz^2=1(abc\neq0)$相切的三个互相垂直的平面的交点在球面$x^2+y^2+z^2=\frac1a+\frac1b+\frac1c$上.

证:(幸子)椭球面方程为$ax^2+by^2+cz^2=1(abc\neq0)$,则法向量${n_i} = \left( {a{x_i},b{y_i},c{z_i}} \right)$,切平面方程为$a{x_i}x + b{y_i}y + c{z_i}z = 1$.

设三个切点分别为${\alpha _i}\left( {{x_i},{y_i},{z_i}} \right)\left( {i = 1,2,3} \right)$,三平面交点为$(x,y,z)$.由三平面垂直可知

\[\overrightarrow {{n_i}} \cdot \overrightarrow {{n_j}} = {a^2}{x_i}{x_j} + {b^2}{y_i}{y_j} + {c^2}{z_i}{z_j} = 0\left( {i \ne j} \right).\]

原点到三切平面的距离分别为

\[\frac{1}{{\sqrt {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} }}\left( {i = 1,2,3} \right).\]

由几何关系(考虑长方体对角线)可知

\[{x^2} + {y^2} + {z^2} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} .\]

设$\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)$,则对任意的$i = 1,2,3$,有

\[1 = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right) = {t_i}\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right).\]

并且

\begin{align*}&{x^2} + {y^2} + {z^2} = \left( {\begin{array}{*{20}{c}}x&y&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{t_1}}&{{t_2}}&{{t_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{b{y_1}}&{c{z_1}}\\{a{x_2}}&{b{y_2}}&{c{z_2}}\\{a{x_3}}&{b{y_3}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)\\=& \sum\limits_{i = 1}^3 {{t_i}^2\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} = \sum\limits_{i = 1}^3 {\frac{{a{x_i}^2 + b{y_i}^2 + c{z_i}^2}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} \\=& a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right).\end{align*}

注意到对任意的$w$,记$w = \sum\limits_{i = 1}^3 {{s_i}{n_i}} $,则${s_i} = \frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}$,即\[w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} .\]

分别令$w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} $,得

\begin{align*}a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{a},\\b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{b},\\c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{c}.\end{align*}

因此有\[{x^2} + {y^2} + {z^2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.\]

解法二.记$d_i=\sqrt{a^2x_i^2+b^2y_i^2+c^2z_i^2},\ i=1,2,3$,则$\begin{pmatrix}\frac{ax_1}{d_1}&\frac{by_1}{d_1}&\frac{cz_1}{d_1}\\\frac{ax_2}{d_2}&\frac{by_2}{d_2}&\frac{cz_2}{d_2}\\\frac{ax_3}{d_3}&\frac{by_3}{d_3}&\frac{cz_3}{d_3}\end{pmatrix} $是正交矩阵,从而有

\[\frac{a^2x_1^2}{d_1^2}+\frac{a^2x_2^2}{d_2^2}+\frac{a^3x_3^2}{d_3^2}=1\Rightarrow \frac{ax_1^2}{d_1^2}+\frac{ax_2^2}{d_2^2}+\frac{ax_3^2}{d_3^2}=\frac{1}{a}.\]

类似得到另外两式, 相加便有

\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ax_1^2+by_1^2+cz_1^2}{d_1^2}+\frac{ax_2^2+by_2^2+cz_2^2}{d_2^2}+\frac{ax_3^2+by_3^2+cz_3^2}{d_3^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.\]

第四届全国大学生数学竞赛决赛（数学组）解析几何试题

设$A$为正整数，直线$L$与双曲线$x^2-y^2=2(x>0)$所围成的面积为$A$，证明：

(1)上述$L$被双曲线$x^2-y^2=2(x>0)$所截线段的中点的轨迹为双曲线；

(2)$L$总是(1)中轨迹曲线的切线.

证明. (1)不妨设直线$L$的方程为$x=my+l(m^2<1)$，直线$L$与双曲线$x^2-y^2=2(x>0)$的交点$P,Q$分别为$(x_1,y_1),(x_2,y_2)$$(\text{其中}y_1<y_2)$,

联立方程，有\[\left\{ \begin{array}{l}x = my + l\\{x^2} - {y^2} = 2\end{array} \right. \Rightarrow \left( {{m^2} - 1} \right){y^2} + 2mly + {l^2} - 2 = 0.\]

由韦达定理，我们有：\[{y_1} + {y_2} = \frac{{2ml}}{{1 - {m^2}}},{y_1}{y_2} = \frac{{{l^2} - 2}}{{{m^2} - 1}},{y_2} - {y_1} = \sqrt {{{\left( {{y_2} + {y_1}} \right)}^2} - 4{y_2}{y_1}}  = \frac{{2\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.\]

由题意得：

即\[x^2-y^2=C(C>2).\]为双曲线轨迹.

(2)再之，由\[2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y} = \frac{1}{m} = {k_L}\]及直线$L$经过点$M$可知，直线$L$总为$M$的轨迹曲线的切线.