方企勤 第七章 典型综合题分析 第9题
📝 题目
例 9 求证: $t$ 的方程
$$ \frac{{x}^{2}}{a - t} + \frac{{y}^{2}}{b - t} + \frac{{z}^{2}}{c - t} = 1\;\left( {a > b > c}\right) $$
( 1 )有三个不同实根 ${t}_{1},{t}_{2},{t}_{3}$ ,分别属于区间 $\displaystyle{- \infty < {t}_{1} < c,c < {t}_{2} <}$ $b,b < {t}_{3} < a$ ,其中(x, y, z)不在坐标平面上;
(2)过任意点的三个曲面 ${t}_{i}\left( {x,y,z}\right) =$ 常数 $\left( {i = 1,2,3}\right)$ 互相正交.
💡 答案解析
证 (1) 令 $f\left( t\right) = \frac{{x}^{2}}{a - t} + \frac{{y}^{2}}{b - t} + \frac{{z}^{2}}{c - t} - 1$ ,则
$$ \mathop{\lim }\limits_{{t \rightarrow {a}^{ - }}}f\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow {b}^{ - }}}f\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow {c}^{ - }}}f\left( t\right) = + \infty ; $$
$$ \mathop{\lim }\limits_{{t \rightarrow {b}^{ + }}}f\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow {c}^{ + }}}f\left( t\right) = - \infty ,\;\mathop{\lim }\limits_{{t \rightarrow - \infty }}f\left( t\right) = - 1; $$
$$ {f}^{\prime }\left( t\right) = \frac{{x}^{2}}{{\left( a - t\right) }^{2}} + \frac{{y}^{2}}{{\left( b - t\right) }^{2}} + \frac{{z}^{2}}{{\left( c - t\right) }^{2}} > 0\;\left( {t \neq a,b,c}\right) . $$
由连续函数中间值定理及单调性,存在惟一的 ${t}_{1} \in \left( {-\infty ,c}\right) ,{t}_{2} \in$ $\left( {c,b}\right) ,{t}_{3} \in \left( {b,a}\right)$ ,使得 $f\left( {t}_{1}\right) = 0,f\left( {t}_{2}\right) = 0,f\left( {t}_{3}\right) = 0$ .
(2)根据(1)我们有
$$ \frac{{x}^{2}}{a - {t}_{i}} + \frac{{y}^{2}}{b - {t}_{i}} + \frac{{z}^{2}}{c - {t}_{i}} = 1\;\left( {i = 1,2,3}\right) . \tag{7.24} $$
对 (7.24) 式微分得到
$$ \mathrm{d}{t}_{i} = \frac{{2x}\mathrm{\;d}x}{{T}_{i}\left( {a - {t}_{i}}\right) } + \frac{{2y}\mathrm{\;d}y}{{T}_{i}\left( {b - {t}_{i}}\right) } + \frac{{2z}\mathrm{\;d}z}{{T}_{i}\left( {c - {t}_{i}}\right) }, $$
其中 ${T}_{i}\overset{\text{ 定义 }}{ = } - \left\lbrack {\frac{{x}^{2}}{{\left( a - {t}_{i}\right) }^{2}} + \frac{{y}^{2}}{{\left( b - {t}_{i}\right) }^{2}} + \frac{{z}^{2}}{{\left( c - {t}_{i}\right) }^{2}}}\right\rbrack \left( {i = 1,2,3}\right)$ ,由此推出
$$ \frac{\partial {t}_{i}}{\partial x} = \frac{2x}{{T}_{i}\left( {a - {t}_{i}}\right) },\;\frac{\partial {t}_{i}}{\partial y} = \frac{2y}{{T}_{i}\left( {b - {t}_{i}}\right) }, $$
$$ \frac{\partial {t}_{i}}{\partial z} = \frac{2z}{{T}_{i}\left( {c - {t}_{i}}\right) }\;\left( {i = 1,2,3}\right) . $$
于是当 $i \neq j$ 时,
$$ \frac{\partial {t}_{i}}{\partial x}\frac{\partial {t}_{j}}{\partial x} + \frac{\partial {t}_{i}}{\partial y}\frac{\partial {t}_{j}}{\partial y} + \frac{\partial {t}_{i}}{\partial z}\frac{\partial {t}_{j}}{\partial z} $$
$$ = \frac{4}{{T}_{i}{T}_{j}}\left\lbrack {\frac{{x}^{2}}{\left( {a - {t}_{i}}\right) \left( {a - {t}_{j}}\right) } + \frac{{y}^{2}}{\left( {b - {t}_{i}}\right) \left( {b - {t}_{j}}\right) }}\right. $$
$$ \left. {+\frac{{z}^{2}}{\left( {c - {t}_{i}}\right) \left( {c - {t}_{j}}\right) }}\right\rbrack \text{ . } \tag{7.25} $$
再由 (7.24) 式可得
$$ \frac{{x}^{2}\left( {{t}_{i} - {t}_{j}}\right) }{\left( {a - {t}_{i}}\right) \left( {a - {t}_{j}}\right) } + \frac{{y}^{2}\left( {{t}_{i} - {t}_{j}}\right) }{\left( {b - {t}_{i}}\right) \left( {b - {t}_{j}}\right) } + \frac{{z}^{2}\left( {{t}_{i} - {t}_{j}}\right) }{\left( {c - {t}_{i}}\right) \left( {c - {t}_{j}}\right) } = 0. $$
(7.26)
(7.26)式除以 ${t}_{i} - {t}_{j} \neq 0$ ,并将所得结果代入 (7.25) 式,即得
$$ \frac{\partial {t}_{i}}{\partial x}\frac{\partial {t}_{j}}{\partial x} + \frac{\partial {t}_{i}}{\partial y}\frac{\partial {t}_{j}}{\partial y} + \frac{\partial {t}_{i}}{\partial z}\frac{\partial {t}_{j}}{\partial z} = 0\;\left( {i,j = 1,2,3;i \neq j}\right) , $$
这意味着过任意点的三曲面 ${t}_{i}\left( {x,y,z}\right) =$ 常数 $\left( {i = 1,2,3}\right)$ 的法向量 $\left( {\frac{\partial {t}_{i}}{\partial x},\frac{\partial {t}_{i}}{\partial y},\frac{\partial {t}_{i}}{\partial z}}\right)$ 两两正交,即三曲面互相正交.