方企勤 第五章 多元函数微分学 第7题
📝 题目
例 7 求椭圆 $5{x}^{2} + {4xy} + 2{y}^{2} = 1$ 的长半轴 $a$ 和短半轴 $b$ .
💡 答案解析
解 问题可转化为求函数 $f\left( {x,y}\right) = {x}^{2} + {y}^{2}$ 在条件 $5{x}^{2} + {4xy} +$ $2{y}^{2} = 1$ 下的极值. 令
$$ \Phi \left( {x,y}\right) = {x}^{2} + {y}^{2} - \lambda \left( {5{x}^{2} + {4xy} + 2{y}^{2} - 1}\right) , $$
则
$$ \left\{ \begin{array}{l} {\Phi }_{x}^{\prime } = {2x} - {10\lambda x} - {4\lambda y} = 0, \\ {\Phi }_{y}^{\prime } = {2y} - {4\lambda x} - {4\lambda y} = 0, \end{array}\right. $$
化简得
$$ \left\{ \begin{array}{l} \left( {1 - {5\lambda }}\right) x - {2\lambda y} = 0, \\ - {2\lambda x} + \left( {1 - {2\lambda }}\right) y = 0. \end{array}\right. \tag{4.14} $$ (4. 15)
要上述方程组有非零解 (因从实际意义看条件极值存在, 即方程组一定有解), 其系数行列式必须为零, 即
$$ \left| \begin{matrix} 1 - {5\lambda } & - {2\lambda } \\ - {2\lambda } & 1 - {2\lambda } \end{matrix}\right| = 0, $$
由此得 $1 - {7\lambda } + 6{\lambda }^{2} = 0$ ,解出 ${\lambda }_{1} = 1,{\lambda }_{2} = 1/6$ . 设对应于 ${\lambda }_{i}$ ,方程组的解为 $\left( {{x}_{i},{y}_{i}}\right) \left( {i = 1,2}\right)$ ,把它们代入方程组,且 (4.14) 式 $\times {x}_{i} + \left( {4.15}\right)$ 式 $\times {y}_{i}$ ,得
$$ {x}_{i}^{2} + {y}_{i}^{2} - {\lambda }_{i}\left( {5{x}_{i}^{2} + 4{x}_{i}{y}_{i} + 2{y}_{i}^{2}}\right) = 0, $$
即得
$$ \left\{ \begin{array}{l} {x}_{1}^{2} + {y}_{1}^{2} = {\lambda }_{1} = 1, \\ {x}_{2}^{2} + {y}_{2}^{2} = {\lambda }_{2} = 1/6. \end{array}\right. $$
所以长半轴 $a = \sqrt{{x}_{1}^{2} + {y}_{1}^{2}} = 1,b = \sqrt{{x}_{2}^{2} + {y}_{2}^{2}} = 1/\sqrt{6}$ .