第七章 典型综合题分析 · 第30题

解法 1 由曲线方程解出

$$ {y}_{ \pm } = - \frac{Bx}{C} \pm \frac{1}{C}\sqrt{C - \left( {{AC} - {B}^{2}}\right) {x}^{2}}. $$

再由 ${y}_{ + } = {y}_{ - }$ ,得出曲线 $y = {y}_{ + }\left( x\right)$ 与曲线 $y = {y}_{ - }\left( x\right)$ 的两个交点的横坐标为

$$ {x}_{ \pm } = \pm \sqrt{C}/D,\;D\overset{\text{ 记为 }}{ = }\sqrt{{AC} - {B}^{2}}. $$

于是平面曲线 $A{x}^{2} + {2Bxy} + C{y}^{2} = 1$ 所围图形的面积,就是曲线 $y =$ ${y}_{ + }\left( x\right)$ 与曲线 $y = {y}_{ - }\left( x\right)$ 所围图形的面积,此面积

$$ S = {\int }_{{x}_{ - }}^{{x}_{ + }}\left( {{y}_{ + }\left( x\right) - {y}_{ - }\left( x\right) }\right) \mathrm{d}x $$

$$ = \frac{2}{C}{\int }_{-\sqrt{c}/D}^{\sqrt{c}/D}\sqrt{C - {D}^{2}{x}^{2}}\mathrm{\;d}x\;\left( {\text{ 令 }u = {Dx}}\right) $$

$$ = \frac{4}{CD}{\int }_{0}^{\sqrt{c}}\sqrt{C - {u}^{2}}\mathrm{\;d}u = \frac{4}{CD} \cdot \frac{\pi }{4}{\left( \sqrt{C}\right) }^{2} $$

$$ = \frac{\pi }{D} = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$

解法 2 所给的椭圆在极坐标下的方程为

$$ {r}^{2} = \frac{1}{A{\cos }^{2}\theta + {2B}\cos \theta \sin \theta + C{\sin }^{2}\theta }, $$

所以椭圆的面积为

$$ S = \frac{1}{2}{\int }_{0}^{2\pi }{r}^{2}\mathrm{\;d}\theta = \frac{1}{2}{\int }_{0}^{2\pi }\frac{\mathrm{d}\theta }{A{\cos }^{2}\theta + {2B}\cos \theta \sin \theta + C{\sin }^{2}\theta } $$

$$ = \frac{1}{2}\left\lbrack {{\int }_{0}^{\pi /2} + {\int }_{\pi /2}^{\pi } + {\int }_{\pi }^{{3\pi }/2} + {\int }_{{3\pi }/2}^{2\pi }}\right\rbrack \frac{\mathrm{{dt}}\tan \theta }{A + {2B}\tan \theta + C{\tan }^{2}\theta } $$

$$ = \frac{1}{2}\left\lbrack {{\int }_{0}^{\pi /2} + {\int }_{\pi /2}^{\pi } + {\int }_{\pi }^{{3\pi }/2} + {\int }_{{3\pi }/2}^{2\pi }}\right\rbrack \times \mathrm{d}\frac{1}{\sqrt{{AC} - {B}^{2}}}{\tan }^{-1}\frac{C\tan \theta + B}{\sqrt{{AC} - {B}^{2}}} $$

$$ = \frac{1}{2\sqrt{{AC} - {B}^{2}}}\left\lbrack {\frac{\pi }{2} - \arctan \frac{B}{\sqrt{{AC} - {B}^{2}}} + \arctan \frac{B}{\sqrt{{AC} - {B}^{2}}}}\right. $$

$$ \left. {+\frac{\pi }{2} + \frac{\pi }{2} - \arctan \frac{B}{\sqrt{{AC} - {B}^{2}}} + \arctan \frac{B}{\sqrt{{AC} - {B}^{2}}} + \frac{\pi }{2}}\right\rbrack $$

$$ = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$

解法 3 设所给椭圆上的点(x, y)到原点的距离为 $d$ ,则 ${d}^{2} =$ ${x}^{2} + {y}^{2}$ ,考虑在条件 $A{x}^{2} + {2Bxy} + C{y}^{2} = 1$ 下求 $d$ 的极值. 令

$$ F\left( {x,y}\right) \overset{\text{ 定义 }}{ = }{x}^{2} + {y}^{2} + \lambda \left( {A{x}^{2} + {2Bxy} + C{y}^{2} - 1}\right) , $$

$$ \left\{ \begin{array}{l} {F}_{x}^{\prime } = \left( {2 + {2A\lambda }}\right) x + {2B\lambda y} = 0, \\ {F}_{y}^{\prime } = {2B\lambda x} + \left( {2 + {2C\lambda }}\right) y = 0. \end{array}\right. \tag{7.52} $$ (7.53)

将 (7.52) 和 (7.53) 式看成未知数为 $x,y$ 的齐次线性方程组,它有非零解,必须有系数行列式为 0 ,即得

$$ \left( {{AC} - {B}^{2}}\right) {\lambda }^{2} + \left( {A + C}\right) \lambda + 1 = 0. \tag{7.54} $$

这是 $\lambda$ 的二次方程,设它的两根为 ${\lambda }_{1},{\lambda }_{2}$ ,显然 ${\lambda }_{1} < 0,{\lambda }_{2} < 0$ . 方程 (7.52) 乘以 $x$ 加上方程 (7.53) 乘以 $y$ ,得到

$$ {x}^{2} + {y}^{2} + \lambda \left( {A{x}^{2} + {2Bxy} + C{y}^{2}}\right) = 0, $$

即得 $d = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{-\lambda }$ . 于是由 (7.54) 式,有

$$ \min d \cdot \max d = \sqrt{{\lambda }_{1}{\lambda }_{2}} = 1/\sqrt{{AC} - {B}^{2}}. $$

因为 $\displaystyle{\max d}$ 和 $\displaystyle{\min d}$ 分别表示椭圆的长半轴和短半轴,所以所求的椭圆面积为

$$ S = \pi /\sqrt{{AC} - {B}^{2}}. $$

解法 4 用极坐标: $x = r\cos \theta ,y = r\sin \theta$ . 所给椭圆在极坐标下的方程为

$$ {r}^{2} = \frac{1}{A{\cos }^{2}\theta + B\sin {2\theta } + C{\sin }^{2}\theta } = \frac{1}{\left( {A + C}\right) /2 + h\sin \left( {{2\theta } + \varphi }\right) }, $$

(7.55)

其中 $h = \sqrt{{B}^{2} + {\left( \frac{A - C}{2}\right) }^{2}},\varphi \overset{\text{ 定义 }}{ = }\arccos \frac{B}{h}$ . 由 (7.55) 式易见

$$ \max {r}^{2} = \frac{1}{\left( {A + C}\right) /2 - h},\;\min {r}^{2} = \frac{1}{\left( {A + C}\right) /2 + h}. $$

于是所求椭圆面积

$$ S = \pi \max r \cdot \min r = \pi /\sqrt{{\left( \frac{A + C}{2}\right) }^{2} - {h}^{2}} = \pi /\sqrt{{AC} - {B}^{2}}. $$

解法 5 将所给曲线方程配方, 得

$$ A{\left( x + \frac{B}{A}y\right) }^{2} + \left( {C - \frac{{B}^{2}}{A}}\right) {y}^{2} = 1. $$

作变换 $u = \sqrt{A}\left( {x + \frac{B}{A}y}\right) ,v = \sqrt{C - \frac{{B}^{2}}{A}}y$ ,则有

$$ \frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) } = \sqrt{{AC} - {B}^{2}} \Rightarrow \frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) } = \frac{1}{\sqrt{{AC} - {B}^{2}}}. $$

于是所求面积

$$ S = {\iint }_{A{x}^{2} + {2Bxy} + C{y}^{2} \leq 1}\mathrm{\;d}x\mathrm{\;d}y = {\iint }_{{u}^{2} + {v}^{2} \leq 1}\frac{1}{\sqrt{{AC} - {B}^{2}}}\mathrm{\;d}u\mathrm{\;d}v = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$

解法 6 考虑过原点的直线 $y = {kx}$ ,它与所给椭圆有两个交点, 设它们的坐标分别为 $\left( {{x}_{1},k{x}_{1}}\right) ,\left( {{x}_{2},k{x}_{2}}\right)$ ,这两个交点之间的距离为

$$ d = \sqrt{1 + {k}^{2}}\left| {{x}_{2} - {x}_{1}}\right| . $$

将 $y = {kx}$ 代入椭圆方程,得 ${x}^{2}\left( {A + {2Bk} + C{k}^{2}}\right) = 1$ . 由此解得

$$ {x}_{1} = - {x}_{2} = \frac{1}{\sqrt{A + {2Bk} + C{k}^{2}}}, $$

从而

$$ d = 2\sqrt{1 + {k}^{2}}/\sqrt{A + {2Bk} + C{k}^{2}}. $$

对任意实数 $k$ ,令

$$ \lambda \left( k\right) = {\left( \frac{d}{2}\right) }^{2} = \frac{1 + {k}^{2}}{A + {2Bk} + C{k}^{2}}, \tag{7.56} $$

$$ {\lambda }^{\prime }\left( k\right) = \frac{{2k}\left( {A + {2Bk} + C{k}^{2}}\right) - \left( {1 + {k}^{2}}\right) \left( {{2B} + {2Ck}}\right) }{{\left( A + 2Bk + C{k}^{2}\right) }^{2}} $$

$$ = \frac{2\left\lbrack {B{k}^{2} + \left( {A - C}\right) k - B}\right\rbrack }{{\left( A + 2Bk + C{k}^{2}\right) }^{2}}\overset{\text{ 令 }}{ = }0. $$

由于 $B = 0$ 时结论显然成立,我们只需讨论 $B \neq 0$ 的情形. 这时上式变为

$$ B{k}^{2} + \left( {A - C}\right) k - B = 0. \tag{7.57} $$

设方程 (7.57) 的解为 ${k}_{1},{k}_{2}$ . 由 (7.56) 式及 ${\lambda }^{\prime }\left( {k}_{i}\right) = 0$ ,我们得到

$$ \lambda \left( {k}_{i}\right) = \frac{1 + {k}_{i}^{2}}{A + {2B}{k}_{i} + C{k}_{i}^{2}} = \frac{{k}_{i}}{B + C{k}_{i}}\;\left( {i = 1,2}\right) . $$

进一步再由 (7.57) 式得到

$$ S = \pi \sqrt{\lambda \left( {k}_{1}\right) \lambda \left( {k}_{2}\right) } = \pi \sqrt{\frac{{k}_{1} \cdot {k}_{2}}{{B}^{2} + {CB}\left( {{k}_{1} + {k}_{2}}\right) + {C}^{2}{k}_{1} : {k}_{2}}} $$

$$ = \pi \sqrt{\frac{-1}{{B}^{2} + {CB} \cdot \frac{C - A}{B} - {C}^{2}}} = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$

解法 7 因为 $A > 0,{AC} - {B}^{2} > 0$ ,所以 $\left\lbrack \begin{array}{ll} A & B \\ B & C \end{array}\right\rbrack$ 是正定矩阵,从而存在正定矩阵 $\left\lbrack \begin{array}{ll} p & q \\ q & r \end{array}\right\rbrack$ ,满足

$$ {\left\lbrack \begin{array}{ll} p & q \\ q & r \end{array}\right\rbrack }^{2} = \left\lbrack \begin{array}{ll} A & B \\ B & C \end{array}\right\rbrack $$

令 $\left\lbrack \begin{array}{l} u \\ v \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} p & q \\ q & r \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack$ ,则有

$$ A{x}^{2} + {2Bxy} + C{y}^{2} = \left( \begin{array}{ll} x & y \end{array}\right) \left\lbrack \begin{array}{ll} A & B \\ B & C \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack $$

$$ = \left( \begin{array}{ll} x & y \end{array}\right) \left\lbrack \begin{array}{ll} p & q \\ q & r \end{array}\right\rbrack \left\lbrack \begin{array}{ll} p & q \\ q & r \end{array}\right\rbrack \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack = {u}^{2} + {v}^{2}, $$

$$ \text{ 且 }\;\frac{\partial \left( {x,y}\right) }{\partial \left( {u,v}\right) } = {\left\lbrack \frac{\partial \left( {u,v}\right) }{\partial \left( {x,y}\right) }\right\rbrack }^{-1} = {\left| \begin{array}{ll} p & q \\ q & r \end{array}\right| }^{-1} = {\left| \begin{array}{ll} A & B \\ B & C \end{array}\right| }^{-1/2} $$

$$ = \frac{1}{\sqrt{{AC} - {B}^{2}}}. $$

于是所求椭圆面积

$$ S = {\iint }_{A{x}^{2} + {2Bxy} + C{y}^{2} \leq 1}\mathrm{\;d}x\mathrm{\;d}y = {\iint }_{{u}^{2} + {v}^{2} \leq 1}\frac{1}{\sqrt{{AC} - {B}^{2}}}\mathrm{\;d}u\mathrm{\;d}v = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$

解法 8 前半部分同解法 6 一样,考虑过原点的直线 $y = {kx}$ ,它与椭圆 $A{x}^{2} + {2Bxy} + C{y}^{2} = 1$ 有两个交点,它们之间距离为 $d$ ,则

$$ \lambda \left( k\right) = {\left( \frac{d}{2}\right) }^{2} = \frac{1 + {k}^{2}}{A + {2Bk} + C{k}^{2}}. \tag{7.58} $$

如果 $\lambda \left( k\right)$ 不是极值,那么由椭圆图形的对称性容易看出,一定有两个不同的 $k$ 取到同一个 $\lambda \left( k\right)$ 的值; 如果 $\lambda \left( k\right)$ 是极值,那么就只有一个 $k$ 取到该 $\lambda \left( k\right)$ 的值. 由此可见,对应于极值的 $\lambda$ ,由 (7.58) 式改写成的 $k$ 的二次方程

$$ \left( {{C\lambda } - 1}\right) {k}^{2} + {2B\lambda k} + \left( {{A\lambda } - 1}\right) = 0 $$

的判别式应等于零,即 $- \left( {{AC} - {B}^{2}}\right) {\lambda }^{2} + \left( {A + C}\right) \lambda - 1 = 0$ . 这是 $\lambda$ 的二次方程,设它的解为 ${\lambda }_{1},{\lambda }_{2}$ ,显然它们都是 (7.58) 式所定义的 $\lambda$ 的极值. 于是求出椭圆的长、短半轴为 $\sqrt{{\lambda }_{1}},\sqrt{{\lambda }_{2}}$ ,因此所求面积为

$$ S = \pi \sqrt{{\lambda }_{1} \cdot {\lambda }_{2}} = \pi \sqrt{\frac{-1}{-\left( {{AC} - {B}^{2}}\right) }} = \frac{\pi }{\sqrt{{AC} - {B}^{2}}}. $$