提示, 本篇回答可能不是特别好理解, 考试(考研数一)不会考这么难, 可以直接看后面的例子

椭圆的一般方程如下

\begin{align} Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \end{align}

实际上, 这是所有圆锥曲线的一般方程.

现在, 假设该圆锥曲线描述的是一个椭圆, 我们要如何得到我们熟悉的椭圆标准方程

\begin{align} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{align} 的形式呢?

一种熟悉的思路, 是配方法, 但是因为配方法有时候比较难算, 我们这里采用线性代数的方法.

正交变换.

所谓正交变换, 对于本回答探讨的内容来说, 从直观上理解, 就是在不改变向量长度, 内积, 夹角和图形面积, 形状的情况下, 将非标准曲线化为标准曲线的过程.

那我们应该怎么做呢?

注意到(令 x=x_1,y=x_2 )

\begin{align} Ax^2+Bxy+Cy^2&=X^T\left[\begin{matrix}A& \frac{B}{2}\\ \frac{B}{2}&C\end{matrix}\right]X\\\\ &=X^TMX \end{align}

现欲寻找 X=QY

使得 P=Y^TQ^TMQY 为标准型

那么, 将椭圆非标准曲线化为椭圆标准曲线, 就变成了寻找正交矩阵 Q 的过程.

该过程是固定的.

先求 M 的特征值.

\begin{align} |M-\lambda E|=0 \end{align} , 解出 \lambda_1 , \lambda_2

因为 M 是实对称矩阵, 其一定有 2 个线性无关的特征向量. 分别是 a_1,a_2

因此, 将 a_1,a_2 正交化(如果 \lambda_1=\lambda_2 ), 再单位化.

得到 a_1',a_2'

得到矩阵 Q=[a_1',a_2']

所以 P=Y^TQ^TMQY 为标准型

所以正交变换后的半标准椭圆方程为 \begin{align} &\color{gray}{\text{(请左右滑动该公式)}}\\\\&\displaystyle P(x',y')+D'x'+E'y'+F'=A'x'^2+C'y'^2+D'x'+E'y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text{不要给我换行}\\ \end{align}

为什么是半标准呢? 因为目前还不是完全标准的椭圆方程.

因为是椭圆方程, P 的秩一定为 2 .

因此, 我们进一步配方. \begin{align} &\color{gray}{\text{(请左右滑动该公式)}}\\\\&\displaystyle A'\Big(x'+\frac{D'}{2}\Big)^2+C'\Big(y'+\frac{E'}{2}\Big)^2+\Big(F'-\frac{D'^2}{4}-\frac{E'^2}{2}\Big)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{不要给我换行}\\ \end{align}

令

\begin{align} F''=\Big(F'-\frac{D'^2}{4}-\frac{E'^2}{2}\Big) \end{align}

\begin{align} x''=x'+\frac{D'}{2} \end{align}

\begin{align} y''=y'+\frac{E'}{2} \end{align}

则有

\begin{align} \frac{x''}{\frac{A'}{-F''}}+\frac{y''}{\frac{C'}{-F''}}=1 \end{align}

这就是标准的椭圆方程.

(当然, 如果系数小于 0 , 就是双曲线或者虚椭圆)

目前很少见这样考的, 主要是这样考没什么考点.

但是经常见类似如下题型.

例: 求椭圆2x^2+4xy+5y^2=1的面积.

解, 设

\begin{align} A=\left[\begin{matrix}2&2\\2&5\end{matrix}\right] \end{align}

对 A 进行正交变换.

|A-\lambda E|=0\Rightarrow \lambda _1=1,\lambda _2=6

求出对应特征向量

\begin{align} a_1=[-2,1]^T \end{align}

\begin{align} a_2=[1,2]^T \end{align}

必定正交

所以单位化

\begin{align} a_1=[\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}}]^T \end{align}

\begin{align} a_2=[\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}]^T \end{align}

所以 \begin{align} Q=\left[\begin{matrix}\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\end{matrix}\right] \end{align}

所以 \begin{align} Q^TAQ=\left[\begin{matrix}1&0\\0&6\end{matrix}\right] \end{align}

所以椭圆曲线化为 x_1^2+6y_1^2

所以有椭圆面积

\begin{align} S=\pi ab=\pi\frac{\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\pi \end{align}