已知圆锥曲线方程求焦点的方法

hejoseph

以下令
\begin{align*}
f(x,y)&=a_{11}x^2+2a_{12}xy+a_{22}y^2+2a_{13}x+2a_{23}y+a_{33}，\\
f_1(x,y)&=a_{11}x+a_{12}y+a_{13}，\\
f_2(x,y)&=a_{12}x+a_{22}y+a_{23}，\\
f_3(x,y)&=a_{13}x+a_{23}y+a_{33}，\\
I_1&=a_{11}+a_{22}，\\
I_2&=a_{11}a_{22}-a_{12}^2，\\
I_3&=-a_{11}a_{23}^2-a_{22}a_{13}^2-a_{12}^2a_{33}+a_{11}a_{22}a_{33}+2a_{12}a_{13}a_{23}。
\end{align*}

若已知圆锥曲线方程 $f(x,y)=0$，设焦点是 $(x_0,y_0)$，则过点 $(x_0,y_0)$ 圆锥曲线 $f(x,y)=0$ 的切线（此时是虚直线）方程是
\begin{align*}
&f(x_0,y_0)f(x,y)-(f_1(x_0,y_0)x+f_2(x_0,y_0)y+f_3(x_0,y_0))^2\\
={}&\left(a_{11}f(x_0,y_0)-f_1^2(x_0,y_0)\right)x^2+2\left(a_{12}f(x_0,y_0)-f_1(x_0,y_0)f_2(x_0,y_0)\right)xy\\
&{}+\left(a_{22}f(x_0,y_0)-f_2^2(x_0,y_0)\right)y^2\\
&{}+2\left(a_{13}f(x_0,y_0)-f_1(x_0,y_0)f_3(x_0,y_0)\right)x+2\left(a_{23}f(x_0,y_0)-f_2(x_0,y_0)f_3(x_0,y_0)\right)y\\
&{}+a_{33}f(x_0,y_0)-f_3^2(x_0,y_0)，
\end{align*}
这个方程是一个圆点的方程，解方程组
\[
\left\{
\begin{aligned}
&a_{11}f(x,y)-f_1^2(x,y)=a_{22}f(x,y)-f_2^2(x,y)，\\
&a_{12}f(x,y)-f_1(x,y)f_2(x,y)=0，
\end{aligned}
\right.
\]
求出的点 $(x,y)$ 即为圆锥曲线的焦点。求出焦点后，再求焦点的极线方程，极线方程就是圆锥曲线的准线方程。下面讨论方程组
\[
\left\{
\begin{aligned}
&a_{11}f(x,y)-f_1^2(x,y)=a_{22}f(x,y)-f_2^2(x,y)，\\
&a_{12}f(x,y)-f_1(x,y)f_2(x,y)=0，
\end{aligned}
\right.
\]
的解法。由
\[
a_{12}f(x,y)-f_1(x,y)f_2(x,y)=0
\]
求得
\begin{equation}
y=-\frac{(a_{11}a_{23}-a_{12}a_{13})x-(a_{12}a_{33}-a_{13}a_{23})}{(a_{11}a_{22}-a_{12}^2)x-(a_{12}a_{23}-a_{13}a_{22})}，\label{eqnjdzjzby}
\end{equation}
代入
\[
a_{11}f(x,y)-f_1^2(x,y)=a_{22}f(x,y)-f_2^2(x,y)
\]
整理得，当 $I_2=0$ 时
\begin{align*}
&2I_1(a_{12}a_{23}-a_{13}a_{22})(a_{11}a_{23}^2+a_{22}a_{13}^2-2a_{12}a_{13}a_{23})x\\
&{}-\left(I_3^2+I_1(a_{22}a_{33}-a_{23}^2)(a_{11}a_{23}^2+a_{22}a_{13}^2-2a_{12}a_{13}a_{23})\right)=0，
\end{align*}
当$I_2\neq 0$时
\[
\left(x-\frac{a_{12}a_{23}-a_{13}a_{22}}{I_2}\right)^4-\frac{I_3(a_{11}-a_{22})}{I_2^2}\left(x-\frac{a_{12}a_{23}-a_{13}a_{22}}{I_2}\right)^2-\frac{a_{12}^2I_3^2}{I_2^4}=0，
\]
求出 $x$ 的值后代入 \eqref{eqnjdzjzby} 便可得 $y$ 值，这样就能求得焦点坐标。

hbghlyj

参见圆环点，迷向直线和拉格尔定理“例 2 利用圆环点的性质求二次曲线的焦点公式.”

hejoseph

将上面的方法转化成重心坐标的方法：

以下设 $\triangle ABC$ 中 $BC=a$，$CA=b$，$AB=c$，面积是$S$，$\overline{S_{\triangle XYZ}}$ 表示 $\triangle XYZ$ 的有向面积，轨迹方程上动点的重心或三线坐标是 $\alpha:\beta:\gamma$，
\begin{align*}
f(\alpha,\beta,\gamma)&=t_{11}\alpha^2+t_{22}\beta^2+t_{33}\gamma^2+2t_{12}\alpha\beta+2t_{13}\alpha\gamma+2t_{23}\beta\gamma，\\
f_1(\alpha,\beta,\gamma)&=t_{11}\alpha+t_{12}\beta+t_{13}\gamma，\\
f_2(\alpha,\beta,\gamma)&=t_{12}\alpha+t_{22}\beta+t_{23}\gamma，\\
f_3(\alpha,\beta,\gamma)&=t_{13}\alpha+t_{23}\beta+t_{33}\gamma，
\end{align*}
容易验证有如下关系：
\begin{align*}
&\alpha f_1(\alpha,\beta,\gamma)+\beta f_2(\alpha,\beta,\gamma)+\gamma f_3(\alpha,\beta,\gamma)=f(\alpha,\beta,\gamma)，\\
&\alpha_1 f_1(\alpha_2,\beta_2,\gamma_2)+\beta_1 f_2(\alpha_2,\beta_2,\gamma_2)+\gamma_1 f_3(\alpha_2,\beta_2,\gamma_2)\\
={}&\alpha_2 f_1(\alpha_1,\beta_1,\gamma_1)+\beta_2 f_2(\alpha_1,\beta_1,\gamma_1)+\gamma_2 f_3(\alpha_1,\beta_1,\gamma_1)。
\end{align*}

重心坐标方程是 $f(\alpha,\beta,\gamma)=0$ 的二次曲线的焦点的重心坐标是 $\alpha_0:\beta_0:\gamma_0$，则 $\alpha_0:\beta_0:\gamma_0$ 是方程组
\begin{align*}
&\frac{(t_{11}+t_{22}-2t_{12})f(\alpha,\beta,\gamma)-(f_1(\alpha,\beta,\gamma)-f_2(\alpha,\beta,\gamma))^2}{c^2}\\
={}&\frac{(t_{11}+t_{33}-2t_{13})f(\alpha,\beta,\gamma)-(f_1(\alpha,\beta,\gamma)-f_3(\alpha,\beta,\gamma))^2}{b^2}\\
={}&\frac{(t_{22}+t_{33}-2t_{23})f(\alpha,\beta,\gamma)-(f_2(\alpha,\beta,\gamma)-f_3(\alpha,\beta,\gamma))^2}{a^2}
\end{align*}
的解，从上面的方程组消去 $f(\alpha,\beta,\gamma)$，得到的方程就
\begin{align*}
&\left((t_{11}+t_{33}-2t_{13})a^2-(t_{11}+t_{22}-2t_{12})b^2\right)(f_1(\alpha,\beta,\gamma)-f_2(\alpha,\beta,\gamma))^2\\
&{}+\left((t_{11}+t_{22}-2t_{12})b^2-(t_{11}+t_{33}-2t_{13})c^2\right)(f_2(\alpha,\beta,\gamma)-f_3(\alpha,\beta,\gamma))^2\\
&{}+\left((t_{22}+t_{33}-2t_{23})c^2-(t_{11}+t_{22}-2t_{12})a^2\right)(f_3(\alpha,\beta,\gamma)-f_1(\alpha,\beta,\gamma))^2=0
\end{align*}
是二次曲线 $f(\alpha,\beta,\gamma)=0$ 主轴的重心坐标方程。

hejoseph

用这个结论可以推得以前很久都得不到的结论：四边形的外接圆锥曲线的焦点的轨迹一般是六次曲线。平行四边形外接圆锥曲线的焦点的轨迹一般是三次曲线。