设
\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},
\end{align*}
点 $P(x_0,y_0)$ 是定点,点 $A$、$B$ 是给定的圆锥曲线 $f(x,y)=0$ 上的两动点,且满足 $PA\perp PB$,点 $Q$ 是直线 $AB$ 上的点,满足 $PQ\perp AB$,则 $Q$ 的轨迹方程是
\[
(a_{11}+a_{22})((x-x_0)^2+(y-y_0)^2)+2(f_1(x_0,y_0)x+f_2(x_0,y_0)y+f_3(x_0,y_0))-f(x_0,y_0)=0。
\]
设
\begin{align*}
f(x,y,z)&=a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+2a_{23}yz+2a_{14}x+2a_{24}y+2a_{34}z+a_{44},\\
f_1(x,y,z)&=a_{11}x+a_{12}y+a_{13}z+a_{14},\\
f_2(x,y,z)&=a_{12}x+a_{22}y+a_{23}z+a_{24},\\
f_3(x,y,z)&=a_{13}x+a_{23}y+a_{33}z+a_{34},\\
f_4(x,y,z)&=a_{14}x+a_{24}y+a_{34}z+a_{44},
\end{align*}
点 $P(x_0,y_0,z_0)$ 是定点,点 $A$、$B$、$C$ 是二次曲面 $f(x,y,z)=0$ 上的动点,且 $AP$、$BP$、$CP$ 两两互相垂直,点 $Q$ 是过点 $P$ 作平面 $ABC$ 垂线的垂足,则点 $Q$ 的轨迹方程是
\begin{align*}
&(a_{11}+a_{22}+a_{33})((x-x_0)^2+(y-y_0)^2+(z-z_0)^2)\\
&{}+2(f_1(x_0,y_0,z_0)x+f_2(x_0,y_0,z_0)y+f_3(x_0,y_0,z_0)z+f_4(x_0,y_0,z_0))-f(x_0,y_0,z_0)=0。
\end{align*} |