从$O$点发出两条射线${l_1},{l_2}$,已知直线$l$分别交${l_1},{l_2}$于$A,B$两点,且${S_{\triangle OAB}} = c$($c$为定值),记$AB$中点为$D$,$D$随着$A,B$的运动构成轨迹$\Gamma$.
求证:(1)$D$的轨迹$\Gamma$关于$\angle AOB$的角平分线反射对称;
(2)轨迹$\Gamma$为双曲线.
分析与解 以${l_1},{l_2}$的角平分线所在直线为$x$轴建立如图所示的直角坐标系.
设$\angle AOx = \angle BOx = \alpha $,$OA = a,OB = b$,$X\left( {x,y} \right)$,
则$${S_{\Delta OAB}} = \dfrac{1}{2}ab\sin 2\alpha = c,$$所以$ab = \dfrac{{2c}}{{\sin 2\alpha }}$.
因为$ A\left( {a\cos \alpha ,a\sin \alpha } \right),B\left( {b\cos \alpha , - b\sin \alpha } \right)$,所以$$\begin{cases}x = \dfrac{{a\cos \alpha + b\cos \alpha }}{2}, \\ y = \dfrac{{a\sin \alpha - b\sin \alpha }}{2}, \end{cases}$$所以$$\begin{cases}\dfrac{x}{{\cos \alpha }} = \dfrac{{a + b}}{2}\cdots (1) \\ \dfrac{y}{{\sin \alpha }} = \dfrac{{a - b}}{2}\cdots (2) .\end{cases}$$${(1)^2} - {(2)^2}$得$$\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{{{\sin }^2}\alpha }} = ab = \dfrac{{2c}}{{\sin 2\alpha }},$$所以,$X$的轨迹为双曲线(的一支),且关于$\angle AOB$的角平分线反射对称.
更多轨迹相关的问题见每日一题[494]滑动的线段.