\((08\text{年江苏卷})\)满足条件\(AB=2,AC=\sqrt{2}BC\)的三角形\(ABC\)的面积的最大值是\(\underline{\qquad\qquad}\).
解析:
建立如图所示坐标系,设$$A(-1,0),B(1,0).$$设\(C(x,y),\)则由\(AC=\sqrt{2} BC\)
![](https://img2018.cnblogs.com/blog/1793042/201911/1793042-20191116131446249-1077812523.png)
可得$C$点的轨迹方程$$\sqrt{\left(x+1\right)^2+y^2}=\sqrt{2}\cdot \sqrt{\left(x-1\right)^2+y^2},y\neq 0.$$ 整理可得$$(x-3)^2+y^2=8,y\neq 0.$$从而$\triangle ABC$的面积最大值为$$ \dfrac{1}{2}\cdot |AB|\cdot 2\sqrt{2}=2\sqrt{2}.$$