在△ABC中,已知AB=10,∠CAB=2∠CBA,求点C的轨迹方程

设AB在X轴上,AB的中点为原点,并设A(-5,0),B(5,0),C(x,y),则
k(AC)=y/(x+5)
k(BC)=y/(x-5)
∠CAB=2∠CBA
tan∠CAB=-tan(2∠CBA)=-2tan∠CBA/(1-tan^2∠CBA)
k(AC)=-2k(BC)/[1-k^2(BC)]
y/(x+5)=-[2y/(x-5)]/{1-[y/(x-5)]^2}
3x^2-10x-y^2-25=0