圆锥曲线题

先建系
以AB中点为原点A(-1,0),B(1,0)
设C(x,y)
tanA=y/(x+1)
tanB=y/(1-x)
tanAtanB=y^2/(1-x^2)
y^2/(1-x^2)=4
整理得y^2+4x^2=4

先建系
以AB中点为原点A(-1,0),B(1,0)
设C(x,y)
tanA=y/(x+1)
tanB=y/(1-x)
tanAtanB=y^2/(1-x^2)
y^2/(1-x^2)=4
整理得y^2+4x^2=4
故该顶点C的轨迹方程为y^2/4 + x^2 = 1