Question

在双曲线x62/13-y^2/12=-1的一支上有三个不同的点,A(x1,y1),B(x2,6),C(x3,y3),与焦

Answer 1

1、双曲线x²/13-y²/12=-1--->a=2√3,c=5

--->离心率e=c/a=5√3/6,上准线方程：y=a²/c=12/5

∵|FA|=e(y1-a²/c)=ey1-a,|FB|=ey2-a,|FC|=ey3-a

|FA|、|FB|、|FC|成等差数列--->|FA|+|FC|=2|FB|

--->(ey1-a)+(ey3-a)=2(ey2-a)--->y1+y3=2y2=12

2、AC的中点M,xM=(x1+x3)/2,yM=(y1+y3)/2 = 6

x1²/13-y1²/12=-1,x3²/13-y3²/12=-1

相减：(x1+x3)(x1-x3)/13-(y1+y3)(y1-y3)/12=0

--->AC的斜率k =(y1-y3)/(x1-x3)=12(x1+x3)/13(y1+y3)=2xM/13

--->xM/k=13/2

--->AC垂直平分线方程：y-yM = (-1/k)(x-xM)

--->y-6 = -x/k + 13/2,令x=0--->y=25/2

--->AC的垂直平分线经过定点(0,25/2)