椭圆（X^2）/9+（Y^2）/25=1上不同三点A(x1,y1),B(9/5,4),C(x2,y2)与焦点F(0,4)的距离成等_数学解答

椭圆（X^2）/9+（Y^2）/25=1上不同三点A(x1,y1),B(9/5,4),C(x2,y2)与焦点F(0,4)的距离成等差数列,则Y1+Y2=?

人气：248 ℃　时间：2020-04-03 23:09:19

解答

a=5,b=3 ,c=4 e=c/a=4/5
FA=a-ey1=5-(4/5)y1
FB=a-e*4=5-(16/5)=9/5
FC=a-ey2=5-(4/5)y2
FA,FB,FC成等差数列
则有2*(9/5)=5-(4/5)y1+5-(4/5)y2
18/5=10-(4/5)(y1+y2)
所以y1+y2=8