Question

已知F1、F2是椭圆x^2/a^2+y^2/b^2=1的左右焦点O为坐标原点点P(2,-√5/5)在椭圆上,线段PF1与

Answer 1

(1) OF=(-c,0) OP=(2, -√5/5) OP+OF=2ON  c-2=0  c=2

4/a^2+1/5(a^2+4)=1 (无解) or 4/a^2+1/5(a^2-4)=1

(a^2-5)(5a^2-16)=0

a^2>4 a^2=5 方程：x^2/5+y^2=1

(2) 设A点横坐标x1,B点横坐标x2 AB方程y=k(x-2)

则M横坐标0, x2横坐标2

λ1=(x1-0)/(2-x1) λ2=(x2-0)/(2-x2)

λ1+λ2=x1/(2-x1)+x2/(2-x2)=2(x1+x2-x1x2)/(4-2x1-2x2+x1x2)

根据y= k(x-2) ,x^2/5+y^2=1 用韦达定理

X1+x2=4k^2/(k^2+0.2) x1x2=(4k^2-1)/(k^2+0.2)

λ1+λ2=2(4k^2-4k^2+1)/(4K^2+0.8-8k^2+4k^2-1)=2*1/-0.2=-10