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一椭圆方程E为(x^2/a^2)+(y^2/b^2)=1(a>b>0)有一斜率为1的直线过F1交E于A,B两点,直线AF2,AB,BF2成等差数列,求椭圆的离心率
人气:133 ℃ 时间:2020-06-10 19:42:21
解答
奥林匹克高手告诉你,高考不会这样出题的.
F1(-c,0),F2(c,0)
2AB=AF2+BF2
2AB+AF1+BF1=AF2+AF1+BF1+BF2=4a
3AB=4a
AB=4a/3
AB:y=x+c
b^2=a^2-c^2
x^2/a^2+(x+c)^2/(a^2-c^2)=1
(2a^2-c^2)x^2+2ca^2*x+2a^2*c^2-a^4=0
xA+xB=-2ca^2/(2a^2-c^2)
xA*xB=(2a^2*C^2-a^4)/(2a^2-c^2)
(xA-xB)^2=(yA-yB)^2=(xA+xB)^2-4xA*xB
AB^2=2(xA-xB)^2=2(xA+xB)^2-8xA*xB
(4a/3)^2=2*[-2ca^2/(2a^2-c^2)]^2-8*(2a^2*C^2-a^4)/(2a^2-c^2)
8c^4-14a^2*c^2+5a^4=0
8e^4-14e^2+5=0
(4e^2-5)*(2e^2-1)=0
e
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