Question

设F1,F2是椭圆E:X^2/a^2+Y^2/b^2=1(a>b>0)d的左右焦点，P为直线x=3a/2上一点，三角形F2PF1是底角为30度的

等腰三角形，则E的离心率为？

Answer 1

设直线交x轴于D,则角PF2D为60度,因为PF2=F1F2=2c,所以F2D=(1/2)PF2=c。OF2+F2D=2c=3/2a。e=c/a=3/4

Answer 2

答案正确3/4
P为直线x=3a/2上的一点，三角形F2PF1是底角为30度的等腰三角形
设M为直线x=3a/2与x轴的交点
可知只有∠PF1F2=∠F1PF2=30°
所以∠PF2F1=60°
则在RT△PF2M中
F2M=1/2*PF2=1/2*2C=c
而F2M=3a/2-c
所以3a/2-c=c
因此3a=4c
e=c/a=3/4

Answer 3

3／4

Answer 4

7^（1/2)/4

追问

要全过程

Answer 5

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