已知双曲线x^2⼀a^2-y^2⼀b^2=1,(a>b>0)的半焦距为c,直线l的方程为bx+ay-ab=o,

2024-11-26 01:02:44
推荐回答(5个)
回答1:

原点(0,0)到直线bx+ay=ab的距离是d=|ab|/√(a²+b²)=(√3/4)c,
两边平方得:a²b²=(3/16)c²(a²+b²)=(3/16)(c²)²,
即:16a²(c²-a²)=3(c²)²,展开,16a²c²-16c^4=3c^4,
两边除以a^4,得:3e^4-16e²+16=0,
解得e²=4或e²=4/3。

因为a>b,所以a^2>b^2,
即a^2>c^2-a^2, 2a^2>c^2,
所以c^2/a^2<2,即e^2<2.
所以e²=4舍去。
只能e²=4/3,
所以离心率e=2√3/3.

回答2:

原点O到直线l的距离为d=|-ab|/根号(a^2+b^2)=(根号3/4)c
ab/c=根号3/4 c
ab=√3c^2/4,......(1)
b=√(c^2-a^2),
代入(1)式,
16a^4-16a^2c^2+3c^4=0,
(4a^2-3c^2)(4a^2-c^2)=0,
a=c/2,或a=√3c/2,
e=c/a=2或2/根号3=2/3根号3.

回答3:

原点O到直线l的距离d=ab/√(a^2+b^2)=ab/c=√3c/4
√3c^2=4ab √3(a^2+b^2)=4ab a=√3b c=2b
e=c/a=2/√3=2√3/3

回答4:

4b/根号3

回答5:

。。。。。。

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