试题与答案

若F1，F2是双曲线x2a2-y2b2=1(a＞0，b＞0)与椭圆x225+y2

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题型：选择题

题目：

若F1，F2是双曲线
x2
a2
-
y2
b2
=1(a＞0，b＞0)与椭圆
x2
25
+
y2
9
=1的共同焦点，点P是两曲线的一个交点，且△PF1F2为等腰三角形，则该双曲线的渐近线方程是（　　）
A．3x±
2
y=0		B．
2
x±3y=0		C．3x±
7
y=0		D．
7
x±3y=0

答案：

∵椭圆

x2
25
+
y2
9
=1的焦点坐标为（±4，0），

∴双曲线

x2
a2
-
y2
b2
=1(a＞0，b＞0)中c=4，a2-b2=16  ①

设P为两曲线在第一象限的交点，则在椭圆中，△PF1F2为等腰三角形，∴PF1=F1F2=8，∴PF2=10-8=2

在双曲线中，2a=PF1-PF2=6，∴a=3  ②

由①②得，双曲线中a=3，b=

7

∴该双曲线的渐近线方程是y=±

7
3
x

故选 D

考点：椭圆的性质（顶点、范围、对称性、离心率）双曲线的性质（顶点、范围、对称性、离心率）
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