题目
题型:济南一模难度:来源:
x2 |
a2 |
y2 |
b2 |
3 |
答案
∴抛物线焦点坐标为F(1,0),因此双曲线中a=1
又∵双曲线
x2 |
a2 |
y2 |
b2 |
3 |
∴
b |
a |
3 |
3 |
3 |
由此可得双曲线方程为x2-
y2 |
3 |
故答案为:x2-
y2 |
3 |
核心考点
举一反三
x2 |
a2 |
y2 |
b2 |
| ||
2 |
(I)求双曲线的方程及k的取值范围;
(II)是否存在常数k,使得向量
OP |
OQ |
AB |
5 |
x2 |
a2 |
y2 |
b2 |
3 |
(1)求双曲线的方程;
(2)若
OP |
OQ |
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024011430-69943.jpg)
x2 |
a2 |
y2 |
b2 |
3 |
x2 |
a2 |
y2 |
b2 |
3 |
b |
a |
3 |
3 |
3 |
y2 |
3 |
y2 |
3 |
x2 |
a2 |
y2 |
b2 |
| ||
2 |
OP |
OQ |
AB |
5 |
x2 |
a2 |
y2 |
b2 |
3 |
OP |
OQ |