题目
题型:不详难度:来源:
x2 |
m+2 |
y2 |
m+1 |
答案
x2 |
m+2 |
y2 |
m+1 |
∴将方程化为标准形式,得
x2 |
m+2 |
y2 |
-m-1 |
可得
|
3 |
2 |
∴m∈(-2,-
3 |
2 |
3 |
2 |
故答案为:(-2,-
3 |
2 |
3 |
2 |
核心考点
举一反三
x2 |
a2 |
y2 |
b2 |
(Ⅰ)求椭圆的离心率e;
(Ⅱ)设直线PF2与椭圆相交于A,B两点,若直线PF2与圆(x+1)2+(y-
3 |
5 |
8 |
![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023103835-87836.jpg)
x2 |
m+2 |
y2 |
m+1 |
x2 |
m+2 |
y2 |
m+1 |
x2 |
m+2 |
y2 |
-m-1 |
|
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
x2 |
a2 |
y2 |
b2 |
3 |
5 |
8 |