题目
题型:不详难度:来源:
x2 |
4 |
y2 |
16 |
答案
x2 |
4 |
y2 |
16 |
纵坐标不变,横坐标变为原来的2倍后的曲线上与P对应的点P′(x,y),
则
|
∴x0=
1 |
2 |
∵P(x0,y0)为椭圆
x2 |
4 |
y2 |
16 |
将P(
1 |
2 |
x2 |
4 |
y2 |
16 |
x2 |
16 |
y2 |
16 |
故答案为:
x2 |
16 |
y2 |
16 |
核心考点
举一反三
x2 |
a2 |
y2 |
b2 |
x2 |
a2+1 |
1 |
4 |
x2 |
4 |
y2 |
16 |
x2 |
4 |
y2 |
16 |
|
1 |
2 |
x2 |
4 |
y2 |
16 |
1 |
2 |
x2 |
4 |
y2 |
16 |
x2 |
16 |
y2 |
16 |
x2 |
16 |
y2 |
16 |
x2 |
a2 |
y2 |
b2 |
x2 |
a2+1 |
1 |
4 |