Question

已知直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)相交于AB两点,M是线段

已知直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)相交于AB两点,M是线段AB上一点,向量AM=-向量BM，且点M在直线y=1/2x上，求椭圆离心率

Answer 1

解：直线x+y-1=0
y=1-x代入x�0�5/a�0�5+y�0�5/b�0�5=1
b�0�5x�0�5+a�0�5(1-x)�0�5=a�0�5b�0�5
b�0�5x�0�5+a�0�5x�0�5-2a�0�5x+a�0�5-a�0�5b�0�5=0
(a�0�5+b�0�5)x�0�5-2a�0�5x+a�0�5-a�0�5b�0�5=0
x1+x2=2a�0�5/(a�0�5+b�0�5)
y1+y2=1-x1+1-x2=2-(x1+x2)=2-(2a�0�5)/(a�0�5+b�0�5)=2b�0�5/(a�0�5+b�0�5)
那么点M的坐标（a�0�5/(a�0�5+b�0�5),b�0�5/(a�0�5+b�0�5))
代入y=x/2
b�0�5/(a�0�5+b�0�5)=a�0�5/2(a�0�5+b�0�5)
a�0�5=2b�0�5
a�0�5=b�0�5+c�0�5
a�0�5=1/2a�0�5+c�0�5
c�0�5=1/2a�0�5
c�0�5/a�0�5=1/2
e�0�5=1/2
e=√2/2