题目

已知椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的一个顶点为A(2,0)离心率为√2/2,直线y=k(x-1）与椭圆c交予不同的两
点M,N,其中已由第一问得出椭圆C的方程为：x^2/4+y^2/2=1
求当△AMN得面积为√10/3时,k的值

提问时间：2020-11-21

答案

设M(x1,y1),N(x2,y2)
联立直线椭圆,得：
(1+2k²)x² - 4k²x+2k²-4=0
x1+x2=4k²/(1+2k²),x1x2=(2k²-4)/(1+2k²)
|MN|=√[(x1-x2)²+(y1-y2)²]
=√{ (x1-x2)² + [k(x1-1) - k(x2-1)]² }
=√[(x1-x2)² + k²(x1-x2)²]
=√[(1+k²)(x1-x2)²]
=√{ (1+k²)[(x1+x2)² - 4x1x2]
=√{ (1+k²)[16k^4/(1+2k²)² - 4(2k²-4)/(1+2k²) ] }
=√[(1+k²)(24k²+16)/(1+2k²)² ]
A点到直线距离为
h=|k|/√(1+k²)
∴S=(1/2)·h·|MN|
=(1/2)·[|k|/√(1+k²)] ·√[(1+k²)(24k²+16)/(1+2k²)² ]
=(1/2)·|k|·√[(24k²+16)/(1+2k²)²]
=√10/3
即：|k|·√[(24k²+16)/(1+2k²)²] = 2√10/3
两边平方,得：(24k^4 + 16k²)/(1+2k²)² = 40/9
即：7k^4 - 2k² - 5=0
解得：k²=1或-5/7 (舍去)
∴k²=1
∴k=±1