Question

已知椭圆C:x²/4+y²=1,直线l与椭圆C相交于A.B两点,OA•OB=0(其中O为

Answer 1

OA•OB=0
向量OA与向量OB垂直
设A(a,b),B(kb,-ka)
则a^2/4+b^2=1,k^2*b^2/4+K^2*a^2=1
a^2+4b^2=4,k^2=4/(4a^2+b^2)
k^2+1=(4+4a^2+b^2)/(4a^2+b^2)=(a^2+4b^2+4a^2+b^2)/(4a^2+b^2)=5(a^2+b^2)/(4a^2+b^2)
直线AB：(ka+b)x-(a-kb)y-k(a^2+b^2)=0
1.
点O到直线AB的距离：
d=|k|*(a^2+b^2)/√[(ka+b)^2+(a-kb)^2]
=|k|*(a^2+b^2)/√[(k^2+1)*(a^2+b^2)]
=√[k^2*(a^2+b^2)/(k^2+1)]
=√[4*(a^2+b^2)/(5^2+5b^2)]
=2√5/5
2.
解法1
令a=2cosθ,b=sinθ
m=|OA|•|OB|
=√[(a^2+b^2)*k^2(a^2+b^2)]
=√[4(a^2+b^2)^2/(4a^2+b^2)]
=√[4(3cosθ^2+1)^2/(15cosθ^2+1)]
m^2/4=(9cosθ^4+6cosθ^2+1)/(15cosθ^2+1)
=9/15*cosθ^2+81/225+144/225/(15cosθ^2+1)
=9/225*(15cosθ^2+1)+72/225+144/225/(15cosθ^2+1)
=36/225*[(15cosθ^2+1)/4+4/(15cosθ^2+1)]+72/225
>=72/225+72/225=16/25
|OA|•|OB|最小值为8/5
此时(15cosθ^2+1)/4=4/(15cosθ^2+1)
15cosθ^2+1=4,cosθ^2=1/5
a^2=b^2=4/5,k^2=1
解法2
|OA|•|OB|=|AB|*d
=2√5/5*√[(a-kb)^2+(b+ka)^2]
=2√5/5*√[(k^2+1)(a^2+b^2)]
=2√[(a^2+b^2)^2/(4a^2+b^2)]……
检查结束
解法3(特简单)
m=|OA|•|OB|=|AB|*d
=d√(|OA|^2+|OB|^2)
>=d√2|OA|•|OB|
m^2>=2md^2
m>=2d^2=8/5
此时|OA|=|OB|,k^2=1……