已知两点M(-1,0),N(1,0),且动点P使向量MN^2=2向量PM•向量PN,求向量PM与PN的夹角的取值范围
问题描述：
已知两点M(-1,0),N(1,0),且动点P使向量MN^2=2向量PM•向量PN,求向量PM与PN的夹角的取值范围
答
设P(x,y)
PM=(-1-x,-y),PN=(1-x,-y)
PM*PN=(1-x^2)+y^2=MN^2/2=2
y^2-x^2=1
P点轨迹为以y轴为对称轴的双曲线.
设PM与PN夹角为α
PM*PN=|PM|*|PN|cosα
cosα =2/√[(1+x)^2+y^2][(1-x)^2+y^2]
整理得：cosα =1/√[x^4+x^2+1]
设f(x)=1/√[x^4+x^2+1]
f'(x)=(-1/2)(4x^3+2x)√[x^4+x^2+1]^3
令f'(x)=0,x=0
x>0,f'(x)