Question

急 已知A(-√ 3,0),B(√ 3,0),动点P满足|向量PA|+|向量PB|=4

Answer 1

1.已知A(-√ 3,0),B(√ 3,0),动点P满足|向量PA|+|向量PB|=4

P为动点,A,B为定点,则P的轨迹是以A,B为焦点,2a=4 c=-√ 3 b^2=a^2-c^2=1

动点P的轨迹方程 x^2/4+y^2=1

2.设M(x1,y1) N>(x2,y2)

点(1,0)作直线l x=ty+1 x1x2=t^2*y1y2+t(y1+y2)+1

代x=ty+1入 x^2/4+y^2=1 得 (t^2+4)y^2+2ty-3=0 y1y2=-3/(t^2+4) y1+y2=-2t/(t^2+4)

向量OM*向量ON=x1x2+y1y2=-3(t^2+1)/(t^2+4)-2t^2/(t^2+4)+1=(-4t^2+1)/(t^2+4)

=-4+15/(t^2+4) ,当t=0时,最大值为-1/4