Question

已知椭圆x2/a2+y2/b2=1(a>b>0)的一个焦点为(√2,0),且椭圆过点A(√2,1)

Answer 1

一个焦点坐标是F1(根号2,0),则另一个坐标是F2(-根号2,0)
那么有AF1+AF2=根号[0+1]+根号[(根号2+根号2)^2+1]=1+3=4=2a
即有a=2,c=根号2,则有b^2=a^2-c^2=4-2=2
即椭圆方程是x^2/4+y^2/2=1

(2)设P坐标是(x,y)
PM^2=x^2+(y-m)^2=4-2y^2+y^2-2my+m^2=-y^2-2my+m^2+4=-(y+m)^2+2m^2+4
由于-根号2<=y<=根号2
(1)对称轴y=-m<-根号2时,即有m>根号2时,PM的最大值是当y=-根号2时取得,即有PM=根号(m^2+2根号2m+2)=m+根号2
(2)-根号2<=-m<=根号2,即有-根号2<=m<=根号2,又m>0,即有0<m<=根号2时,当y=-m时有最大值是PM=根号(2m^2+4)