1.由已知得两轴为两个腰,所以k=±1
k=1时,y=x+5;
k=-1时,y=-x+3
2.设y=kx-2k+1
S=1/2*|(2k-1)/k|*|-2k+1|
=2|k|+1/2|k|-2
4k^2=1
k=±1/2
所以y=0.5x或y=-0.5x+2
3.(1)设y=kx+k+3(k≠0),则两焦点为A=[(4k+24)/(3-4k),(15k+9)/(3-4k)]
和B=[(4k+9)/(3-4k),9/(3-4k)]
|AB|=√{[15/(3-4k)]^2+[15k/(3-4k)]^2}=3√2
解得k1=-1/7,k2=7
y1=-1/7x+20/7
y2=7x+10
(2)设y=3,A=(8,3),B=(3,3)
|AB|=5≠3√2,舍去
(3)设x=-1,A(-1,-15/4),B(-1,0)
|AB|=15/4≠3√2,舍去
所以综上所述
y1=-1/7x+20/7
y2=7x+10