y+1=ax-10a
ax-y-10a-1=0
依题意点(2,3)到直线的距离为
d=|2a-3-10a-1|/根号(a^2+1)=|-8a-4|/根号(a^2+1)=4|2a+1|/根号(a^2+1)
设y=4|2a+1|/根号(a^2+1)
y根号(a^2+1)=|2a+1|
y^2(a^2+1)=4a^2+4a+1
(4-y^2)a^2+4a+1-y^2=0
所以△=16-4(4-y^2)(1-y^2)>=0
4-(4-4y^2-y^2+y^4)>=0
5y^2-y^4>=0
y^4-5y^2