已知实数X、Y满足(x-6)^2+y^2＝9,则u＝y/(x-1)的最大值是

题目

已知实数X、Y满足(x-6)^2+y^2＝9,则u＝y/(x-1)的最大值是
刚才有个人便宜了他 竟然按采纳了

答案

u＝y/(x-1)
y=u(x-1)代入(x-6)^2+y^2=9
x^2-12x+36+u^2*x^2-2u^2x+u^2-9=0
(1+u^2)x^2-(12+2u^2)x+27+u^2=0
x为实数,则方程有解
(12+2u^2)^2-4(1+u^2)(27+u^2)≥0
u^4+12u^2+36-u^4-28u^2-27≥0
-16u^2+9≥0
16u^2-9≤0
(4u+3)(4u-3)≤0
解得-3/4≤u≤3/4
所以u＝y/(x-1)的最大值是3/4