(已知实数x，y满足(x-2)^2+y^2=3,求frac{y}{x}的最大值)
(设frac{y}{x}=t)
(则y=tx)
((x-2)^2+t^2x^2=3)
(整理为：(1+t^2)x^2-4x+1=0)
(上式的判别式delta为16-4(1+t^2)≥0)
(则:1+t^2≤4)
(即：t^2≤3)
(即：|t|≤sqrt{3})
(则：-sqrt{3}≤t≤sqrt{3})
(即frac{y}{x}的最大值是sqrt{3})
或
(t^2≤3)
(t^2-3≤0)
(t^2-sqrt{3}≤0)
(t^2-(-sqrt{3})≤0)
(即(t+sqrt{3})(t-sqrt{3})≤0)
(即：t+sqrt{3}与t-sqrt{3}异号)
(即t+sqrt{3}>0且t-sqrt{3}<0或者：t+sqrt{3}<0且t-sqrt{3}>0)
(即：t>-sqrt{3}且t<sqrt{3},或者t<-sqrt{3}且t>sqrt{3})
(第二种情况显然不成立，故t>-sqrt{3}且t<sqrt{3})
(故：frac{y}{x}最大值为sqrt{3})