Question

已知P(x,y)为圆C:(x+3)^2+(y-4)^2=1上任意一点,求y-6/x的最值

Answer 1

设（y-6）/x =T

则Y=TX+6

如图X＜0，故T最大时Y最小为3且X=-3，T最小时Y最大5且X=-3

即T最大为1最小时为1/3

Answer 2

(y-6)/(x-0)=k
所以k是过P(x,y).B(0,6)的斜率
P在C上则圆心到直线距离小于等于半径

kx-y+6=0
所以|-3k-4+6|/√(k²+1)<=1
平方
9k²-12k+4<=k²+1
8k²-12k+3<=0
(3-√3)/4<=k<=(3+√3)/4
所以最大(3+√3)/4
最小(3-√3)/4

Answer 3

是y- (6/x)
还是 (y-6)/x
???

追问

（y-6）/x
不好意思~~~~~~~~~~~~

追答

let x' = x+3
y' = y-4

C:(x+3)^2+(y-4)^2=1
x'^2+y'^2=1 (1)

let
x'= cosa
y'= sina

S= (y-6)/x
= (y'-2)/(x'-3)
= (sina-2)/(cosa-3)
dS/da = [(cosa-3)cosa +(sina-2)sina] /(cosa-3)^2=0
(cosa)^2-3cosa +(sina)^2-2sina =0
1-3cosa= 2sina
1-6cosa +9(cosa)^2 = 4-4(cosa)^2
13(cosa)^2-6cosa-3=0
cosa = (3+4√3)/13 or (3-4√3)/13

cosa = (3+4√3)/13 , sina = (2-6√3)/13 (max)
cosa = (3-4√3)/13 , sina = (2+6√3)/13 (min)

min S
=(2-sina)/(3-cosa)
=[2- (2+6√3)/13 ]/[3- (3-4√3)/13]
= (24-6√3)/(36+4√3)

max S
= (2-sina)/(3-cosa)
=[2- (2-6√3)/13 ]/[3- (3+4√3)/13]
= (24+6√3)/(36-4√3)