回覆 4# mathguy 的帖子
第 16 題
在坐標平面上，\(A\)點坐標為\((8,0)\)，\(B\)點坐標為\((0,6)\)，\(P\)為圓：\(x^2+y^2=16\)上的動點，求\(3\overline{PA}+2\overline{PB}\)的最小值=
[解答]
P(x，y)，M(0，m)
令 PM = (2/3)PB
9PM^2 = 4PB^2
9x^2 + 9(y - m)^2 = 4x^2 + 4(y - 6)^2
5x^2 + 5y^2 - 6(3m - 8)y + (9m^2 - 144) = 0
80 - 6(3m - 8)y + (9m^2 - 144) = 0
6(3m - 8)y - (9m^2 - 64) = 0
(3m - 8)(6y - 3m - 8) = 0
m = 8/3，M(0，8/3)
3PA + 2PB = 3[PA + (2/3)PB] = 3(PA + PM) ≧ 3AM = 8√10
