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:
x2+y2=16上的動點,求
3PA+2PB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