p點在y軸上移動 a(5,-8) pa段與兩軸形成三角形面積不大於1 求po線段的長度範圍 o為原點?

設P(0,y), PA段交x軸於 (x,0)
(0-y)/(x-0)=(0+8)/(x-5)
8x+xy-5y=0
x=5y/(y+8)
PA與兩軸形成三角形面積=(1/2)(5y^2)/(y+8)<=1
5y^2-2y-16<=0
(5y+8)(y-2)<=0
-8/5<=y<=2
-8/5<=PO<= 2 ....Ans

Sol
A(5,-8)到y軸距離=5
0<=PO*5/2<=1
0<=PO<=2/5