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?
回答 (2)
2017-05-17 12:39 pm
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
(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
2017-05-16 3:28 pm
Sol
A(5,-8)到y軸距離=5
0<=PO*5/2<=1
0<=PO<=2/5
A(5,-8)到y軸距離=5
0<=PO*5/2<=1
0<=PO<=2/5
收錄日期: 2021-04-30 22:24:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170516055718AArfXdk