Picture: http://postimg.org/image/6jnytptth/
The figure shows a circle x^2 +y^2=9. A(0,3) and P(2a,2b-3) are points on the
circle. Let M(a,b) be the mid point of AP.
If P moves along on the circle, find the equation of the locus of M.
The answer is a^2 +b^2-3b=0. It is found by putting P(2a,2b-3) into x^2 +y^2=9.
But I don't know why we should find the equation in this way. We put the
coordinates of P to find the equation of M. I think it is a little bit strange.
Can anyone explain this method? Thank you!!
F.5 Maths equation of circle
2015-03-21 9:03 am
回答 (1)
2015-03-21 10:47 am
✔ 最佳答案
Please Read：圖片參考：https://s.yimg.com/rk/HA07344914/o/1301084698.png
2015-03-21 16:57:22 補充：
x²+y²=9 ... (1)
你係唔係講緊1條不會變嘅直線L穿過(1)
其中(1)及L相交2點?
咁即係P只係有2個情況會on L [ 當L與(1)相交時 ]
當P is on L嘅時候，
就可以係L到搵到P嘅座標。
因為P嘅座標可以用M嘅座標黎寫，
所以就可以搵埋M嘅座標。
但咁做就只係可以搵到2個M嘅座標(x1,y1) and (x2,y2)
所以如果P係on (1)但唔係on L嘅時候。
咁樣就唔會搵到P嘅座標。
因為將P嘅座標放入L只係搵到(x1,y1) and (x2,y2)
所以就唔可以搵整個locus of M
2015-03-21 18:10:40 補充：
係
呢題入面P(r,s) is on x²+y²=9
It implies r²+s²=9 ... (2)
(2)係講緊r,s嘅關係
因為r,s同M嘅a,b係有關
所以P(r,s)可以寫成P(2a,2b-3)
將(2a,2b-3)放入(2)裏
(2a)²+(2b-3)²=9 ... (3)
(3)仍然係講緊r同s嘅關係
不過今次r,s係用a,b黎表示
但(3)亦可以話講緊a同b嘅關係
其實有左a同b嘅關係
咁就可以寫到1條方程出黎
因為M嘅locus就係講緊a同b嘅關係
所以a同b嘅關係系講緊M嘅locus
M嘅locus就係講緊a同b嘅關係
2015-03-21 18:20:47 補充：
好似x²+y²=9...(4)
呢條方程講緊x同y嘅關係
而呢個關係就限制x同y
所以有左呢個限制就可以由(4)可以畫到1個圓形出黎
而呢個圓形就係講緊locus of P
收錄日期: 2021-04-15 18:55:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150321000051KK00003