標題: [Core] Locus Problem [打印本頁] 作者: 數痴夢王 時間: 2016-3-11 02:59 PM 標題: Locus Problem
Given two fixed points A and B. P is a variable point such that PAB=k:1.
Prove that the locus of P is a circle geometrially. 作者: chman0216 時間: 2016-3-11 04:00 PM
提示: 作者被禁止或刪除 內容自動屏蔽作者: 數痴夢王 時間: 2016-3-14 07:36 AM
not the stuff i want
Thanks anyway 作者: 數痴夢王 時間: 2016-3-15 08:28 AM
Solved but need cosine law 作者: erectus 時間: 2016-3-16 12:21 AM
See attached. 作者: 數痴夢王 時間: 2016-3-16 07:22 AM
Do not think they are similar
as <POA = ,<POB
and <PAO less than <PBO 作者: erectus 時間: 2016-3-16 12:48 PM
I don't see why you made such a simple mistake. 作者: 數痴夢王 時間: 2016-3-17 07:44 AM
Sorry Proved
2 sides proportional and included angle
so PA/PB = r/(r-1) = k 作者: 數痴夢王 時間: 2016-3-17 08:46 AM
erectus
不過這個証明假設了個軌跡是圓形,不能証明沒有其他圓形符合條件。 作者: 數痴夢王 時間: 2016-3-17 11:26 AM
放棄用幾何的方法証肥了,改用了解析幾何做
設 P(x,y) A(0,0) B(a,0)
PA=k PB
x^2+y^2=k^2[(x-a)^2+y^2]
(k^2-1)x^2+(k^2-1)y^2-2k^2ax+k^2a^2=0
x^2+y^2-2k^2a/(k^2-1)x+k^2a^2/(k^2-1)=0
Proved r^2 = k^4a^2/(k^2-1)^2-k^2a^2/(k^2-1) = k^2a^2/(k^2-1)^2 >0
so it is a real circle. 作者: erectus 時間: 2016-3-17 04:39 PM
No, it's already a proof. But with some details to be filled in.
In my diagram, it is assumed that k > 1. (if 0 < k < 1, the circle will be around point A instead).
And it has also to be proved when P is on the far end of the diameter, though it's quite easy.
This circle satisfies the locus of point P. Can some point F lies inside or outside the circle still satisfies the condition?
Now AP:BP = k : 1. Can FA be k(d-x) so that AF:BF= k : 1.? No, that should be FG which is parallel to AP.
So F cannot lie inside the circle. It can be proved impossible outside the circle similarly. So the circle is the only locus. 作者: 數痴夢王 時間: 2016-3-21 07:47 AM
WHY FG is parallel to PA? 作者: erectus 時間: 2016-3-22 02:34 AM
I mean FG parallel to AP in the diagram equals k(d-x) in length and cannot be equal to FA. 作者: 數痴夢王 時間: 2016-3-22 09:18 AM