本區搜索:
Yahoo!字典
打印

[Core] Locus Problem

[隱藏]

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.
   

TOP

https://youtu.be/YD_1_IVdqUA
有講Locus例題解答 ,可以參考一下

TOP

not the stuff i want
Thanks anyway

TOP

Solved but need cosine law

TOP

See attached.

TOP

Do not think they are similar
as <POA = ,<POB
and <PAO less than <PBO

TOP

I don't see why you made such a simple mistake.

TOP

Sorry Proved
2 sides proportional and included angle
so PA/PB = r/(r-1) = k

TOP

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。

TOP

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.

TOP

重要聲明:小卒資訊論壇 是一個公開的學術交流及分享平台。 論壇內所有檔案及內容 都只可作學術交流之用,絕不能用商業用途。 所有會員均須對自己所發表的言論而引起的法律責任負責(包括上傳檔案或連結), 本壇並不擔保該等資料之準確性及可靠性,且概不會就因有關資料之任何不確或遺漏而引致之任何損失或 損害承擔任何責任(不論是否與侵權行為、訂立契約或其他方面有關 ) 。