Re: [中學] 軌跡問題

[Honor1984 (希望願望成真)]

※ 引述《sincere617 (頂著鋼盔往前衝)》之銘言：
: 過原點的一直線與  x+y=2 x-y=2 交於 P Q 兩點
: 試求  PQ中點所成圖形之方程式為?
: 謝謝大大解答囉

P(t,2-t)

Q = k(t,2-t) => k(t-(2-t))=2 => k = 1/(t-1)

=> Q = (t/(t-1),(2-t)/(t-1)) = ( 1+1/(t-1) , -1+1/(t-1) )

M為PQ的中點 = (x,y) = (1/2)( (t+1)+1/(t-1) , (1-t)+1/(t-1) )

            2x-2y = 2t => t = x-y

            2x+2y = 2 + 2/(t-1) = 2 + 2/(x-y-1) = (2x-2y)/(x-y-1)

            => (x^2 - y^2 - x - y) =  x - y

            => x^2 - y^2 - 2x  = 0

            => (x-1)^2 - y^2 = 1

中心在(1,0)的雙曲線

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※ 編輯: Honor1984       來自: 128.220.147.163      (06/29 22:01)
→ sincere617  :非常謝謝你喔 ~^^