Re: [問題] 數學

[akitohayama (細雨紛飛)]

※ 引述《pipibjc (　　)》之銘言：
: x^2=8y 上有兩點p(x1,y1)，q(x2,y2)
: pq 線段長 12 且連線通過焦點
: 問 y1 + y2 = ?

線段長^2 = (x1-x2)^2 + (y1-y2)^2
　　　　 = (x1+x2)^2 + (y1+y2)^2 - 4x1x2 - 4y1y2

焦點為 (0,2) 假設斜率為 m => y-2 = mx
跟 x^2 = 8y 聯立 =>

x^2 - 8(mx+2) = 0  --> x^2 - 8mx - 16 = 0

(x1+x2) = 8m & x1x2 = -16

代入 y = mx + 2

(y1+y2) = m(x1+x2) + 4 = 8m^2 + 4
y1y2 = (mx1+2)(mx2+2) = m^2(x1x2) + 2m(x1+x2) + 4
                      = -16m^2 + 16m^2 + 4 = 4

線段長^2 = 144 = (8m)^2 + (8m^2+4)^2 - 4*(-16) - 4*(4)
               = 64m^2 + 64m^4 + 64m^2 + 16 + 64 - 16
               = 64m^4 + 128m^2 + 64

--> 9 = 4m^4 + 8m^2 + 4 --> 4m^4 + 8m^2 - 5 = 0

(2m^2 + 5)(2m^2 - 1) = 0

取 2m^2 = 1

y1+y2 = 8m^2 + 4 = 4 + 4 = 8


大概就這樣了, 有錯請指正一下 :p

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