1樓：匿名使用者
如圖:由對稱性,不妨設k>0.
設 m為(x1,y1),n為(x2,y2)
∴2/(|oq|^2) = (1/|om|^2) + (1/|on|^2)
<=> 2/(m^2 + n^2) = 1/(x1^2 + y1^2) + 1/(x2^2 + y2^2)
又∵x1^2 + (y1 - 4)^2 = 4
<=> x1^2 + y1^2 - 8y1 + 12 = 0
<=> x1^2 + y1^2 = 8y1 - 12
x2^2 + (y2 - 4)^2 = 4
<=> x2^2 + y2^2 - 8y2 + 12 = 0
<=> x2^2 + y2^2 = 8y2 - 12
∴1/(x1^2 + y1^2) + 1/(x2^2 + y2^2)
= 1/(8y1 - 12) + 1/(8y2 - 12)
= (1/4)(1/(2y1 - 3) + 1/(2y2 - 3))
= (1/4)( ((2y2 - 3) + (2y1 - 3))/((2y1 - 3)*(2y2 - 3)) )
= (1/4)( (2y1 + 2y2 - 6)/((2y1 - 3)*(2y2 - 3)) )
= (1/2)( (y1 + y2 - 3)/(4y1y2 - 6y1 - 6y2 + 9) )
代入 y=kx 得:
1/(x1^2 + y1^2) + 1/(x2^2 + y2^2)
= (1/2)( (y1 + y2 - 3)/(4y1y2 - 6y1 - 6y2 + 9) )
= (1/2)( (kx1 + kx2 - 3)/(4k^2x1x2 - 6kx1 - 6kx2 + 9) )
=(1/2)( (k(x1+x2) - 3)/(4k^2x1x2 - 6k(x1+x2) + 9) )
這裡 x1+x2 , x1x2 是未知的,下面來嘗試求出 x1+x2:
已知圓方程: x^2 + (y-4)^2 = 4
代入 m,n 得:
x1^2 + (y1 - 4)^2 = 4 ---------------------------(一),
x2^2 + (y2 - 4)^2 = 4 ---------------------------(二).
(一)(二)式相減得:
x1^2 - x2^2 + (y1-4)^2 - (y2-4)^2 = 0
<=> (x1+x2)(x1-x2) + (y1+y2-8)(y1-y2) = 0
<=> k = (y1-y2)/(x1-x2) = -(x1+x2)/(y1+y2-8)
<=> x1+x2 = -(y1+y2-8)k = -(kx1+kx2-8)k = -(k^2(x1+x2)-8k) = -(x1+x2)k^2 + 8k
<=> (x1+x2)(1+k^2) = 8k
<=> x1+x2 = 8k/(1+k^2)
下面來嘗試求出 x1x2:
(一)(二)式相加得:
x1^2 + x2^2 + (y1-4)^2 + (y2-4)^2 = 8
<=> x1^2 + x2^2 + y1^2 + y2^2 - 8y1 - 8y2 + 32 = 8
<=> x1^2 + x2^2 + k^2x1^2 + k^2x2^2 - 8kx1 - 8kx2 + 24 = 0
<=> (k^2+1)(x1^2 + x2^2) - 8k(x1+x2) + 24 = 0
<=> (k^2+1)(x1^2 + x2^2) - 64k^2/(1+k^2) + 24 = 0
<=> (k^2+1)(x1^2 + x2^2) - (64k^2 - 24 - 24k^2)/(1+k^2) = 0
<=> (k^2+1)(x1^2 + x2^2) - (40k^2 - 24)/(1+k^2) = 0
<=> (k^2+1)(x1^2 + x2^2) = (40k^2 - 24)/(1+k^2)
<=> x1^2 + x2^2 = (40k^2 - 24)/(1+k^2)^2
∵ x1x2 = ( (x1+x2)^2 - (x1^2 + x2^2) )/2
∴ x1x2 = ( 64k^2/(1+k^2)^2 - (40k^2 - 24)/(1+k^2)^2 )/2
= ( (64k^2 - 40k^2 + 24)/(1+k^2)^2 )/2
= ( (24k^2 + 24)/(1+k^2)^2 )/2
= ( 24/(1+k^2) )/2
= 12/(1+k^2)
現在有等式:
x1+x2 = 8k/(1+k^2)
x1x2 = 12/(1+k^2)
x1^2 + x2^2 = (40k^2 - 24)/(1+k^2)^2
代入 1/(x1^2 + y1^2) + 1/(x2^2 + y2^2) = (1/2)( (k(x1+x2) - 3)/(4k^2x1x2 - 6k(x1+x2) + 9) ) 得:
(1/2)( (k(x1+x2) - 3)/(4k^2x1x2 - 6k(x1+x2) + 9) )
=(1/2)( (8k^2/(1+k^2) - 3)/(48k^2/(1+k^2) - 48k^2/(1+k^2) + 9) )
=(1/2)( ((8k^2 - 3 - 3k^2)/(1+k^2))/9 )
=(5k^2-3)/18(1+k^2)
即2/(m^2 + n^2) = (5k^2-3)/18(1+k^2)
<=> m^2 + n^2 = 36(1+k^2)/(5k^2-3)
<=> n^2 = 36(1+k^2)/(5k^2-3) - m^2
<=> n = √(36(1+k^2)/(5k^2-3) - m^2)
n = √(36(1+k^2)/(5k^2-3) - m^2) 即為所求.
下面計算各變數取值範圍:
由第一小題的 k<=-2π/3 或 k>=2π/3
∴ k^2 >= (2π/3)^2 >= 2^2 = 4
∴ 5k^2-3 >= 17
∴ 36(1+k^2)/(5k^2-3)
= 36(1+k^2)/5(k^2-3/5)
= 36(k^2-3/5 + 8/5)/5(k^2-3/5)
= (36(k^2-3/5) + 36*8/5)/5(k^2-3/5)
= 36(k^2-3/5)/5(k^2-3/5) + 36*8/25(k^2-3/5)
= 36/5 + 36*8/25(k^2-3/5)
<= 36/5 + 36*8/25(4-3/5)
= 36/5 + 36*8/85
= (36*17+36*8)/(5*17)
= 36*25/(5*17)
= 36*5/17
< 10.58
且有36(1+k^2)/(5k^2-3)
= 36/5 + 36*8/25(k^2-3/5)
> 36/5
= 7.2
∵ m∈(-2,0)∪(0,2)
∴ m^2∈(0,4)
∴ 36(1+k^2)/(5k^2-3) - m^2 > 7.2 - 4 = 3.2 > 0
∴ k 保持原範圍 (-∞,-2π/3)∪(2π/3,+∞)
m∈(-2,0)∪(0,2)
2樓：匿名使用者
第二問不就是帶入計算麼
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