Q1. The coordinates of the centre of the circle Ｃ are (4, 6). It is given that the
horizontal line ｙ=1 is a tangent to Ｃ.
(a) Find the equation of Ｃ. 我知道這條是這樣做: (ｘ- 4)^2 + (ｙ- 6)^2 = 5^2
ｘ^2 + ｙ^2 - 8ｘ- 12ｙ +27= 0
(b) Ｌ is a straight line with slope -1/2 and x-intercept ｋ. Ｌ cuts Ｃ at Ａ and
Ｂ. Express the mid-point of Ａ and Ｂ in terms of ｋ. 我不知道這條如何做?
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Q2. Consider a quadratuc equation (x - r)(x - 2) = 3x - 2, where r is a constant.
(a) Find the discriminant fi the equation in terms of r. 我知道這條是這樣做:
- r^2 + 6r + 7
(b) If the quadratic equation has tew equal real roots, find the possible values of r.
我知道這條是這樣做: r = - 1
(c) Using the result in (b), solve the equation (x - r)(x + 2)(x - 1) = (3x + 2)(x - 1)
if r < - 5. 我不明白也不知道這條如何做?
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Thanks
math Q x2
2013-12-28 1:10 am
回答 (2)
2013-12-28 6:51 am
✔ 最佳答案
將 y = (-1/2)x + k 代入 ｘ² + ｙ² - 8ｘ- 12ｙ +27= 0得到 x² + (k-x/2)² - 8x -12(k-x/2) + 2 = 0
x² + k² -kx + x²/4 - 8x - 12k + 6x + 2 = 0
(5/4)x² + (6-8-k)x + k² - 12k + 2 = 0
Let roots be X1 and X2
X1 + X2 = -(6-8-k)/[(2)(5/4}]
(X1 + X2)/2 = (k+2)/5
2013-12-27 22:14:26 補充：
這便是交點中點的 X 座標，你自己再計 Y 座標啦！
2013-12-27 22:44:04 補充：
(x - r)(x - 2) = 3x - 2
x² -2x - rx + 2r - 3x + 2 = 0
x² -(r+2+3)x + 2r + 2 = 0
∆ = (r+5)² - 4(2r+2)
r² + 10r + 25 - 8r - 8 = 0
r² + 2r + 17 = 0 (冇實數解)
原式應該為 (x - r)(x + 2) = (3x + 2) [ 是不是你打錯 (x + 2) 呢？ ]
2013-12-27 22:48:06 補充：
(x - r)(x + 2) = (3x + 2)
x² -rx +2x - 2r - 3x -2 =0
x² -(r-2+3)x -2(r+1) = 0
∆ = (r+1)² - (4)[-2(r+1)]
=r²+2r+1+[8(r+1)]
=r²+2r+1+8r+8
=r²+10r+9
if ∆ = 0
r²+10r+9 = 0
(r+9)(r+1) = 0
r = -1 or r = -9
2013-12-27 22:48:31 補充：
(x - r)(x + 2) = (3x + 2)
x² -rx +2x - 2r - 3x -2 =0
x² -(r-2+3)x -2(r+1) = 0
∆ = (r+1)² - (4)[-2(r+1)]
=r²+2r+1+[8(r+1)]
=r²+2r+1+8r+8
=r²+10r+9
if ∆ = 0
2013-12-27 22:51:22 補充：
原式應該為 (x - r)(x + 2) = (3x + 2) [ 是不是你打錯 (x + 2) 呢？ ]
(x - r)(x + 2) = (3x + 2)
x² -rx +2x - 2r - 3x -2 =0
x² -(r-2+3)x -2(r+1) = 0
∆ = (r+1)² - (4)[-2(r+1)]
=r²+2r+1+[8(r+1)]
=r²+2r+1+8r+8
=r²+10r+9
if ∆ = 0
r²+10r+9 = 0
(r+9)(r+1) = 0
r = -1 or r = -9
(x - r)(x + 2)(x - 1) = (3x + 2)(x - 1)
x = 1 or (x - r)(x + 2) = (3x + 2)
if r< -5 then r = -9
then (x+9)(x+2)=(3x+2)
do it yourself.
2013-12-28 2:07 am
我想問 Q1(b) 是否要找它的 coordinate？
收錄日期: 2021-04-13 19:52:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131227000051KK00138