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，可否列式。
--------------------------------------------------------------------------------------------------------------------
我不知道這條如何做，可否列式。
Q2. In the beginning of 2011, Penny bought a cat. Her cat ate 30 cans of food
every month in the first year. In the beginning of every succeding year, her cat
requires 5 more cans of food every month than the previous year.
(a) How many cans of food does she buy in ｎ years? (Assume the life of the
cat is longer than ｎ years.)
(b) Penny wants to buy one more cat in the beginning of 2014. She expects that the new cat eats 20 cans of food every month in the first year and 3 more cans
every month than the previous year. She wants to find the budget on feeding the cats. If the cost of a can of food is $15, and she assumes that the cats will not
die before the end of 2020 and the costs of food wil not change, how much will
she spend on feeding the two cats from the beginning of 2011 to the end of
2020?
S6 math (help me!, thanks)
2013-12-30 11:34 pm
回答 (2)
2013-12-31 12:48 am
✔ 最佳答案
Q.1(a)
Radius of C
= 6 - 1
= 5
Equation of C :
(x - 4)² + (y - 6)² = 5²
x² - 8x + 16 + y² - 12y + 36 = 25
x² + y² - 8x - 12y + 27 = 0
(b)
L passes through (k, 0) and has a slope -1/2.
The equation of L:
y - 0 = (-1/2)(x - k)
2y = -x + k
L : x = k - 2y ...... [1]
C : x² + y² - 8x - 12y + 27 = 0 ...... [2]
Put [1] into [2] :
(k - 2y)² + y² - 8(k - 2y) - 12y + 27 = 0
4y² - 4ky + k² + y² + 16y - 8k - 12y + 27 = 0
5y² + (4 - 4k)y + (k² - 8k + 27) = 0 ...... [3]
Let (x1, y1) and (x2, y2) be thecoordinates of A and B respectively.
Then y1 and y2 are the roots of [3].
Sum of roots : y1 + y2 = -(4 - 4k)/5
y1 + y2 = (4k - 4)/5
Let M(p, q) be the coordinates of the mid-point of A and B.
q = (y1 + y2)/2
q = [(4k - 4)/5]/2
q = (2k - 2)/5
Put q = (2k - 2)/5 into [1]
p = k - [2(2k - 2)/5]
p = (5k/5) + [(-4k + 4)/5]
p = (k + 4)/5
Hence, the mid-point of A and B = ((k + 4)/5, (2k- 2)/5))
Q.2
(a)
Number of cans of food that Penny buys :
1st year = 30 * 12 = 360
2nd year = (30 + 5)*12 = 360 + 60
3nd year = (30 + 5*2)*12 = 360 + 60*2
nth year = [30 + 5(n - 1)]*12 = 360 + 60(n - 1)
The sum is an arithmetic sequence.
1st term, a = 360
common difference, d = 60
Total number of cans of food that Penny buys in n years
= n[2a + (n - 1)d]/2
= n[2*360 + (n - 1)*60]/2
= n[720 + 60n - 60]/2
= 30n(n+ 11)
= 30n² + 330n
(b)
Similarly, total number of cans of food that Penny buys for the second cat in nyears
= n[2*(20*12) + (n - 1)(3*12)]/2
= n[2*240 + (n - 1)*36]/2
= 18n² + 222n
Number of years for her to feed the first cat
= 2020 - 2011 + 1
= 10
Number of years for her to feed the second cat
= 2020 - 2014 + 1
= 7
Amount required for her to spend on feeding the two cats
= $15 * [(30 * 10² + 330 * 10) + (18 * 7² + 222 * 7)]
= $131,040
參考: 賣女孩的火柴
2013-12-30 11:53 pm
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
這便是交點中點的 X 座標
2013-12-30 16:05:09 補充：
y = mx + c 是一條直線，由題意得 y = -x/2 + k
2y = -x + 2k
x = 2(k-y)
代入 C :ｘ² + ｙ² - 8ｘ- 12ｙ +27= 0
[2(k-y)]² + y² - 8[2(k-y)] -12y + 27 = 0
4(k-y)² + y² - 16k + 16y -12y + 27 = 0
4k² - 8ky + 4y² + y² - 16k + 4y + 27 = 0
5y² + (4-8k)y + (4k²-16k+27) = 0
2013-12-30 16:08:40 補充：
這是一條對於 y 的二次方程，两根之和 = Y1+Y2=-(4-8k)/5 ..... [ -b/a ]
(Y1+Y2)/2=-2(1-2k)/5
交點的中點座標為 ((k+2)/5 , -2(1-2k)/5 )
2013-12-30 16:21:34 補充：
Sorry X 座標 計算上 有錯誤
(5/4)x² + (6-8-k)x + k² - 12k + 2 = 0
Let roots be X1 and X2
X1 + X2 = -(6-8-k)/(5/4)
X1 + X2 = -(-2-k)/(5/4)
X1 + X2 = (2+k)/(5/4)
X1 + X2 = 4(2+k)/5
(X1 + X2)/2 = 2(2+k)/5
交點的中點座標為 ( 2(k+2)/5 , 2(2k-1)/5 )
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
這便是交點中點的 X 座標
2013-12-30 16:05:09 補充：
y = mx + c 是一條直線，由題意得 y = -x/2 + k
2y = -x + 2k
x = 2(k-y)
代入 C :ｘ² + ｙ² - 8ｘ- 12ｙ +27= 0
[2(k-y)]² + y² - 8[2(k-y)] -12y + 27 = 0
4(k-y)² + y² - 16k + 16y -12y + 27 = 0
4k² - 8ky + 4y² + y² - 16k + 4y + 27 = 0
5y² + (4-8k)y + (4k²-16k+27) = 0
2013-12-30 16:08:40 補充：
這是一條對於 y 的二次方程，两根之和 = Y1+Y2=-(4-8k)/5 ..... [ -b/a ]
(Y1+Y2)/2=-2(1-2k)/5
交點的中點座標為 ((k+2)/5 , -2(1-2k)/5 )
2013-12-30 16:21:34 補充：
Sorry X 座標 計算上 有錯誤
(5/4)x² + (6-8-k)x + k² - 12k + 2 = 0
Let roots be X1 and X2
X1 + X2 = -(6-8-k)/(5/4)
X1 + X2 = -(-2-k)/(5/4)
X1 + X2 = (2+k)/(5/4)
X1 + X2 = 4(2+k)/5
(X1 + X2)/2 = 2(2+k)/5
交點的中點座標為 ( 2(k+2)/5 , 2(2k-1)/5 )
收錄日期: 2021-04-13 19:52:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131230000051KK00089