Yahoo Answers: Answers and Comments for Maths(Coordinate Treatment of Simple Locus Problems) [數學]
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Thu, 05 Apr 2007 23:51:19 +0000
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Yahoo Answers: Answers and Comments for Maths(Coordinate Treatment of Simple Locus Problems) [數學]
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From chaoseng2000: The equation of line L is x+3y2+k(xy6)=0
(a...
https://hk.answers.yahoo.com/question/index?qid=20070405000051KK05272
https://hk.answers.yahoo.com/question/index?qid=20070405000051KK05272
Fri, 06 Apr 2007 00:58:18 +0000
The equation of line L is x+3y2+k(xy6)=0
(a)Express the slope of L interms of k.
由L:x+3y2+k(xy6)=0
x+3y2+kxky6k=0
ky3y=kx+x(6k+2)
(k3)y=(k+1)x(6k+2)
y=[(k+1)/(k3)]x(6k+2)/(k3)
即L的斜率為(k+1)/(k3)
(b)FInd the value of k if L is parallel to x+2y4=0.
∵L is parallel to x+2y4=0
x+2y4=0的斜率為1/2
即(k+1)/(k3)=1/2
2k+2=k+3
3k=1
∴k=1/3
(c)If L passes through a fixed point P for any value of k,find the coorsinates of P
由L : x+3y2+k(xy6)=0與k無關得
{x+3y2=0(1)
{xy6=0(2)
(1)(2)
4y+4=0
y=1
代入(2)
x+16=0
x=5
∴P的坐標為(5,1)

From WM: (a) L: x+3y2+k(xy6) = 0
x+3y2+k(xy6) = ...
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https://hk.answers.yahoo.com/question/index?qid=20070405000051KK05272
Fri, 06 Apr 2007 00:43:22 +0000
(a) L: x+3y2+k(xy6) = 0
x+3y2+k(xy6) = 0
x + 3y  2 + kx  ky  6k = 0
ky  3y = kx + x  6k  2
(k  3)y = (k + 1)x +  2(3k +1) =0
y = [(k + 1)/(k  3)]x  [2(3k +1)/(k  3)]
therefore, slope of L = (k + 1)/(k  3)
(b) the line x+2y4=0 can be expressed as:
2y = x + 4
y = (1/2)x + (4/2)
therefore slope of the line x+2y4=0 is 1/2
if L is parallel to the line x+2y4=0
slope of L = 1/2
(k + 1)/(k  3) = 1/2
2k + 2 = k + 3
3k = 1
k = 1/3
(c) L can be written as (k+ 1)x + (3 k)y  2(3k + 1) = 0
Let the coordinates of P be (a, b), and substitue (a,b) into L, it becomes
(k+ 1)a + (3  k)b  2(3k + 1) = 0 .................(1)
L passes through a fixed point P for any value of k, so
put k = 1, (1) becomes
(1 + 1)a + [3(1)]b  2[3(1) + 1] = 0
4b + 4 = 0
b = 1
put k = 3, (1) becomes
(3+ 1)a + (3  3)b  2[3(3) + 1] = 0
4a  20 = 0
a = 5
therefore, the coordinates of P is (5, 1)