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樂仔 2007-07-05 20:56

A.Maths Locus...

1. Q is a variable point on the curve xy=4. OQ is joined and produced to P such that PQ=0.5OQ, where O is the origin. Find the equation of the locus of P.

2. From a variable point P, perpendiculars are drawn to the line y=2x and the x-axis cutting them at M and N respectively. If PM=5^1/2PN, find the equation of the locus of P.

5^1/2即係5 square root

唉...識做既幫下我
就黎灰到喊...

kylau 2007-07-05 21:44
Q2

Let P = (x,y), M = (b, 2b), N = (x,0)

Slope of OM =2, So Slope of PM = -1/2
=> y-2b / x-b = -1/2
=> x+2y-5b =0 .........(1)

PM=root 5 x PN

So PM square = 5x PN square

=> (x-b)^2 + (y-2b)^2 = 5y^2 ...........(2)

Solving (1) and (2), get b=0 (rej) or y=b

Hence x=3b, and since y=b, get LOCUS of P is x=3y



kylau 2007-07-05 21:49
Q1

Let P= (x,y) , then Q= (2x/3, 2y/3)

Since Q lies on xy=4

So (2x/3)(2y/3)=4, get xy=9

So locus of P is xy=9

樂仔 2007-07-05 21:59
萬分感謝
好在有你...

其實除左呢兩條我仲有幾條locus都係唔識做...er...介唔介紹我係度問埋?
(如果你唔得閒既話唔緊要ga..唔駛理我 )

樂仔 2007-07-05 22:03
Q2果度...
Solving (1) and (2)果PART...係唔係要將(1)果條式轉做b係subject...再代入(2)果條式既b入面?

kylau 2007-07-05 22:05
引用第4樓樂仔2007-07-05 22:03發表的“”:
Q2果度...
Solving (1) and (2)果PART...係唔係要將(1)果條式轉做b係subject...再代入(2)果條式既b入面?


no need, just take y-2b / x-b = -1/2

Hence x-b = -2(y-2b) 就ok

咁代x-b 做-2(y-2b) 入(2), much faster

樂仔 2007-07-05 22:10
引用第5樓kylau2007-07-05 22:05發表的“”:
no need, just take y-2b / x-b = -1/2
Hence x-b = -2(y-2b) 就ok
.......

係唔係會變左咁樣:
(4b-2y)^2+(y-2b)^2=5y^2

樂仔 2007-07-05 22:14
引用第2樓kylau2007-07-05 21:49發表的“”:
Q1
Let P= (x,y) , then Q= (2x/3, 2y/3)
Since Q lies on xy=4
.......

做到啦

樂仔 2007-07-05 22:21
del

樂仔 2007-07-05 22:24
引用第1樓kylau2007-07-05 21:44發表的“”:
Q2
Let P = (x,y), M = (b, 2b), N = (x,0)
Slope of OM =2, So Slope of PM = -1/2
.......

我answer key度話有兩個答案...x-3y=0同x+2y=0...點解既
我淨係跟你方法...計到x-3y=0

kylau 2007-07-05 22:25
引用第6樓樂仔2007-07-05 22:10發表的“”:
係唔係會變左咁樣:
(4b-2y)^2+(y-2b)^2=5y^2


yes, 一抽就可以有common term, 約左個5, 快好多


其他的方法都一樣, 唔識再問

kylau 2007-07-05 22:26
引用第9樓樂仔2007-07-05 22:24發表的“”:
我answer key度話有兩個答案...x-3y=0同x+2y=0...點解既 [表情]
我淨係跟你方法...計到x-3y=0


x+2y-5b =0

如果b=0 不rej, 會出x+2y=0, though it doesn't make sense when u draw a graph to see

樂仔 2007-07-05 22:27
哦!!!
會唔會係...y=0唔駛reject?!

樂仔 2007-07-05 22:28
引用第11樓kylau2007-07-05 22:26發表的“”:
x+2y-5b =0
如果b=0 不rej, 會出x+2y=0, though it doesn't make sense when u draw a graph to see

係喎!
M果個point係(b,2b)丫嘛...如果b係0...咁應該做唔到

樂仔 2007-07-05 22:29
呢條我聽日返學問阿sir丫...
你得唔得閒幫我解多幾條

kylau 2007-07-05 22:33
just ask

樂仔 2007-07-05 22:36
Q3. A variable straight line which is parallel to the x-axis cuts the curves y^2=8x and 2x-y-5=0 at P and Q respectively. Find the equation of the locus of the mid-point of PQ

Q4. A line passing through the point (1,0) cuts the curve y^2=4x at two points A and B. Find the equation of the locus of the mid-point of AB as the line moves.

樂仔 2007-07-05 22:42
真係好唔掂
我覺得我應該要練到一見到題目就諗到個graph出黎...
但好明顯我真係做唔到...連A.Maths都做到咁...其他科目更加做到更差

kylau 2007-07-05 23:13
Q3

Let the variable straight line which is parallel to the x-axis be y=c

Let P= (a,c), Q=(b,c), Mid-pt M = ((a+b/2),c)

For P, c square =8a.......(1)

For Q, c=2b-5.......(2)


We need to find a+b first

from (1) and (2), get
a+b= (c^2)/8 + (c+5)/2
    = (c^2+4c+20)/8
Hence a+b/2     =   (c^2+4c+20)/16

Now M = ((c^2+4c+20)/16, c)

Hence get LOCUS of M is x=(y^2+4y+20)/16


樂仔 2007-07-05 23:24
引用第18樓kylau2007-07-05 23:13發表的“”:
Q3
Let the variable straight line which is parallel to the x-axis be y=c
Let P= (a,c), Q=(b,c), Mid-pt M = ((a+b/2),c)
.......

哦...原來係咁做...

點解你可以咁快諗到既
如果得我自己一個諗既話諗到聽朝都未諗到點做

kylau 2007-07-05 23:36
Q4

Let the line passing through the point (1,0) be y=mx+c
M be the mid-pt of A and B

As it passes through (1,0), 0=m+c, get c=-m,
hence y=mx-m = m(x-1).......(*)

Consider y=m(x-1) and y^2 =4x
Get (m^2)(x^2)- (2m+4)x +m =0, with roots x = a and b

By property of quad equation, get Sum of roots =a+b =(2m+4)/(m^2)

Hence (a+b)/2 = (m+2)/(m^2)
This is x-coordinate of M


Similarly, Consider y=m(x-1) and y^2 =4x
Get my^2-4y-4m=0, with roots y = c and d

By property of quad equation, get Sum of roots =c+d =4/m

Hence (c+d)/2 = 2/m
This is y-coordinate of M

Hence we can see that when y=2/m,
get Locus of M is x = (y/2) (1+y) = (y^2+y)/2


kylau 2007-07-05 23:37
引用第19樓樂仔2007-07-05 23:24發表的“”:
哦...原來係咁做...
點解你可以咁快諗到既 [表情]
如果得我自己一個諗既話諗到聽朝都未諗到點做 [表情]


u check check the sol to see whether this is correct

I am not 100% sure

樂仔 2007-07-05 23:44
Consider y=m(x-1) and y^2 =4x
Get (m^2)(x^2)- (2m+4)x +m =0, with roots x = a and b

呢步我計到(m^2)x^2 -(2m^2+4)x+m^2既

樂仔 2007-07-05 23:44
引用第21樓kylau2007-07-05 23:37發表的“”:
u check check the sol to see whether this is correct
I am not 100% sure

我冇solution...得answer key咋

kylau 2007-07-05 23:50
引用第22樓樂仔2007-07-05 23:44發表的“”:
Consider y=m(x-1) and y^2 =4x
Get (m^2)(x^2)- (2m+4)x +m =0, with roots x = a and b
呢步我計到(m^2)x^2 -(2m^2+4)x+m^2既 [表情]


you are correct, i mis-typed

Then follow the same idea to derive the sol

樂仔 2007-07-06 00:03
Hence we can see that when y=2/m,
get Locus of M is x = (y/2) (1+y) = (y^2+y)/2
呢步唔明

樂仔 2007-07-06 00:11
引用第20樓kylau2007-07-05 23:36發表的“”:
Q4
Let the line passing through the point (1,0) be y=mx+c
M be the mid-pt of A and B
.......

佩服佩服...條條計到既答案都同answer key一樣

樂仔 2007-07-06 00:15
我依家計到個M係[(m^2+2)/m^2, 2/m]

x係m square加2 成個over m square

y係2 over m

樂仔 2007-07-06 00:23
哦~~!!
我諗到喇


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