A.Maths Locus... [復制鏈接]

just ask

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.

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

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

引用第18樓kylau於2007-07-05 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

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

引用第19樓樂仔於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

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

引用第21樓kylau於2007-07-05 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

引用第22樓樂仔於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

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

引用第20樓kylau於2007-07-05 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

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

哦~~!!
我諗到喇