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)
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kylau於2007-07-05 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.
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
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