Locus

Let the line passing through (1,0) be y =m(x-1)
Put y = m(x-1) into y^2 = 4x
m^2(x-1)^2 = 4x
m^2x^2 -(2m^2+4)x + m^2 = 0 ...(*)

Let (x1, y1) and (x2, y2) be the coordinates of A and B.
Then, x1 and x2 are the roots of the equation (*).
x1+x2 = (2m^2+4)/m^2

Let (x,y) be the midpoint of AB.
x = (x1+x2)/2 = (m^2+2)/m^2 ...(1)
y = (y1+y2)/2 = (m(x1-1)+m(x2-1))/2 = m(x1+x2)/2 - m
y = (m^2+2)/m - m = 2/m
Therefore, m = 2/y ...(2)
Substitute (2) into (1), x = (4/y^2 + 2)/(4/y^2)
x=(4+2y^2)/4
y^2 = 2(x-1)

Let the line passing through the point (1,0)
be y = mx + c
0 = m(1) + c, thus c = -m
ie the line is y = m(x - 1)

{y = m(x - 1)--------(1)
{y^2 = 4x-------------(2)

sub (1) into (2)
we have [m(x - 1)]^2 = 4x
ie m^2 * (x^2 - 2x +1) = 4x
ie (mx)^2 - (4 + 2m^2) x + m^2 = 0 --------------(**)

x-coor of mid-pt of AB = 0.5(sum of roots of (**))
= (2 + m^2)/m^2
which is on (1)
thus y-coor of mid-pt of AB = 2/m

{x = (2 + m^2)/m^2---------------------(3)
{y = 2/m------------------------------------(4)

rearrange (4) and sub it into (3), we have locus of mid-pt of AB
4x = 2y^2+4
y^2 = 2(x - 1)

2007-09-12 17:39:13 補充：
roots of (**) are x-xoor of A and B