A.maths 高手請答

1.a line L: y=k(x-1) ,k is a real constant .
L cuts the curve y^2 =4x at two points A(x1,y1) and B(x2,y2).
x1 and x2 are the roots of
k^2x^2 -2(k^2+2)x +k^2 =0
M is the mid -point of AB

As k varies ,find the equation od the locus of M.
ANSWER
let the point M (a,b)
Then a=(x1+x2)/2 ; b=(y1+y2)/2
But (x1+x2)=2(k^2+2)/k^2 [sum of the roots]
So 2a=2(k^2+2)/k^2 or a=(k^2+2)/k^2
a=1+2/k^2
2/k^2=(a-1)
k=√[2/(a-1)]
Also y1=k(x1-1) ; y2=k(x2-1)
y1+y2
=k(x1-1+x2-1)
=k(x1+x2-2)
(y1+y2)=k(x1+x2-2)
2b=k(2a-2)
2b=2k(a-1)
b=k(a-1)
substitute the value of k
b={√[2/(a-1)]}(a-1)
b^2=2(a-1)^2/(a-1)
b^2=2(a-1)
change back a,b to x,y
The equation of the locus of M is
y^2=2(x-1)