Let P(x, y)

Then Q(- 1, y)

Vector qp, vector qf = vector FP, vector FQ.

(x+ 1，0)(2，-y)=(x- 1，y)(-2，y)

2(x+ 1)=-2(x- 1)+y^2

y^2=4x

The equation y 2 = 4x of trajectory c is a parabola.

(2)m intersection (-1, 0)

M: y = k (x+ 1) is substituted into y 2 = 4x.

k^2x^2+(2k^2-4)x+k^2=0

k 1=(yA-0)/(xA- 1)

k2=(yB-0)/(xB- 1)

k 1+k2

= yA/(xA- 1)+yB/(x B- 1)

= k(xA+ 1)/(xA- 1)+k(xB+ 1)/(xB- 1)

= k[(xAxB+(-xA+xB)- 1+xAxB+xA-xB- 1)]/[xAxB-(xA+xB)+ 1]

= 2k(xAxB- 1)/[xAxB-(xA+xB)+ 1]

=2k( 1- 1)/( 1+(2k^2-4)/k^2+ 1)

=0

certificate

Let R(x, y)

MA/MB=RA/RB

(xA+ 1)/(x-xA)=(x b+ 1)/(x B- x)

x =(2 AXB+xa+XB)/(xa+XB+2)= 1。

R abscissa constant = 1

F( 1，0)

∴RF⊥MF

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