(1) Find the trajectory equation of the center of motion c;

(2) Let A and B be two different points on the trajectory C different from the origin O, and the inclination angles of the straight lines OA and OB are α and β respectively. When α and β change and α+β is constant θ (0 < θ < π), it is proved that the straight line AB passes through a fixed point and the coordinates of the fixed point are obtained.

Thinking analysis: This topic is entitled Conic curve synthesis problem. When solving the trajectory equation of (1) moving point, it is easy to analyze by combining its geometric characteristics and the definition of commonly used conic curve, that is, the method of combining numbers and shapes is often used. (2) The problem of a straight line passing through a fixed point must be solved by introducing an equation with parameters representing a straight line.

(1) solution: as shown in the figure, let m be the center of the motion circle, (? 0) is marked as f, and the intersection point m is marked as a straight line x=? The vertical line, the vertical foot is n, from the meaning of the question |MF|=|MN|, that is, the moving point M to the fixed point F and the alignment x=? According to the definition of parabola, it is known that the trajectory of point M is a parabola, where f (? , 0) as the focus, x=-? It is a directrix, so the ballistic equation is y2 = 2px (P > 0).

(2) Proof: As shown in the figure, let A(x 1, y 1) and B(x2, y2),

X 1≠x2 (otherwise α+β = π) and x1,x2≠0.

So the slope of straight line AB exists, and its equation is y = kx+b. 。

Clearly x 1=? ，x2=？ .

Y=kx+b and y2 = 2px (p > 0) together to eliminate x, ky2-2py+2pb=0.

Vieta theorem knows that y 1+y2=? ，y 1 y2=？ . ①

When θ =? That is α+β =? ，tan α tan β = 1。

So what? = 1，x 1x2-y 1y2=0，

-y 1y2=0, so y 1y2=4p2.

Through 1 know? =4p2, so b=2pk.

So the equation of straight line AB can be expressed as y=kx+2pk,

That is, k(x+2p)-y=0. So the straight line AB passes through the fixed point (-2p, 0).

When θ ≦? From α+β = θ, tanθ=tan(α+β)=? .

Substituting Formula ① into the above formula and simplifying it, we can get tanθ=? ,

So b=? +2pk。

At this point, the equation of straight line AB can be expressed as y=kx+? +2pk，

K(x+2p)-(y-？ )=0.

So the straight line AB passes through a fixed point (-2p,? ).

So, when θ =? When the straight line AB continuously passes through the fixed point (-2p, 0),

When θ ≦? The timeline AB keeps passing through the fixed point (-2p,? ).