Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is at p. when the Q point moves on the circle, we can find Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is p. when Q moves on the circle, the trajectory of P is obtained

Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is at p. when the Q point moves on the circle, we can find Q is the moving point on the circle x ^ 2 + y ^ 2 = 4, and there is another point a (radical 3,0). The intersection radius OQ of the vertical bisector of the line segment AQ is p. when Q moves on the circle, the trajectory of P is obtained

Let P (x, y), Q be written as x = 2Sin Θ, y = 2cos Θ
From the vertical bisector of AQ, we can see that if the midpoint of vector AQ is m, PM and AQ are vertical, an equation can be obtained
The module of vector PQ is equal to the module of PA: | Po | = | PA|
These two equations can get the trajectory of P, you try it yourself

Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

From the equation of the circle, we can see that the center of the circle is C (- 1,0), and the radius is equal to 5. Let the coordinates of the point m be (x, y), and the vertical bisector of ∵ AQ intersects CQ at m, then | MQ | + | MC | = radius 5, and | MC | + | Ma | = 5 | AC |. According to the definition of ellipse, the locus of the point m is an ellipse with a and C as the focus