The equation of circle C (x-1) ^ 2 + y ^ 2 = 9, point P is a moving point on the circle, the coordinate of fixed point a is (a, 0), and the vertical bisector of line AP and straight line CP intersect at point M (1) If a = - 1, find the trajectory equation of point M; (2) When a changes, try to discuss the type of locus of point M

The equation of circle C (x-1) ^ 2 + y ^ 2 = 9, point P is a moving point on the circle, the coordinate of fixed point a is (a, 0), and the vertical bisector of line AP and straight line CP intersect at point M (1) If a = - 1, find the trajectory equation of point M; (2) When a changes, try to discuss the type of locus of point M

(1) It can be seen from the drawing that | am | = | PM |, | PM | + | MC | = r = 3, that is, the sum of the distances from point m to point a and point C is 3, so the trajectory equation of M is ellipse
(2) In the discussion, with the help of drawing, a moves on the x-axis when A3, - 1

Given the point a (5,0) and the circle B: (x + 5) ^ 2 + y ^ 2 = 36, P is the moving point on the circle B, and the intersection of the straight line BP and the vertical bisector of the segment AP at the point Q, then the point Q (x, y) is the moving point
Given the points a (5,0) and ⊙ B: (x + 5) ^ 2 + y ^ 2 = 36, P is the moving point on ⊙ B, and the intersection of the line BP and the vertical bisector of the line AP at point Q, then the trajectory equation satisfied by point Q (x, y) is ()

The line BP intersects the vertical bisector of the segment AP at point Q
So QA = QP
So | qa-qb | = | qp-qb | = | BP | = 6 (radius of circle)
So the distance difference between Q and fixed point a (5,0) B (- 5,0) is fixed value 6, so the trajectory of Q point is hyperbola
2a=6 a=3 c=5
So B = 4
So the trajectory equation is
x^2/9-y^2/16=1

Given points a (2, - 5) and B (4, - 7), try to find a point P on the y-axis to minimize the absolute value PA plus the absolute value Pb

If a '(- 2, - 5) of point a is symmetric about y axis, then PA = PA'. If the absolute value PA plus absolute value Pb is the minimum, then PA '+ Pb is also the minimum. Only when a', P and B are on the same route, PA '+ Pb is also the minimum. If the linear equation passing through point P is y = KX + B, then y passes through a' (- 2. - 5), and B (4, - 7) is substituted to obtain the linear equation: y = - 1 / 3x -

Given that a (1,3) B (5, - 2) P is a point on the x-axis, if the absolute value of | PA | - | Pb | is the largest, the coordinates of point P can be obtained?

If P is not on the line AB, then ABP forms a triangle, so | PA | - | Pb ||