There are two fixed points a and B in the plane, and | ab | = 2. The ratio of the distance between a moving point m and two points a and B in the plane is 2:1

There are two fixed points a and B in the plane, and | ab | = 2. The ratio of the distance between a moving point m and two points a and B in the plane is 2:1

Taking the line AB as the x-axis and the vertical bisector of the line AB as the y-axis to establish the rectangular coordinate system, then a (- 1,0), B (1,0)
Let the coordinates of M be (x, y), then | Ma | = √ [(x + 1) &# 178; + Y & # 178;], and | MB | = √ [(x-1) &# 178; + Y & # 178;]
It is known that √ [(x + 1) &# 178; + Y & # 178;] = 2 √ [(x-1) &# 178; + Y & # 178;]
It is reduced to 3x & # 178; + 3Y & # 178; - 10x + 3 = 0
The formula is (X-5 / 3) &# 178; + Y & # 178; = 16 / 9
This is a circle with (5 / 3,0) as the center and 4 / 3 as the radius

The ratio of the distance between a moving point m and AB on the plane is 2:1. To solve the trajectory equation of a moving point m, it is known that AB has a length of 2a and ab is a fixed point

B. set up the coordinate system with the midpoint of AB as the origin
A(-a,0),B(a,0)
M(x,y)
Then √ [(x + a) & sup2; + (y-0) & sup2;]: √ [(x-a) & sup2; + (y-0) & sup2;] = 2:1
square
x²+2ax+a²+y²=4x²-8ax+4a²+4y²
So 3x & sup2; - 10ax + 3A & sup2; + 3Y & sup2; = 0
That is, (x-5a / 3) & sup2; + Y & sup2; = 34a & sup2; / 9

"The trajectory equation of a moving point whose distance to the point (1,1) is equal to the distance to the y-axis is"

Let P (x, y) be a point on the trajectory equation
√((x-1)²+(y-1)²)=|x|
(x-1)²+(y-1)²=x²
（y-1)²=2x-1
So the trajectory equation is (Y-1) & sup2; = 2x-1 (x ≥ 1 / 2)

The distance ratio between a fixed point F (C, 0) and a fixed line L: x = a ^ 2 / C is constant E = C / a (0)

Let m (x, y)
Then √ [(x-C) & sup2; + Y & sup2;] / | x-a & sup2; / C | = C / A
square
x²-2cx+c²+y²=(x²-2a²x/c+a^4/c²)(c²/a²)
x²-2cx+c²+y²=c²x²/a²-2cx+a²
(a²-c²)x²/a²+y²=a²-c²
Then x & sup2; / A & sup2; + Y & sup2; / (A & sup2; - C & sup2;) = 1