Given two fixed points F1 (- 5,0), F2 (5,0), find the trajectory equation of point P whose absolute value of the difference between the distances of F1 and F2 is 6

The distance difference between two points is a fixed value
So it's a hyperbola
Distance difference = 2A = 6
a=3
Focal length = 2C = 5 - (- 5)
So C = 5
b²=c²-a²=16
The focus is on the x-axis
x²/9-y²/16=1

1. The locus of point m whose sum of distances from two fixed points F1 (- 2,0) and F2 (2,0) in the plane is 4 is

The trajectory of a point whose distance from two known points F1 (- 2,0) and F2 (2,0) is equal to 4 is a line segment
The center coordinate is (0,0), the length of major axis is 4, so a = 2, the focal length is 4, so C = 2, so B ^ 2 = a ^ 2-C ^ 2 = 0
In conclusion, it is a line segment

If the absolute value of the distance difference between two fixed points F1 (- 7,0) and F2 (7,0) on the plane is equal to 10, the trajectory equation is ()

It is known that the curve is Hyperbola by "the absolute value of the distance difference between two fixed points F 1 (- 7,0) and F 2 (7,0) is equal to 10"
Because the distance difference is 10;
So 2A = 10, a = 5;
Because fixed point F1 (- 7,0), F2 (7,0);
So 2C = 14, C = 7;
So B & # 178; = C & # 178; - A & # 178; = 49-25 = 24
So B = 2 × root 6
So the equation is: X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1

Given that the difference between the distance from the moving point P to F1 (5,0) and its F2 (- 5,0) is equal to 6, then the trajectory equation of P is

Let P (x, y)
√[(x-5)^2+y^2]-√[(x+5)^2+y^2]=6
Transfer term, square: (X-5) ^ 2 + y ^ 2 = 36 + (x + 5) ^ 2 + y ^ 2 + 12 √ [(x + 5) ^ 2 + y ^ 2]
-5/3x-3=√[(x+5)^2+y^2]
Then we get the trajectory equation: (5x / 3 + 3) ^ 2 = (x + 5) ^ 2 + y ^ 2

Given two fixed points F1 (- 5,0), F2 (5,0), find the trajectory equation of point P whose absolute value of the difference between the distances of F1 and F2 is 6