It is known that the absolute value of the difference between two points F 1 (- 5,0) f 2 (5,0) and them is the moving point trajectory equation of G

It is known that the absolute value of the difference between two points F 1 (- 5,0) f 2 (5,0) and them is the moving point trajectory equation of G

The title is not very clear

It is known that F1F2 is the left and right focus of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b = 1 (a > 0, b > 0), the line passing through F1 and perpendicular to X axis intersects hyperbola at
A. B two points, if △ abf2 is an acute triangle, then the value range of eccentricity of the hyperbola is () to find the specific process, thank you
Answer: (1,1 + radical 2)

∵ F1 is the left focus
∴F1A>F2A
It must be an acute angle
∵ ab ⊥ X axis
∴F2A=F2B
∠F1AF2=∠F1BF2
∵ triangle abf2 is an acute triangle
Only ∠ af2b is an acute angle
∵∠AF2F1=∠BF2F1=1/2∠AF2B

Given the focus F1F2 of hyperbola X-Y / 2 = 1, the point m is on the hyperbola and the vector MF1 multiplies the vector MF2 = 0 to find the distance from the point m to the x-axis

|MF1-MF2|=2 MF1²+MF2²=(2c)²=4c²=12
Equation 1 is squared, and equation 2 is combined to get MF1 * MF2 = 4
If the distance between M and X axis is h, then 1 / 2 * MF1 * MF2 = 1 / 2 * 2 3 * h, so h = 2 3 / 3

The left and right focus of the ellipse are F1 and F2, the eccentricity is 2 / 2 of the root, and the right quasilinear equation is x = 2
The line L passing through the point F1 intersects the ellipse at two points m and N, and the | vector F2m + vector f2n | = (2 times the root 26) / 3, the linear l equation is solved
(the first problem has been solved as x2 / 2 + y2 = 10

Let the slope of a straight line be K, m (x1, Y1), n (X2, Y2)
If F 1 (- 1,0), F 2 (1,0) and a line L passes through F 1, then the equation of the line is y = K (x + 1)
Linear equation and elliptic equation are combined, and (2k ^ 2 + 1) x ^ 2 + 4K ^ 2x + 2K ^ 2-20 = 0
Then X1 + x2 = - 4K ^ 2 / (2k ^ 2 + 1), Y1 + y2 = K (x1 + x2 + 2) = 2K / (2k ^ 2 + 1)
|Vector F2m + vector f2n | ^ 2 = 4 * 26 / 9, namely | X1 + x2-2, Y1 + Y2 | ^ 2 = 4 * 26 / 9
Take X1 + X2, Y1 + Y2 as a whole, and get 5K ^ 4-4k ^ 2-1 = 0
The solution is k = 1, k = - 1
So the equation of a line is X-Y + 1 = 0 or x + y + 1 = 0