It is known that the focus of the hyperbola x2 / 64-y2 / 36 = 1 is F1 and F2. The left branch of the straight line passing through the intersection of F1 hyperbola is at two points a and B, [AB] = m, and the circumference of the triangle abf2 is calculated All right, score [AB] is the absolute value of ab

It is known that the focus of the hyperbola x2 / 64-y2 / 36 = 1 is F1 and F2. The left branch of the straight line passing through the intersection of F1 hyperbola is at two points a and B, [AB] = m, and the circumference of the triangle abf2 is calculated All right, score [AB] is the absolute value of ab

a=8,b=6,c^2=64+36=100,c=10
|AF2|-|AF1|=2a=16
|BF2|-|BF1|=2a=16
|AF2|+|BF2|-(|AF1|+|BF1|)=4a=32
|AF1|+|BF1|=|AB|=m
So: | af2 | + | BF2 | = 32 + M
Triangle abf2 perimeter = | af2 | + | BF2 | + | ab | = 32 + m + M = 32 + 2m