It is known that F 1 and F 2 are the left and right focal points of the hyperbola X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1, and P (x, y) is a moving point on the right branch of the hyperbola, The minimum value of Pf1 is 8. The minimum value of vector Pf1 and vector PF2 is - 16. Solve hyperbolic equation

It is known that F 1 and F 2 are the left and right focal points of the hyperbola X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1, and P (x, y) is a moving point on the right branch of the hyperbola, The minimum value of Pf1 is 8. The minimum value of vector Pf1 and vector PF2 is - 16. Solve hyperbolic equation

If the minimum value of Pf1 is 8, we can know that a + C = 8 (1)
The minimum value of vector Pf1 and vector PF2 is - 16, which is obtained when point P is at the vertex of the right branch
So C-A = 2 (2)
So a = 2, C = 6
So B = root 32
So the hyperbolic equation is
x²/4-y²/32=1

Let p be the moving point on the hyperbola x ^ 2 / 16-y ^ 2 / 9 = 1 with F1 and F2 as the focus, then the trajectory equation of the center of gravity of the triangle f1f2p is?

Let the center of gravity be (x, y)
Then F1 (- 5,0) F2 (5,0)
So there is P point (3x, 3Y)
And P is on the hyperbola, so
(3x)^2/9+(3y)^2/16=1
x^2+9y^2/16=1