It is known that the hyperbola has a point P on one branch, the left and right focus is F1, F2, and | Pf1 | = 4 | PF2 |, so find the maximum value of eccentricity! Known hyperbola has a point P on a branch, the left and right focus is F1, F2, and | Pf1 | = 4 | PF2 |, find the maximum centrifugal rate, to complete the whole process of solving the problem!

It is known that the hyperbola has a point P on one branch, the left and right focus is F1, F2, and | Pf1 | = 4 | PF2 |, so find the maximum value of eccentricity! Known hyperbola has a point P on a branch, the left and right focus is F1, F2, and | Pf1 | = 4 | PF2 |, find the maximum centrifugal rate, to complete the whole process of solving the problem!

Let Pf1 = ex + A, PF2 = ex-a, so ex + a = 4 (ex-a), and E = 5A / 3x. Because P is on the right branch of hyperbola, X is greater than or equal to a, so e is greater than or equal to 5 / 3, that is, the maximum value of E is 5 / 3

Point a is on the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0). F1 and F2 are the two focuses of the hyperbola. The trajectory equation of the center of gravity g of the triangle af1f2 is obtained

Let g (x, y), a (x1, Y1) = = = > be derived from the formula of barycenter coordinates, 3x = x1.3y = Y1. = = = > be substituted into hyperbolic equation, and barycenter trajectory equation: (9x ^ 2 / A ^ 2) - (9y ^ 2 / b ^ 2) = 1