P is any point on the hyperbola X / 16-y / 9 = 1 which is different from the vertex. F1F2 is the two focuses of the hyperbola. The trajectory equation of the center of gravity of △ pf1f2 is obtained

P is any point on the hyperbola X / 16-y / 9 = 1 which is different from the vertex. F1F2 is the two focuses of the hyperbola. The trajectory equation of the center of gravity of △ pf1f2 is obtained

It is known that a ^ 2 = 16, B ^ 2 = 9, so C ^ 2 = a ^ 2 + B ^ 2 = 25,
So F1 (- 5,0), F2 (5,0), let g (x, y) be the center of gravity,
From 3G = P + F1 + F2, the P coordinate is (3x, 3Y),
And because P is on hyperbola, so (3x) ^ 2 / 16 - (3Y) ^ 2 / 9 = 1,
It is reduced to x ^ 2 / (16 / 9) - y ^ 2 = 1. Because P is different from vertex, y ≠ 0,
So the trajectory equation of the center of gravity is x ^ 2 / (16 / 9) - y ^ 2 = 1 (Y ≠ 0)

(1 / 2) it is known that the center of ellipse C is at the origin, the focus is on the x-axis, the left and right focus are F1F2, and the absolute value of F1F2 is 2, and the point (3 / 2) is at
(1 / 2) it is known that the center of ellipse C is at the origin, the focus is on the x-axis, the left and right focus are F1F2, and the absolute value of F1F2 is 2, and the point (3 / 1,2) is on ellipse C
(2) Straight line passing through point F1

2c=2
c=1
Let the equation x & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1
（1） 1/a²+9/4b²=1
（2）a²-b²=1
The solution is a & # 178; = 4, B & # 178; = 3
Equation x & # 178 / 4 + Y & # 178 / 3 = 1

The left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) are F1 and F2 respectively, and O is the coordinate origin,
The slope of OP and F2P are - 24 / 7 and - 3 / 4 respectively
(1) Verification: Pf1 ⊥ PF2
(2) If the area of △ opf1 is 3, find the elliptic equation

According to the formula of angle between two straight lines, Tan & lt; opf2 = (k2-k1) / (1 + k1k2) = (- 3 / 4 + 24 / 7) / [1 + (- 24 / 7) * (- 3 / 4) = 3 / 4, Tan & lt; PF2 = - Tan & lt; pf2x = 3 / 4, 〈 opf2 = 〈 pf2o, 〉△ opf2 is isosceles △, 〈 op | = | of 2 | = | of 1 |, F1, P and F2 take o as the center and half focal length as half focal length

It is known that F1 and F2 are the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the point P is on the ellipse, O is the coordinate origin, and the area of △ pof2 is root 3 and is an equilateral triangle, then the value of B ^ 2 is____

Opf2 is an equilateral triangle, so of 2 = C
S=1/2OF2*OPsin60
Substituting C = 2, pof1 is an isosceles triangle, and Pf1 = 2 times root sign 3 can be obtained
Pf1 + PF2 = 2A, so a = radical 3 + 1
The values of a and C all know that B "2 = 2 times the root sign 3