It is known that a focus of hyperbola C: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) is F2 (2,0) and B = root 3a. (1) the equation for solving hyperbola C (2) let a normal vector of line L passing through focus F2 be (m, 1). When line L and the right branch of hyperbola C intersect at two different points a and B, then, Find the value range of the real number m: and prove that the key point m of AB is on the curve 3 (x-1) ^ 2-y ^ 2 = 3. (3) let the right branch of the straight line L and the hyperbola C in (2) intersect at two points a and B. ask if there is a real number m, so that the angle AOB is an acute angle? If there is, find the range of M. if not, explain the reason

It is known that a focus of hyperbola C: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) is F2 (2,0) and B = root 3a. (1) the equation for solving hyperbola C (2) let a normal vector of line L passing through focus F2 be (m, 1). When line L and the right branch of hyperbola C intersect at two different points a and B, then, Find the value range of the real number m: and prove that the key point m of AB is on the curve 3 (x-1) ^ 2-y ^ 2 = 3. (3) let the right branch of the straight line L and the hyperbola C in (2) intersect at two points a and B. ask if there is a real number m, so that the angle AOB is an acute angle? If there is, find the range of M. if not, explain the reason

（1）c=2c^2=a^2+b^2
The hyperbola is x ^ 2-y ^ 2 / 3 = 1
(2) L: m (X-2) + y = 0 by {y = - MX + 2m
x^2-y^2/3=1
Then (3-m ^ 2) x ^ 2 + 4m ^ 2x-4m ^ 2-3 = 0
From △ 0 to 4m ^ 4 + (3-m ^ 2) (4m ^ 2 + 3) > 0
12m ^ 2 + 9-3m ^ 2 ＞ 0 means m ^ 2 + 1 ＞ 0
And {X1 + x2 ＞ 0
x1•x2＞0
4m^2/(m^2-3)＞0
(4m^2+3)/(m^2-3)＞0
∪ m ^ 2 ＞ 3 ∪ m ∈ (- ∞, - radical 3) ∪ (radical 3, + ∞)
Let a (x1, Y1), B (X2, Y2),
Then (x1 + x2) / 2 = (2m ^ 2 / m ^ 2-3) (Y1 + Y2) / 2 = - 2m ^ 3 / (m ^ 2-3) + 2m = - 6m (m ^ 2-3)
The midpoint m of AB (2m2-3, - 6mm2-3)
∵ 3[(2m^2)/(m^2-3)-1]^2-36m^2/[(m^2-3)^2]=3
M is on the curve 3 (x-1) ^ 2-y ^ 2 = 3
(3) A (x1, Y1), B (X2, Y2), let there be a real number m, so that ∠ AOB is an acute angle, then OA → & # 8226; ob → > 0
∴x1x2+y1y2＞0
Because y1y2 = (- MX1 + 2m) (- MX2 + 2m) = m ^ 2x1x2-2m ^ 2 (x1 + x2) + 4m ^ 2
∴（1+m^2）x1x2-2m^2（x1+x2）+4m^2＞0
(1 + m ^ 2) (4m ^ 2 + 3) - 8m ^ 4 + 4m ^ 2 (m ^ 2-3) > 0, that is 7m ^ 2 + 3-12m ^ 2 > 0
The contradiction between m ^ 2 ＜ 35 and m ^ 2 ＞ 3
That doesn't exist

If 0 〈 K 〈 a, the hyperbola (x ^ 2 / A ^ 2-k) - (y ^ 2 / b ^ 2 + k) = 1 and the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 have the same?

(a^2-k)+(b^2+k)=a^2+b^2,
They have a common focus
That condition should be 0