In hyperbola C: X & # 178; / A & # 178; - Y & # 178; / B & # 178; = 1, the chord length perpendicular to the real axis through the focus is 2 √ 3 / 3 The distance from the focus to an asymptote is 1, find C

In hyperbola C: X & # 178; / A & # 178; - Y & # 178; / B & # 178; = 1, the chord length perpendicular to the real axis through the focus is 2 √ 3 / 3 The distance from the focus to an asymptote is 1, find C

An asymptote equation is y = BX / A,
Let the chord be AB, passing through the right focus F2, ∵ be symmetric about the X axis, | F2a | = | ab | / 2 = √ 3 / 3, and the right focus coordinate F2 (C, 0),
c^2/a^2-(1/3)/b^2=1,(1)
Asymptote equation: BX ay = 0,
Let d be the distance from the right focus to the asymptote,
According to the formula of point line distance, d = bc-0 | / √ (a ^ 2 + B ^ 2) = BC / C = b = 1,
b=1,
Substituting (1) into the formula,
c^2/√（c^2-1)-1/3=1,
c^2=4,
∴c=2.

Make a chord with an inclination angle of 45 ° through the right focus of the hyperbola x &# 178; - y &# 178; = 1, find the distance from the midpoint C of the chord AB to the right focus F2, and find the length of the chord ab

From the title: C ^ 2 = 16 + 9 = 25, C = 5
Therefore, the coordinate of the right focus is F2 (5,0). Because a straight line with an inclination angle of 45 ° is made through it, the straight line is:
y=x-5
(1) The linear equation is substituted into the curve equation
The result is: 16x ^ 2-9y ^ 2 = 144
16x^2-9(x-5)^2=144
The result is: 7x ^ 2 + 90x-369 = 0
According to Weida's theorem:
x1+x2= -90/7
y1+y2=(x1-5)+(x2-5)=-160/7
Because C is the midpoint of AB, XC = (x1 + x2) / 2 = - 45 / 7
Yc=(y1+y2)/2= -160/7
So: | CF2 | = √ [(- 45 / 7-5) ^ 2 + (- 160 / 7-0) ^ 2] = (80 √ 5) / 7
(2).7x^2+90x-369=0
According to Weida's theorem: x1x2 = - 369 / 7
|AB|=√[(x1-x2)^2+(y1-y2)^2]=√[(x1-x2)^2+(x1-5-x2+5)^2]=√[2(x1-x2)^2]
=√[2(x1+x2)^2-8x1x2]=√[2*(-90/7)^2+4*369/7]=192/7
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If F1 and F2 are the left and right focal points of hyperbola = 1, AB is the chord passing F1 on the left branch of hyperbola, | ab | = m, then the perimeter of △ abf2 is
From the definition of hyperbola, we know that | af2 | - | AF1 | = 2A, | BF2 | - | BF1 | = 2A,
∴|AF2|-|AF1|+|BF2|-|BF1|=4a,①
That is | af2 | + | BF2 | - | ab | = 4A, ②
∴|AF2|+|BF2|=4a+m,
So the perimeter of △ abf2 is | af2 | + | BF2 | + | ab | = 4A + m + M = 4A + 2m
∴ 4a+2m
What is the reason of how ① to ② (- | AF1 | - | BF1 | = - | ab |

Let P (x0, Y0) according to the focal radius formula PF2 = ex0-a = 2C. ① because F1F2 = PF2, the triangle pf1f2 is an isosceles triangle. According to the graph, the distance between F2 and Pf1 is 2a, then half of Pf1 is 2B, Pf1 = 4b, so a + ex0 = 4B. ② eliminate ex0 of Formula 1 and formula 2, and get C = 2b-a, because C * 2 = a * 2 + B

If F2 is the right focus, / AB / = m, find the circumference of triangle abf2
Urgent need, problem solving process

Firstly, hyperbola is defined as f2a-f1a = 2A, f2b-f1b = 2A, and then special value method is used. When the chord AB is perpendicular to the X axis, M + 4A = F2a + f2b plus m is the total perimeter: 4A + 2m