It is known that the two focuses of hyperbola x ^ 2 - (y ^ 2 / 3) = 1 are F1 and F2 respectively. The slope of a straight line AB passing through the left focus F1 is 3. The area of rtabf2 is calculated Brother, your answer is wrong.
S = 2C * (y1-y2) / 2
c=2
Let the linear equation be y = 3x + 6
Substituting into hyperbolic equation
3x^2-9(x^2+4x+4)=3
-6x^2-36x-39=0
x1+x2=-6 x1x2=13/2
So Y1 + y2 = - 6, y1y2 = 9x1x2 + 18 (x1 + x2) + 36
Y1-y2 = √ [(Y1 + Y2) ^ 2-4y1y2] = 3 √ 10
So s = 6 √ 10
This time, it's right. I'm really sorry that the computing power has decreased