It is known that the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, (a > 0, b > 0) f1.f2 is the two focal points of the hyperbola, and the point P is on the hyperbola. Find the minimum value of | Pf1 | * | PF2 |

It is known that the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, (a > 0, b > 0) f1.f2 is the two focal points of the hyperbola, and the point P is on the hyperbola. Find the minimum value of | Pf1 | * | PF2 |

Let the angle f1pf2 = t, then in the triangle pf1f2, from the cosine theorem, we obtain that:
PF1^2+PF2^2-2PF1*PF2*cost=(PF1-PF2)^2+2PF1*PF2(1-cost)=F1F2^2
Because (pf1-pf2) ^ 2 = (2a) ^ 2 = 4A ^ 2, F1F2 ^ 2 = (2C) ^ 2 = 4C ^ 2 (the first definition of ellipse)
Substituting into the above formula, we get: 4A ^ 2 + 2pf1 * PF2 (1-cost) = 4C ^ 2
That is, Pf1 * PF2 (1-cost) = 2 (C ^ 2-A ^ 2) = 2B ^ 2
So Pf1 * PF2 = 2B ^ 2 / (1-cost)
Because 0 degrees