Without losing generality, let the equation of equilateral hyperbola be: x 2/a 2-y 2/a 2 = 1,

Let the left focus be F 1, the right focus be F2, and the hyperbola center be O (the origin of the coordinate axis).

In △PF 1F2, OP is the median line of F 1F2, which is obtained by the median line theorem:

pf 1^2+pf2^2=2op^2+2of 1^2=2op^2+4a^2①

According to the definition of hyperbola:

|PF 1-PF2|=2a

(PF 1-PF2)^2=4a^2

PF 1^2+PF2^2-2PF 1？ PF2=4a^2 ②

Substitute ① into ② to get:

2OP^2+4a^2-2PF 1？ PF2=4a^2

Simplify and get the conclusion:

PF 1*PF2=OP^2=d^2.

Solution 2

Without losing generality, let the hyperbolic equation be: x 2/a 2-y 2/a 2 = 1, and we get:

The focal coordinates of this hyperbola are f1(-√ 2a,0) and F2 (√ 2a,0), and the central coordinate of this hyperbola is O (0 0,0).

Let P(m, n) be a point on the hyperbola. Then:

|pf 1|=√[(m+√2a)^2+n^2],|pf2|=√[(m-√2a)^2+n^2]。

∴|pf 1||pf2|=√{[(m+√2a)^2+n^2][(m-√2a)^2+n^2]}

=√[(m^2-2a^2)^2+(m^2+2a^2+2√2am+m^2+2a^2-√2am)n^2+n^4]

=√[m^4-4a^2m^2+4a^4+(2m^2+4a^2)n^2+n^4]

=√(m^4+2m^2n^2+n^4+4a^4-4a^2m^2+4a^2n^2)

=√[(m^2+n^2)^2+4a^4-4a^2(m^2-n^2)]

Obviously, P(m, n) satisfies hyperbolic equation, ∴ m 2/a 2-n 2/a 2 = 1, ∴ m 2-n 2 = a 2.

∴|pf 1||pf2|=√[(m^2+n^2)^2+4a^4-4a^2(a^2)]=m^2+n^2。

And | po | 2 = (m-0) 2+(n-0) 2 = m 2+n 2.

∴|PF 1||PF2|=|PO|^2=d^2。