It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, P is one of their foci, then It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, and P is one of their foci, then the shape of triangle pf1f 2 is

It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, P is one of their foci, then It is known that the ellipse x ^ 2 / M-Y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) with the same two foci F 1 and F 2, and P is one of their foci, then the shape of triangle pf1f 2 is

The ellipse should be x ^ 2 / M + y ^ 2 = 1, A1 = √ m, B1 = 1, C = √ (m-1), where a1 and B1 are the major and minor axes of the ellipse,
According to the definition of ellipse, | Pf1 | + | PF2 | = 2A1 = 2 √ m, (1)
Hyperbolic real semiaxis A2 = √ n, imaginary semiaxis B2 = 1, C = √ (n + 1)
According to the definition of hyperbola, | - Pf1 | - | - PF2 | = 2A2 = 2 √ n,
Let | Pf1 | > | PF2 |, | Pf1 | - | PF2 | = 2 √ n, (2)
(1) And (2),
|PF1|=（√m+√n),
|PF2|=(√m-√n),
PF1^2+PF2^2=m+n-2√mn+m+n+2√mn
=2(m+n),
F1F2^2=(2c)^2=4(m-1)=4(n+1),
2F1F2^2=4m-4+4n-4=4m+4n,
F1F2^2=2(m+n)=PF1^2+PF2^2,
According to the Pythagorean inverse theorem,
The triangle pf1f2 is a right triangle