选项B.
设,点P坐标为(X1,Y1),
x^2/49+y^2/24=1,
a=7,b=√24=2√6,c=√(a^2-b^2)=5,有
x1^2/49+Y1^2/24=1,
24X^2+49Y1^2=49*24.
令,直线PF1的斜率为Kpf1,直线PF2的斜率为KPf2.
因为Kpf1*kpf2=-1,
而,KPF1*KPF2=y1*y1/[(x1+c)*(x1-c)]=-1.
y1^2=-(x1^2-c^2).代入x1^2/49+Y1^2/24=1,中得
X1^2=49/25.
而,三角形PF1F2的面积是
=1/2*|PF1|*|PF2|
=1/2*(e^2)*[(a^2/c)^2-x1^2)]
=1/2*25/49*[(49*49)/24-49/25]
=24.
选项B.