椭圆x^2/m^2+y^2=1(m>1)与双曲线x^2/n^2-y^2=1(n>0)有公共焦点F1,F2,P是他们的一个交点,求三角形F1PF2的S

有共同焦点c^2=m^2-1=n^2+1m^2-n^2=2两边加上2n^2m^2+n^2=2n^2+2求交点x^2=m^2(1-y^2)=n^2(y^2+1)m^2-m^2y^2=n^2y^2+n^2y^2=(m^2-n^2)/(m^2+n^2)=2/(2n^2+2)=1/(n^2+1)所以三角形的高=|y|=1/√(n^2+1)底边F1F2=2c=2...