设双曲线x^2/a^2-y^2/b^2=1
m,n (x1,y1) (x2,y2)
则y1=x1-1
y2=x2-1
x1^2/a^2-y1^2/b^2=1
x2^2/a^2-y2^2/b^2=1
(x1+x2)(x1-x2)/a^2-(y1+y2)(y1-y2)/b^2=0
又因为x1+x2=-4/3
y1+y2=x1+x2-2=-10/3
y1-y2=x1-x2
所以4/a^2=10/b^2
点c为根号7
a^2+b^2=7
所以
x^2/2-y^2/5=1