The shape of the curve looks like an hyperbola (non orthogonal).
I suggest to fit the curve (least squares fitting method, involving 5 parameters) to the general quadratic equation and then, compute the characteristic parameters of the hyperbola.
It might be much simpler if you already know the gradient of the asymtotes m1 and m2 . In this case, the equation of the hyperbola is :
C1, C2 and C3 can be computed thanks to a linear regression involving 3 parameters (to be defined in relation with C1, C2, C3).