Calculating the Activity Coefficient parameters form Henry's Law

Here is the question, I am stuck on part B:


[![Question](https://i.stack.imgur.com/RAxBr.png)](https://i.stack.imgur.com/RAxBr.png)


Basically, this is my thought process so far, as $x\_a \rightarrow 0$, the Henry's law line becomes tangential to the actual partial pressure curve..
[![Graphs](https://i.stack.imgur.com/tVoz0.png)](https://i.stack.imgur.com/tVoz0.png) Henry's law states that:  

$$
p\_{\alpha} = Hx\_{\alpha}
$$


Similarly, using the activity coefficient model:
$$
p\_{\alpha} = \gamma\_{\alpha}x\_{\alpha}p^\*\_{\alpha}
$$


Where H is Henry's constant and $p^\*\_{\alpha}$ is the vapour pressure of the pure component. So at small $x\_{\alpha}$ values one can assume:
$$
H = \gamma\_{\alpha}p^\*\_{\alpha}
$$


From this I can get values for the $\gamma$ (I've been given values for pure component pressure). Then I can get values for $\ln \gamma\_1, \ln \gamma\_2$. My next step would be to substitute in values of $x\_2$ and $x\_1$ into the given correlations. So, for $\gamma\_1$ to be correct, $x\_1$ must be very small, assume it is zero therefore $x\_2 \approx 1$, therefore:


$$
\ln \gamma\_1 = \frac{A \times 1^2}{\bigg( (1-1)^2\frac{A}{B} + 1 \bigg)^2} \implies \ln \gamma\_1 = A
$$


Clearly, this is wrong because if I run through the same process for $\ln \gamma\_2$, I get a different A value. Can someone please advise, thanks.


