Let  C and D are corresponding points of  A and B on the auxiliary circle  x² + y² = 16 respectively.
∴     A (4 cos θ₁, 2 sin θ₁) ,
        B(4 cos θ₂, 2 sin θ₂) ,
        C(4 cos θ₁, 4 sin θ₁) ,
        D(4 cos θ₂, 4 sin θ₂)

$\text{ Also, }\mathrm{ar}.\left(\mathrm{△}\mathrm{AOB}\right)=\frac{1}{2}\mathrm{ar}.\left(\mathrm{△}\mathrm{COD}\right)$

∴   For maximum area of  ΔAOB,  area (ΔCOD) must be maximum

$\Rightarrow \mathrm{\angle }\mathrm{COD}={90}^{\circ }$

$\therefore \text{ Maximum area of }\mathrm{△}\mathrm{AOB}=\frac{1}{2}\times \left(\text{ Maximum area of }\mathrm{△}\mathrm{COD}\right)=\frac{1}{2}\left(\frac{1}{2}\times 4\times 4\right)=4$

Let slope of line  CD be  m (m < 0)

$\therefore \text{ The equation of }\mathrm{CD}\text{ is }\mathrm{mx}-\mathrm{y}-\mathrm{m}\sqrt{264}=0$

$\text{ As, }\mathrm{OM}=2\sqrt{2}\Rightarrow \frac{\left|\mathrm{m}\sqrt{264}\right|}{\sqrt{1+{\mathrm{m}}^2}}=2\sqrt{2}\Rightarrow \mathrm{m}=\frac{-2\sqrt{2}}{16}\Rightarrow \mathrm{m}=\frac{-2}{8\sqrt{2}}$

$\therefore \text{ The slope of }\mathrm{AB}=\frac{\mathrm{m}}{2}=\frac{-1}{8\sqrt{2}}$

Aliter :

Equation of chord AB is $\frac{\mathrm{x}}{4}\cos \left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)+\frac{\mathrm{y}}{2}\sin \left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)=\cos \left(\frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_2}{2}\right)$      ……(1)

$\text{ Area of }\mathrm{△}\mathrm{AOB}=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 4\cos {\mathrm{\theta }}_1& 2\sin {\mathrm{\theta }}_1& 1\\ 4\cos {\mathrm{\theta }}_2& 2\sin {\mathrm{\theta }}_2& 1\end{array}\right|$

$\mathrm{\Delta }=4\sin \left({\mathrm{\theta }}_2-{\mathrm{\theta }}_1\right)$

$\text{ Hence, }{\mathrm{\Delta }}_{max.}=4\text{, when }\sin \left({\mathrm{\theta }}_2-{\mathrm{\theta }}_1\right)=1\text{. }$

$\therefore {\mathrm{\theta }}_2=\left(\frac{\mathrm{\pi }}{2}+{\mathrm{\theta }}_1\right)$

$\text{ Now, since }\mathrm{P}(\sqrt{264},0)\text{ lies on }\mathrm{AB}$

$\text{ Hence, }\cos \left(\frac{\mathrm{\pi }}{4}+{\mathrm{\theta }}_1\right)=\frac{1}{\sqrt{33}}$

$\text{ Slope of }\mathrm{AB}=\frac{-2}{4}\cot \left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)=\frac{-1}{2}\cot \left(\frac{\mathrm{\pi }}{4}+{\mathrm{\theta }}_1\right)=\frac{-1}{2}\times \frac{1}{\sqrt{32}}=\frac{-1}{8\sqrt{2}}$