I was actually trying to find a partial differential equation for the surface, but I got two equations I don't know which is correct. Here are my steps:
The surface is $\log(au-1)=x+ay+b$, so $au = 1+e^{x+ay+b}$. In the picture below you can see I have got two equations, please help
Actually both the PDEs are same, just eliminate $u_{y}$ and $u_{yy}$ from second PDE. Because $u_{y}^{2}=a^{2}u_{x}^{2}$ and $ u_{yy}=a^{2}u_{xx}. $
I will give you simple geometrical explanation, consider two differential equations $$ \frac{dy}{dx}=1 ~ \text{and} ~ \left ( \frac{dy}{dx} \right)^{2}=1 .$$ Then we need to find curve with slope $1$ and for second one, curve with slope $1$ or $-1$. Both the differential equations are different, one is linear and the other is nonlinear, but still have common solution $y=x.$ In other way you can say that for curve $y=x$, we have two different differential equations.