ap.analysis of pdes – Beltrami equation

The second order linear PDE

$$a(x,y)u_{xx}+ 2b(x,y)u_{xy}+c(x,y)u_{yy}= 0$$

was transformed by introducing new independent variable $$ξ=φ(x,y)$$ and $$η=ψ(x,y)$$, transforming the PDE to the form

$$A(ξ,η)u_{ξξ}+ 2B(ξ,η)u_{ξη}+C(ξ,η)u_{ηη}= 0$$

with

$$A=aphi_x^2 + 2bphi_xphi_y+cphi_y^2$$

$$B=aphi_xpsi_x+b(phi_xpsi_y+phi_ypsi_x)+cphi_ypsi_y$$

$$C=apsi_x^2+2bpsi_xpsi_y+cpsi_y^2$$

Assume we are in the elliptic case.
Show that in this case we can obtain the situation $$A=C$$ and $$B=0$$ in case that $$ψ$$ and $$φ$$ are solutions of

$$φ_x=frac{bψ_x+cφ_y}{sqrt{ac−b^2}},;;;φ_y=−frac{aψ_x+bφ_y}{sqrt{ac−b^2}}$$

and $$ψ$$ is a solution of the Beltrami equation:

$$left(frac{aψ_x+bφ_y}{sqrt{ac−b^2}}right)_x+ left(frac{bψ_x+cφ_y}{sqrt{ac−b^2}}right)_y= 0$$.

I’ve tried to calculate $$A-C=0$$ and $$B=0$$ and set them even. But I’m stuck at

$$frac{phi_x+ipsi_x}{phi_y+ipsi_y} = frac{-bpmsqrt{b^2-ac}}{a}$$.