Gaussian beam reflection derivation

In this PhD thesis I found a good treatment of refraction and reflection of Gaussian beams at a curved surface.
My doubt is from where does it come the cube in the cosine at the denominator of $x^2$ in eq. 4.28 (pag. 58):
$$z_iapprox x_i tan theta_i+frac{x_i^2}{2 R_I cos^3 theta_i}+frac{y_i^2}{2 R_S cos theta_i}$$
Following that approach (actually from the paper "Reflection and Refraction of Gaussian Light Beams at Tilted Ellipsoidal Surfaces") I have to find an approximated solution for z on the ellipsoid of revolution from eq. 4.27:
$$ (x_i+A sin theta_i)^2+(A y_i/B)^2+(z_i-A cos theta_i)^2=A^2$$
The exact solution is
$$z_i=A cos theta_i pm sqrt{A^2-(x_i+A sin theta_i)^2+(A y_i/B)^2} = = A cos theta_i pm A cos theta_i sqrt{1- frac{x_i^2+2 A x_i sin theta_i+ (A y_i/B)^2}{A^2 cos^2 theta_i}}$$
Where the minus solution is to be selected. The first order approximation is then
$$z_i approx A cos theta_i left( frac{x_i^2+2A x_i sin theta_i+ (A y_i/B)^2}{2 A^2 cos^2 theta_i}right) = x_i tan theta_i + frac{x_i^2}{2 A cos theta_i}+ frac{A y_i^2}{2 B^2 cos theta_i}$$
It adds up, except for the $x^2$ term.

Moreover, I don’t understand how it gets to eq. 4.31:
$$psi_r(x_i, y_i)= k_i x_i tan theta_i+ frac{k_i x_i^2}{2} left( frac{1}{q_{1r}}+ frac{1-2 cos^2 theta_i}{R_I cos^3 theta_i} right) + frac{k_i y_i^2}{2} left( frac{1}{q_{2r}}+ frac{1-2 cos^2 theta_i}{R_S cos^3 theta_i} right)$$