Equation for shape of water drop on hydrophobic surface

Initially, before the water droplet meets any other material surface, it's (ideal) initial spherical shape sphere contains and maintains an isotropic pressure distribution. Surface tension, due to cohesive forces, tends keeps the water molecules bound in the spherical shape. 

In your question, you then introduce a hydrophobic surface, in this case wax paper, which will obviously minimise the conversion of cohesion to adhesion, but also obviously will distort the ideal spherical shape towards one with an elliptical cross section.

Luckily for me, the Young Laplace Equation, as taken from Wikipedia, is described as the equation involved in determining the resulting shape:

The Young–Laplace equation relates the pressure difference to the shape of the surface or wall and it is fundamentally important in the study of static capillary surfaces. It is a statement of normal stress balance for static fluids meeting at an interface, where the interface is treated as a surface (zero thickness):

$${displaystyle {begin{aligned}Delta p&=-gamma {bar {nabla }}cdot {hat {n}}&=2gamma H&=gamma left({frac {1}{R_{1}}}+{frac {1}{R_{2}}}right)end{aligned}}}$$

where ${displaystyle Delta p}$ is the pressure difference across the fluid interface,

$γ$ is the surface tension (or wall tension), 

${displaystyle {hat {n}}}$ is the unit normal pointing out of the surface, 

${displaystyle H}$ is the mean curvature,

${displaystyle R_{1}}$ and ${displaystyle R_{2}}$ are the principal radii of curvature.

Note that the ${displaystyle {bar {nabla }}}$ is the divergence on surface, which is defined by $${displaystyle {bar {nabla }}=nabla -{hat {n}}({hat {n}}cdot nabla )}$$

Note that only normal stress is considered, this is because it can be shown that a static interface is possible only in the absence of tangential stress.