I have a point $P'$ given by known Cartesian coordinates $(x',y',z')$ and an oblate spheroidal surface parameterized by \begin{align} x &= a \cosh{\mu} \cos{\nu} \cos{\phi} \\ y &= a \cosh{\mu} \cos{\nu} \sin{\phi} \\ z &= a \sinh{\mu} \sin{\nu} \end{align} where $a,\mu$ are known constants. For an oblate spheroid, $\mu$ is similar to a radial coordinate, $\nu$ is similar to a latitude, and $\phi$ is the same azimuthal angle from spherical coordinates. So for an oblate spheroidal surface, $\mu$ is fixed.
When you draw a line from the point $P'$ perpendicular to the oblate spheroid surface ($P'$ is outside the surface), it will intersect the surface at some point $P$ at coordinates $(x,y,z)$. I need to find the location of this point (it will intersect the surface at 2 points, but I want the intersection closest to $P'$ and normal to the surface).
To do this, I write the equation of that line containing both $P'$ and $P$ as: $$ \mathbf{x}' = \mathbf{x} + t \hat{e}_{\mu} $$ where $\mathbf{x}$ is the point I want to find, $t$ is unknown, and $\hat{e}_{\mu}$ is the outward normal vector to the oblate spheroid surface at the point $\mathbf{x}$: $$ \hat{e}_{\mu} = {1 \over \sqrt{\sinh^2{\mu} + \sin^2{\nu}}} \begin{pmatrix} \sinh{\mu} \cos{\nu} \cos{\phi} \\ \sinh{\mu} \cos{\nu} \sin{\phi} \\ \cosh{\mu} \sin{\nu} \end{pmatrix} $$ So we have 3 equations and 2 unknowns: $t$ and $\nu$. As I mentioned before, $a$ and $\mu$ are known constants, $\mathbf{x}'$ is given, and the angle $\phi$ can be determined from symmetry, since both points $P$ and $P'$ will share the same azimuth angle.
Since $\mu$ and $\phi$ are known, the goal is to solve for $\nu$ since that would immediately yield the coordinates $(x,y,z)$. I don't see an easy way to solve this equation analytically - I could use numerical methods but it seems to me that there should be an analytical solution to this. Does anyone see how?