The radius of curvature of a surface of revolution The radii of curvature $R_1$ and $R_2$ at point $A$ are $R_1=AM$ ($M$ is the center of curvature of the meridian curve in the plane of the figure) and $R_2=AN$ (in the perpendicular plane). $\vec{n}$ is the normal to the surface at point $A$. x is the symmetry axis.

My question is whether $N$ must lies on the symmetry axis.Is the length of $AN$ equal to the radius of curvature parellel to the latitude? If not, is there any  condition we need to add in order to make sure that the point $N$ lies on the symmetry axis?
 A: First, let's define a surface of revolution with $z$ as the symmetry axis:
$$f(t,\varphi) = \left ( r(t) \cos \varphi, \; r(t) \sin \varphi, \; h(t) \right )$$
For simplicity of notation, let's use
$$r = r(t), \quad \dot{r} = \frac{d r(t)}{d t}, \quad \ddot{r} = \frac{d^2 r(t)}{d t^2} \\
h = h(t), \quad \dot{h} = \frac{d h(t)}{d t}, \quad \ddot{h} = \frac{d^2 h(t)}{d t^2}$$
The surface normal unit vector is
$$\vec{n}(t, \varphi) = \left ( \frac{-\dot{h}\cos\varphi}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}}, \; \frac{-\dot{h}\sin\varphi}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}}, \; \frac{\dot{r}}{\sqrt{{\dot{r}}^2 + {\dot{h}}^2}} \right )$$
If $h\ne0$, and we have a point $\vec{p}(t, \varphi) = \left ( r\cos\varphi, r\sin\varphi, h \right)$ on the surface, and the unit normal $\vec{n}(t, \varphi)$ at that point, their slopes on the $xy$ plane are the same. This means that a line extending the unit normal will always intersect the symmetry axis ($z$ axis), as long as $h\ne0$. (If $h = 0$, then the surface normal is parallel to the symmetry axis.)
The principal curvatures -- maximum and minimum curvature at a given point $(t,\varphi)$ -- are
$$\begin{cases}
\kappa_1 = \frac{\dot{r}\ddot{h} - \ddot{r}\dot{h}}{({\dot{r}}^2+{\dot{h}}^2)^{3/2}} \\
\kappa_2 = \frac{\dot{h}}{r \sqrt{ {\dot{r}}^2 + {\dot{h}}^2 }}
\end{cases}$$
and the Gaussian curvature is
$$K = \kappa_1 \kappa_2 = - \frac{\ddot{r}}{r}$$
These hold as long as $r(t)$ and $h(t)$ are twice differentiable.

Radius of curvature is defined as the inverse of curvature. In that sense, the principal radii of curvature are
$$\begin{cases}
R_1 = \frac{1}{\kappa_1} = \frac{({\dot{r}}^2+{\dot{h}}^2)^{3/2}}{\dot{r}\ddot{h} - \ddot{r}\dot{h}} \\
R_2 = \frac{1}{\kappa_2} = \frac{r \sqrt{ {\dot{r}}^2 + {\dot{h}}^2 }}{\dot{h}}
\end{cases}$$
In three dimensions, the centers of their respective circles are always on the line extending the surface normal. (They do differ in their orientation, rotation of the respective circle around the surface normal.)
Alternatively, we can think of the radii as the radii of two spheres, one defining the minimum curvature at that point on the surface, and the other the maximum curvature. The centers of the spheres are still on the line extending the surface normal.
If we look at the original illustration,

then, given that this is a surface of revolution, a line passing through $A$ and $M$, or through $A'$ and $M'$, will always pass through the symmetry axis.
In all honesty, I fail to see any relevance in the location of $M$, really. Since 3D surfaces have two principal curvatures at each point on the surface, using a single point to describe (a part of) it seems.. counterintuitive to me. It is much easier to use the surface normal and curvature parameters, in my opinion.
A: I think the answer to your question is in the equation that was given for $R_2$ in the original answer. If you pass the h' in the denominator to inside the square root, you'll end up with $R_2 = r\sqrt{1+(dr/dt)^2}$ which is the same as $r/cos(\theta)$, which gives the radius you were asking about. 
I had the same question a few years ago, since I was taught this with no proof as an engineering undergrad. I had forgotten the proof by now, and it was very helpful to be reminded of the main steps in the first answer to your question. Thanks for posting the question so properly.
A: (Clarifying remarks as the topic is still current).
Every doubly curved surface has two principal curvatures. Their reciprocal radii of curvature are: 
$$ \kappa_1== \dfrac {1}{R_1},\,\kappa_2== \dfrac {1}{R_2}\,;$$
For the meridional section $ x-z $ shown in the plane of paper considering neighborhood meridional arc,
$$ AM= R_1$$
Next consider plane orthogonal to the above. If the cone with slant heights is cut and removed, the slant height $R_2$ between  base of cone and symmetry axis is:
$$ AN= R_2 $$
Centers of curvature $(M,N)$ lie on same or opposite sides of meridian according as curvature product is positive or negative, as shown in sketch.
There is a smooth transition from $\kappa_1$  to $\kappa_2$ as $ \psi $ varies from $0$ to $\pi/2$  in the tangent plane when normal intersection plane rotates around normal line axis $ANM$ of the normal vector $\vec n$  at any point $A$ on the surface.
$$ \kappa_n = \kappa_1\cos^2 \psi +\kappa_2\ \sin ^2 \psi $$
according to Euler's relation of curvatures.
