Radius vs Radius of curvature of an ellipse I am a bit confused by the physical meaning of radius vs radius of curvature, with regard to an ellipse.
For a standard ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
In this case, the $a$ and $b$ refer to the "radius of curvature" of the ellipse in the $x$ and $y$ direction respectively.
In contrast to the radius of curvature for an ellipse:
$$ \frac{(a^2 \sin^2t + b^2 \cos^2t)^\frac{3}{2}}{ab} $$
Let's say that at $t = 0$, we get a radius of curvature of $\frac{b^2}{a}$.
How does this value relate to the original equation, where I should get 
$x = b$ at $t = 0$ (since b is the "radius" in the x-direction) instead?
 A: Let's say we're given a point $P_0$ on the ellipse. We can pick two other points, $P_1$ and $P_2$ on either side of $P_0$.
It is well known that three points uniquely define a circle. So, we can find a unique circle passing through $P_0$, $P_1$, and $P_2$.
Now imagine moving $P_1$ and $P_2$ towards $P_0$, until they are infinitesimally close. The radius of the resulting circle is the curvature radius. The circle itself is called the oscillating circle. Note that the radius of curvature is different from the lengths of the semi-major and semi-minor axes as you have pointed out.
How can we find the radius of curvature?
For the sake of generalization, let's say we have a random parametric curve, $(X(t),Y(t))$. In our case, $X(t)=a\cos t$ and $Y(t)=b\sin t$. Pick a random point on this curve, $(X(t_0), Y(t_0))$. Our goal is to find the radius of curvature, $R(t_0)$, at this point. Consider the point $(X(t_0+dt),Y(t_0+dt))$. The arc length between the these two points, $ds$, can be can be written like so:
$$ds=\sqrt{dX(t_0)^2+dY(t_0)^2}$$
However, since the arc length of a circle is $\theta r$, we can also write the arc length like so:
$$ds=
d\phi(t_0)\cdot R(t_0)$$
Where $\phi(t_0)$ is the tangential angle at $t_0$. Try to convince yourself that this is true by drawing pictures (it took me a while at first).
So, we can deduce the following:
$$R(t_0)=\frac{\sqrt{dX(t_0)^2+dY(t_0)^2}}{d\phi(t_0)}=\frac{\sqrt{\left(\frac{dX(t_0)}{dt}\right)^2+\left(\frac{dY(t_0)}{dt}\right)^2}}{\frac{d\phi(t_0)}{dt}}=\frac{\sqrt{X'(t_0)^2+Y'(t_0)^2}}{\phi(t_0)}$$
We also know the following:
$$\tan\left(\phi(t_0)\right)=\frac{Y'(t)}{X'(t)}$$
So…
$$\frac{d}{dt}\tan\left(\phi(t_0)\right)=\sec^2(\phi(t_0))\cdot\phi'(t_0)=\frac{X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)}{\left(X'(t_0)\right)^2}$$
From this, we deduce the following:
\begin{equation} \label{eq1}
\begin{split}
\phi'(t_0) & = \frac{1}{\sec^2\theta}\cdot\frac{X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)}{\left(X'(t_0)\right)^2} \\
 & = \frac{1}{1+\tan^2\theta}\cdot\frac{X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)}{\left(X'(t_0)\right)^2} \\
&=\frac{1}{1+\frac{Y'(t_0)}{X'(t_0)}}\cdot\frac{X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)}{\left(X'(t_0)\right)^2}\\
&=\frac{X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)}{X'(t_0)^2+Y'(t_0)^2}
\end{split}
\end{equation}
Finally, we come to the conclusion that
$$R(t_0)=\frac{\left(X'(t_0)^2+Y'(t_0)^2\right)^\frac{3}{2}}{\left|X'(t_0)Y''(t_0)-Y'(t_0)X''(t_0)\right|}$$
The absolute value in the denominator comes from the fact that $R(t_0)$ must be positive.
Now plug in $X(t)=a\cos t$ and $Y(t)=b\sin t$. You will get the same result mentioned in the OP.
I got the derivation from this website: http://mathworld.wolfram.com/Curvature.html
[2]: http://istack.imgur.com/ypzxQ.png
A: The radius of curvature $R$ of a curve at a point is the radius of the circular arc which ''best'' approximates the curve at that point.
Here ''best'' means that the system given by the curve equation and the circle equation have a double root in the point of contact.
Properly $a$ is not a ''radius'' of the ellipse, but it is the radius of the circle with center $(0,0)$ and that pass in the vertex $(a,0)$ of the ellipse.
This is not the circle that ''best approximates'' the ellipse in this point, because the system has also the solution $(-a,0)$.
The ''best'' circle is
$$
\left[x-\left( a-\dfrac{b^2}{a}\right)  \right]^2+y^2=\left(\dfrac{b^2}{a}\right)^2
$$
that has as radius the curvature radius $R=\dfrac{b^2}{a} $ as found in OP.
And you can see that the system of this equation and the equation of the ellipse has double root in $(a,0)$.
