412. Determine the radii of curvature at the vertices of the ellipse $x=a \cos t, y=b \sin t$.
Solution. Let's find the derivatives $\dot{x}=-a \sin t, \ddot{x}=-a \cos t$, $\dot{y}=b \cos t, \ddot{y}=-b \sin t$ and determine the radius of curvature of the ellipse at any of its points:

$$
R(t)=\frac{1}{K(t)}=\frac{\left(\dot{x}^{2}+\dot{y}^{2}\right)^{\frac{3}{2}}}{|\dddot{x} \ddot{y}-\dot{y}| \dot{x} \mid}=\frac{\left(a^{2} \sin ^{2} t+b^{2} \cos ^{2} t\right)^{\frac{3}{2}}}{a b}
$$

For the vertices of the ellipse lying on its axis $2 a$, the parameter $t$ is 0 or $\pi$. Therefore, the radius of curvature of the ellipse at these vertices $R(0)=R(\pi)=\frac{b^{2}}{a}$.

At the other two vertices of the ellipse, lying on the axis $2 b$, $t=\frac{\pi}{2}$ or $t=\frac{3 \pi}{2}$. At these vertices, the radius of curvature of the ellipse $R\left(\frac{\pi \pi}{2}\right)=R\left(\frac{3 \pi}{2}\right)=\frac{a^{2}}{b}$.