38.16. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ find the focal point corresponding to the point $\left(a \cos t_{0}, b \sin t_{0}\right)$; find the radius of curvature of the ellipse at this point.
38.16. For a fixed point ( $x_{0}, y_{0}$ ), consider the function $F(t)=\left(x_{0}-a \cos t\right)^{2}+\left(y_{0}-b \sin t\right)^{2}$ on the ellipse. We are interested in the point $\left(x_{0}, y_{0}\right)$ for which $F^{\prime}\left(t_{0}\right)=0$ and $F^{\prime \prime}\left(t_{0}\right)=0$, i.e.,

$$
\begin{aligned}
x_{0} a \sin t_{0}-y_{0} b \cos t_{0}+\left(b^{2}-a^{2}\right) \sin t_{0} \cos t_{0} & =0 \\
x_{0} a \cos t_{0}+y_{0} b \sin t_{0}+\left(b^{2}-a^{2}\right)\left(\cos ^{2} t_{0}-\sin ^{2} t_{0}\right) & =0
\end{aligned}
$$

Solving this system of equations, we find

$$
x_{0}=\frac{a^{2}-b^{2}}{a} \cos ^{3} t_{0}, \quad y_{0}=-\frac{a^{2}-b^{2}}{b} \sin ^{3} t_{0}
$$

The radius of curvature $R$ is then computed using the formula

$$
R^{2}=\left(x_{0}-a \cos t_{0}\right)^{2}+\left(y_{0}-b \sin t_{0}\right)^{2}=\frac{\left(a^{2} \sin ^{2} t_{0}+b^{2} \cos ^{2} t_{0}\right)^{3}}{a^{2} b^{2}}
$$