38.14. Calculate the curvature of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at each point.

## 38.4. Osculating Circle

Consider two curves on a plane given by the equations $y=f(x)$ and $y=g(x)$. These curves intersect if $f\left(x_{0}\right)=g\left(x_{0}\right)$. The curves touch (have a contact of order 1) at the point of intersection if $f^{\prime}\left(x_{0}\right)=g^{\prime}\left(x_{0}\right)$. Curves that touch have a contact of order $n$ if $f\left(x_{0}\right)=g\left(x_{0}\right), f^{\prime}\left(x_{0}\right)=g^{\prime}\left(x_{0}\right), \ldots, \ldots, f^{(n)}\left(x_{0}\right)=g^{(n)}\left(x_{0}\right)$. Curves that have a contact of order 2 are called osculating.

If the curvature of a curve at some point is non-zero, then there is a unique circle that is osculating to the curve at that point (the osculating circle). The equation of the osculating circle in the case where the curve is given by the equation $y=f(x)$ is found as follows. Let $y=g(x)$ be the function locally defining a circle of radius $R$ with center $(a, b)$, i.e., $g$ satisfies the relation

$$
(x-a)^{2}+(g(x)-b)^{2}=R^{2}
$$

Differentiating this relation, we successively obtain

$$
\begin{array}{r}
2(x-a)+2 g^{\prime}(x)(g(x)-b)=0 \\
2+2 g^{\prime \prime}(x)(g(x)-b)+2\left(g^{\prime}(x)\right)^{2}=0
\end{array}
$$

If the curve $y=f(x)$ is osculating to the specified circle at the point $\left(x_{0}, y_{0}\right)$, then

$$
\begin{aligned}
\left(x_{0}-a\right)^{2}+\left(y_{0}-b\right)^{2} & =R^{2} \\
\left(x_{0}-a\right)+y_{0}^{\prime}\left(y_{0}-b\right) & =0 \\
1+y_{0}^{\prime \prime}\left(y_{0}-b\right)+\left(y_{0}^{\prime}\right)^{2} & =0
\end{aligned}
$$

where $y_{0}^{\prime}=f^{\prime}\left(x_{0}\right)$ and $y_{0}^{\prime \prime}=f^{\prime \prime}\left(x_{0}\right)$. Moreover, if this system of equations for $a, b, R$ has a solution, then the converse is also true: the curve is osculating to the circle. It is easy to see that if $y_{0}^{\prime \prime} \neq \mathbf{0}$, then this system of equations has a unique solution. For example, if $y_{0}^{\prime}=0$ (which can always be achieved by a suitable choice of coordinate system), then $a=x_{0}, b=y_{0}+\frac{1}{y_{0}^{\prime \prime}}, R^{2}=\frac{1}{\left(y_{0}^{\prime \prime}\right)^{2}}$.

The center of the osculating circle is called the center of curvature of the curve. According to problem 38.13 in the considered situation, $k^{2}=\frac{\left(y_{0}^{\prime \prime}\right)^{2}}{\left(1+y_{0}^{\prime 2}\right)^{3}}=\left(y_{0}^{\prime \prime}\right)^{2}$. Thus, the center of curvature lies on the normal to the curve at the given point and is at a distance of $\pm 1 / k$ from this point.
38.14. Consider the following parameterization of an ellipse: $x(t) = a \cos t, y(t) = b \sin t$. Then

$$
\begin{gathered}
x'' y' - y'' x' = -ab(\cos^2 t + \sin^2 t) = -ab, \\
x'^2 + y'^2 = a^2 \sin^2 t + b^2 \cos^2 t
\end{gathered}
$$

Therefore, according to problem 38.13,

$$
k^2 = \frac{a^2 b^2}{(a^2 \sin^2 t + b^2 \cos^2 t)^3}
$$

Remark. Another method for calculating the curvature of an ellipse is given in problem 38.16.