Answer
From the figure we see that the ellipse bends minimally at $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$. Whereas at $t=0$ and $t = \pi $, it bends maximally.
Applying Eq. (10) confirms our prediction.
Work Step by Step
Without loss of generality, let $a>b$, where $a$ is the semimajor axis located in the $x$-axis; and $b$ is the semiminor axis located in the $y$-axis (see the figure attached). Note that this assumption conforms to the notation in Eq. (10). Since curvature is a measure of how much a curve bends, from the figure we see that the ellipse bends minimally at $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$. Whereas at $t=0$ and $t = \pi $, it bends maximally. Thus, we predict that the points of minimal curvature occurs at $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$, and the points of maximal curvature occurs at $t=0$ and $t = \pi $.
Now, we use Eq. (10) of Exercise 25 for the ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$ to confirm our prediction:
$\kappa \left( t \right) = \frac{{ab}}{{{{\left( {{b^2}{{\cos }^2}t + {a^2}{{\sin }^2}t} \right)}^{3/2}}}}$
Since $a$ and $b$ are constants, $\kappa \left( t \right)$ is minimal if the denominator is maximal. Whereas, $\kappa \left( t \right)$ is maximal if the denominator is minimal.
Since $0 \le {\cos ^2}t \le 1$, $0 \le {\sin ^2}t \le 1$ and $a>b$, the denominator is maximal at the points where $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$. Thus, the curvature is minimal at $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$.
Similarly, the denominator is minimal at the points where $t=0$ and $t = \pi $. Thus, the curvature is maximal at $t=0$ and $t = \pi $.
