Answer
Using Eq. (10) we obtain the minimal curvature ${\kappa _{\min }} = \frac{b}{{{a^2}}}$ and the maximal curvature ${\kappa _{\max }} = \frac{a}{{{b^2}}}$. Therefore, we conclude that
$\frac{b}{{{a^2}}} \le \kappa \left( t \right) \le \frac{a}{{{b^2}}}$ for all $t$.
Work Step by Step
We have Eq. (10) of Exercise 25 for the ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$:
$\kappa \left( t \right) = \frac{{ab}}{{{{\left( {{b^2}{{\cos }^2}t + {a^2}{{\sin }^2}t} \right)}^{3/2}}}}$
In Exercise 26, we have shown that the curvature is minimal at $t = \frac{\pi }{2}$ and $t = \frac{{3\pi }}{2}$.
Substituting either $t = \frac{\pi }{2}$ or $t = \frac{{3\pi }}{2}$ in $\kappa \left( t \right)$ gives the minimum of $\kappa \left( t \right)$:
${\kappa _{\min }} = \frac{{ab}}{{{{\left( {{a^2}} \right)}^{3/2}}}} = \frac{b}{{{a^2}}}$
Also in Exercise 26, we have shown that the curvature is maximal at $t=0$ and $t=\pi$.
Substituting either $t=0$ or $t=\pi$ in $\kappa \left( t \right)$ gives the maximum of $\kappa \left( t \right)$:
${\kappa _{\max }} = \frac{{ab}}{{{{\left( {{b^2}} \right)}^{3/2}}}} = \frac{a}{{{b^2}}}$
Since ${\kappa _{\min }} = \frac{b}{{{a^2}}}$ and ${\kappa _{\max }} = \frac{a}{{{b^2}}}$, we conclude that $\frac{b}{{{a^2}}} \le \kappa \left( t \right) \le \frac{a}{{{b^2}}}$ for all $t$.
