Change to polar coordinates the equation $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.
The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is
$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$
$\begin{aligned}
\therefore \; \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 & \implies \dfrac{r^2 \cos^2 \theta}{a^2} + \dfrac{r^2 \sin^2 \theta}{b^2} = 1 \\\\
& i.e. \;\; r^2 \left(\dfrac{b^2 \cos^2 \theta + a^2 \sin^2 \theta}{a^2 b^2}\right) = 1 \\\\
& i.e. \;\; b^2 \cos^2 \theta + a^2 \sin^2 \theta = \dfrac{a^2 b^2}{r^2}
\end{aligned}$
$\therefore \;$ The polar form of the given Cartesian equation $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ is $\;$ $b^2 \cos^2 \theta + a^2 \sin^2 \theta = \dfrac{a^2 b^2}{r^2}$.