\[\frac{x^2}{a^2} + \frac{y^2}{b^2} =1, \; a lt b\]
the perimeter is given by \[P=\int_C \sqrt{(dx)^2+(dy)^2} = \]
& where \[x=acos \theta , \; y=bsin \theta \rightarrow dx=-asin \theta d \theta , \; dy=bcos \theta d \theta\]
, where \[0 \le \theta \lt 2 \pi\]
.Hence
\[\begin{equation} \begin{aligned} P &= \int^{2 \pi}_0 \sqrt{(a^2sin^2 \theta (d \theta )^2+ b^2 cos^2 \theta (d \theta )^2} \\ &= b \int^{2 \pi}_0 \sqrt{\frac{a^2}{b^2}sin^2 \theta + cos^2 \theta} d \theta \\ &= b \int^{2 \pi}_0 \sqrt{-sin^2 \theta + \frac{a^2}{b^2}sin^2 \theta +cos^2 \theta +sin^2 \theta } d \theta \\ &= \int^{2 \pi }_0 \sqrt{1- (1- \frac{a^2}{b^2} sin^2 \theta } d \theta \end{aligned} \end{equation}\]
.This is an elliptic integral of the second kind (
\[ \int^{\theta}_0 \sqrt{1-k^2 sin^2 \theta } d \theta \]
).