How to show $A=\{(x,y)\in R^2:4x^2+9y^2=36\}$ is path connected and compact?

Question

let $A=\{(x,y)\in R^2:4x^2+9y^2=36\}$ . Show that A is path connected and compact.

my attempt: since $\frac {x^2}{9}+\frac{y^2}{4}=1$ is elips. A is bounded and closed. so is compact. (by heine borel) How do we show it is path connected?

Answer 1

Find a path between two points, and you will have shown that the set is path connected.

Note that the parameterization $$ x = 3 \cos t; \quad y = 2 \sin t; \quad t \in \mathbb{R} $$ Yields a path through the entire space. Use this to "connect" any two points.

Alternatively: $A$ is the continuous image of $S^1$ under some map (which you should find). Since continuous maps preserve path connectedness, it suffices to note (or show) that $S^1$ is path connected.