Answer
The set includes all the points inside the ellipse $\frac{x^2}{4}+y^2 = 1$
Note that the points on the ellipse are also included in the set.
Work Step by Step
$\{(x,y)\vert~x^2+4y^2 \leq 4 ~\}$
$x^2+4y^2 = 4$
$\frac{x^2}{4}+y^2 = 1$
We can write the general form of an ellipse:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$
The equation in the question is an equation of an ellipse, where $a = 2, b = 1, h = 0,$ and $k = 0$
The set includes all the points inside the ellipse $\frac{x^2}{4}+y^2 = 1$
Note that the points on the ellipse are also included in the set.
