Answer
$(0,1)$ and $(2,0)$ are on the graph
Work Step by Step
In order to determine if a point is on the graph of the equation, we have to substitute into the equation: $x^2+4y^2=4$
For the point of $(0,1)$ substitute $x=0$ and $y=1$ into the equation
$0^2+4(1)^2=4\\
4=4$
The equation is satisfied, therefore the point is on the graph.
For $(2,0)$:
$x=2$ and $y=0$
$2^2+4(0)^2=4\\
4=4$
The equation is satisfied, therefore the point is on the graph.
For $(2,\frac{1}{2})$:
$x=2$ $y=\frac{1}{2}$
$2^2+4(\frac{1}{2}^2=4\\
5\ne4$
The equation is not satisfied, therefore the point is not on the graph.
