Answer
See graph
$(0,-2),(1,0)$
Work Step by Step
We are given the equations:
$\begin{cases}
4x^2+y^2=4\\
2x-y=2
\end{cases}$
$\begin{cases}
\dfrac{4x^2}{4}+\dfrac{y^2}{4}=1\\
y=2x-2
\end{cases}$
$\begin{cases}
\dfrac{x^2}{1}+\dfrac{y^2}{4}=1\\
y=2x-2
\end{cases}$
The first equation represents an ellipse. Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$:
$h=0$
$k=0$
$a^2=4\Rightarrow a=2$
$b^2=1\Rightarrow b=1$
Determine the foci:
$c^2=a^2-b^2$
$c^2=2^2-1^2$
$c^2=3$
$c=\pm\sqrt 3$
The foci are:
$F_1(h,k-c)=(0,-\sqrt 3)$
$F_2(h,k+c)=(0,\sqrt 3)$
The second equation represents a line with slope 2 and $y$-intersection $2$.
Draw the two equations.
There are 2 intersections:
$(0,-2),(1,0)$