#### Answer

$(-1,0)$, $(1,0)$

#### Work Step by Step

Here, we have $(4x^2-y^2)+(4x^2+y^2)=4+4$
Now, $4x^2-y^2+4x^2+y^2=8$
This gives: $8x^2=8$
This yields : $x^2=1 \implies x=1,-1$
when $x=1$ then we have $4 (1)^2+y^2=4 \implies y^2=0$
and $y=0$
when $x=-1$ then we have $4 (-1)^2+y^2=4 \implies y^2=0$
and $y=0$
Hence, our answers are: $(-1,0)$, $(1,0)$