So I need to find the intersection between these two:
$4x^2+16y^2-64=0$ and $x^2+y^2=9$
Following a similar approach of substitution, I was able to get 4 points: $(\sqrt {20/3}, \sqrt{7/3}), (\sqrt{-20/3}, \sqrt{7/3}), (\sqrt{-20/3},\sqrt{-7/3}), (\sqrt{20/3}, \sqrt{-7/3})$. But the answer says $(3,0) and (-3,0)$... What went wrong?
$4x^2+16y^2-64=0$ and $x^2+y^2=9$
One little picture says more than a long speech!
In your answer : $$(\sqrt {20/3}, \sqrt{7/3}), (\sqrt{-20/3}, \sqrt{7/3}), (\sqrt{-20/3},\sqrt{-7/3}), (\sqrt{20/3}, \sqrt{-7/3})$$ the signs $-$ are not at right places. The correct typing is : $$(\sqrt {20/3}, \sqrt{7/3}), (-\sqrt{20/3}, \sqrt{7/3}), (-\sqrt{20/3},-\sqrt{7/3}), (\sqrt{20/3}, -\sqrt{7/3})$$ Obviously $(3,0)$ and $(-3,0)$ are not correct. Probably there is a typo in the wording of the problem and/or given solution.
If they are only two points of intersection $(3,0)$ and $(-3,0)$ one possibility is : $4x^2+16y^2=36$.
Another possibility with non-concentric circle and ellipse : $4x^2+16(y+c)^2=64 \quad,\quad c=\pm\frac{\sqrt{7}}{2}$.
They are other possibilities with non-concentric circle and ellipse (bigger ellipses).