<s>[INST] Find $\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\}\cap\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$ 
Find $A \cap  B$, where
$A=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2=3\},\quad B=\{(a,b) \in \mathbb{R}^2 : 2a^2 + 3 b^2=6\}$

I know that $(\sqrt{3},0)$ and $(-\sqrt{3},0)\in A \cap B$ so $A\cap B\neq\emptyset$, but how can I find all of elements of $A \cap B$ ?
 [/INST] An.option:
Ellipse $2x^2+3y^2=6$, or
$3(x^2+y^2) =6+x^2$.
With $d^2$ the (distance)$^2$ from the origin:
$d^2:= x^2+y^2= 2 +(1/3)x^2 \le 3$,
since  $|x| \le √3$ for points on the ellipse.
$d^2 = 3$ , for  $x = ^{+}_{-}√3$, else $d^2<3$, i.e
only $2$ points on the circle with $r^2=3$.
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