Is the set $\{(x, y) : 3x^2 − 2y^ 2 + 3y = 1\}$ connected? Is the set $\{(x, y)\in\mathbb{R}^2 : 3x^2 − 2y^ 2 + 3y = 1\}$ connected?
I have checked that it is an hyperbola, hence disconnected am i right?
 A: Your set $S$ contains the points $(0,{1\over2})$ and $(0,1)$, but does not contain any points on the line $y={3\over4}$. Therefore $$S\cap\{(x,y)|y<{3\over4}\},\quad S\cap\{(x,y)|y>{3\over4}\}$$
is a partition of $S$ into disjoint (relatively) open subsets of ${\mathbb R}^2$. This shows that $S$ is not connected.
A: Assuming, you mean 
$$S=\{(x,y)\in\mathbb R^2:3x^2-2y^2+3y=1\},  $$
observe that $f\colon(x,y)\mapsto y-\frac23$ is a continuos map. There is no point $(x,y)\in S$ with $f(x,y)=0$ because $3x^2-2\cdot \frac49+3\cdot \frac23=1$ (i.e. $3x^2=-\frac19$) has no real solution. On the other hand, $(0,1)\in S$ and $(0,\frac12)\in S$, so that $f$ does take both positive ans negative values. Therefore we can write $S=f^{-1}((-\infty,0))\cup f^{-1}((0,\infty))$ as disjoint union of nonempty open sets, i.e. $S$ is not connected.
A: For any $x$, you can solve for exactly two solutions in $y$. One solution set has $y \geq 1$, the other $y \leq \frac{1}{2}$. This will give you the decomposition that witnesses that the set is disconnected.
A: Per definition hyperbola is an unbounded set, consisting of two disconnected parts. Proof is however to raise that this is a hyperbola. And you say you have that already.
