3. Let the bivariate curve $C$ be given by the equation: $3 y^{2}-4(x+1) \cdot y+12(x-2)=0$. Then the graph of $C$ is (.
A. Ellipse
B. Hyperbola
C. Circle
D. None of the above
3. $B$

The original equation can be transformed into: $(x-1)^{2}+y^{2}=(x-2 y+5)^{2} \Rightarrow \frac{\sqrt{(x-1)^{2}+y^{2}}}{|x-2 y+5 / \sqrt{5}|}=\sqrt{5}$, which means $t=\sqrt{5}$.
According to the definitions of hyperbolas and ellipses,
curve $C$ should be a hyperbola.