4. Planar vectors $\vec{u}=(x, y), \vec{b}=\left(x^{2}, y^{2}\right), \vec{c}=(1,1), \vec{a}=\left(\frac{1}{4}, \frac{1}{9}\right)$, if $\vec{a} \cdot \vec{c}$ $=1, \vec{b} \cdot \vec{d}=1$, then the number of such vectors $\vec{a}$ is ( ).
A. 1
B. 2
C. More than 2
D. Does not exist
4. B

From $\vec{a} \cdot \vec{c}=1 \Rightarrow x+y=1, \vec{b} \cdot \vec{d}=1 \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$, it can be seen from the graph that there are two such $\vec{a}$