4. Let $S_{n}$ be the sum of the products of the elements of all 3-element subsets of the set $A=\left\{1, \frac{1}{2}, \frac{1}{4}, \cdots, \frac{1}{2^{n-1}}\right\}$, and $\lim _{n \rightarrow \infty} \frac{4 S_{n}}{n^{2}}=a$. Then the line described by the polar equation $\rho=\frac{1}{2-a \cos \theta}$ is $\qquad$ .
4. The right branch of a hyperbola.

It is known that $S_{n}=C_{n-1}^{2}\left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right)$
$$
\begin{aligned}
& =(n-1)(n-2)\left(1-\frac{1}{2^{n}}\right) . \\
\therefore a & =\lim _{n \rightarrow \infty} \frac{4 S_{n}}{n^{2}}=4 \lim _{n \rightarrow \infty} \frac{(n-1)(n-2)\left(1-\frac{1}{2^{n}}\right)}{n^{2}} \\
& =4 \lim _{n \rightarrow \infty}\left[\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{1}{2^{n}}\right)\right]=4 .
\end{aligned}
$$
$\therefore \rho=\frac{1}{2-a \cos \theta}=\frac{\frac{1}{2}}{1-2 \cos \theta}$ represents the right branch of a hyperbola.