1. Identify the two special points in the elliptical orbit

For a satellite moving in an elliptical orbit around a planet:

• At the closest point (perigee), distance from planet is minimum, say $r_{\,c}$, and speed is maximum, say $v_{\,c}$.
• At the farthest point (apogee), distance from planet is maximum, say $r_{\,f}$, and speed is minimum, say $v_{\,f}$.

Given: $v_{\,f} \text{ is 6 times less than } v_{\,c}$ so, $v_{\,f} = \frac{v_{\,c}}{6}$

2. Use conservation of angular momentum

For motion under a central force, angular momentum is conserved.

At closest and farthest points, velocity is perpendicular to radius vector, so angular momentum magnitude is: $L = mrv$

Hence, $m r_{\,c} v_{\,c} = m r_{\,f} v_{\,f}$ Cancelling $m$, $r_{\,c} v_{\,c} = r_{\,f} v_{\,f}$

Substitute $v_{\,f} = \dfrac{v_{\,c}}{6}$: $r_{\,c} v_{\,c} = r_{\,f}\left(\frac{v_{\,c}}{6}\right)$

Cancel $v_{\,c}$: $r_{\,c} = \frac{r_{\,f}}{6}$

Therefore, $r_{\,c} : r_{\,f} = 1 : 6$

3. Match with the given options

The required ratio of distances at closest and farthest points is: $\boxed{1:6}$

So the correct option is C.