Is it possible to have a non-escape orbit that is equidistant to two celestial bodies (for example, Earth and Mars) at all times? If possible, how much does it cost to keep an object (which is much smaller than both of the large bodies) in such a trajectory?

Detailed in response to comments:

As @BMF says, all points that are equidistant to two distinct given points is a plane perpendicular to the line segment connecting them passing through its midpoint. However, there is a peculiar orbit in that medial plane such that you make a near-right angle between lines of sight to said bodies. That orbit enables a 3-axis stabilised satellite to have its antenna assembly and solar (panel and sail) assembly in the same side and at the same time enabling very tightly focussed (narrower than arc-second) communication beams, greatly simplifying the satellite design. Which is the initial motive of asking this question.

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    $\begingroup$ Are you asking specifically about an object orbiting the Sun, or an object in space more or less equidistant from two similar-sized bodies (but no massive third-body nearby), such as a binary star system? $\endgroup$ Jul 27 '20 at 13:19
  • $\begingroup$ I don't think there is such an orbit but there are a heck of a lot crazy-shaped three-body orbits that might do something like this! Have a look through Elemental periodic orbits associated with the libration points in the circular restricted 3-body problem (cited here $\endgroup$
    – uhoh
    Jul 27 '20 at 13:21
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    $\begingroup$ @ErkinAlpGüney your title does state that the midpoint is what you're asking about. $\endgroup$ Jul 27 '20 at 19:47
  • $\begingroup$ I assume you are not looking for Lagrange points? $\endgroup$ Jul 28 '20 at 9:52
  • $\begingroup$ @OscarLanzi No. Edited the question to clarify. Not the trajectory of midpoints (not useful for an artificial satellite), but not an arbitrary orbit in the medial plane either. $\endgroup$ Jul 28 '20 at 19:01

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