The military tested ballistic flare projectile?

Question

Curious amateur photographer located at (0E, 30N) noticed a bright object rising over the horizon, and made a picture of its trajectory.
He directed his camera horizotally due south, and let exposition continue until the object fell under horizon. The trace of the projectile on the film is

a) parabola
b) ellipse
c) hyperbola
d) impossible to predict

I will draw a picture on Tuesday

I posted a picture here:
http://alexandersemenov.tripod.com/ya/hpb/index.htm


"Skewed ellipses" do not exist.
"Skewed ellipse" is simply another perfect ellipse.
Central projection of conic section is always conic section. But ellipse can become hyperbola, and visa versa, depending on relative positions of center of projection, original ellipse, and projection plane.

Answer 1

This is a repost (except now you don't specify the launch and impact points and you don't say to ignore Earth's rotation) and I believe my first answer, repeated below, is correct if the trajectory is still in the equatorial plane and the camera longitude is still midway between the launch and impact points:
b) (part of an) ellipse. Ballistic trajectories are simply partial orbits, which are elliptical. Since the launch and impact points are 160 deg apart, the semiminor axis of this ellipse is shorter than earth's radius. The camera is directed 30 degrees above the normal to the (equatorial) plane of the ellipse, resulting in some "flattening" of the ellipse in the image.
I'm assuming that "ballistic" is literally true; the projectile was unpowered, in free fall for the entire trajectory, otherwise it's unpredictable. So if the object was "bright" because of a rocket plume, the problem is self-contradictory.
EDIT: Your comment on my previous answer is mistaken. You said the ellipse is seen as a hyperbola because that's the intersection between the film plane and a cone that extends from the lens (tip of cone) to the orbit (base). I have to disagree. First, the cone is imaginary. (I assume it's a projection of the camera's - actually rectangular - field of view.) The ellipse is already formed and you're not making a new conic section, you're viewing the existing one. And an ellipse is topologically different from a hyperbola. Second, the film plane would have to be parallel with the cone's axis of symmetry, thus perpendicular to the ellipse plane, to form a hyperbolic intersection, and how would you image the ellipse that way? The film plane is in fact normal to the cone's axis of symmetry.
When you step away from an ellipse and view it from out of plane, if you are located in a plane of symmetry of the ellipse (a plane defined by one of the ellipse's axes of symmetry and perpendicularity to the ellipse plane), you will never see a hyperbola, you simply see an ellipse, possibly with different major/minor axis ratios. (If you're not in that plane of symmetry you see a skewed ellipse.) And of course the camera sees what you do.
EDIT (after seeing picture): OK, I get it. There's more to the problem than meets the eye. Literally, because in looking at the ellipse you're changing the retinal plane orientation so it doesn't intersect the ellipse, whereas the camera film plane is fixed per the problem statement. Very nice!
BTW skewed ellipse is a common term describing an ellipse projected such that its original axes of symmetry are no longer perpendicular.