If gravity is uniform - the force has the same magnitude and direction everywhere, the trajectory is a parabola. This is a very good approximation for trajectories that don't go very far.
But in fact the force is not perfectly uniform. It actually does point to the center of the earth. It is stronger nearer the center. The trajectory for this case is an ellipse.
A typical parabolic trajectory hits the ground before it gets very far. If it didn't, it would be a very long skinny ellipse.
A parabola is an infinitely long ellipse.
For typical trajectories that hit the ground quickly, parabolic and elliptical trajectories are almost identical.
Edited to respond to comments.
The rotation of the earth does have an effect. From the point of view of an inertial observer floating in space, the initial velocity of a thrown rock is about the speed of rotation of the earth's surface at that latitude.
The earth rotates $360$ degrees in $1$ sidereal day = $85604.1$ sec, or about $0.0042$ degrees/sec. So gravity isn't quite uniform. It has tilted a bit by the end of the trajectory. But in a flight lasting only a couple seconds, it isn't enough to notice.
The observer in space sees an observer on the ground moving sideways at the speed of the earth's surface. The ground observer is following a circular path. In those few seconds, he deviates from a straight line at $0.0042$ degrees/sec. To a good approximation, he is moving at uniform speed on a straight line. This gives the same result as if he wasn't moving.
If the rock fell through the earth and orbited, it would follow an ellipse as seen by the space observer. It would not be as skinny as I had thought. To the observer on earth, the motion would look complicated.
So thanks to Peter for pointing this out.
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