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# Parabolic envelope of fireworks

The envelope of parabolic trajectories from a common launch point is itself a parabola. In the U.S. this weekend many will have a chance to observe this fact direcly, as the 4th of July is traditionally celebrated with fireworks.

If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows the parabola $$y = x \tan \theta - [x^2 /(2 v^2)] (1 + \tan^2 \theta)$$ and the envelope of all such trajectories is another parabola: $$y = v^2 /2 - x^2 / (2v^2)$$

These equations are not difficult to derive. I have two questions. First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations? Is there a purely geometric argument? Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.

Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity. Is the envelope of these elliptical trajectories also an ellipse? (I didn't try to work out the equations.) Perhaps the same geometric viewpoint (if it exists) could apply, e.g., by slightly tilting the sections.

I ask these questions in a weekend recreational spirit.