Envelope of Projectile Trajectories 

For a given launch velocity $v$ and launch angle $\theta$, the trajectory of a projectile may be described by the standard formula
  $$y=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$
For different values of $\theta$ what is the envelope of the different trajectories? Is it a parabola itself?

The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in $\tan\theta$ and set the discriminant to zero. The resulting relationship between $x,y$ is the envelope. 
This question is posted to see if there are other approaches to the solution.

Edit 1
Thanks for the nice solutions from Jack and Blue, received so far.
From the solution of the envelope it can be worked out that the envelope itself corresponds to the right half of the trajectory of a projectile launched at $(-\frac{v^2}{g^2},0)$ at a launch angle $\alpha=\frac{\pi}4$ and a launch velocity $V=v\sqrt2$. This means that both vertical and horizontal components of the launch velociy are equal to $v$. It would be interesting to see if these conclusions can be inferred from the problem itself by inspection and without first solving it. If so, then this would form another solution.

See also this other question posted subsequently. 

Although there are several very nice solutions provided above, here is another approach. No calculus is needed.
Consider a projectile launched with angle $\theta$ with relation to an inclined
plane with tilt angle $\varphi$. The initial speed is $v_0$.
We want to know at what distance from the launch point the projectile will
hit the inclined plane.
We use a Cartesian coordinate system rotated by the angle $\varphi$ (see
figure 1).
The trajectory is given by
\begin{aligned}
    x'(t) &=v_{0x'}t+\frac{g_{x'}t^2}2,\\
    y'(t) &= v_{0y'}t+\frac{g_{y'}t^2}2,
\end{aligned}
where $x'$ and $y'$ indicate the coordinates of the projectile in the rotated
coordinate frame.
Furthermore, we have $v_{0x'}=v_0\cos\theta$, $v_{0y'}=v_0\sin\theta$, $g_{x'}= -g\sin\varphi$
and $g_{y'}=-g\cos\varphi$.
At the point where the projectile hits the inclined plane $y'=0$.
Hence, we obtain that the time span for the collision is
$t_{hit}=-2v_{0y'}/g_{y'}$.
Replacing this time, we find
\begin{aligned}
x'_{hit} &=-2\frac{v_{0x'}v_{0y'}}{g_{y'}}+2\frac{g_{x'}v^2_{0y'}}{g^2_{y'}}\\
&=\frac{v_0^2}{g\cos^2\varphi}\left[\cos\varphi\sin(2\theta)-\sin\varphi(1-\cos(2\theta))\right]\\
&=\frac{v_0^2}{g\cos^2\varphi}\left[\sin(2\theta+\varphi)-\sin\varphi\right]
\leq\frac{v_0^2}{g\cos^2\varphi}\left(1-\sin\varphi\right)=\frac{v_0^2}{g(1+\sin\varphi)}.
\end{aligned}
If we now turn to polar coordinates, the envelope of the projectile trajectories
is given by
$$
r(\varphi)=\frac{v_0^2}{g(1+\sin\varphi)},\qquad(1)
$$
with $0\leq\varphi\leq\pi$.
See in figure 2 that this curve is the envelope.
Projectile trajectory hitting an inclined plane

Envelope given by equation (1)

