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. 

Here's another approach. Not as elegant as Jack's though. Perhaps similar to Blue's. (Edit: Revised slightly to use substitution of $H=v^2/2g$ instead of $\lambda=g/v^2$ for more convenient reference, per solution by Narasimham)
This approach maximises the value of $y$ for any given value of $x$.  
For a given value of $x$, say $x=k$, 
$$\begin{align}
y&=kT-\frac {k^2}{4H}(1+T^2)\qquad\text{where $T=\tan\theta, H=\frac {v^2}{2g}$}\\
\frac{dy}{dT}&=k-\frac {k^2}{4H}\cdot 2T=0\qquad\text{when
$T=\frac {2H}k$}\\
\text{At $T=\frac {2H}k$}:\qquad\qquad
y&=k\cdot\frac{2H}k-\frac{k^2} {4H}\left(1+\frac {4H^2}{k^2}\right)
\color{lightgrey}{=2H-\frac {k^2}{4H}-H}\\
&=H-\frac {k^2}{4H}
\end{align}$$
Hence the equation of the envelope is
$$y=\frac {v^2}{2g}-\frac {gx^2}{2v^2}\quad\blacksquare$$
