<s>[INST] Find envelope of $x \sin \theta - y \cos \theta + z = a \theta$, where $\theta$ is a parameter. I have tried to find envelope for
$$x \sin \theta - y \cos \theta + z = a \theta$$
First I find derivative w.r.t. $\theta$
$$F(\theta)=x \sin \theta - y \cos \theta + z - a \theta = 0$$
$$\frac{\partial F(\theta)}{\partial \theta}=y \sin \theta + x \cos \theta - a = 0$$
Then by solving these two above equation in order to eliminate parameter $\theta$, I find values of $\sin \theta$ and $\cos \theta$
$$\sin \theta = \frac{ax\theta + ay - xz}{x^2 + y^2}$$
and
$$\cos \theta = \frac{ax - ay\theta + yz}{x^2 + y^2}$$
Finally to find value of $\theta$ I squares and add above two equations then I get a quadratic equation in $\theta$ as
$$a^2\theta^2 - 2az\theta + a^2 + z^2 - x^2 - y^2=0$$
Then solving this for $\theta$ by applying quadratic formula, I get condition for real values of $\theta$ i.e.
$\theta$ is real only if
$$x^2+y^2 \ge a^2$$
Now here is my question. Either $x^2+y^2 \ge a^2$ is required envelope or something more to do?
Because $x^2+y^2 \ge a^2$ is an equality in which we have eliminated parameter $\theta$.
 [/INST] Hint:
You get rid of the annoying trigonometric functions by
$$(z-a\theta)^2+a^2=x^2+y^2.$$
From this you draw $\theta$, which you plug in one of the initial equations. Not a really nice expression.

Alternatively, in cylindrical coordinates $(\phi,\rho)$ we rewrite
$$\begin{cases}\rho\sin(\phi-\theta)=z-a\theta,\\\rho\cos(\phi-\theta)=a.\end{cases}$$
Then
$$\phi=\theta+\arccos\frac a\rho=\frac{z\pm\sqrt{\rho^2-a^2}}a+\arccos\frac a\rho.$$
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