How do I remove the parameter $t$ from the $z$-function in the following: $$\begin{align}x&=a\cos{t}-a\\ y&=a\sin{t}\\ z&=nt\end{align}$$ (where $n,a$ are arbitrary coefficients) So far I have: $$\begin{align} x&=(a^2-y^2)^{0.5} - a\\ y&=(a^2 -(x+a))^{0.5}\end{align}$$ (though I haven't used $a^2=x^2+y^2+z^2$, which I am thinking that I should replace $a$ with it.) But $z$ has me confused on how to eliminate $t$.
Any help would be appreciated! Thanks.
On
The first two equations give you $$ \cos t = \frac{x+a}{a} \quad ; \quad \sin t = \frac{y}{a} $$ Dividing the second by the first gives $$ \tan t = \frac{\sin t}{\cos t} = \frac{y}{x+a} $$ and so $$ t = \arctan\frac{y}{x+a} + 2k\pi $$ for some integer $k$. Substituting this in the equation for $z$ gives $$ z = n \arctan\frac{y}{x+a} + 2k\pi $$ The equations represent a helix, if you're interested.
$\cos ^2{t}-\sin^2{t}=\cos{2t}=\dfrac{y^2-(x-a)^2}{a^2} \implies t=\dfrac{1}{2}(\arccos{\dfrac{y^2-(x-a)^2}{a^2}}+2k\pi)\implies z=\dfrac{n}{2}(\arccos{\dfrac{y^2-(x-a)^2}{a^2}}+2k\pi)$