Let L be the line of intersection of the planes $cx + y + z = c$ and $x - cy + cz = -1$, where c is a real number. 
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*Find symmetric equations for $L$

*As the number $c$ varies, the line $L$ sweeps out a surface $S$. Find an equation for the curve of intersection of $S$ with the horizontal plane $z = t$ (the trace of S in the plane $z = t$)

I got the symmetric equations as $$\frac{x+1}{-2c}=\frac{y-c}{c^2-1}=\frac{z-c}{c^2+1}$$
I got part one.
 A: Hint: Show that $x^2+y^2=1+t^2$.
A: The equations for $L$ that you got are correct as can be easily verified.
From this we can write
$x = -1 - 2 c \lambda $
$y = c + (c^2 - 1) \lambda $
$ z = t = c + (c^2 + 1) \lambda$
From the third equation, we get
$ \lambda = \dfrac{t - c }{c^2 + 1 } $
Substitute this in the first two equations
$ x = -1 - 2 c \dfrac{t - c}{c^2 + 1} $
$ y = c  + (c^2 - 1) \dfrac{t - c}{c^2 + 1 } $
And this simplifies to
$ x = \dfrac{ c^2 - 1 }{c^2 + 1 } - \dfrac{2 c}{c^2 + 1} t $
$ y = \dfrac{2 c}{c^2 + 1} + \dfrac{c^2 - 1}{c^2 + 1} t $
Note that $ \bigg( \dfrac{c^2 - 1}{c^2 + 1} \bigg)^2 + \bigg( \dfrac{2c}{c^2 + 1} \bigg) ^2 = 1 $
Therefore, we can let $ \dfrac{c^2-1}{c^2 + 1} = \cos(\theta) $ and $ \dfrac{2c}{c^2 + 1} = \sin(\theta) $
Then
$ x = \cos(\theta) - \sin(\theta) t $
$ y = \sin(\theta) + \cos(\theta) t $
Hence $(x, y)$ is a rotation of $(1, t)$ about the origin by a variable angle $\theta$, or by simply squaring $x$ and $y$ and adding we get
$ x^2 + y^2 = 1 + t^2 $
Thus the trace of the surface at $z = t$ is a circle of radius $\sqrt{1 + t^2} $
And this also means the surface $S$ is the hyperboloid of one sheet
$ x^2 + y^2 - z^2 = 1 $
