Answer
\(a)\) Hyperboloid of one sheet directed along \(z\)-axis.
\(b)\) Hyperboloid of one sheet directed along \(y\)-axis. The change will rotate the figure.
\(c)\) The figure will move it's center to \((0,-1,0)\).
Work Step by Step
\(a)\) The function \(x^2+y^2-z^2=1\) can be written as \(\frac{x^2}{1} + \frac{y^2}{1}- \frac{z^2}{1}=1\) that match \(\textbf{exactly}\) witg the hyperboloid of one sheet formula:
\[
\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} - \frac{(z-z_0)^2}{c^2} =1
\]
where \(a\),\(b\) and \(c\) are constants, and \((x_0,y_0,z_0)\) is the center of the hyperboloid.
\(b)\) The figure will be the same (Hyperboloid of one sheet), but it will be directed along the \(y\)-axis, since it is always directed along the axis of the \(\textbf{neagtive}\) variable.
\(c)\)
\[
x^2+y^2+2y-z^2=0
\]
Add \(1\) on both sides:
\[
x^2+y^2+2y+1-z^2=1 \\
x^2+(y+1)^2-z^2=1 \\
x^2+(y-(-1))^2-z^2=1
\]
Following standard formula:
\[
\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} - \frac{(z-z_0)^2}{c^2} =1
\]
where \(a\),\(b\) and \(c\) are constants, and \((x_0,y_0,z_0)\) is the center of the hyperboloid.
The figure will move it's center to \((0,-1,0)\).
