Answer
\(\textbf{(a)}\) Hyperboloid of one sheet directed along \(z\)-axis
\(\textbf{(b)}\) Hyperboloid of one sheet directed along \(y\)-axis. The change will rotate the figure
\(\textbf{(c)}\) The figure will move it's center to \((0,-1,0)\)
Work Step by Step
\(\textbf{(a)}\) 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 \(\textbf{exactly}\) with 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.
\(\textbf{(b)}\) The figure will be the same (Hyperboloid of one sheet), but it will be directed along de \(y\)-axis, since it is always directed along de axis of the \( \textbf{negative}\) variable.
\(\textbf{(c)}\) \[
x^2+y^2+2y-z^2=0 \\
\text{Add 1 on both sides:} \\
x^2+y^2+2y+1-z^2=1 \\
x^2+(y+1)^2-z^2=1 \\
x^2+(y-(-1))^2-z^2=1 \\
\]
Following the 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 \((0,-1,0)\)
