Answer
(a) $a<0$, $b<0$
There is no quadric surface that satisfies the equation.
(b) $a>0$, $b>0$
The quadric surface is a hyperboloid of one sheet.
(c) $a>0$, $b<0$
The quadric surface is a hyperboloid of two sheets.
Work Step by Step
(a) $a<0$, $b<0$
Since $a<0$, $b<0$, we can write the quadric surface $a{x^2} + b{y^2} - {z^2} = 1$ as
$ - \left| a \right|{x^2} - \left| b \right|{y^2} - {z^2} = 1$
Since the left-hand side is always negative, but the right-hand side is positive, we conclude that there is no such quadric surface.
(b) $a>0$, $b>0$
Since $a>0$, $b>0$, we can write the quadric surface $a{x^2} + b{y^2} - {z^2} = 1$ as
${\left( {\frac{x}{{1/\sqrt a }}} \right)^2} + {\left( {\frac{y}{{1/\sqrt b }}} \right)^2} = {\left( {\frac{z}{1}} \right)^2} + 1$
Referring to Eq. (2) of Section 13.6, this is the equation of the hyperboloid of one sheet.
(c) $a>0$, $b<0$
Since $a>0$, $b<0$, we can write the quadric surface $a{x^2} + b{y^2} - {z^2} = 1$ as
$a{x^2} - \left| b \right|{y^2} - {z^2} = 1$
$\left| b \right|{y^2} + {z^2} = a{x^2} - 1$
${\left( {\frac{y}{{1/\sqrt {\left| b \right|} }}} \right)^2} + {\left( {\frac{z}{1}} \right)^2} = {\left( {\frac{x}{{1/\sqrt a }}} \right)^2} - 1$
Referring to Eq. (2) of Section 13.6, this is the equation of the hyperboloid of two sheets.
