Identify each quadric surface. List its intercepts and its traces in the coordinate planes.

Solution (a) We begin by dividing the equation by $9$ to put it in standard form. $$\dfrac{x^{2}}{9}-\dfrac{y^{2}}{9}+\dfrac{z^{2}}{9}=-1$$

This is a hyperboloid of two sheets. It has two intercepts $( 0,3,0)$ and $(0,-3,0).$ If $y=0$, the equation has no real solution, so there is no trace in the $xz$-plane. Traces parallel to the $xz$-plane are ellipses and are defined for $\left\vert y\right\vert \gt3.$ The trace in the $xy$-plane is the hyperbola $\dfrac{y^{2}}{9}-\dfrac{x^{2}}{ 9}=1$ and the trace in the $yz$-plane is the hyperbola $\dfrac{y^{2}}{9}- \dfrac{z^{2}}{9}=1.$ Traces parallel to the $xy$- and $yz$-plane are also hyperbolas.

The $y$-axis is the axis of this hyperboloid of two sheets. See Figure 67 for the graph.

(b) This equation defines a hyperbolic paraboloid. The only intercept is at the origin. The trace in the $xy$-plane is the pair of lines $y=\pm \dfrac{2}{3}x,$ which intersect at the origin. The traces parallel to the $xy$-plane are hyperbolas given by the equations $\dfrac{x^{2}}{9}- \dfrac{y^{2}}{4}=k,$ where $z=k.$

The trace in the $xz$-plane is the parabola $z=\dfrac{x^{2}}{9}$, which opens up. Traces parallel to the $xz$-plane are also parabolas that open up. The trace in the $yz$-plane is the parabola $z=-\dfrac{y^{2}}{4},$ which opens down. Traces parallel to the $yz$-plane are also parabolas that open down.

The origin is the saddle point of the hyperbolic paraboloid. See Figure 68 for the graph.