Question

Use traces to sketch and identify the surface.

\( - {x^2} + 4{y^2} - {z^2} = 4\)

Short answer

The surface \( - {x^2} + 4{y^2} - {z^2} = 4\)is a hyperboloid of two sheets.

Step-by-step solution

the given surface

Consider the given surface is \( - {x^2} + 4{y^2} - {z^2} = 4\).

Substitute \(x = k\) in the surface equation

Case (i)

Let \(x = k\).

Substitute \(x = k\) in \( - {x^2} + 4{y^2} - {z^2} = 4\).

\(\begin{array}{l} - {k^2} + 4{y^2} - {z^2} = 4\\4{y^2} - {z^2} = 4 + {k^2}\end{array}\)

The surface area represents the trace of hyperbola.

Substitute \(y = k\) in surface equation

Case (ii):

Let \(y = k\)

Substitute \(y = k\) in \( - {x^2} + 4{y^2} - {z^2} = 4\).

\(\begin{array}{l} - {x^2} + 4{k^2} - {z^2} = 4\\ - {x^2} - {z^2} = 4 - 4{k^2}\\{x^2} + {z^2} = 4{k^2} - 4\end{array}\)

The surface area represents the trace of ellipse when \({k^2} > 1\).

Substitute \(z = k\) in surface equation

Case (iii):

Let \(z = k\).

Substitute \(z = k\) in \( - {x^2} + 4{y^2} - {z^2} = 4\).

\(\begin{array}{l} - {x^2} + 4{y^2} - {k^2} = 4\\ - {x^2} + 4{y^2} = 4 + {k^2}\end{array}\)

The surface area represents the trace of hyperbola.

observe the above cases and plot graph

From the above information, it is observed that the graph of \( - {x^2} + 4{y^2} - {z^2} = 4\) is combined behaviour of hyperbola and circle.

Since the two minus signs in the given equation indicates the two sheets in the hyperboloid, the surface equation \( - {x^2} + 4{y^2} - {z^2} = 4\) is a hyperboloid of two sheets.

Plot the hyperboloid of two sheets surface \( - {x^2} + 4{y^2} - {z^2} = 4\) as shown below in Figure 1.

From Figure 1, it is observed that the surface is a hyperboloid of two sheets.